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**Unformatted text preview: **Instructor Solutions ManualforPhysicsbyHalliday, Resnick, and KranePaul StanleyBeloit CollegeVolume 2A Note To The Instructor...The solutions here are somewhat brief, as they are designed for the instructor, not for the student.Check with the publishers before electronically posting any part of these solutions; website, ftp, orserver access must be restricted to your students.I have been somewhat casual about subscripts whenever it is obvious that a problem is onedimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.Although this does not change the validity of the answer, it will sometimes obfuscate the approachif viewed by a novice.There are some traditional formula, such as22vx = v0x + 2ax x,which are not used in the text. The worked solutions use only material from the text, so there maybe times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know aneasier approach existed. But if it was not in the text, I did not use it here.I also tried to avoid reinventing the wheel. There are some exercises and problems in the textwhich build upon previous exercises and problems. Instead of rederiving expressions, I simply referyou to the previous solution.I adopt a dierent approach for rounding of signicant gures than previous authors; in particular, I usually round intermediate answers. As such, some of my answers will dier from those inthe back of the book.Exercises and Problems which are enclosed in a box also appear in the Students Solution Manualwith considerably more detail and, when appropriate, include discussion on any physical implicationsof the answer. These student solutions carefully discuss the steps required for solving problems, pointout the relevant equation numbers, or even specify where in the text additional information can befound. When two almost equivalent methods of solution exist, often both are presented. You areencouraged to refer students to the Students Solution Manual for these exercises and problems.However, the material from the Students Solution Manual must not be copied.Paul StanleyBeloit Collegestanley@clunet.edu1E25-1The charge transferred isQ = (2.5 104 C/s)(20 106 s) = 5.0 101 C.E25-2 Use Eq. 25-4:(8.99109 Nm2 /C2 )(26.3106 C)(47.1106 C)= 1.40 m(5.66 N)r=E25-3 Use Eq. 25-4:F =(8.99109 Nm2 /C2 )(3.12106 C)(1.48106 C)= 2.74 N.(0.123 m)2E25-4 (a) The forces are equal, so m1 a1 = m2 a2 , orm2 = (6.31107 kg)(7.22 m/s2 )/(9.16 m/s2 ) = 4.97107 kg.(b) Use Eq. 25-4:q=E25-5(6.31107 kg)(7.22 m/s2 )(3.20103 m)2= 7.201011 C(8.99109 Nm2 /C2 )(a) Use Eq. 25-4,F =1 q1 q21(21.3 C)(21.3 C)== 1.77 N212 C2 /N m2 )4 0 r124(8.8510(1.52 m)2(b) In part (a) we found F12 ; to solve part (b) we need to rst nd F13 . Since q3 = q2 andr13 = r12 , we can immediately conclude that F13 = F12 .We must assess the direction of the force of q3 on q1 ; it will be directed along the line whichconnects the two charges, and will be directed away from q3 . The diagram below shows the directions.FF 12F2323F netF 12From this diagram we want to nd the magnitude of the net force on q1 . The cosine law isappropriate here:F net 2F net22= F12 + F13 2F12 F13 cos ,2= (1.77 N) + (1.77 N)2 2(1.77 N)(1.77 N) cos(120 ),= 9.40 N2 ,= 3.07 N.2E25-6 Originally F0 = CQ2 = 0.088 N, where C is a constant. When sphere 3 touches 1 the0charge on both becomes Q0 /2. When sphere 3 the touches sphere 2 the charge on each becomes(Q0 + Q0 /2)/2 = 3Q0 /4. The force between sphere 1 and 2 is thenF = C(Q0 /2)(3Q0 /4) = (3/8)CQ2 = (3/8)F0 = 0.033 N.0E25-7 The forces on q3 are F31 and F32 . These forces are given by the vector form of CoulombsLaw, Eq. 25-5,F31=F32=1 q3 q11 q3 q1 ,r2 r31 = 42 314 0 r310 (2d)1 q3 q21 q3 q2 .r2 r32 = 42 324 0 r320 (d)These two forces are the only forces which act on q3 , so in order to have q3 in equilibrium the forcesmust be equal in magnitude, but opposite in direction. In short,F311 q3 q131r4 0 (2d)2q131r4= F32 ,1 q3 q232 ,= r4 0 (d)2q2= 32 .r1Note that 31 and 32 both point in the same direction and are both of unit length. We then getrrq1 = 4q2 .E25-8 The horizontal and vertical contributions from the upper left charge and lower right chargeare straightforward to nd. The contributions from the upper left charge require slightly more work.The diagonal distance is 2a; the components will be weighted by cos 45 = 2/2. The diagonalcharge will contribute1 (q)(2q) 22 q2 Fx =i=i,4 0 ( 2a)2 28 0 a21 (q)(2q) 2 2 q2 Fy =j=j.4 0 ( 2a)2 28 0 a2(a) The horizontal component of the net force is then1 (2q)(2q)2 q2 Fx =i+i,4 0a28 0 a24 + 2/2 q 2 =i,4 0 a2=(4.707)(8.99109 N m2 /C2 )(1.13106 C)2 /(0.152 m)2 = 2.34 N ii.(b) The vertical component of the net force is then1 (q)(2q) 2 q2 j+j,Fy = 24 0 a8 0 a22 + 2/2 q 2 =j,8 0a2=(1.293)(8.99109 N m2 /C2 )(1.13106 C)2 /(0.152 m)2 = 0.642 N jj.3E25-9 The magnitude of the force on the negative charge from each positive charge isF = (8.99109 N m2 /C2 )(4.18106 C)(6.36106 C)/(0.13 m)2 = 14.1 N.The force from each positive charge is directed along the side of the triangle; but from symmetryonly the component along the bisector is of interest. This means that we need to weight the aboveanswer by a factor of 2 cos(30 ) = 1.73. The net force is then 24.5 N.E25-10 Let the charge on one sphere be q, then the charge on the other sphere is Q = (52.6 106 C) q. Then1 qQ= F,4 0 r2(8.99109 Nm2 /C2 )q(52.6106 C q) = (1.19 N)(1.94 m)2 .Solve this quadratic expression for q and get answers q1 = 4.02105 C and q2 = 1.24106 N.E25-11 This problem is similar to Ex. 25-7. There are some additional issues, however. It iseasy enough to write expressions for the forces on the third chargeF31=F32=1 q3 q12 r31 ,4 0 r311 q3 q22 r32 .4 0 r32ThenF311 q3 q12 r314 0 r31q12 r31r31= F32 ,1 q3 q2= 2 r32 ,4 0 r32q2= 2 32 .rr32The only way to satisfy the vector nature of the above expression is to have 31 = 32 ; this meansrrthat q3 must be collinear with q1 and q2 . q3 could be between q1 and q2 , or it could be on eitherside. Lets resolve this issue now by putting the values for q1 and q2 into the expression:(1.07 C)31r2r312r32 31r= (3.28 C)32 ,r2r322= (3.07)r31 32 .rSince squared quantities are positive, we can only get this to work if 31 = 32 , so q3 is not betweenrrq1 and q2 . We are then left with22r32 = (3.07)r31 ,so that q3 is closer to q1 than it is to q2 . Then r32 = r31 + r12 = r31 + 0.618 m, and if we take thesquare root of both sides of the above expression,r31 + (0.618 m) =(3.07)r31 ,(0.618 m) =(3.07)r31 r31 ,(0.618 m) = 0.752r31 ,0.822 m = r314E25-12 The magnitude of the magnetic force between any two charges is kq 2 /a2 , where a =0.153 m. The force between each charge is directed along the side of the triangle; but from symmetryonly the component along the bisector is of interest. This means that we need to weight the aboveanswer by a factor of 2 cos(30 ) = 1.73. The net force on any charge is then 1.73kq 2 /a2 .The length of the angle bisector, d, is given by d = a cos(30 ).The distance from any charge to the center of the equilateral triangle is x, given by x2 =(a/2)2 + (d x)2 . Thenx = a2 /8d + d/2 = 0.644a.The angle between the strings and the plane of the charges is , given bysin = x/(1.17 m) = (0.644)(0.153 m)/(1.17 m) = 0.0842,or = 4.83 .The force of gravity on each ball is directed vertically and the electric force is directed horizontally.The two must then be related bytan = F E /F G ,so1.73(8.99109 N m2 /C2 )q 2 /(0.153 m)2 = (0.0133 kg)(9.81 m/s2 ) tan(4.83 ),or q = 1.29107 C.E25-13 On any corner charge there are seven forces; one from each of the other seven charges.The net force will be the sum. Since all eight charges are the same all of the forces will be repulsive.We need to sketch a diagram to show how the charges are labeled.21467385The magnitude of the force of charge 2 on charge 1 isF12 =1 q22 ,4 0 r12where r12 = a, the length of a side. Since both charges are the same we wrote q 2 . By symmetry weexpect that the magnitudes of F12 , F13 , and F14 will all be the same and they will all be at rightangles to each other directed along the edges of the cube. Written in terms of vectors the forces5would beF12F13=F141 q2 i,4 0 a21 q2 j,4 0 a21 q2 k.4 0 a2==The force from charge 5 is1 q22 ,4 0 r15and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonaldistance, and can be found from2r15 = a2 + a2 = 2a2 ,F15 =then1 q2.4 0 2a2By symmetry we expect that the magnitudes of F15 , F16 , and F17 will all be the same and they willall be directed along the diagonals of the faces of the cube. In terms of components we would haveF15 =F15=F16=F17=1 q24 0 2a21 q24 0 2a21 q24 0 2a2 2 + k/ 2 ,j/ 2 + k/ 2 ,i/ 2 + 2 .i/j/The last force is the force from charge 8 on charge 1, and is given byF18 =1 q22 ,4 0 r18and is directed along the cube diagonal away from charge 8. The distance r18 is also the cubediagonal distance, and can be found from2r18 = a2 + a2 + a2 = 3a2 ,then in term of componentsF18 =1 q2 i/ 3 + 3 + k/ 3 .j/24 0 3aWe can add the components together. By symmetry we expect the same answer for each components, so well just do one. How about This component has contributions from charge 2, 6, 7,i.and 8:11 q2 12,+ + 24 0 a1 2 2 3 3or1 q2(1.90)4 0 a2The three components add according to Pythagoras to pick up a nal factor of 3, soF net = (0.262)6q2.20aE25-14 (a) Yes. Changing the sign of y will change the sign of Fy ; since this is equivalent toputting the charge q0 on the other side, we would expect the force to also push in the otherdirection.(b) The equation should look Eq. 25-15, except all ys should be replaced by xs. ThenFx =1q0 q.4 0 x x2 + L2 /4(c) Setting the particle a distance d away should give a force with the same magnitude asF =14 0 dq0 qd2 + L2 /4.This force is directed along the 45 line, so Fx = F cos 45 and Fy = F sin 45 .(d) Let the distance be d = x2 + y 2 , and then use the fact that Fx /F = cos = x/d. ThenFx = F1x q0 qx=.2 + y 2 + L2 /4)3/2d4 0 (xFy = F1y q0 qy=.2 + y 2 + L2 /4)3/2d4 0 (xandE25-15 (a) The equation is valid for both positive and negative z, so in vector form it would readF = Fz k =1q0 q zk.4 0 (z 2 + R2 )3/2(b) The equation is not valid for both positive and negative Reversing the sign of z shouldz.reverse the sign of Fz , and one way to x this is to write 1 = z/ z 2 . Then1 2q0 qz4 0 R2F = Fz k =11 2zz2k.E25-16 Divide the rod into small dierential lengths dr, each with charge dQ = (Q/L)dr. Eachdierential length contributes a dierential force1 q dQ1 qQ=dr.24 0 r4 0 r2 LdF =Integrate:x+LF==1 qQdr,4 0 r2 Lx1 qQ 114 0 L x x + LdF =E25-17 You must solve Ex. 16 before solving this problem! q0 refers to the charge that had beencalled q in that problem. In either case the distance from q0 will be the same regardless of the signof q; if q = Q then q will be on the right, while if q = Q then q will be on the left.Setting the forces equal to each other one gets1 qQ4 0 L11x x+L=orr=x(x + L).71 qQ,4 0 r2E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem.If all charges are positive then moving q0 o axis will result in a net force away from the axis.Thats unstable.If q = Q then both q and Q are on the same side of q0 . Moving q0 closer to q will result in theattractive force growing faster than the repulsive force, so q0 will move away from equilibrium.E25-19 We can start with the work that was done for us on Page 577, except since we areconcerned with sin = z/r we would have1q0 dz4 0 (y 2 + z 2 )dFx = dF sin =zy2+ z2.We will need to take into consideration that changes sign for the two halves of the rod. ThenFx====0q0 4 0q0 2 0L/2L/20q0 2 0q0 2 0(y 2L/2z dz++ z 2 )3/20+z dz(y 2 + z 2 )3/2,z dz,(y 2 + z 2 )3/2L/21y2 + z21y,01y2+ (L/2)2.E25-20 Use Eq. 25-15 to nd the magnitude of the force from any one rod, but write it asF =14 0 rqQr2 + L2 /4,where r2 = z 2 + L2 /4. The component of this along the z axis is Fz = F z/r. Since there are 4 rods,we have1qQz1qQzF =,=,22 + L2 /42 + L2 /4) z 2 + L2 /2 0r r 0 (zEquating the electric force with the force of gravity and solving for Q,Q= 0 mg 2(z + L2 /4) z 2 + L2 /2;qzputting in the numbers,(8.851012 C2 /Nm2 )(3.46107 kg)(9.8m/s2 )((0.214m)2+(0.25m)2 /4) (0.214m)2 +(0.25m)2 /2(2.451012 C)(0.214 m)so Q = 3.07106 C.E25-21 In each case we conserve charge by making sure that the total number of protons is thesame on both sides of the expression. We also need to conserve the number of neutrons.(a) Hydrogen has one proton, Beryllium has four, so X must have ve protons. Then X must beBoron, B.(b) Carbon has six protons, Hydrogen has one, so X must have seven. Then X is Nitrogen, N.(c) Nitrogen has seven protons, Hydrogen has one, but Helium has two, so X has 7 + 1 2 = 6protons. This means X is Carbon, C.8E25-22 (a) Use Eq. 25-4:(8.99109 Nm2 /C2 )(2)(90)(1.601019 C)2= 290 N.(121015 m)2F =(b) a = (290 N)/(4)(1.661027 kg) = 4.41028 m/s2 .E25-23 Use Eq. 25-4:(8.99109 Nm2 /C2 )(1.601019 C)2= 2.89109 N.(2821012 m)2F =E25-24 (a) Use Eq. 25-4:q=(3.7109 N)(5.01010 m)2= 3.201019 C.(8.99109 Nm2 /C2 )(b) N = (3.201019 C)/(1.601019 C) = 2.E25-25Use Eq. 25-4,F =( 1 1.6 1019 C)( 1 1.6 1019 C)1 q1 q233== 3.8 N.24 0 r124(8.85 1012 C2 /N m2 )(2.6 1015 m)2E25-26 (a) N = (1.15107 C)/(1.601019 C) = 7.191011 .(b) The penny has enough electrons to make a total charge of 1.37105 C. The fraction is then(1.15107 C)/(1.37105 C) = 8.401013 .E25-27 Equate the magnitudes of the forces:1 q2= mg,4 0 r2sor=(8.99109 Nm2 /C2 )(1.601019 C)2= 5.07 m(9.111031 kg)(9.81 m/s2 )E25-28 Q = (75.0 kg)(1.601019 C)/(9.111031 kg) = 1.31013 C.3E25-29 The mass of water is (250 cm3 )(1.00 g/cm ) = 250 g. The number of moles of water is(250 g)/(18.0 g/mol) = 13.9 mol. The number of water molecules is (13.9 mol)(6.021023 mol1 ) =8.371024 . Each molecule has ten protons, so the total positive charge isQ = (8.371024 )(10)(1.601019 C) = 1.34107 C.E25-30 The total positive charge in 0.250 kg of water is 1.34107 C. Marys imbalance is thenq1 = (52.0)(4)(1.34107 C)(0.0001) = 2.79105 C,while Johns imbalance isq2 = (90.7)(4)(1.34107 C)(0.0001) = 4.86105 C,The electrostatic force of attraction is then1 q1 q2(2.79105 )(4.86105 )F == (8.99109 N m2 /C2 )= 1.61018 N.4 0 r2(28.0 m)29E25-31(a) The gravitational force of attraction between the Moon and the Earth isFG =GM E M M,R2where R is the distance between them. If both the Earth and the moon are provided a charge q,then the electrostatic repulsion would beFE =1 q2.4 0 R2Setting these two expression equal to each other,q2= GM E M M ,4 0which has solutionq=4 0 GM E M M ,=4(8.851012 C2/Nm2 )(6.671011 Nm2/kg2 )(5.981024 kg)(7.361022 kg),=5.71 1013 C.(b) We need(5.71 1013 C)/(1.60 1019 C) = 3.57 1032protons on each body. The mass of protons needed is then(3.57 1032 )(1.67 1027 kg) = 5.97 1065 kg.Ignoring the mass of the electron (why not?) we can assume that hydrogen is all protons, so weneed that much hydrogen.P25-1them isAssume that the spheres initially have charges q1 and q2 . The force of attraction betweenF1 =1 q1 q2= 0.108 N,24 0 r12where r12 = 0.500 m. The net charge is q1 + q2 , and after the conducting wire is connected eachsphere will get half of the total. The spheres will have the same charge, and repel with a force ofF2 =1 1 (q1 + q2 ) 1 (q1 + q2 )22= 0.0360 N.24 0r12Since we know the separation of the spheres we can nd q1 + q2 quickly,2q1 + q2 = 2 4 0 r12 (0.0360 N) = 2.00 CWell put this back into the rst expression and solve for q2 .0.108 N=1 (2.00 C q2 )q2,24 0r123.00 1012 C2 = (2.00 C q2 )q2 ,20 = q2 + (2.00 C)q2 + (1.73 C)2 .The solution is q2 = 3.0 C or q2 = 1.0 C. Then q1 = 1.0 C or q1 = 3.0 C.10P25-2 The electrostatic force on Q from each q has magnitude qQ/4 0 a2 , where a is the lengthof the side of the square. The magnitude of the vertical (horizontal) component of the force of Q onQ is 2Q2 /16 0 a2 .(a) In order to have a zero net force on Q the magnitudes of the two contributions must balance,so 22QqQ=,216 0 a4 0 a2or q = 2Q/4. The charges must actually have opposite charge.(b) No.P25-3 (a) The third charge, q3 , will be between the rst two. The net force on the third chargewill be zero if1 q q31 4q q32 = 42 ,4 0 r310 r32which will occur if12=r31r32The total distance is L, so r31 + r32 = L, or r31 = L/3 and r32 = 2L/3.Now that we have found the position of the third charge we need to nd the magnitude. Thesecond and third charges both exert a force on the rst charge; we want this net force on the rstcharge to be zero, so1 q q31 q 4q2 = 42 ,4 0 r130 r12orq34q= 2,(L/3)2Lwhich has solution q3 = 4q/9. The negative sign is because the force between the rst and secondcharge must be in the opposite direction to the force between the rst and third charge.(b) Consider what happens to the net force on the middle charge if is is displaced a small distancez. If the charge 3 is moved toward charge 1 then the force of attraction with charge 1 will increase.But moving charge 3 closer to charge 1 means moving charge 3 away from charge 2, so the force ofattraction between charge 3 and charge 2 will decrease. So charge 3 experiences more attraction toward the charge that it moves toward, and less attraction to the charge it moves away from. Soundsunstable to me.P25-4(a) The electrostatic force on the charge on the right has magnitudeF =q2,4 0 x2The weight of the ball is W = mg, and the two forces are related byF/W = tan sin = x/2L.Combining, 2Lq 2 = 4 0 mgx3 , sox=q2 L2 01/3.(b) Rearrange and solve for q,q=2(8.851012 C2 /N m2 )(0.0112 kg)(9.81 m/s2 )(4.70102 m)3= 2.28108 C.(1.22 m)11P25-5 (a) Originally the balls would not repel, so they would move together and touch; aftertouching the balls would split the charge ending up with q/2 each. They would then repel again.(b) The new equilibrium separation isP25-61/3(q/2)2 L2 0 mgx ==141/3x = 2.96 cm.Take the time derivative of the expression in Problem 25-4. Thendx2 x dq2 (4.70102 m)==(1.20109 C/s) = 1.65103 m/s.dt3 q dt3 (2.28108 C)P25-7The force between the two charges isF =1 (Q q)q.24 0r12We want to maximize this force with respect to variation in q, this means nding dF/dq and settingit equal to 0. Thend1 (Q q)q1 Q 2qdF==.22dqdq 4 0r124 0 r12This will vanish if Q 2q = 0, or q = 1 Q.2P25-8Displace the charge q a distance y. The net restoring force on q will be approximatelyF 2qQ1yqQ 16=y.4 0 (d/2)2 (d/2)4 0 d3Since F/y is eectively a force constant, the period of oscillation isT = 2m=k0 m3 3d1/2.qQP25-9 Displace the charge q a distance x toward one of the positive charges Q. The net restoringforce on q will beF=qQ1124 0 (d/2 x)(d/2 + x)2qQ 32x.4 0 d3Since F/x is eectively a force constant, the period of oscillation isT = 2m=k0 m3 32qQ12d1/2.,P25-10 (a) Zero, by symmetry.(b) Removing a positive Cesium ion is equivalent to adding a singly charged negative ion at thatsame location. The net force is thenF = e2 /4 0 r2 ,where r is the distance between the Chloride ion and the newly placed negative ion, orr=3(0.20109 m)2The force is thenF =(1.61019 C)2= 1.92109 N.4(8.851012 C2 /N m2 )3(0.20109 m)2P25-11 We can pretend that this problem is in a single plane containing all three charges. Themagnitude of the force on the test charge q0 from the charge q on the left isFl =q q01.4 0 (a2 + R2 )A force of identical magnitude exists from the charge on the right. we need to add these two forcesas vectors. Only the components along R will survive, and each force will contribute an amountF l sin = F l R,+ a2R2so the net force on the test particle will beq q0R2.4 0 (a2 + R2 ) R2 + a2We want to nd the maximum value as a function of R. This means take the derivative, and set itequal to zero. The derivative is2q q04 0(a213R2 22 )3/2+R(a + R2 )5/2which will vanish whena2 + R2 = 3R2 ,a simple quadratic equation with solutions R = a/ 2.13,E26-1 E = F/q = ma/q. ThenE = (9.111031 kg)(1.84109 m/s2 )/(1.601019 C) = 1.05102 N/C.E26-2 The answers to (a) and (b) are the same!F = Eq = (3.0106 N/C)(1.601019 C) = 4.81013 N.E26-3F = W , or Eq = mg, soE=mg(6.64 1027 kg)(9.81 m/s2 )== 2.03 107 N/C.q2(1.60 1019 C)The alpha particle has a positive charge, this means that it will experience an electric force whichis in the same direction as the electric eld. Since the gravitational force is down, the electric force,and consequently the electric eld, must be directed up.E26-4 (a) E = F/q = (3.0106 N)/(2.0109 C) = 1.5103 N/C.(b) F = Eq = (1.5103 N/C)(1.601019 C) = 2.41016 N.(c) F = mg = (1.671027 kg)(9.81 m/s2 ) = 1.61026 N.(d) (2.41016 N)/(1.61026 N) = 1.51010 .E26-5 Rearrange E = q/4 0 r2 ,q = 4(8.851012 C2 /N m2 )(0.750 m)2 (2.30 N/C) = 1.441010 C.E26-6 p = qd = (1.601019 C)(4.30109 ) = 6.881028 C m.E26-7Use Eq. 26-12 for points along the perpendicular bisector. ThenE=1 p(3.56 1029 C m)= (8.99 109 N m2 /C2 )= 1.95 104 N/C.4 0 x3(25.4 109 m)3E26-8 If the charges on the line x = a where +q and q instead of +2q and 2q then at thecenter of the square E = 0 by symmetry. This simplies the problem into nding E for a charge +qat (a, 0) and q at (a, a). This is a dipole, and the eld is given by Eq. 26-11. For this exercise wehave x = a/2 and d = a, so1qaE=,4 0 [2(a/2)2 ]3/2or, putting in the numbers, E = 1.11105 N/C.E26-9 The charges at 1 and 7 are opposite and can be eectively replaced with a single charge of6q at 7. The same is true for 2 and 8, 3 and 9, on up to 6 and 12. By symmetry we expect theeld to point along a line so that three charges are above and three below. That would mean 9:30.E26-10 If both charges are positive then Eq. 26-10 would read E = 2E+ sin , and Eq. 26-11would look like1qx,E = 22 + (d/2)22 + (d/2)24 0 xx1 q x 24 0 x2 x2when xd. This can be simplied to E = 2q/4 0 x2 .14E26-11 Treat the two charges on the left as one dipole and treat the two charges on the right asa second dipole. Point P is on the perpendicular bisector of both dipoles, so we can use Eq. 26-12to nd the two elds.For the dipole on the left p = 2aq and the electric eld due to this dipole at P has magnitudeEl =2aq14 0 (x + a)3and is directed up.For the dipole on the right p = 2aq and the electric eld due to this dipole at P has magnitudeEr =12aq4 0 (x a)3and is directed down.The net electric eld at P is the sum of these two elds, but since the two component elds pointin opposite directions we must actually subtract these values,E= Er El,2aq11=,34 0 (x a)(x + a)3aq 111=332 0 x(1 a/x)(1 + a/x)3.We can use the binomial expansion on the terms containing 1 a/x,E==aq 1((1 + 3a/x) (1 3a/x)) ,2 0 x3aq 1(6a/x) ,2 0 x33(2qa2 ).2 0 x4E26-12 Do a series expansion on the part in the parentheses111 + R2 /z 21 11 R22 z2=R2.2z 2Substitute this in,Ez R2 Q=.2 2 0 2z4 0 z 2E26-13 At the surface z = 0 and Ez = /2 0 . Half of this value occurs when z is given by1z,=1 2z 2 + R2which can be written as z 2 + R2 = (2z)2 . Solve this, and z = R/ 3.E26-14 Look at Eq. 26-18. The electric eld will be a maximum when z/(z 2 + R2 )3/2 is amaximum. Take the derivative of this with respect to z, and get132z 2z 2 + R2 3z 2=.2 (z 2 + R2 )5/2(z 2 + R2 )3/2(z 2 + R2 )5/2This will vanish when the numerator vanishes, or when z = R/ 2.15E26-15 (a) The electric eld strength just above the center surface of a charged disk is given byEq. 26-19, but with z = 0,E=2 0The surface charge density is = q/A = q/(R2 ). Combining,q = 2 0 R2 E = 2(8.85 1012 C2 /N m2 )(2.5 102 m)2 (3 106 N/C) = 1.04 107 C.Notice we used an electric eld strength of E = 3 106 N/C, which is the eld at air breaks downand sparks happen.(b) We want to nd out how many atoms are on the surface; if a is the cross sectional area ofone atom, and N the number of atoms, then A = N a is the surface area of the disk. The numberof atoms is(0.0250 m)2A== 1.31 1017N=a(0.015 1018 m2 )(c) The total charge on the disk is 1.04 107 C, this corresponds to(1.04 107 C)/(1.6 1019 C) = 6.5 1011electrons. (We are ignoring the sign of the charge here.) If each surface atom can have at most oneexcess electron, then the fraction of atoms which are charged is(6.5 1011 )/(1.31 1017 ) = 4.96 106 ,which isnt very many.E26-16 Imagine switching the positive and negative charges. The electric eld would also needto switch directions. By symmetry, then, the electric eld can only point vertically down. Keepingonly that component,/2E1 dsin ,4 0 r2=2=2 .4 0 r20But = q/(/2), so E = q/ 2 0 r2 .E26-17We want to t the data to Eq. 26-19,Ez =2 01 z2z+ R2.There are only two variables, R and q, with q = R2 .We can nd very easily if we assume that the measurements have no error because then at thesurface (where z = 0), the expression for the electric eld simplies toE=.2 0Then = 2 0 E = 2(8.854 1012 C2 /N m2 )(2.043 107 N/C) = 3.618 104 C/m2 .Finding the radius will take a little more work. We can choose one point, and make that thereference point, and then solve for R. Starting withEz =2 01 16z2z+ R2,and then rearranging,2 0 Ez2 0 Ez11 + (R/z)2=1 = 1=1 + (R/z)2=Rzz,z 2 + R211 + (R/z)22 0 Ez1,12,(1 2 0 Ez /)=12(1 2 0 Ez /), 1.Using z = 0.03 m and Ez = 1.187 107 N/C, along with our value of = 3.618 104 C/m2 , wendRz=R=1(1 22(8.8541012 C2/Nm2 )(1.187107 N/C)/(3.618104 C/m2 ))2.167(0.03 m) = 0.065 m.(b) And now nd the charge from the charge density and the radius,q = R2 = (0.065 m)2 (3.618 104 C/m2 ) = 4.80 C.E26-18 (a) = q/L.(b) Integrate:L+aE===1 dxx2 ,4 0a11,4 0 a L + aq1,4 0 a(L + a)since = q/L.(c) If aL then L can be replaced with 0 in the above expression.E26-19 A sketch of the eld looks like this.17 1,E26-20 (a) F = Eq = (40 N/C)(1.601019 C) = 6.41018 N(b) Lines are twice as far apart, so the eld is half as large, or E = 20N/C.E26-21 Consider a view of the disk on edge.E26-22 A sketch of the eld looks like this.18E26-23 To the right.E26-24 (a) The electric eld is zero nearer to the smaller charge; since the charges have oppositesigns it must be to the right of the +2q charge. Equating the magnitudes of the two elds,2q5q=,4 0 x24 0 (x + a)2or5x =which has solution2(x + a),x= 2a = 2.72a.5 2E26-25 This can be done quickly with a spreadsheet.ExdE26-26 (a) At point A,E=14 0q2qd2(2d)219=1 q,4 0 2d2where the negative sign indicates that E is directed to the left.At point B,1q2q1 6qE==,224 0 (d/2)(d/2)4 0 d2where the positive sign indicates that E is directed to the right.At point C,q2q1 7q1+ 2=,E=24 0 (2d)d4 0 4d2where the negative sign indicates that E is directed to the left.E26-27 (a) The electric eld does (negative) work on the electron. The magnitude of this workis W = F d, where F = Eq is the magnitude of the electric force on the electron and d is the distancethrough which the electron moves. Combining,W = F d = q E d,which gives the work done by the electric eld on the electron. The electron originally possessed a1kinetic energy of K = 2 mv 2 , since we want to bring the electron to a rest, the work done must benegative. The charge q of the electron is negative, so E and d are pointing in the same direction,and E d = Ed.By the work energy theorem,1W = K = 0 mv 2 .2We put all of this together and nd d,d=mv 2(9.111031 kg)(4.86 106 m/s)2W=== 0.0653 m.qE2qE2(1.601019 C)(1030 N/C)(b) Eq = ma gives the magnitude of the acceleration, and v f = v i + at gives the time. Butv f = 0. Combining these expressions,t=mv i(9.111031 kg)(4.86 106 m/s)== 2.69108 s.Eq(1030 N/C)(1.601019 C)(c) We will apply the work energy theorem again, except now we dont assume the nal kineticenergy is zero. Instead,W = K = K f K i ,and dividing through by the initial kinetic energy to get the fraction lost,WKf Ki== fractional change of kinetic energy.KiKiBut K i = 1 mv 2 , and W = qEd, so the fractional change is2W(1.601019 C)(1030 N/C)(7.88103 m)qEd== 12.1%.= 11231 kg)(4.86 106 m/s)2Ki2 mv2 (9.1110E26-28 (a) a = Eq/m = (2.16104 N/C)(1.601019 C)/(1.671027 kg) = 2.071012 m/s2 .(b) v = 2ax = 2(2.071012 m/s2 )(1.22102 m) = 2.25105 m/s.20E26-29 (a) E = 2q/4 0 r2 , orE=(1.88107 C)= 5.85105 N/C.2(8.851012 C2 /N m2 )(0.152 m/2)2(b) F = Eq = (5.85105 N/C)(1.601019 C) = 9.361014 N.E26-30 (a) The average speed between the plates is (1.95102 m)/(14.7109 s) = 1.33106 m/s.The speed with which the electron hits the plate is twice this, or 2.65106 m/s.(b) The acceleration is a = (2.65106 m/s)/(14.7109 s) = 1.801014 m/s2 . The electric eldthen has magnitude E = ma/q, orE = (9.111031 kg)(1.801014 m/s2 )/(1.601019 C) = 1.03103 N/C.E26-31 The drop is balanced if the electric force is equal to the force of gravity, or Eq = mg.The mass of the drop is given in terms of the density by4m = V = r3 .3Combining,q=mg4r3 g4(851 kg/m3 )(1.64106 m)3 (9.81 m/s2 )=== 8.111019 C.E3E3(1.92105 N/C)We want the charge in terms of e, so we divide, and getq(8.111019 C)== 5.07 5.e(1.601019 C)E26-32 (b) F = (8.99109 N m2 /C2 )(2.16106 C)(85.3109 C)/(0.117m)2 = 0.121 N.(a) E2 = F/q1 = (0.121 N)/(2.16106 C) = 5.60104 N/C.E1 = F/q2 = (0.121 N)/(85.3109 C) = 1.42106 N/C.E26-33 If each value of q measured by Millikan was a multiple of e, then the dierence betweenany two values of q must also be a multiple of q. The smallest dierence would be the smallestmultiple, and this multiple might be unity. The dierences are 1.641, 1.63, 1.60, 1.63, 3.30, 3.35,3.18, 3.24, all times 1019 C. This is a pretty clear indication that the fundamental charge is on theorder of 1.6 1019 C. If so, the likely number of fundamental charges on each of the drops is shownbelow in a table arranged like the one in the book:45781011121416The total number of charges is 87, while the total charge is 142.69 1019 C, so the average chargeper quanta is 1.64 1019 C.21E26-34 Because of the electric eld the acceleration toward the ground of a charged particle isnot g, but g Eq/m, where the sign depends on the direction of the electric eld.(a) If the lower plate is positively charged then a = g Eq/m. Replace g in the pendulum periodexpression by this, and thenLT = 2.g Eq/m(b) If the lower plate is negatively charged then a = g + Eq/m. Replace g in the pendulumperiod expression by this, and thenT = 2L.g + Eq/mE26-35 The ink drop travels an additional time t = d/vx , where d is the additional horizontaldistance between the plates and the paper. During this time it travels an additional vertical distancey = vy t , where vy = at = 2y/t = 2yvx /L. Combining,y =2yvx t2yd2(6.4104 m)(6.8103 m)=== 5.44104 m,LL(1.6102 m)so the total deection is y + y = 1.18103 m.E26-36 (a) p = (1.48109 C)(6.23106 m) = 9.221015 C m.(b) U = 2pE = 2(9.221015 C m)(1100 N/C) = 2.031011 J.E26-37 Use = pE sin , where is the angle between p and E. For this dipole p = qd = 2edor p = 2(1.6 1019 C)(0.78 109 m) = 2.5 1028 C m. For all three casespE = (2.5 1028 C m)(3.4 106 N/C) = 8.5 1022 N m.The only thing we care about is the angle.(a) For the parallel case = 0, so sin = 0, and = 0.(b) For the perpendicular case = 90 , so sin = 1, and = 8.5 1022 N m..(c) For the anti-parallel case = 180 , so sin = 0, and = 0.E26-38 (c) Equal and opposite, or 5.221016 N.(d) Use Eq. 26-12 and F = Eq. Thenp===E26-394 0 x3 F,q4(8.851012 C2 /N m2 )(0.285m)3 (5.221016 N),(3.16106 C)4.251022 C m.The point-like nucleus contributes an electric eldE+ =1 Ze,4 0 r2while the uniform sphere of negatively charged electron cloud of radius R contributes an electriceld given by Eq. 26-24,1 ZerE =.4 0 R322The net electric eld is just the sum,E=Ze4 01r 3r2RE26-40 The shell theorem rst described for gravitation in chapter 14 is applicable here since bothelectric forces and gravitational forces fall o as 1/r2 . The net positive charge inside the sphere ofradius d/2 is given by Q = 2e(d/2)3 /R3 = ed3 /4R3 .The net force on either electron will be zero wheneQ4e2 d3e2 de2== 2= 3,d2(d/2)2d 4R3Rwhich has solution d = R.P26-1 (a) Let the positive charge be located closer to the point in question, then the electriceld from the positive charge isq1E+ =4 0 (x d/2)2and is directed away from the dipole.The negative charge is located farther from the point in question, soE =1q4 0 (x + d/2)2and is directed toward the dipole.The net electric eld is the sum of these two elds, but since the two component elds point inopposite direction we must actually subtract these values,E= E + E ,1q1q,=4 0 (z d/2)24 0 (z + d/2)2111 q=4 0 z 2 (1 d/2z)2(1 + d/2z)2We can use the binomial expansion on the terms containing 1 d/2z,E=1 q((1 + d/z) (1 d/z)) ,4 0 z 21 qd2 0 z 3(b) The electric eld is directed away from the positive charge when you are closer to the positivecharge; the electric eld is directed toward the negative charge when you are closer to the negativecharge. In short, along the axis the electric eld is directed in the same direction as the dipolemoment.P26-2The key to this problem will be the expansion of11 2(x2 + (z d/2)2 )3/2(x + z 2 )3/22313 zd2 x2 + z 2.for dx2 + z 2 . Far from the charges the electric eld of the positive charge has magnitudeE+ =1q,4 0 x2 + (z d/2)2the components of this areEx,+=q12 + z24 0 xx2xEz,+=q14 0 x2 + z 2x2 + (z d/2)2+ (z d/2)2(z d/2),.Expand both according to the rst sentence, thenEx,+Ez,+=1xq4 0 (x2 + z 2 )3/21 (z d/2)q4 0 (x2 + z 2 )3/23 zd2 x2 + z 23 zd1+2 x2 + z 21+,.Similar expression exist for the negative charge, except we must replace q with q and the + in theparentheses with a , and z d/2 with z + d/2 in the Ez expression. All that is left is to add theexpressions. ThenEx==Ez===1xq3 zd1xq1++4 0 (x2 + z 2 )3/22 x2 + z 24 0 (x2 + z 2 )3/23xqzd1,4 0 (x2 + z 2 )5/23 zd1 (z d/2)q1 (z + d/2)q1++2 + z22 + z 2 )3/24 0 (x2x4 0 (x2 + z 2 )3/213z 2 dq1dq,2 + z 2 )5/24 0 (x4 0 (x2 + z 2 )3/21 (2z 2 x2 )dq.4 0 (x2 + z 2 )5/213 zd2 x2 + z 2,13 zd2 x2 + z 2,P26-3 (a) Each point on the ring is a distance z 2 + R2 from the point on the axis in question.Since all points are equal distant and subtend the same angle from the axis then the top half of thering contributesq1zE1z =,2 + R2 )2 + R24 0 (xzwhile the bottom half contributes a similar expression. Add, andEz =q1 + q2zq=4 0 (z 2 + R2 )3/240(z 2z,+ R2 )3/2which is identical to Eq. 26-18.(b) The perpendicular component would be zero if q1 = q2 . It isnt, so it must be the dierenceq1 q2 which is of interest. Assume this charge dierence is evenly distributed on the top half ofthe ring. If it is a positive dierence, then E must point down. We are only interested then in thevertical component as we integrate around the top half of the ring. ThenE==1 (q1 q2 )/cos d,4 0 z 2 + R20q1 q21.22 + R22 0 z24P26-4 Use the approximation 1/(z d)2 (1/z 2 )(1Add the contributions:E==2d/z + 3d2 /z 2 ).q2qq1 2 +,24 0 (z + d)z(z d)2q2d 3d22d 3d21+ 2 2+1++ 224 0 zzzzz2q 6d3Q=,4 0 z 2 z 24 0 z 4,where Q = 2qd2 .P26-5 A monopole eld falls o as 1/r2 . A dipole eld falls o as 1/r3 , and consists of twooppositely charge monopoles close together. A quadrupole eld (see Exercise 11 above or readProblem 4) falls o as 1/r4 and (can) consist of two otherwise identical dipoles arranged with antiparallel dipole moments. Just taking a leap of faith it seems as if we can construct a 1/r6 eldbehavior by extending the reasoning.First we need an octopole which is constructed from a quadrupole. We want to keep things assimple as possible, so the construction steps are1. The monopole is a charge +q at x = 0.2. The dipole is a charge +q at x = 0 and a charge q at x = a. Well call this a dipole atx = a/23. The quadrupole is the dipole at x = a/2, and a second dipole pointing the other way atx = a/2. The charges are then q at x = a, +2q at x = 0, and q at x = a.4. The octopole will be two stacked, oset quadrupoles. There will be q at x = a, +3q atx = 0, 3q at x = a, and +q at x = 2a.5. Finally, our distribution with a far eld behavior of 1/r6 . There will be +q at x = 2a, 4q atx = a, +6q at x = 0, 4q at x = a, and +q at x = 2a.P26-6 The vertical component of E is simply half of Eq. 26-17. The horizontal component isgiven by a variation of the work required to derive Eq. 26-16,dEz = dE sin =1 dz4 0 y 2 + z 2zy2+ z2,which integrates to zero if the limits are to +, but in this case,Ez =dEz =0 1.4 0 zSince the vertical and horizontal components are equal then E makes an angle of 45 .P26-7 (a) Swap all positive and negative charges in the problem and the electric eld must reversedirection. But this is the same as ipping the problem over; consequently, the electric eld mustpoint parallel to the rod. This only holds true at point P , because point P doesnt move when youip the rod.25(b) We are only interested in the vertical component of the eld as contributed from each pointon the rod. We can integrate only half of the rod and double the answer, so we want to evaluateL/2Ez= 20=1 dz4 0 y 2 + z 2zy2+ z2,(L/2)2 + y 2 y2.4 0 y (L/2)2 + y 2(c) The previous expression is exact. If yexpansion toEz =L, then the expression simplies with a Taylor L2,4 0 y 3which looks similar to a dipole.P26-8EvaluateRE=01z dq,4 0 (z 2 + r2 )3/2where r is the radius of the ring, z the distance to the plane of the ring, and dq the dierentialcharge on the ring. But r2 + z 2 = R2 , and dq = (2r dr), where = q/2R2 . ThenRqR2 r2 r drE =,R50 4 0q1=.4 0 3R2P26-9 The key statement is the second italicized paragraph on page 595; the number of eldlines through a unit cross-sectional area is proportional to the electric eld strength. If the exponentis n, then the electric eld strength a distance r from a point charge isE=kq,rnand the total cross sectional area at a distance r is the area of a spherical shell, 4r2 . Then thenumber of eld lines through the shell is proportional toEA =kq4r2 = 4kqr2n .rnNote that the number of eld lines varies with r if n = 2. This means that as we go farther fromthe point charge we need more and more eld lines (or fewer and fewer). But the eld lines can onlystart on charges, and we dont have any except for the point charge. We have a problem; we reallydo need n = 2 if we want workable eld lines.P26-10 The distance traveled by the electron will be d1 = a1 t2 /2; the distance traveled by theproton will be d2 = a2 t2 /2. a1 and a2 are related by m1 a1 = m2 a2 , since the electric force is thesame (same charge magnitude). Then d1 + d2 = (a1 + a2 )t2 /2 is the 5.00 cm distance. Divide bythe proton distance, and thend1 + d2a1 + a2m2==+ 1.d2a2m1Thend2 = (5.00102 m)/(1.671027 /9.111031 + 1) = 2.73105 m.26P26-11 This is merely a fancy projectile motion problem. vx = v0 cos while vy,0 = v0 sin . Thex and y positions are x = vx t andy=1 2ax2at + vy,0 t = 2+ x tan .22v0 cos2 The acceleration of the electron is vertically down and has a magnitude ofa=FEq(1870 N/C)(1.61019 C)=== 3.2841014 m/s2 .mm(9.111031 kg)We want to nd out how the vertical velocity of the electron at the location of the top plate. Ifwe get an imaginary answer, then the electron doesnt get as high as the top plate.vy=vy,0 2 + 2ay,=(5.83106 m/s)2 sin(39 )2 + 2(3.2841014 m/s2 )(1.97102 m),= 7.226105 m/s.This is a real answer, so this means the electron either hits the top plate, or it misses both plates.The time taken to reach the height of the top plate ist=vy(7.226105 m/s) (5.83106 m/s) sin(39 )== 8.972109 s.a(3.2841014 m/s2 )In this time the electron has moved a horizontal distance ofx = (5.83106 m/s) cos(39 )(8.972109 s) = 4.065102 m.This is clearly on the upper plate.P26-12Near the center of the ring zR, so a Taylor expansion yieldsE= z.2 0 R2The force on the electron is F = Ee, so the eective spring constant is k = e/2 0 R2 . This means=P26-13k=me=2 0 mR2eq.4 0 mR3U = pE cos , so the work required to ip the dipole isW = pE [cos(0 + ) cos 0 ] = 2pE cos 0 .P26-14 If the torque on a system is given by | | = , where is a constant, then the frequencyof oscillation of the system is f = /I/2. In this case = pE sin pE, sof=pE/I/2.27P26-15 Use the a variation of the exact result from Problem 26-1. The two charge are positive,but since we will eventually focus on the area between the charges we must subtract the two eldcontributions, since they point in opposite directions. ThenEz =q411(z a/2)2(z + a/2)20and then take the derivative,dEzq=dz211(z a/2)3(z + a/2)30Applying the binomial expansion for points zdEzdz.a,8q 111332 0 a(2z/a 1)(2z/a + 1)38q 1 ((1 + 6z/a) (1 6z/a)) ,2 0 a38q 1=. 0 a3= ,There were some fancy sign ips in the second line, so review those steps carefully!(b) The electrostatic force on a dipole is the dierence in the magnitudes of the electrostaticforces on the two charges that make up the dipole. Near the center of the above charge arrangementthe electric eld behaves asEz Ez (0) +dEzdzz + higher ordered terms.z=0The net force on a dipole isF+ F = q(E+ E ) = q Ez (0) +dEzdzz+ Ez (0) z=0dEzdzzz=0where the + and - subscripts refer to the locations of the positive and negative charges. Thislast line can be simplied to yieldqdEzdz(z+ z ) = qdz=028dEzdz.z=0E27-1 E = (1800 N/C)(3.2103 m)2 cos(145 ) = 7.8103 N m2 /C.E27-2 The right face has an area element given by A = (1.4 m)2j.2 = 0.(a) E = A E = (2.0 m )j (6 N/C)i(b) E = (2.0 m2 ) (2 N/C) = 4N m2 /C.jj(c) E = (2.0 m2 ) [(3 N/C) + (4 N/C)k] = 0.ji(d) In each case the eld is uniform so we can simply evaluate E = E A, where A has sixparts, one for every face. The faces, however, have the same size but are organized in pairs withopposite directions. These will cancel, so the total ux is zero in all three cases.E27-3(a) The at base is easy enough, since according to Eq. 27-7,E =E dA.There are two important facts to consider in order to integrate this expression. E is parallel to theaxis of the hemisphere, E points inward while dA points outward on the at base. E is uniform, soit is everywhere the same on the at base. Since E is anti-parallel to dA, E dA = E dA, thenE dA = E =E dA.Since E is uniform we can simplify this asE = dA = EA = R2 E.E dA = EThe last steps are just substituting the area of a circle for the at side of the hemisphere.(b) We must rst sort out the dot productEdARWe can simplify the vector part of the problem with E dA = cos E dA, soE =E dA =cos E dAOnce again, E is uniform, so we can take it out of the integral,E =cos E dA = Ecos dAFinally, dA = (R d)(R sin d) on the surface of a sphere centered on R = 0.29Well integrate around the axis, from 0 to 2. Well integrate from the axis to the equator,from 0 to /2. Then2E = E/2R2 cos sin d d.cos dA = E00Pulling out the constants, doing the integration, and then writing 2 cos sin as sin(2),/2/2cos sin d = R2 EE = 2R2 E0sin(2) d,0Change variables and let = 2, then we haveE = R2 E01sin d = R2 E.2E27-4 Through S1 , E = q/ 0 . Through S2 , E = q/ 0 . Through S3 , E = q/ 0 . Through S4 ,E = 0. Through S5 , E = q/ 0 .E27-5By Eq. 27-8,E =q0=(1.84 C)= 2.08105 N m2 /C.(8.851012 C2 /N m2 )E27-6 The total ux through the sphere isE = (1 + 2 3 + 4 5 + 6)(103 N m2 /C) = 3103 N m2 /C.The charge inside the die is (8.851012 C2 /N m2 )(3103 N m2 /C) = 2.66108 C.E27-7 The total ux through a cube would be q/ 0 . Since the charge is in the center of the cubewe expect that the ux through any side would be the same, or 1/6 of the total ux. Hence the uxthrough the square surface is q/6 0 .E27-8 If the electric eld is uniform then there are no free charges near (or inside) the net. Theux through the netting must be equal to, but opposite in sign, from the ux through the opening.The ux through the opening is Ea2 , so the ux through the netting is Ea2 .E27-9 There is no ux through the sides of the cube. The ux through the top of the cube is(58 N/C)(100 m)2 = 5.8105 N m2 /C. The ux through the bottom of the cube is(110 N/C)(100 m)2 = 1.1106 N m2 /C.The total ux is the sum, so the charge contained in the cube isq = (8.851012 C2 /N m2 )(5.2105 N m2 /C) = 4.60106 C.E27-10 (a) There is only a ux through the right and left faces. Through the right faceR = (2.0 m2 ) (3 N/C m)(1.4 m) = 8.4 N m2 /C.jjThe ux through the left face is zero because y = 0.30E27-11 There are eight cubes which can be wrapped around the charge. Each cube has threeexternal faces that are indistinguishable for a total of twenty-four faces, each with the same uxE . The total ux is q/ 0 , so the ux through one face is E = q/24 0 . Note that this is the uxthrough faces opposite the charge; for faces which touch the charge the electric eld is parallel tothe surface, so the ux would be zero.E27-12 Use Eq. 27-11, = 2 0 rE = 2(8.851012 C2 /N m2 )(1.96 m)(4.52104 N/C) = 4.93106 C/m.E27-13 (a) q = A = (2.0106 C/m2 )(0.12 m)(0.42 m) = 3.17107 C.(b) The charge density will be the same! q = A = (2.0 106 C/m2 )(0.08 m)(0.28 m) =1.41107 C.E27-14 The electric eld from the sheet on the left is of magnitude E l = /2 0 , and points directlyaway from the sheet. The magnitude of the electric eld from the sheet on the right is the same,but it points directly away from the sheet on the right.(a) To the left of the sheets the two elds add since they point in the same direction. This meansthat the electric eld is E = (/ 0 )i.(b) Between the sheets the two electric elds cancel, so E = 0.(c) To the right of the sheets the two elds add since they point in the same direction. Thismeans that the electric eld is E = (/ 0 )i.E27-15 The electric eld from the plate on the left is of magnitude E l = /2 0 , and points directlytoward the plate. The magnitude of the electric eld from the plate on the right is the same, but itpoints directly away from the plate on the right.(a) To the left of the plates the two elds cancel since they point in the opposite directions. Thismeans that the electric eld is E = 0.(b) Between the plates the two electric elds add since they point in the same direction. Thismeans that the electric eld is E = (/ 0 )i.(c) To the right of the plates the two elds cancel since they point in the opposite directions.This means that the electric eld is E = 0.E27-16 The magnitude of the electric eld is E = mg/q. The surface charge density on the platesis = 0 E = 0 mg/q, or=(8.851012 C2 /N m2 )(9.111031 kg)(9.81 m/s2 )= 4.941022 C/m2 .(1.601019 C)E27-17 We dont really need to write an integral, we just need the charge per unit length in thecylinder to be equal to zero. This means that the positive charge in cylinder must be +3.60nC/m.This positive charge is uniformly distributed in a circle of radius R = 1.50 cm, so=3.60nC/m3.60nC/m== 5.09C/m3 .R2(0.0150 m)231E27-18 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.The E eld will be perpendicular to the surface, so Gauss law will simplify toq enc /0=E dA =E dA = EdA = 4r2 E.(a) For point P1 the charge enclosed is q enc = 1.26107 C, soE=(1.26107 C)= 3.38106 N/C. m2 )(1.83102 m)24(8.851012 C2 /N(b) Inside a conductor E = 0.E27-19 The proton orbits with a speed v, so the centripetal force on the proton is FC = mv 2 /r.This centripetal force is from the electrostatic attraction with the sphere; so long as the proton isoutside the sphere the electric eld is equivalent to that of a point charge Q (Eq. 27-15),E=1 Q.4 0 r2If q is the charge on the proton we can write F = Eq, or1 Qmv 2=qr4 0 r2Solving for Q,Q ==4 0 mv 2 r,q4(8.851012 C2 /N m2 )(1.671027 kg)(294103 m/s)2 (0.0113 m),(1.601019 C)= 1.13109 C.E27-20 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.The E eld will be perpendicular to the surface, so Gauss law will simplify toq enc /0=E dA =E dA = EdA = 4r2 E.(a) At r = 0.120 m q enc = 4.06108 C. ThenE=(4.06108 C)= 2.54104 N/C. m2 )(1.20101 m)24(8.851012 C2 /N(b) At r = 0.220 m q enc = 5.99108 C. ThenE=(5.99108 C)= 1.11104 N/C. m2 )(2.20101 m)24(8.851012 C2 /N(c) At r = 0.0818 m q enc = 0 C. Then E = 0.32E27-21 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E eld will be perpendicular to the curved surface and parallel to the end surfaces, soGauss law will simplify toq enc /0E dA ==E dA = EdA = 2rLE,where L is the length of the cylinder. Note that = q/2rL represents a surface charge density.(a) r = 0.0410 m is between the two cylinders. ThenE=(24.1106 C/m2 )(0.0322 m)= 2.14106 N/C.(8.851012 C2 /N m2 )(0.0410 m)It points outward.(b) r = 0.0820 m is outside the two cylinders. ThenE=(24.1106 C/m2 )(0.0322 m) + (18.0106 C/m2 )(0.0618 m)= 4.64105 N/C.(8.851012 C2 /N m2 )(0.0820 m)The negative sign is because it is pointing inward.E27-22 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E eld will be perpendicular to the curved surface and parallel to the end surfaces, soGauss law will simplify toq enc /0E dA ==E dA = EdA = 2rLE,where L is the length of the cylinder. The charge enclosed isq enc =dV = L r2 R2The electric eld is given byE= r 2 R2L r2 R2=.2 0 rL2 0rAt the surface,Es = (2R)2 R23R=.2 0 2R4 0Solve for r when E is half of this:3Rr 2 R2=,82r3rR = 4r2 4R2 ,0 = 4r2 3rR 4R2 .The solution is r = 1.443R. Thats (2R 1.443R) = 0.557R beneath the surface.E27-23 The electric eld must do work on the electron to stop it. The electric eld is given byE = /2 0 . The work done is W = F d = Eqd. d is the distance in question, sod=2 0K2(8.851012 C2 /N m2 )(1.15105 eV)== 0.979 mq(2.08106 C/m2 )e33E27-24 Let the spherical Gaussian surface have a radius of R and be centered on the origin.Choose the orientation of the axis so that the innite line of charge is along the z axis. The electriceld is then directed radially outward from the z axis with magnitude E = /2 0 , where is theperpendicular distance from the z axis. Now we want to evaluateE dA,E =over the surface of the sphere. In spherical coordinates, dA = R2 sin d d, = R sin , andE dA = EA sin . Then2Rsin R d d =.E =2 00E27-25 (a) The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E eld will be perpendicular to the curved surface and parallel to the end surfaces, soGauss law will simplify toq enc /0=E dA =E dA = EdA = 2rLE,where L is the length of the cylinder. Now for the q enc part. If the (uniform) volume charge densityis , then the charge enclosed in the Gaussian cylinder isq enc =dV = V = r2 L.dV = Combining, r2 L/ 0 = E2rL or E = r/2 0 .(b) Outside the charged cylinder the charge enclosed in the Gaussian surface is just the chargein the cylinder. Thenq enc =dV = V = R2 L.dV = andR2 L/0= E2rL,and then nallyE=R2 .2 0rE27-26 (a) q = 4(1.22 m)2 (8.13106 C/m2 ) = 1.52104 C.(b) E = q/ 0 = (1.52104 C)/(8.851012 C2 /N m2 ) = 1.72107 N m2 /C.(c) E = / 0 = (8.13106 C/m2 )/(8.851012 C2 /N m2 ) = 9.19105 N/CE27-27 (a) = (2.4106 C)/4(0.65 m)2 = 4.52107 C/m2 .(b) E = / 0 = (4.52107 C/m2 )/(8.851012 C2 /N m2 ) = 5.11104 N/C.E27-28 E = /E27-290= q/4r2 0 .(a) The near eld is given by Eq. 27-12, E = /2 0 , soE(6.0106 C)/(8.0102 m)2= 5.3107 N/C.2(8.851012 C2 /N m2 )(b) Very far from any object a point charge approximation is valid. ThenE=1 q1(6.0106 C)== 60N/C.4 0 r24(8.851012 C2 /N m2 ) (30 m)234P27-1For a spherically symmetric mass distribution choose a spherical Gaussian shell. Theng dA =Theng dA = gdA = 4r2 g.ggr2== m,4GGorGm.r2The negative sign indicates the direction; g point toward the mass center.g=P27-2 (a) The ux through all surfaces except the right and left faces will be zero. Through theleft face,l = Ey A = b aa2 .Through the right face,r = Ey A = b 2aa2 .The net ux is then = ba5/2 ( 2 1) = (8830 N/C m1/2 )(0.130 m)5/2 ( 2 1) = 22.3 N m2 /C.(b) The charge enclosed is q = (8.851012 C2 /N m2 )(22.3 N m2 /C) = 1.971010 C.P27-3 The net force on the small sphere is zero; this force is the vector sum of the force of gravityW , the electric force FE , and the tension T .TFEWThese forces are related by Eq = mg tan . We also have E = /2 0 , so===2 0 mg tan ,q2(8.851012 C2 /N m2 )(1.12106 kg)(9.81 m/s2 ) tan(27.4 ),(19.7109 C)5.11109 C/m2 .35P27-4 The materials are conducting, so all charge will reside on the surfaces. The electric eldinside any conductor is zero. The problem has spherical symmetry, so use a Gaussian surface whichis a spherical shell. The E eld will be perpendicular to the surface, so Gauss law will simplify toq enc /0=E dA =E dA = EdA = 4r2 E.Consequently, E = q enc /4 0 r2 .(a) Within the sphere E = 0.(b) Between the sphere and the shell q enc = q. Then E = q/4 0 r2 .(c) Within the shell E = 0.(d) Outside the shell q enc = +q q = 0. Then E = 0.(e) Since E = 0 inside the shell, q enc = 0, this requires that q reside on the inside surface. Thisis no charge on the outside surface.P27-5 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E eld will be perpendicular to the curved surface and parallel to the end surfaces, soGauss law will simplify toq enc /0=E dA =E dA = EdA = 2rLE,where L is the length of the cylinder. Consequently, E = q enc /2 0 rL.(a) Outside the conducting shell q enc = +q 2q = q. Then E = q/2 0 rL. The negative signindicates that the eld is pointing inward toward the axis of the cylinder.(b) Since E = 0 inside the conducting shell, q enc = 0, which means a charge of q is on theinside surface of the shell. The remaining q must reside on the outside surface of the shell.(c) In the region between the cylinders q enc = +q. Then E = +q/2 0 rL. The positive signindicates that the eld is pointing outward from the axis of the cylinder.P27-6Subtract Eq. 26-19 from Eq. 26-20. ThenzE=.2 0 z 2 + R2P27-7 This problem is closely related to Ex. 27-25, except for the part concerning q enc . Wellset up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered onthe axis of the physical cylinder. For Gaussian surfaces of radius r < R, there is no charge enclosedwhile for Gaussian surfaces of radius r > R, q enc = l.Weve already worked out the integralE dA = 2rlE,tubefor the cylinder, and then from Gauss law,q enc =E dA = 2 0 rlE.0tube(a) When r < R there is no enclosed charge, so the left hand vanishes and consequently E = 0inside the physical cylinder.(b) When r > R there is a charge l enclosed, soE=.2 0 r36P27-8 This problem is closely related to Ex. 27-25, except for the part concerning q enc . Well setup the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on theaxis of the physical cylinders. For Gaussian surfaces of radius r < a, there is no charge enclosedwhile for Gaussian surfaces of radius b > r > a, q enc = l.Weve already worked out the integralE dA = 2rlE,tubefor the cylinder, and then from Gauss law,q enc =E dA = 2 0 rlE.0tube(a) When r < a there is no enclosed charge, so the left hand vanishes and consequently E = 0inside the inner cylinder.(b) When b > r > a there is a charge l enclosed, soE=.2 0 rP27-9 Uniform circular orbits require a constant net force towards the center, so F = Eq =q/2 0 r. The speed of the positron is given by F = mv 2 /r; the kinetic energy is K = mv 2 /2 =F r/2. Combining,K===P27-10q,4 0(30109 C/m)(1.61019 C),4((8.85 1012 C2 /N m2 )4.311017 J = 270 eV. = 2 0 rE, soq = 2(8.851012 C2 /N m2 )(0.014 m)(0.16 m)(2.9104 N/C) = 3.6109 C.P27-11 (a) Put a spherical Gaussian surface inside the shell centered on the point charge. Gausslaw statesq encE dA =.0Since there is spherical symmetry the electric eld is normal to the spherical Gaussian surface, andit is everywhere the same on this surface. The dot product simplies to E dA = E dA, while sinceE is a constant on the surface we can pull it out of the integral, and we end up withEdA =q,0where q is the point charge in the center. Now dA = 4r2 , where r is the radius of the Gaussiansurface, soqE=.4 0 r2(b) Repeat the above steps, except put the Gaussian surface outside the conducting shell. Keepit centered on the charge. Two things are dierent from the above derivation: (1) r is bigger, and37(2) there is an uncharged spherical conducting shell inside the Gaussian surface. Neither change willaect the surface integral or q enc , so the electric eld outside the shell is stillE=q,4 0 r2(c) This is a subtle question. With all the symmetry here it appears as if the shell has no eect;the eld just looks like a point charge eld. If, however, the charge were moved o center the eldinside the shell would become distorted, and we wouldnt be able to use Gauss law to nd it. Sothe shell does make a dierence.Outside the shell, however, we cant tell what is going on inside the shell. So the electric eldoutside the shell looks like a point charge eld originating from the center of the shell regardless ofwhere inside the shell the point charge is placed!(d) Yes, q induces surface charges on the shell. There will be a charge q on the inside surfaceand a charge q on the outside surface.(e) Yes, as there is an electric eld from the shell, isnt there?(f) No, as the electric eld from the outside charge wont make it through a conducting shell.The conductor acts as a shield.(g) No, this is not a contradiction, because the outside charge never experienced any electrostaticattraction or repulsion from the inside charge. The force is between the shell and the outside charge.P27-12ThenThe repulsive electrostatic forces must exactly balance the attractive gravitational forces.1 q2m2=G 2 ,4 0 r2ror m = q/ 4 0 G. Numerically,(1.601019 C)m== 1.86109 kg.24(8.851012 C2 /N m2 )(6.671011 N m2 /kg )P27-13 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.The E eld will be perpendicular to the surface, so Gauss law will simplify toq enc /0=E dA =E dA = EdA = 4r2 E.Consequently, E = q enc /4 0 r2 .q enc = q + 4 r2 dr, orrAr dr = q + 2A(r2 a2 ).q enc = q + 4aThe electric eld will be constant if q enc behaves as r2 , which requires q = 2Aa2 , or A = q/2a2 .P27-14 (a) The problem has spherical symmetry, so use a Gaussian surface which is a sphericalshell. The E eld will be perpendicular to the surface, so Gauss law will simplify toq enc /0=E dA =E dA = EConsequently, E = q enc /4 0 r2 .q enc = 4 r2 dr = 4r3 /3, soE = r/3380dA = 4r2 E.and is directed radially out from the center. Then E = r/3 0 .(b) The electric eld in the hole is given by Eh = E Eb , where E is the eld from part (a) andEb is the eld that would be produced by the matter that would have been in the hole had the holenot been there. ThenEb = b/3 0 ,where b is a vector pointing from the center of the hole. ThenEh =br=(r b).3 03 03 0But r b = a, so Eh = a/3 0 .P27-15 If a point is an equilibrium point then the electric eld at that point should be zero.If it is a stable point then moving the test charge (assumed positive) a small distance from theequilibrium point should result in a restoring force directed back toward the equilibrium point. Inother words, there will be a point where the electric eld is zero, and around this point there will bean electric eld pointing inward. Applying Gauss law to a small surface surrounding our point P ,we have a net inward ux, so there must be a negative charge inside the surface. But there shouldbe nothing inside the surface except an empty point P , so we have a contradiction.P27-16 (a) Follow the example on Page 618. By symmetry E = 0 along the median plane. Thecharge enclosed between the median plane and a surface a distance x from the plane is q = Ax.ThenE = Ax/ 0 A = A/ 0 .(b) Outside the slab the charge enclosed between the median plane and a surface a distance xfrom the plane is is q = Ad/2, regardless of x. TheE = Ad/2/ 0 A = d/2 0 .P27-17(a) The total charge is the volume integral over the whole sphere,Q= dV.This is actually a three dimensional integral, and dV = A dr, where A = 4r2 . ThenQ = dV,RS r4r2 dr,R04S 1 4=R ,R 4= S R3 .=(b) Put a spherical Gaussian surface inside the sphere centered on the center. We can use Gausslaw here because there is spherical symmetry in the entire problem, both inside and outside theGaussian surface. Gauss law statesq encE dA =.039Since there is spherical symmetry the electric eld is normal to the spherical Gaussian surface, andit is everywhere the same on this surface. The dot product simplies to E dA = E dA, while sinceE is a constant on the surface we can pull it out of the integral, and we end up withEdA =q enc,0NowdA = 4r2 , where r is the radius of the Gaussian surface, soq enc.4 0 r2E=We arent done yet, because the charge enclosed depends on the radius of the Gaussian surface. Weneed to do part (a) again, except this time we dont want to do the whole volume of the sphere, weonly want to go out as far as the Gaussian surface. Thenq enc= dV,rS r4r2 dr,R04S 1 4=r ,R 4r4= S .R=Combine these last two results andE===S r4,4 0 r2 RS r2,4 0 RQ r2.4 0 R4In the last line we used the results of part (a) to eliminate S from the expression.P27-18 (a) Inside the conductor E = 0, so a Gaussian surface which is embedded in the conductorbut containing the hole must have a net enclosed charge of zero. The cavity wall must then have acharge of 3.0 C.(b) The net charge on the conductor is +10.0 C; the charge on the outer surface must then be+13.0 C.P27-19 (a) Inside the shell E = 0, so the net charge inside a Gaussian surface embedded in theshell must be zero, so the inside surface has a charge Q.(b) Still Q; the outside has nothing to do with the inside.(c) (Q + q); see reason (a).(d) Yes.40Throughout this chapter we will use the convention that V () = 0 unless explicitly statedotherwise. Then the potential in the vicinity of a point charge will be given by Eq. 28-18,V = q/4 0 r.E28-1 (a) Let U12 be the potential energy of the interaction between the two up quarks. ThenU12 = (8.99109 N m2 /C2 )(2/3)2 e(1.601019 C)= 4.84105 eV.(1.321015 m)(b) Let U13 be the potential energy of the interaction between an up quark and a downquark. ThenU13 = (8.99109 N m2 /C2 )(1/3)(2/3)e(1.601019 C)= 2.42105 eV(1.321015 m)Note that U13 = U23 . The total electric potential energy is the sum of these three terms, or zero.E28-2 There are six interaction terms, one for every charge pair. Number the charges clockwisefrom the upper left hand corner. ThenU12U23U34U41U13U24====q 2 /4q 2 /4q 2 /4q 2 /40 a,0 a,0 a,0 a,= (q) /4 0 ( 2a),= q 2 /4 0 ( 2a).2Add these terms and get2 42The amount of work required is W = U .U=q2q2= 0.2064 0 a0aE28-3 (a) We build the electron one part at a time; each part has a charge q = e/3. Moving therst part from innity to the location where we want to construct the electron is easy and takes nowork at all. Moving the second part in requires work to change the potential energy toU12 =1 q1 q2,4 0 rwhich is basically Eq. 28-7. The separation r = 2.82 1015 m.Bringing in the third part requires work against the force of repulsion between the third chargeand both of the other two charges. Potential energy then exists in the form U13 and U23 , where allthree charges are the same, and all three separations are the same. Then U12 = U13 = U12 , so thetotal potential energy of the system isU =31 (e/3)23(1.601019 C/3)2== 2.721014 J4 0 r4(8.851012 C2 /N m2 ) (2.821015 m)(b) Dividing our answer by the speed of light squared to nd the mass,m=2.72 1014 J= 3.02 1031 kg.(3.00 108 m/s)241E28-4 There are three interaction terms, one for every charge pair. Number the charges from theleft; let a = 0.146 m. ThenU12=U13=U23=(25.5109 C)(17.2109 C),4 0 a(25.5109 C)(19.2109 C),4 0 (a + x)(17.2109 C)(19.2109 C).4 0 xAdd these and set it equal to zero. Then(25.5)(17.2)(25.5)(19.2) (17.2)(19.2)=+,aa+xxwhich has solution x = 1.405a = 0.205 m.E28-5 The volume of the nuclear material is 4a3 /3, where a = 8.01015 m. Upon dividing inhalf each part will have a radius r where 4r3 /3 = 4a3 /6. Consequently, r = a/ 3 2 = 6.351015 m.Each fragment will have a charge of +46e.(a) The force of repulsion isF =(46)2 (1.601019 C)2= 3000 N4(8.851012 C2 /N m2 )[2(6.351015 m)]2(b) The potential energy isU=(46)2 e(1.601019 C)= 2.4108 eV4(8.851012 C2 /N m2 )2(6.351015 m)E28-6 This is a work/kinetic energy problem:v0 =122 mv0= qV . Then2(1.601019 C)(10.3103 V)= 6.0107 m/s.(9.111031 kg)E28-7 (a) The energy released is equal to the charges times the potential through which thecharge was moved. ThenU = qV = (30 C)(1.0 109 V) = 3.0 1010 J.(b) Although the problem mentions acceleration, we want to focus on energy. The energy willchange the kinetic energy of the car from 0 to K f = 3.0 1010 J. The speed of the car is thenv=2K=m2(3.0 1010 J)= 7100 m/s.(1200 kg)(c) The energy required to melt ice is given by Q = mL, where L is the latent heat of fusion.ThenQ(3.0 1010 J)m=== 90, 100kg.L(3.33105 J/kg)42E28-8 (a) U = (1.601019 C)(1.23109 V) = 1.971010 J.(b) U = e(1.23109 V) = 1.23109 eV.122 mvE28-9 This is an energy conservation problem:bining,=q22 0 m=v(3.1106 C)22(8.851012 C2 /N m2 )(18106 kg)== qV ; V = q/4 0 (1/r1 1/r2 ). Com-11,r1r211,(0.90103 m) (2.5103 m)2600 m/s.E28-10 This is an energy conservation problem:1q21m(2v)2 = mv 2 .24 0 r2Rearrange,=q2,6 0 mv 2=r(1.601019 C)2= 1.42109 m.6(8.851012 C2 /N m2 )(9.111031 kg)(3.44105 m/s)2 )E28-11 (a) V = (1.601019 C)/4(8.851012 C2 /N m2 )(5.291011 m) = 27.2 V.(b) U = qV = (e)(27.2 V) = 27.2 eV.(c) For uniform circular orbits F = mv 2 /r; the force is electrical, or F = e2 /4 0 r2 . Kineticenergy is K = mv 2 /2 = F r/2, soK=e2(1.601019 C)== 13.6 eV.12 C2 /N m2 )(5.291011 m)8 0 r8(8.8510(d) The ionization energy is (K + U ), orE ion = [(13.6 eV) + (27.2 eV)] = 13.6 eV.E28-12 (a) The electric potential at A isVA =14 0q1q2+r1r2= (8.99109 N m2 /C)(5.0106 C) (2.0106 C)+(0.15 m)(0.05 m)= 6.0104 V.The electric potential at B isVB =14 0q1q2+r2r1= (8.99109 N m2 /C)(5.0106 C) (2.0106 C)+(0.05 m)(0.15 m)= 7.8105 V.(b) W = qV = (3.0106 C)(6.0104 V 7.8105 V) = 2.5 J.(c) Since work is positive then external work is converted to electrostatic potential energy.43E28-13(a) The magnitude of the electric eld would be found fromE=F(3.90 1015 N)== 2.44 104 N/C.q(1.60 1019 C)(b) The potential dierence between the plates is found by evaluating Eq. 28-15,bV = E ds.aThe electric eld between two parallel plates is uniform and perpendicular to the plates. ThenE ds = E ds along this path, and since E is uniform,bV = bE ds = abE ds = Eads = Ex,awhere x is the separation between the plates. Finally, V = (2.44 104 N/C)(0.120 m) = 2930 V.E28-14 V = Ex, sox =2 02(8.851012 C2 /N m2 )V =(48 V) = 7.1103 m(0.12106 C/m2 )E28-15 The electric eld around an innitely long straight wire is given by E = /2 0 r. Thepotential dierence between the inner wire and the outer cylinder is given bybV = (/2 0 r) dr = (/2 0 ) ln(a/b).aThe electric eld near the surface of the wire is then given byE=V(855 V)= 1.32108 V/m.==2 0 aa ln(a/b)(6.70107 m) ln(6.70107 m/1.05102 m)The electric eld near the surface of the cylinder is then given byE=V(855 V)=== 8.43103 V/m.2 m) ln(6.70107 m/1.05102 m)2 0 aa ln(a/b)(1.0510E28-16 V = Ex = (1.92105 N/C)(1.50102 m) = 2.88103 V.E28-17 (a) This is an energy conservation problem:K=1 (2)(79)e2(2)(79)e(1.601019 C)= (8.99109 N m2 /C)= 3.2107 eV4 0r(7.01015 m)(b) The alpha particles used by Rutherford never came close to hitting the gold nuclei.E28-18 This is an energy conservation problem: mv 2 /2 = eq/4 0 r, orv=(1.601019 C)(1.761015 C)= 2.13104 m/s m2 )(1.22102 m)(9.111031 kg)2(8.851012 C2 /N44E28-19(a) We evaluate VA and VB individually, and then nd the dierence.VA =1 q1(1.16C)== 5060 V,12 C2 /N m2 ) (2.06 m)4 0 r4(8.85 10VB =1 q1(1.16C)== 8910 V,4 0 r4(8.85 1012 C2 /N m2 ) (1.17 m)andThe dierence is then VA VB = 3850 V.(b) The answer is the same, since when concerning ourselves with electric potential we only careabout distances, and not directions.E28-20 The number of excess electrons on each grain isn=4 0 rV4(8.851012 C2 /N m)(1.0106 m)(400 V)== 2.8105e(1.601019 C)E28-21 The excess charge on the shuttle isq = 4 0 rV = 4(8.851012 C2 /N m)(10 m)(1.0 V) = 1.1109 CE28-22 q = 1.37105 C, soV = (8.99109 N m2 /C2 )E28-23(1.37105 C)= 1.93108 V.(6.37106 m)The ratio of the electric potential to the electric eld strength isV=E1 q4 0 r/1 q4 0 r2= r.In this problem r is the radius of the Earth, so at the surface of the Earth the potential isV = Er = (100 V/m)(6.38106 m) = 6.38108 V.E28-24 Use Eq. 28-22:V = (8.99109 N m2 /C2 )(1.47)(3.341030 C m)= 1.63105 V.(52.0109 m)2E28-25 (a) When nding VA we need to consider the contribution from both the positive andthe negative charge, so1qVA =qa +4 0a+dThere will be a similar expression for VB ,VB =14 0qa +45qa+d.Now to evaluate the dierence.V A VB====q11qa +4 0a+d4 0q11,2 0 a a + dqa+da2 0 a(a + d) a(a + d)qd.2 0 a(a + d)qa+dqa +,,(b) Does it do what we expect when d = 0? I expect it the dierence to go to zero as thetwo points A and B get closer together. The numerator will go to zero as d gets smaller. Thedenominator, however, stays nite, which is a good thing. So yes, Va VB 0 as d 0.E28-26 (a) Since both charges are positive the electric potential from both charges will be positive.There will be no nite points where V = 0, since two positives cant add to zero.(b) Between the charges the electric eld from each charge points toward theother, so E willvanish when q/x2 = 2q/(d x)2 . This happens when d x = 2x, or x = d/(1 + 2).E28-27 The distance from C to either charge is 2d/2 = 1.39102 m.(a) V at C is2(2.13106 C)V = (8.99109 N m2 /C2 )= 2.76106 V(1.39102 m)(b) W = qV = (1.91106 C)(2.76106 V) = 5.27 J.(c) Dont forget about the potential energy of the original two charges!U0 = (8.99109 N m2 /C2 )(2.13106 C)2= 2.08 J(1.96102 m)Add this to the answer from part (b) to get 7.35 J.E28-28 The potential is given by Eq. 28-32; at the surface V s = R/2 0 , half of this occurs whenR2 + z 2 z = R/2,R2 + z 2 = R2 /4 + Rz + z 2 ,3R/4 = z.E28-29We can nd the linear charge density by dividing the charge by the circumference,Q,2Rwhere Q refers to the charge on the ring. The work done to move a charge q from a point x to theorigin will be given by=WW= qV,= q (V (0) V (x)) ,1Q1Q= q2 + x24 0 R4 0 R2qQ11.=2 + x24 0 RR46,Putting in the numbers,(5.931012 C)(9.12109 C)4(8.851012 C2 /N m2 )11.48m1(1.48m)2 + (3.07m)2= 1.861010 J.E28-30 (a) The electric eld strength is greatest where the gradient of V is greatest. That isbetween d and e.(b) The least absolute value occurs where the gradient is zero, which is between b and c andagain between e and f .E28-31 The potential on the positive plate is 2(5.52 V) = 11.0 V; the electric eld between theplates is E = (11.0 V)/(1.48102 m) = 743 V/m.E28-32 Take the derivative: E = V /z.E28-33eld,The radial potential gradient is just the magnitude of the radial component of the electricEr = VrThenVr= =1 q,4 0 r279(1.60 1019 C),4(8.85 1012 C2 /N m2 ) (7.0 1015 m)21= 2.321021 V/m.E28-34 Evaluate V /r, andE=Ze4 01r+2 32r2R.E28-35 Ex = V /x = 2(1530 V/m2 )x. At the point in question, E = 2(1530 V/m2 )(1.28102 m) = 39.2 V/m.E28-36 Draw the wires so that they are perpendicular to the plane of the page; they will thencome out of the page. The equipotential surfaces are then lines where they intersect the page,and they look like47E28-37 (a) |VB VA | = |W/q| = |(3.94 1019 J)/(1.60 1019 C)| = 2.46 V. The electric elddid work on the electron, so the electron was moving from a region of low potential to a region ofhigh potential; or VB > VA . Consequently, VB VA = 2.46 V.(b) VC is at the same potential as VB (both points are on the same equipotential line), soVC VA = VB VA = 2.46 V.(c) VC is at the same potential as VB (both points are on the same equipotential line), soVC VB = 0 V.E28-38 (a) For point charges r = q/4 0 V , sor = (8.99109 N m2 /C2 )(1.5108 C)/(30 V) = 4.5 m(b) No, since V 1/r.E28-39 The dotted lines are equipotential lines, the solid arrows are electric eld lines. Note thatthere are twice as many electric eld lines from the larger charge!48E28-40 The dotted lines are equipotential lines, the solid arrows are electric eld lines.E28-41 This can easily be done with a spreadsheet. The following is a sketch; the electric eld isthe bold curve, the potential is the thin curve.49sphere radiusrE28-42 Originally V = q/4 0 r, where r is the radius of the smaller sphere.(a) Connecting the spheres will bring them to the same potential, or V1 = V2 .(b) q1 + q2 = q; V1 = q1 /4 0 r and V2 = q2 /4 0 2r; combining all of the above q2 = 2q1 andq1 = q/3 and q2 = 2q/3.E28-43 (a) q = 4R2 , so V = q/4 0 R = R/ 0 , orV = (1.601019 C/m2 )(6.37106 m)/(8.851012 C2 /N m2 ) = 0.115 V(b) Pretend the Earth is a conductor, then E = /epsilon0 , soE = (1.601019 C/m2 )/(8.851012 C2 /N m2 ) = 1.81108 V/m.E28-44 V = q/4 0 R, soV = (8.99109 N m2 /C2 )(15109 C)/(0.16 m) = 850 V.E28-45 (a) q = 4 0 RV = 4(8.851012 C2 /N m2 )(0.152 m)(215 V) = 3.63109 C(b) = q/4R2 = (3.63109 C)/4(0.152 m)2 = 1.25108 C/m2 .E28-46 The dotted lines are equipotential lines, the solid arrows are electric eld lines.50E28-47 (a) The total charge (Q = 57.2nC) will be divided up between the two spheres so thatthey are at the same potential. If q1 is the charge on one sphere, then q2 = Q q1 is the charge onthe other. ConsequentlyV11 q14 0 r1q 1 r2q1= V2 ,1 Q q1=,4 0 r2= (Q q1 )r1 ,Qr2=.r2 + r1Putting in the numbers, we ndq1 =Qr1(57.2 nC)(12.2 cm)== 38.6 nC,r2 + r1(5.88 cm) + (12.2 cm)and q2 = Q q1 = (57.2 nC) (38.6 nC) = 18.6 nC.(b) The potential on each sphere should be the same, so we only need to solve one. Then1 q11(38.6 nC)== 2850 V.12 C2 /N m2 ) (12.2 cm)4 0 r14(8.85 10E28-48 (a) V = (8.99109 N m2 /C2 )(31.5109 C)/(0.162 m) = 1.75103 V.(b) V = q/4 0 r, so r = q/4 0 V , and thenr = (8.99109 N m2 /C2 )(31.5109 C)/(1.20103 V) = 0.236 m.That is (0.236 m) (0.162 m) = 0.074 m above the surface.51E28-49(a) Apply the point charge formula, but solve for the charge. Then1 q4 0 rqq= V,==4 0 rV,4(8.85 1012 C2 /N m2 )(1 m)(106 V) = 0.11 mC.Now thats a fairly small charge. But if the radius were decreased by a factor of 100, so wouldthe charge (1.10 C). Consequently, smaller metal balls can be raised to higher potentials with lesscharge.(b) The electric eld near the surface of the ball is a function of the surface charge density,E = / 0 . But surface charge density depends on the area, and varies as r2 . For a given potential,the electric eld near the surface would then be given byE==0q40r2=V.rNote that the electric eld grows as the ball gets smaller. This means that the break down eld ismore likely to be exceeded with a low voltage small ball; youll get sparking.E28-50 A Volt is a Joule per Coulomb. The power required by the drive belt is the product(3.41106 V)(2.83103 C/s) = 9650 W.P28-1 (a) According to Newtonian mechanics we want K = 1 mv 2 to be equal to W = qV2which meansmv 2(0.511 MeV)V === 256 kV.2q2emc2 is the rest mass energy of an electron.(b) Lets do some rearranging rst.1= mc2KKmc21 21=1 211 , 1,K,+1 =mc21 21=1 2,Kmc2 + 11= 1 2,2Kmc2 + 1and nally,=11Kmc2+12Putting in the numbers,1(256(5111keV) + 1keV)so v = 0.746c.522= 0.746,P28-2 (a) The potential of the hollow sphere is V = q/4 0 r. The work required to increase thecharge by an amount dq is dW = V /, dq. Integrating,eW =0e2qdq =.4 0 r8 0 rThis corresponds to an electric potential energy ofW =e(1.601019 C)= 2.55105 eV = 4.081014 J. m2 )(2.821015 m)8(8.851012 C2 /N(b) This would be a mass of m = (4.081014 J)/(3.00108 m/s)2 = 4.531031 kg.P28-3The negative charge is held in orbit by electrostatic attraction, ormv 2qQ=.r4 0 r2The kinetic energy of the charge is1qQmv 2 =.28 0 rK=The electrostatic potential energy isU =qQ,4 0 rE=qQ.8 0 rso the total energy isThe work required to change orbit is thenW =P28-4qQ8 011r1r2.(a) V = E dr, sorV =0qrqr2dr = .34 0 R8 0 R3(b) V = q/8 0 R.(c) If instead of V = 0 at r = 0 as was done in part (a) we take V = 0 at r = , thenV = q/4 0 R on the surface of the sphere. The new expression for the potential inside the spherewill look like V = V + Vs , where V is the answer from part (a) and Vs is a constant so that thesurface potential is correct. ThenVs =and thenV =q4 0 R+qR23qR2=,8 0 R38 0 R3qr23qR2q(3R2 r2 )+=.338 0 R8 0 R8 0 R353P28-5 The total electric potential energy of the system is the sum of the three interaction pairs.One of these pairs does not change during the process, so it can be ignored when nding the changein potential energy. The change in electrical potential energy is thenU = 2q2q2q22=4 0 rf4 0 ri2 011rfri.In this case ri = 1.72 m, while rf = 0.86 m. The change in potential energy is thenU = 2(8.99109 N m2 /C2 )(0.122 C)211(0.86 m) (1.72 m)= 1.56108 JThe time required ist = (1.56108 )/(831 W) = 1.87105 s = 2.17 days.P28-6(a) Apply conservation of energy:K=qQqQ, or d =,4 0 d4 0 Kwhere d is the distance of closest approach.(b) Apply conservation of energy:K=1qQ+ mv 2 ,4 0 (2d) 2so, combining with the results in part (a), v =P28-7K/m.(a) First apply Eq. 28-18, but solve for r. Thenr=q4 0 V=(32.0 1012 C)= 562 m.4(8.85 1012 C2 /N m2 )(512 V)(b) If two such drops join together the charge doubles, and the volume of water doubles, but theradius of the new drop only increases by a factor of 3 2 = 1.26 because volume is proportional tothe radius cubed.The potential on the surface of the new drop will beV new===1 q new,4 0 rnew1 2q old,4 0 3 2 rold1 q old(2)2/3= (2)2/3 V old .4 0 roldThe new potential is 813 V.P28-8 (a) The work done is W = F z = Eqz = qz/2 0 .(b) Since W = qV , V = z/2 0 , soV = V0 (/2 0 )z.54P28-9(a) The potential at any point will be the sum of the contribution from each charge,V =1 q11 q2+,4 0 r14 0 r2where r1 is the distance the point in question from q1 and r2 is the distance the point in questionfrom q2 . Pick a point, call it (x, y). Since q1 is at the origin,x2 + y 2 .r1 =Since q2 is at (d, 0), where d = 9.60 nm,r2 =(x d)2 + y 2 .Dene the Stanley Number as S = 4 0 V . Equipotential surfaces are also equi-Stanley surfaces.In particular, when V = 0, so does S. We can then write the potential expression in a sightlysimplied formq1q2S=+ .r1r2If S = 0 we can rearrange and square this expression.q1r12r12q1x2 + y 22q1= ==q2,r22r22,q2(x d)2 + y 2,2q2Let = q2 /q1 , then we can write2 x2 + y 2=2 x2 + 2 y 2( 1)x + 2xd + (2 1)y 222(x d)2 + y 2 ,= x2 2xd + d2 + y 2 ,= d2 .We complete the square for the (2 1)x2 + 2xd term by adding d2 /(2 1) to both sides of theequation. Then2d1(2 1) x + 2+ y 2 = d2 1 + 2. 1 1The center of the circle is at2d(9.60 nm)== 5.4 nm.1(10/6)2 1(b) The radius of the circle is1+d212 12 1,which can be simplied tod|(10/6)|= (9.6 nm)= 9.00 nm.2 1(10/6)2 1(c) No.55P28-10 An annulus is composed of dierential rings of varying radii r and width dr; the chargeon any ring is the product of the area of the ring, dA = 2r dr, and the surface charge density, ordq = dA =k2k2r dr = 2 dr.3rrThe potential at the center can be found by adding up the contributions from each ring. Since weare at the center, the contributions will each be dV = dq/4 0 r. Thenbkk dr=32 0r4 0V =a11 22ab.=k b2 a2.4 0 b2 a2The total charge on the annulus isbQ=a2kdr = 2kr21 1a b= 2kba.baCombining,V =P28-11Add the three contributions, and then do a series expansion for dVr.q111+ +,4 0 r + d rrdq11+1+,4 0 r 1 + d/r1 d/rddq1 + + 1 + 1 +,4 0 rrrq2d1+.4 0 rr==P28-12Q a+b.8 0 ab(a) Add the contributions from each dierential charge: dq = dy. Theny+LV =ydy =ln4 0 y4 0y+Ly.(b) Take the derivative:Ey = Vy LL==.y4 0 y + L y 24 0 y(y + L)(c) By symmetry it must be zero, since the system is invariant under rotations about the axisof the rod. Note that we cant determine E from derivatives because we dont have the functionalform of V for points o-axis!P28-13 (a) We follow the work done in Section 28-6 for a uniform line of charge, starting withEq. 28-26,dV=14 0dV=14 0 dxx2 + y 2L056,kx dxx2 + y 2,==k4 0k4 0x2 + y 2L,0L2 + y 2 y .(b) The y component of the electric eld can be found fromEy = V,ywhich (using a computer-aided math program) isEy =k4 01yL2+ y2.(c) We could nd Ex if we knew the x variation of V . But we dont; we only found the values ofV along a xed value of x.(d) We want to nd y such that the ratiok4 0is one-half. Simplifying,L2 + y 2 y/k(L)4 0L2 + y 2 y = L/2, which can be written asL2 + y 2 = L2 /4 + Ly + y 2 ,or 3L2 /4 = Ly, with solution y = 3L/4.P28-14 The spheres are small compared to the separation distance. Assuming only one sphere ata potential of 1500 V, the charge would beq = 4 0 rV = 4(8.851012 C2 /N m)(0.150 m)(1500 V) = 2.50108 C.The potential from the sphere at a distance of 10.0 m would beV = (1500 V)(0.150 m)= 22.5 V.(10.0 m)This is small compared to 1500 V, so we will treat it as a perturbation. This means that we canassume that the spheres have charges ofq = 4 0 rV = 4(8.851012 C2 /N m)(0.150 m)(1500 V + 22.5 V) = 2.54108 C.P28-15 Calculating the fraction of excess electrons is the same as calculating the fraction ofexcess charge, so well skip counting the electrons. This problem is eectively the same as Exercise28-47; we have a total charge that is divided between two unequal size spheres which are at the samepotential on the surface. Using the result from that exercise we haveq1 =Qr1,r2 + r1where Q = 6.2 nC is the total charge available, and q1 is the charge left on the sphere. r1 is theradius of the small ball, r2 is the radius of Earth. Since the fraction of charge remaining is q1 /Q,we can writeq1r1r1== 2.0 108 .Qr 2 + r1r257P28-16The positive charge on the sphere would beq = 4 0 rV = 4(8.851012 C2 /N m2 )(1.08102 m)(1000 V) = 1.20109 C.The number of decays required to build up this charge isn = 2(1.20109 C)/(1.601019 C) = 1.501010 .The extra factor of two is because only half of the decays result in an increase in charge. The timerequired ist = (1.501010 )/(3.70108 s1 ) = 40.6 s.P28-17 (a) None.(b) None.(c) None.(d) None.(e) No.P28-18 (a) Outside of an isolated charged spherical object E = q/4 0 r2 and V = q/4 0 r.Then E = V /r. Consequently, the sphere must have a radius larger than r = (9.15106 V)/(100106 V/m) = 9.15102 m.(b) The power required is (320106 C/s)(9.15106 V) = 2930 W.(c) wv = (320106 C/s), so=(320106 C/s)= 2.00105 C/m2 .(0.485 m)(33.0 m/s)58E29-1 (a) The charge which ows through a cross sectional surface area in a time t is given byq = it, where i is the current. For this exercise we haveq = (4.82 A)(4.60 60 s) = 1330 Cas the charge which passes through a cross section of this resistor.(b) The number of electrons is given by (1330 C)/(1.60 1019 C) = 8.31 1021 electrons.E29-2 Q/t = (200106 A/s)(60s/min)/(1.601019 C) = 7.51016 electrons per minute.E29-3 (a) j = nqv = (2.101014 /m3 )2(1.601019 C)(1.40105 m/s) = 9.41 A/m2 . Since the ionshave positive charge then the current density is in the same direction as the velocity.(b) We need an area to calculate the current.E29-4 (a) j = i/A = (1231012 A)/(1.23103 m)2 = 2.59105 A/m2 .(b) v d = j/ne = (2.59105 A/m2 )/(8.491028 /m3 )(1.601019 C) = 1.911015 m/s.E29-5The current rating of a fuse of cross sectional area A would beimax = (440 A/cm2 )A,and if the fuse wire is cylindrical A = d2 /4. Thend=4 (0.552 A)= 4.00102 cm. (440 A/m2 )E29-6 Current density is current divided by cross section of wire, so the graph would look like:I (A/mil^2 x10^3)43215010059150200d(mils)E29-7 The current is in the direction of the motion of the positive charges. The magnitude of thecurrent isi = (3.11018 /s + 1.11018 /s)(1.601019 C) = 0.672 A.E29-8 (a) The total current isi = (3.501015 /s + 2.251015 /s)(1.601019 C) = 9.20104 A.(b) The current density isj = (9.20104 A)/(0.165103 m)2 = 1.08104 A/m2 .E29-9 (a) j = (8.70106 /m3 )(1.601019 C)(470103 m/s) = 6.54107 A/m2 .(b) i = (6.54107 A/m2 )(6.37106 m)2 = 8.34107 A.E29-10 i = wv, so = (95.0106 A)/(0.520 m)(28.0 m/s) = 6.52106 C/m2 .E29-11The drift velocity is given by Eq. 29-6,vd =ji(115 A)=== 2.71104 m/s.neAne(31.2106 m2 )(8.491028 /m3 )(1.601019 C)The time it takes for the electrons to get to the starter motor ist=x(0.855 m)== 3.26103 s.v(2.71104 m/s)Thats about 54 minutes.E29-12 V = iR = (50103 A)(1800 ) = 90 V.E29-13The resistance of an object with constant cross section is given by Eq. 29-13,R=L(11, 000 m)= (3.0 107 m)= 0.59 .A(0.0056 m2 )E29-14 The slope is approximately [(8.2 1.7)/1000] cm/ C, so=16.5103 cm/ C 4103 /C1.7 cmE29-15 (a) i = V /R = (23 V)/(15103 ) = 1500 A.(b) j = i/A = (1500 A)/(3.0103 m)2 = 5.3107 A/m2 .(c) = RA/L = (15103 )(3.0103 m)2 /(4.0 m) = 1.1107 m. The material is possiblyplatinum.E29-16 Use the equation from Exercise 29-17. R = 8 ; thenT = (8 )/(50 )(4.3103 /C ) = 37 C .The nal temperature is then 57 C.60E29-17Start with Eq. 29-16, 0 = 0 av (T T0 ),and multiply through by L/A,LL( 0 ) = 0 av (T T0 ),AAto getR R0 = R0 av (T T0 ).E29-18 The wire has a length L = (250)2(0.122 m) = 192 m. The diameter is 0.129 inches; thecross sectional area is thenA = (0.129 0.0254 m)2 /4 = 8.43106 m2 .The resistance isR = L/A = (1.69108 m)(192 m)/(8.43106 m2 ) = 0.385 .E29-19If the length of each conductor is L and has resistivity , thenRA = andRB = L4L=D2 /4D24LL=.(4D2 /4 D2 /4)3D2The ratio of the resistances is thenRA= 3.RBE29-20 R = R, so 1 L1 /(d1 /2)2 = 2 L2 /(d2 /2)2 . Simplifying, 1 /d2 = 2 /d2 . Then12d2 = (1.19103 m)(9.68108 m)/(1.69108 m) = 2.85103 m.E29-21 (a) (750103 A)/(125) = 6.00103 A.(b) V = iR = (6.00103 A)(2.65106 ) = 1.59108 V.(c) R = V /i = (1.59108 V)/(750103 A) = 2.12108 .E29-22 Since V = iR, then if V and i are the same, then R must be the same.2222(a) Since R = R, 1 L1 /r1 = 2 L2 /r2 , or 1 /r1 = 2 /r2 . Thenriron /rcopper =(9.68108 m)(1.69108 m) = 2.39.(b) Start with the denition of current density:j=iVV==.ARALSince V and L is the same, but is dierent, then the current densities will be dierent.61E29-23 Conductivity is given by Eq. 29-8, j = E. If the wire is long and thin, then themagnitude of the electric eld in the wire will be given byE V /L = (115 V)/(9.66 m) = 11.9 V/m.We can now nd the conductivity,=(1.42104 A/m2 )j== 1.19103 ( m)1 .E(11.9 V/m)E29-24 (a) v d = j/en = E/en. Thenv d = (2.701014 / m)(120 V/m)/(1.601019 C)(620106 /m3 + 550106 /m3 ) = 1.73102 m/s.(b) j = E = (2.701014 / m)(120 V/m) = 3.241014 A/m2 .E29-25 (a) R/L = /A, so j = i/A = (R/L)i/. For copper,j = (0.152103 /m)(62.3 A)/(1.69108 m) = 5.60105 A/m2 ;for aluminum,j = (0.152103 /m)(62.3 A)/(2.75108 m) = 3.44105 A/m2 .(b) A = L/R; if is density, then m = lA = l/(R/L). For copper,m = (1.0 m)(8960 kg/m3 )(1.69108 m)/(0.152103 /m) = 0.996 kg;for aluminum,m = (1.0 m)(2700 kg/m3 )(2.75108 m)/(0.152103 /m) = 0.488 kg.E29-26 The resistance for potential dierences less than 1.5 V are beyond the scale.R (Kiloohms)108642126234V(Volts)E29-27(a) The resistance is dened asR=V(3.55 106 V/A2 )i2== (3.55 106 V/A2 )i.iiWhen i = 2.40 mA the resistance would beR = (3.55 106 V/A2 )(2.40 103 A) = 8.52 k.(b) Invert the above expression, andi = R/(3.55 106 V/A2 ) = (16.0 )/(3.55 106 V/A2 ) = 4.51 A.E29-28 First, n = 3(6.021023 )(2700 kg/m3 )(27.0103 kg) = 1.811029 /m3 . Then=m(9.111031 kg)== 7.151015 s.229 /m3 )(1.601019 C)2 (2.75108 m)ne(1.8110E29-29 (a) E = E0 /e = q/4 0 e R2 , soE=(1.00106 C)=4(8.851012 C2 /N m2 )(4.7)(0.10 m)2(b) E = E0 = q/4 0 R2 , soE=(c) ind =0 (E0(1.00106 C)=4(8.851012 C2 /N m2 )(0.10 m)2 E) = q(1 1/e )/4R2 . Then ind =(1.00106 C)4(0.10 m)211(4.7)= 6.23106 C/m2 .E29-30 Midway between the charges E = q/ 0 d, soq = (8.851012 C2 /N m2 )(0.10 m)(3106 V/m) = 8.3106 C.E29-31 (a) At the surface of a conductor of radius R with charge Q the magnitude of the electriceld is given by1E=QR2 ,4 0while the potential (assuming V = 0 at innity) is given byV =1QR.4 0The ratio is V /E = R.The potential on the sphere that would result in sparking isV = ER = (3106 N/C)R.(b) It is easier to get a spark o of a sphere with a smaller radius, because any potential onthe sphere will result in a larger electric eld.(c) The points of a lighting rod are like small hemispheres; the electric eld will be large nearthese points so that this will be the likely place for sparks to form and lightning bolts to strike.63P29-1 If there is more current owing into the sphere than is owing out then there must be achange in the net charge on the sphere. The net current is the dierence, or 2 A. The potential onthe surface of the sphere will be given by the point-charge expression,V =1 q,4 0 rand the charge will be related to the current by q = it. Combining,V =ort=1 it,4 0 r4 0 V r4(8.85 1012 C2 /N m2 )(980 V)(0.13 m)== 7.1 ms.i(2 A)P29-2 The net current density is in the direction of the positive charges, which is to the east. Thereare two electrons for every alpha particle, and each alpha particle has a charge equal in magnitudeto two electrons. The current density is thenjP29-3= q e ne v e + q + n v ,= (1.61019 C)(5.61021 /m3 )(88 m/s) + (3.21019 C)(2.81021 /m3 )(25 m/s),= 1.0105 C/m2 .(a) The resistance of the segment of the wire isR = L/A = (1.69108 m)(4.0102 m)/(2.6103 m)2 = 3.18105 .The potential dierence across the segment isV = iR = (12 A)(3.18105 ) = 3.8104 V.(b) The tail is negative.(c) The drift speed is v = j/en = i/Aen, sov = (12 A)/(2.6103 m)2 (1.61019 C)(8.491028 /m3 ) = 4.16105 m/s.The electrons will move 1 cm in (1.0102 m)/(4.16105 m/s) = 240 s.P29-4 (a) N = it/q = (250109 A)(2.9 s)/(3.21019 C) = 2.271012 .(b) The speed of the particles in the beam is given by v = 2K/m, sov=2(22.4 MeV)/4(932 MeV/c2 ) = 0.110c.It takes (0.180 m)/(0.110)(3.00108 m/s) = 5.45109 s for the beam to travel 18.0 cm. The numberof charges is thenN = it/q = (250109 A)(5.45109 s)/(3.21019 C) = 4260.(c) W = qV , so V = (22.4 MeV)/2e = 11.2 MV.64P29-5(a) The time it takes to complete one turn is t = (250 m)/c. The total charge isq = it = (30.0 A)(950 m)/(3.00108 m/s) = 9.50105 C.(b) The number of charges is N = q/e, the total energy absorbed by the block is thenU = (28.0109 eV)(9.50105 C)/e = 2.66106 J.This will raise the temperature of the block byT = U/mC = (2.66106 J)/(43.5 kg)(385J/kgC ) = 159 C .P29-6(a) i =j dA = 2i = 2jr dr;0R j0 (1 r/R)r dr = 2j0 (R2 /2 R3 /3R) = j0 R2 /6.(b) Integrate, again:i = 2P29-70R j0 (r/R)r dr = 2j0 (R3 /3R) = j0 R2 /3.(a) Solve 20 = 0 [1 + (T 20 C)], orT = 20 C + 1/(4.3103 /C ) = 250 C.(b) Yes, ignoring changes in the physical dimensions of the resistor.P29-8The resistance when on is (2.90 V)/(0.310 A) = 9.35 . The temperature is given byT = 20 C + (9.35 1.12 )/(1.12 )(4.5103 / C) = 1650 C.P29-9 Originally we have a resistance R1 made out of a wire of length l1 and cross sectional areaA1 . The volume of this wire is V1 = A1 l1 . When the wire is drawn out to the new length we havel2 = 3l1 , but the volume of the wire should be constant soA2 l2 = A1 l1 ,A2 (3l1 ) = A1 l1 ,A2 = A1 /3.The original resistance isR1 = The new resistance isR2 = l1.A1l23l1== 9R1 ,A2A1 /3or R2 = 54 .P29-10 (a) i = (35.8 V)/(935 ) = 3.83102 A.(b) j = i/A = (3.83102 A)/(3.50104 m2 ) = 109 A/m2 .(c) v = (109 A/m2 )/(1.61019 C)(5.331022 /m3 ) = 1.28102 m/s.(d) E = (35.8 V)/(0.158 m) = 227 V/m.65P29-11 (a) = (1.09103 )(5.5103 m)2 /4(1.6 m) = 1.62108 m. This is possibly silver.(b) R = (1.62108 m)(1.35103 m)4/(2.14102 m)2 = 6.08108 .P29-12 (a) L/L = 1.7105 for a temperature change of 1.0 C . Area changes are twice this,or A/A = 3.4105 .Take the dierential of RA = L: R dA+A dR = dL+L d, or dR = dL/A+L d/AR dA/A.For nite changes this can be written asRL A=+.RLA/ = 4.3103 . Since this term is so much larger than the other two it is the only signicanteect.P29-13We will use the results of Exercise 29-17,R R0 = R0 av (T T0 ).To save on subscripts we will drop the av notation, and just specify whether it is carbon c oriron i.The disks will be eectively in series, so we will add the resistances to get the total. Lookingonly at one disk pair, we haveRc + Ri= R0,c (c (T T0 ) + 1) + R0,i (i (T T0 ) + 1) ,= R0,c + R0,i + (R0,c c + R0,i i ) (T T0 ).This last equation will only be constant if the coecient for the term (T T0 ) vanishes. ThenR0,c c + R0,i i = 0,but R = L/A, and the disks have the same cross sectional area, soLc c c + Li i i = 0,orLci i(9.68108 m)(6.5103 /C )=== 0.036.Lic c(3500108 m)(0.50103 /C )P29-14 The current entering the cone is i. The current density as a function of distance x fromthe left end is thenij=.[a + x(b a)/L]2The electric eld is given by E = j. The potential dierence between the ends is thenLV =LE dx =00iiLdx =2[a + x(b a)/L]abThe resistance is R = V /i = L/ab.66P29-15The current is found from Eq. 29-5,j dA,i=where the region of integration is over a spherical shell concentric with the two conducting shellsbut between them. The current density is given by Eq. 29-10,j = E/,and we will have an electric eld which is perpendicular to the spherical shell. Consequently,i=1E dA =1E dABy symmetry we expect the electric eld to have the same magnitude anywhere on a spherical shellwhich is concentric with the two conducting shells, so we can bring it out of the integral sign, andthen14r2 Ei = E dA =,where E is the magnitude of the electric eld on the shell, which has radius r such that b > r > a.The above expression can be inverted to give the electric eld as a function of radial distance,since the current is a constant in the above expression. Then E = i/4r2 The potential is given byaV = E ds,bwe will integrate along a radial line, which is parallel to the electric eld, soaV= E dr,baidr,4r2bi a dr= ,4 b ri 1 1=.4 a b= We divide this expression by the current to get the resistance. ThenR=41 1a bP29-16 Since = d , v d . For an ideal gas the kinetic energy is proportional to the/v temperature, so K T .67E30-1We apply Eq. 30-1,q = CV = (50 1012 F)(0.15 V) = 7.5 1012 C;E30-2 (a) C = V /q = (73.01012 C)/(19.2 V) = 3.801012 F.(b) The capacitance doesnt change!(c) V = q/C = (2101012 C)/(3.801012 F) = 55.3 V.E30-3 q = CV = (26.0106 F)(125 V) = 3.25103 C.E30-4 (a) C = 0 A/d = (8.851012 F/m)(8.22102 m)2 /(1.31103 m) = 1.431010 F.(b) q = CV = (1.431010 F)(116 V) = 1.66108 C.E30-5Eq. 30-11 gives the capacitance of a cylinder,C = 20L(0.0238 m)= 2(8.851012 F/m)= 5.461013 F.ln(b/a)ln((9.15mm)/(0.81mm))E30-6 (a) A = Cd/ 0 = (9.701012 F)(1.20103 m)/(8.851012 F/m) = 1.32103 m2 .(b) C = C0 d0 /d = (9.701012 F)(1.20103 m)/(1.10103 m) = 1.061011 F.(c) V = q0 /C = [V ]0 C0 /C = [V ]0 d/d0 . Using this formula, the new potential dierencewould be [V ]0 = (13.0 V)(1.10103 m)/(1.20103 m) = 11.9 V. The potential energy has changedby (11.9 V) (30.0 V) = 1.1 V.E30-7 (a) From Eq. 30-8,C = 4(8.851012 F/m)(b) A = Cd/0(0.040 m)(0.038 m)= 8.451011 F.(0.040 m) (0.038 m)= (8.451011 F)(2.00103 m)/(8.851012 F/m) = 1.91102 m2 .E30-8 Let a = b + d, where d is the small separation between the shells. ThenC(b + d)bab= 4 0,abdb2 4 0 = 0 A/d.d=40E30-9 The potential dierence across each capacitor in parallel is the same; it is equal to 110 V.The charge on each of the capacitors is thenq = CV = (1.00 106 F)(110 V) = 1.10 104 C.If there are N capacitors, then the total charge will be N q, and we want this total charge to be1.00 C. Then(1.00 C)(1.00 C)N=== 9090.q(1.10 104 C)68E30-10 First nd the equivalent capacitance of the parallel part:C eq = C1 + C2 = (10.3106 F) + (4.80106 F) = 15.1106 F.Then nd the equivalent capacitance of the series part:111=+= 3.23105 F1 .6 F)C eq(15.110(3.90106 F)Then the equivalent capacitance of the entire arrangement is 3.10106 F.E30-11 First nd the equivalent capacitance of the series part:111=+= 3.05105 F1 .C eq(10.3106 F) (4.80106 F)The equivalent capacitance is 3.28106 F. Then nd the equivalent capacitance of the parallel part:C eq = C1 + C2 = (3.28106 F) + (3.90106 F) = 7.18106 F.This is the equivalent capacitance for the entire arrangement.E30-12 For one capacitor q = CV = (25.0106 F)(4200 V) = 0.105 C. There are three capacitors, so the total charge to pass through the ammeter is 0.315 C.E30-13(a) The equivalent capacitance is given by Eq. 30-21,111115=+=+=C eqC1C2(4.0F) (6.0F)(12.0F)or C eq = 2.40F.(b) The charge on the equivalent capacitor is q = CV = (2.40F)(200 V) = 0.480 mC. Forseries capacitors, the charge on the equivalent capacitor is the same as the charge on each of thecapacitors. This statement is wrong in the Student Solutions!(c) The potential dierence across the equivalent capacitor is not the same as the potentialdierence across each of the individual capacitors. We need to apply q = CV to each capacitorusing the charge from part (b). Then for the 4.0F capacitor,V =q(0.480 mC)== 120 V;C(4.0F)and for the 6.0F capacitor,q(0.480 mC)== 80 V.C(6.0F)Note that the sum of the potential dierences across each of the capacitors is equal to the potentialdierence across the equivalent capacitor.V =E30-14 (a) The equivalent capacitance isC eq = C1 + C2 = (4.0F) + (6.0F) = (10.0F).(c) For parallel capacitors, the potential dierence across the equivalent capacitor is the same asthe potential dierence across either of the capacitors.(b) For the 4.0F capacitor,q = CV = (4.0F)(200 V) = 8.0104 C;and for the 6.0F capacitor,q = CV = (6.0F)(200 V) = 12.0104 C.69E30-15 (a) C eq = C + C + C = 3C;deq =d0A0A== .C eq3C3(b) 1/C eq = 1/C + 1/C + 1/C = 3/C;deq =0A0A== 3d.C eqC/3E30-16 (a) The maximum potential across any individual capacitor is 200 V; so there must beat least (1000 V)/(200 V) = 5 series capacitors in any parallel branch. This branch would have anequivalent capacitance of C eq = C/5 = (2.0106 F)/5 = 0.40106 F.(b) For parallel branches we add, which means we need (1.2106 F)/(0.40106 F) = 3 parallelbranches of the combination found in part (a).E30-17 Look back at the solution to Ex. 30-10. If C3 breaks down electrically then the circuit iseectively two capacitors in parallel.(b) V = 115 V after the breakdown.(a) q1 = (10.3106 F)(115 V) = 1.18103 C.E30-18 The 108F capacitor originally has a charge of q = (108106 F)(52.4 V) = 5.66103 C.After it is connected to the second capacitor the 108F capacitor has a charge of q = (108 106 F)(35.8 V) = 3.87103 C. The dierence in charge must reside on the second capacitor, so thecapacitance is C = (1.79103 C)/(35.8 V) = 5.00105 F.E30-19 Consider any junction other than A or B. Call this junction point 0; label the four nearestjunctions to this as points 1, 2, 3, and 4. The charge on the capacitor that links point 0 to point 1 isq1 = CV01 , where V01 is the potential dierence across the capacitor, so V01 = V0 V1 , whereV0 is the potential at the junction 0, and V1 is the potential at the junction 1. Similar expressionsexist for the other three capacitors.For the junction 0 the net charge must be zero; there is no way for charge to cross the plates ofthe capacitors. Then q1 + q2 + q3 + q4 = 0, and this meansCV01 + CV02 + CV03 + CV04 = 0orV01 + V02 + V03 + V04 = 0.Let V0i = V0 Vi , and then rearrange,4V0 = V1 + V2 + V3 + V4 ,orV0 =E30-20 U = uV =0EU=21(V1 + V2 + V3 + V4 ) .4V /2, where V is the volume. Then1(8.851012 F/m)(150 V/m)2 (2.0 m3 ) = 1.99107 J.270E30-21 The total capacitance is (2100)(5.0106 F) = 1.05102 F. The total energy stored isU=11C(V )2 = (1.05102 F)(55103 V)2 = 1.59107 J.22The cost is(1.59107 J)$0.033600103 J= $0.133.1E30-22 (a) U = 2 C(V )2 = 1 (0.061 F)(1.0104 V)2 = 3.05106 J.26(b) (3.0510 J)/(3600103 J/kW h) = 0.847kW h.E30-23(a) The capacitance of an air lled parallel-plate capacitor is given by Eq. 30-5,C=0Ad=(8.851012 F/m)(42.0 104 m2 )= 2.861011 F.(1.30 103 m)(b) The magnitude of the charge on each plate is given byq = CV = (2.861011 F)(625 V) = 1.79108 C.(c) The stored energy in a capacitor is given by Eq. 30-25, regardless of the type or shape of thecapacitor, so11U = C(V )2 = (2.861011 F)(625 V)2 = 5.59 J.22(d) Assuming a parallel plate arrangement with no fringing eects, the magnitude of the electriceld between the plates is given by Ed = V , where d is the separation between the plates. ThenE = V /d = (625 V)/(0.00130 m) = 4.81105 V/m.(e) The energy density is Eq. 30-28,u=11212F/m))(4.81105 V/m)2 = 1.02 J/m3 .0 E = ((8.851022E30-24 The equivalent capacitance is given by1/C eq = 1/(2.12106 F) + 1/(3.88106 F) = 1/(1.37106 F).The energy stored is U = 1 (1.37106 F)(328 V)2 = 7.37102 J.2E30-25 V /r = q/4 0 r2 = E, so that if V is the potential of the sphere then E = V /r is theelectric eld on the surface. Then the energy density of the electric eld near the surface isu=1(8.851012 F/m)20E =22(8150 V)(0.063 m)2= 7.41102 J/m3 .E30-26 The charge on C3 can be found from considering the equivalent capacitance. q3 = (3.10106 F)(112 V) = 3.47104 C. The potential across C3 is given by [V ]3 = (3.47104 C)/(3.90106 F) = 89.0 V.The potential across the parallel segment is then (112 V) (89.0 V) = 23.0 V. So [V ]1 =[V ]2 = 23.0 V.Then q1 = (10.3106 F)(23.0 V) = 2.37104 C and q2 = (4.80106 F)(23.0 V) = 1.10104 C..71E30-27 There is enough work on this problem without deriving once again the electric eldbetween charged cylinders. I will instead refer you back to Section 26-4, and stateE=1 q,2 0 Lrwhere q is the magnitude of the charge on a cylinder and L is the length of the cylinders.The energy density as a function of radial distance is found from Eq. 30-28,u=1120E =28 2q22 20 L rThe total energy stored in the electric eld is given by Eq. 30-24,U=1 q2q 2 ln(b/a)=,2C2 2 0 Lwhere we substituted into the last part Eq. 30-11, the capacitance of a cylindrical capacitor.We want to show that integrating a volume integral from r = a to r = ab over the energydensity function will yield U/2. Since we want to do this problem the hard way, we will pretend wedont know the answer, and integrate from r = a to r = c, and then nd out what c is.Then1U2=u dV,c2L=aq=020c28 2 0 L2 a 0cq2dr,4 0 L a rcq2ln .4 0 L a==q22 20 L r18 2L0r dr d dz,drd dz,rNow we equate this to the value for U that we found above, and we solve for c.1 q 2 ln(b/a)2 2 2 0 Lln(b/a)(b/a)abE30-28 (a) d =0 A/C,q2cln ,4 0 L a= 2 ln(c/a),= (c/a)2 ,== c.ord = (8.851012 F/m)(0.350 m2 )/(51.31012 F) = 6.04103 m.(b) C = (5.60)(51.31012 F) = 2.871010 F.E30-29 Originally, C1 =C2 /C1 = d1 /d2 , so0 A/d1 .After the changes, C2 = 0 A/d2 . Dividing C2 by C1 yields = d2 C2 /d1 C1 = (2)(2.571012 F)/(1.321012 F) = 3.89.72E30-30 The required capacitance is found from U = 1 C(V )2 , or2C = 2(6.61106 J)/(630 V)2 = 3.331011 F.The dielectric constant required is = (3.331011 F)/(7.401012 F) = 4.50. Try transformer oil.E30-31Capacitance with dielectric media is given by Eq. 30-31,e 0 A.dThe various sheets have dierent dielectric constants and dierent thicknesses, and we want tomaximize C, which means maximizing e /d. For mica this ratio is 54 mm1 , for glass this ratio is35 mm1 , and for paran this ratio is 0.20 mm1 . Mica wins.C=E30-32 The minimum plate separation is given byd = (4.13103 V)/(18.2106 V/m) = 2.27104 m.The minimum plate area is thenA=dC(2.27104 m)(68.4109 F)== 0.627 m2 . 0(2.80)(8.851012 F/m)E30-33 The capacitance of a cylindrical capacitor is given by Eq. 30-11,C = 2(8.851012 F/m)(2.6)1.0103 m= 8.63108 F.ln(0.588/0.11)E30-34 (a) U = C (V )2 /2, C = e 0 A/d, and V /d is less than or equal to the dielectricstrength (which we will call S). Then V = Sd andU=1e 0 AdS 2 ,2so the volume is given byV = 2U/e 0 S 2 .This quantity is a minimum for mica, soV = 2(250103 J)/(5.4)(8.851012 F/m)(160106 V/m)2 = 0.41 m3 .(b) e = 2U/V0S2, soe = 2(250103 J)/(0.087m3 )(8.851012 F/m)(160106 V/m)2 = 25.E30-35(a) The capacitance of a cylindrical capacitor is given by Eq. 30-11,C = 2 0 eL.ln(b/a)The factor of e is introduced because there is now a dielectric (the Pyrex drinking glass) betweenthe plates. We can look back to Table 29-2 to get the dielectric properties of Pyrex. The capacitanceof our glass is thenC = 2(8.851012 F/m)(4.7)(0.15 m)= 7.31010 F.ln((3.8 cm)/(3.6 cm)(b) The breakdown potential is (14 kV/mm)(2 mm) = 28 kV.73E30-36 (a) C = e C = (6.5)(13.51012 F) = 8.81011 F.(b) Q = C V = (8.81011 F)(12.5 V) = 1.1109 C.(c) E = V /d, but we dont know d.(d) E = E/e , but we couldnt nd E.E30-37 (a) Insert the slab so that it is a distance a above the lower plate. Then the distancebetween the slab and the upper plate is d a b. Inserting the slab has the same eect as havingtwo capacitors wired in series; the separation of the bottom capacitor is a, while that of the topcapacitor is d a b.The bottom capacitor has a capacitance of C1 = 0 A/a, while the top capacitor has a capacitanceof C2 = 0 A/(d a b). Adding these in series,1C eq===11+,C1C2adab+,0A0Adb.0ASo the capacitance of the system after putting the copper slab in is C =(b) The energy stored in the system before the slab is inserted isUi =0 A/(d b).q2 dq2=2C i2 0Awhile the energy stored after the slab is inserted isUf =q2q2 d b=2C f2 0AThe ratio is U i /U f = d/(d b).(c) Since there was more energy before the slab was inserted, then the slab must have gone inwillingly, it was pulled in!. To get the slab back out we will need to do work on the slab equal tothe energy dierence.q2 dq2 d bq2 bUi Uf ==.2 0A2 0A2 0AE30-38 (a) Insert the slab so that it is a distance a above the lower plate. Then the distancebetween the slab and the upper plate is d a b. Inserting the slab has the same eect as havingtwo capacitors wired in series; the separation of the bottom capacitor is a, while that of the topcapacitor is d a b.The bottom capacitor has a capacitance of C1 = 0 A/a, while the top capacitor has a capacitanceof C2 = 0 A/(d a b). Adding these in series,1C eq===11+,C1C2adab+,0A0Adb.0ASo the capacitance of the system after putting the copper slab in is C =740 A/(d b).(b) The energy stored in the system before the slab is inserted isUi =C i (V )2(V )2 0 A=22dwhile the energy stored after the slab is inserted isUf =C f (V )2(V )2 0 A=22 dbThe ratio is U i /U f = (d b)/d.(c) Since there was more energy after the slab was inserted, then the slab must not have gone inwillingly, it was being repelled!. To get the slab in we will need to do work on the slab equal to theenergy dierence.Uf Ui =(V )2 0 A(V )2 0 A(V )2 0 Ab=.2 db2d2 d(d b)E30-39 C = e 0 A/d, so d = e 0 A/C.(a) E = V /d = CV /e 0 A, orE=(1121012 F)(55.0 V)= 13400 V/m.(5.4)(8.851012 F/m)(96.5104 m2 )(b) Q = CV = (1121012 F)(55.0 V) = 6.16109 C..(c) Q = Q(1 1/e ) = (6.16109 C)(1 1/(5.4)) = 5.02109 C.E30-40 (a) E = q/e 0 A, soe =(890109 C)= 6.53(1.40106 V/m)(8.851012 F/m)(110104 m2 )(b) q = q(1 1/e ) = (890109 C)(1 1/(6.53)) = 7.54107 C.P30-1The capacitance of the cylindrical capacitor is from Eq. 30-11,C=2 0 L.ln(b/a)If the cylinders are very close together we can write b = a + d, where d, the separation between thecylinders, is a small number, soC=2 0 L2 0 L=.ln ((a + d)/a)ln (1 + d/a)Expanding according to the hint,C2 0 L2a 0 L=d/adNow 2a is the circumference of the cylinder, and L is the length, so 2aL is the area of a cylindricalplate. Hence, for small separation between the cylinders we haveCwhich is the expression for the parallel plates.750Ad,P30-2(a) C =0 A/x;take the derivative anddCdT0 dA0 A dx 2,x dTx dT1 dA 1 dx= CA dTx dT=.(b) Since (1/A)dA/dT = 2a and (1/x)dx/dT = s , we needs = 2a = 2(23106 /C ) = 46106 /C .P30-3 Insert the slab so that it is a distance d above the lower plate. Then the distance betweenthe slab and the upper plate is abd. Inserting the slab has the same eect as having two capacitorswired in series; the separation of the bottom capacitor is d, while that of the top capacitor is abd.The bottom capacitor has a capacitance of C1 = 0 A/d, while the top capacitor has a capacitanceof C2 = 0 A/(a b d). Adding these in series,1C eq===11+,C1C2dabd+,0A0Aab.0ASo the capacitance of the system after putting the slab in is C =0 A/(a b).P30-4 The potential dierence between any two adjacent plates is V . Each interior plate has acharge q on each surface; the exterior plate (one pink, one gray) has a charge of q on the interiorsurface only.The capacitance of one pink/gray plate pair is C = 0 A/d. There are n plates, but only n 1plate pairs, so the total charge is (n 1)q. This means the total capacitance is C = 0 (n 1)A/d.P30-5 Let V0 = 96.6 V.As far as point e is concerned point a looks like it is originally positively charged, and point d isoriginally negatively charged. It is then convenient to dene the charges on the capacitors in termsof the charges on the top sides, so the original charge on C1 is q 1,i = C1 V0 while the original chargeon C2 is q 2,i = C2 V0 . Note the negative sign reecting the opposite polarity of C2 .(a) Conservation of charge requiresq 1,i + q 2,i = q 1,f + q 2,f ,but since q = CV and the two capacitors will be at the same potential after the switches are closedwe can writeC1 V0 C2 V0 = C1 V + C2 V,(C1 C2 ) V0 = (C1 + C2 ) V,C1 C2V0 = V.C1 + C2With numbers,V = (96.6 V)(1.16 F) (3.22 F)= 45.4 V.(1.16 F) + (3.22 F)76The negative sign means that the top sides of both capacitor will be negatively charged after theswitches are closed.(b) The charge on C1 is C1 V = (1.16 F)(45.4 V) = 52.7C.(c) The charge on C2 is C2 V = (3.22 F)(45.4 V) = 146C.P30-6 C2 and C3 form an eective capacitor with equivalent capacitance Ca = C2 C3 /(C2 + C3 ).The charge on C1 is originally q0 = C1 V0 . After throwing the switch the potential across C1is given by q1 = C1 V1 . The same potential is across Ca ; q2 = q3 , so q2 = Ca V1 . Charge isconserved, so q1 + q2 = q0 . Combining some of the above,V1 =and thenC1q0=V0 ,C1 + CaC1 + Caq1 =22C1C1 (C2 + C3 )V0 =V0 .C1 + CaC1 C2 + C1 C3 + C2 C3q2 =Ca C1V0 =C1 + CaSimilarly,111++C1C2C31V0 .q3 = q2 because they are in series.P30-7 (a) If terminal a is more positive than terminal b then current can ow that will charge thecapacitor on the left, the current can ow through the diode on the top, and the current can chargethe capacitor on the right. Current will not ow through the diode on the left. The capacitors areeectively in series.Since the capacitors are identical and series capacitors have the same charge, we expect thecapacitors to have the same potential dierence across them. But the total potential dierenceacross both capacitors is equal to 100 V, so the potential dierence across either capacitor is 50 V.The output pins are connected to the capacitor on the right, so the potential dierence acrossthe output is 50 V.(b) If terminal b is more positive than terminal a the current can ow through the diode on theleft. If we assume the diode is resistanceless in this conguration then the potential dierence acrossit will be zero. The net result is that the potential dierence across the output pins is 0 V.In real life the potential dierence across the diode would not be zero, even if forward biased. Itwill be somewhere around 0.5 Volts.P30-8 Divide the strip of width a into N segments, each of width x = a/N . The capacitance ofeach strip is C = 0 ax/y. If is small then11d1= 1 x/d).yd + x sin d + x(Since parallel capacitances add,C=C =a0ad(1 x/d)dx =07720ad1a2d.P30-9 (a) When S2 is open the circuit acts as two parallel capacitors. The branch on the left hasan eective capacitance given by1111=+=,Cl(1.0106 F) (3.0106 F)7.5107 Fwhile the branch on the right has an eective capacitance given by1111=+=.Cl(2.0106 F) (4.0106 F)1.33106 FThe charge on either capacitor in the branch on the left isq = (7.5107 F)(12 V) = 9.0106 C,while the charge on either capacitor in the branch on the right isq = (1.33106 F)(12 V) = 1.6105 C.(b) After closing S2 the circuit is eectively two capacitors in series. The top part has an eectivecapacitance ofC t = (1.0106 F) + (2.0106 F) = (3.0106 F),while the eective capacitance of the bottom part isC b = (3.0106 F) + (4.0106 F) = (7.0106 F).The eective capacitance of the series combination is given by1111=+=.6 F)6 F)C eq(3.010(7.0102.1106 FThe charge on each part is q = (2.1106 F)(12 V) = 2.52105 C. The potential dierence acrossthe top part isV t = (2.52105 C)/(3.0106 F) = 8.4 V,and then the charge on the top two capacitors is q1 = (1.0 106 F)(8.4 V) = 8.4 106 C andq2 = (2.0106 F)(8.4 V) = 1.68105 C. The potential dierence across the bottom part isV t = (2.52105 C)/(7.0106 F) = 3.6 V,and then the charge on the top two capacitors is q1 = (3.0 106 F)(3.6 V) = 1.08 105 C andq2 = (4.0106 F)(3.6 V) = 1.44105 C.P30-10 Let V = Vxy . By symmetry V2 = 0 and V1 = V4 = V5 = V3 = V /2.Suddenly the problem is very easy. The charges on each capacitor is q1 , except for q2 = 0. Then theequivalent capacitance of the circuit isC eq =q1 + q4q== C1 = 4.0106 F.V2V178P30-11(a) The charge on the capacitor with stored energy U0 = 4.0 J is q0 , whereU0 =2q0.2CWhen this capacitor is connected to an identical uncharged capacitor the charge is shared equally,so that the charge on either capacitor is now q = q0 /2. The stored energy in one capacitor is thenU=q2q 2 /41= 0 = U0 .2C2C4But there are two capacitors, so the total energy stored is 2U = U0 /2 = 2.0 J.(b) Good question. Current had to ow through the connecting wires to get the charge from onecapacitor to the other. Originally the second capacitor was uncharged, so the potential dierenceacross that capacitor would have been zero, which means the potential dierence across the connecting wires would have been equal to that of the rst capacitor, and there would then have beenenergy dissipation in the wires according toP = i2 R.Thats where the missing energy went.P30-12R = L/A and C =0 A/L.Combining, R = 0 /C, orR = (9.40 m)(8.851012 F/m)/(1101012 F) = 0.756 .1P30-13 (a) u = 2 0 E 2 = e2 /32 2 0 r4 .(b) U = u dV where dV = 4r2 dr. ThenU = 4piRe2e2 1r2 dr =.2 r432 08 0 R(c) R = e2 /8 0 mc2 , orR=P30-14P30-15(1.601019 C)28(8.851012 F/m)(9.111031 kg)(3.00108 m/s)2= 1.401015 m.1U = 2 q 2 /C = q 2 x/2A 0 . F = dU/dx = q 2 /2A 0 .According to Problem 14, the force on a plate of a parallel plate capacitor isF =q2.2 0AThe force per unit area is thenq22F==,A2 0 A22 0where = q/A is the surface charge density. But we know that the electric eld near the surface ofa conductor is given by E = / 0 , soF1= 0E2.A279P30-16 A small surface area element dA carries a charge dq = q dA/4R2 . There are three forceson the elements which balance, sop(V0 /V )dA + q dq/4 0 R2 = p dA,or3pR0 + q 2 /16 2 0 R = pR3 .This can be rearranged as3q 2 = 16 2 0 pR(R3 R0 ).P30-17 The magnitude of the electric eld in the cylindrical region is given by E = /2 0 r,where is the linear charge density on the anode. The potential dierence is given by V =(/2 0 ) ln(b/a), where a is the radius of the anode b the radius of the cathode. Combining, E =V /r ln(b/a), this will be a maximum when r = a, soV = (0.180103 m) ln[(11.0103 m)/(0.180103 m)](2.20106 V/m) = 1630 V.P30-18This is eectively two capacitors in parallel, each with an area of A/2. Then0 A/2C eq = e1d+ e20 A/2d=0Ade1 + e22.P30-19 We will treat the system as two capacitors in series by pretending there is an innitesimally thin conductor between them. The slabs are (I assume) the same thickness. The capacitanceof one of the slabs is then given by Eq. 30-31,C1 =e1 0 A,d/2where d/2 is the thickness of the slab. There would be a similar expression for the other slab. Theequivalent series capacitance would be given by Eq. 30-21,1C eq===C eq=11+,C1C2d/2d/2+,e1 0 A e2 0 Ad e2 + e1,2 0 A e1 e22 0 A e1 e2.d e2 + e1P30-20 Treat this as three capacitors. Find the equivalent capacitance of the series combinationon the right, and then add on the parallel part on the left. The right hand side is1dd2d=+=C eqe2 0 A/2 e3 0 A/20Ae2 + e3e2 e3Add this to the left hand side, andC eq==e1 0 A/2e2 e30A+2d2d e2 + e3A e1e2 e30+.2d2e2 + e380,.P30-21 (a) q doesnt change, but C = C/2. Then V = q/C = 2V .(b) U = C(V )2 /2 = 0 A(V )2 /2d. U = C (V )2 /2 = 0 A(2V )2 /4d = 2U .(c) W = U U = 2U U = U = 0 A(V )2 /2d.P30-22 The total energy is U = qV /2 = (7.021010 C)(52.3 V)/2 = 1.84108 J.(a) In the air gap we haveUa =20 E0 V2=(8.851012 F/m)(6.9103 V/m)2 (1.15102 m2 )(4.6103 m)= 1.11108 J.2That is (1.11/1.85) = 60% of the total.(b) The remaining 40% is in the slab.P30-23 (a) C = 0 A/d = (8.851012 F/m)(0.118 m2 )/(1.22102 m) = 8.561011 F.(b) Use the results of Problem 30-24.C =(4.8)(8.851012 F/m)(0.118 m2 )= 1.191010 F(4.8)(1.22102 m) (4.3103 m)(4.8 1)(c) q = CV = (8.561011 F)(120 V) = 1.03108 C; since the battery is disconnected q = q.(d) E = q/ 0 A = (1.03108 C)/(8.851012 F/m)(0.118 m2 ) = 9860 V/m in the space betweenthe plates.(e) E = E/e = (9860 V/m)/(4.8) = 2050 V/m in the dielectric.(f) V = q/C = (1.03108 C)/(1.191010 F) = 86.6 V.(g) W = U U = q 2 (1/C 1/C )/2, orW =(1.03108 C)2[1/(8.561011 F) 1/(1.191010 F)] = 1.73107 J.2P30-24 The result is eectively three capacitors in series. Two are air lled with thicknesses ofx and d b x, the third is dielectric lled with thickness b. All have an area A. The eectivecapacitance is given by1C==C==xdbxb++,e 0 A0A0A1b(d b) +,Ae00A,d b + b/ee 0 A.e b(e 1)81E31-1 (5.12 A)(6.00 V)(5.75 min)(60 s/min) = 1.06104 J.E31-2 (a) (12.0 V)(1.601019 C) = 1.921018 J.(b) (1.921018 J)(3.401018 /s) = 6.53 W.E31-3If the energy is delivered at a rate of 110 W, then the current through the battery isi=P(110 W)== 9.17 A.V(12 V)Current is the ow of charge in some period of time, sot =q(125 A h)== 13.6 h,i(9.2 A)which is the same as 13 hours and 36 minutes.E31-4 (100 W)(8 h) = 800 W h.(a) (800 W h)/(2.0 W h) = 400 batteries, at a cost of (400)($0.80) = $320.(b) (800 W h)($0.12103 W h) = $0.096.E31-5 Go all of the way around the circuit. It is a simple one loop circuit, and although it doesnot matter which way we go around, we will follow the direction of the larger emf. Then(150 V) i(2.0 ) (50 V) i(3.0 ) = 0,where i is positive if it is counterclockwise. Rearranging,100 V = i(5.0 ),or i = 20 A.Assuming the potential at P is VP = 100 V, then the potential at Q will be given byVQ = VP (50 V) i(3.0 ) = (100 V) (50 V) (20 A)(3.0 ) = 10 V.E31-6 (a) Req = (10 ) + (140 ) = 150 . i = (12.0 V)/(150 ) = 0.080 A.(b) Req = (10 ) + (80 ) = 90 . i = (12.0 V)/(90 ) = 0.133 A.(c) Req = (10 ) + (20 ) = 30 . i = (12.0 V)/(30 ) = 0.400 A.E31-7 (a) Req = (3.0 V 2.0 V)/(0.050 A) = 20 . Then R = (20 ) (3.0 ) (3.0 ) = 14 .(b) P = iV = i2 R = (0.050 A)2 (14 ) = 3.5102 W.E31-8 (5.0 A)R1 = V . (4.0 A)(R1 +2.0 ) = V . Combining, 5R1 = 4R1 +8.0 , or R1 = 8.0 .E31-9 (a) (53.0 W)/(1.20 A) = 44.2 V.(b) (1.20 A)(19.0 ) = 22.8 V is the potential dierence across R. Then an additional potentialdierence of (44.2 V) (22.8 V) = 21.4 V must exist across C.(c) The left side is positive; it is a reverse emf.E31-10 (a) The current in the resistor is (9.88 W)/(0.108 ) = 9.56 A. The total resistance ofthe circuit is (1.50 V)/(9.56 A) = 0.157 . The internal resistance of the battery is then (0.157 ) (0.108 ) = 0.049 .(b) (9.88 W)/(9.56 A) = 1.03 V.82E31-11We assign directions to the currents through the four resistors as shown in the gure.12ab34Since the ammeter has no resistance the potential at a is the same as the potential at b. Consequently the potential dierence (V b ) across both of the bottom resistors is the same, and thepotential dierence (V t ) across the two top resistors is also the same (but dierent from thebottom). We then have the following relationships:V t + V bi1 + i2Vj= E,= i3 + i4 ,= i j Rj ,where the j subscript in the last line refers to resistor 1, 2, 3, or 4.For the top resistors,V1 = V2 implies 2i1 = i2 ;while for the bottom resistors,V3 = V4 implies i3 = i4 .Then the junction rule requires i4 = 3i1 /2, and the loop rule requires(i1 )(2R) + (3i1 /2)(R) = E or i1 = 2E/(7R).The current that ows through the ammeter is the dierence between i2 and i4 , or 4E/(7R) 3E/(7R) = E/(7R).E31-12 (a) Dene the current i1 as moving to the left through r1 and the current i2 as movingto the left through r2 . i3 = i1 + i2 is moving to the right through R. Then there are two loopequations:E1E2= i1 r1 + i3 R,= (i3 i1 )r2 + i3 R.Multiply the top equation by r2 and the bottom by r1 and then add:r2 E1 + r1 E2 = i3 r1 r2 + i3 R(r1 + r2 ),which can be rearranged asi3 =r2 E1 + r1 E2.r1 r2 + Rr1 + Rr2(b) There is only one current, soE1 + E2 = i(r1 + r2 + R),ori=E1 + E2.r1 + r 2 + R83E31-13 (a) Assume that the current ows through each source of emf in the same direction asthe emf. The the loop rule will give us three equationsE1 i1 R1 + i2 R2 E2 i1 R1E2 i2 R2 + i3 R1 E3 + i3 R1E1 i1 R1 + i3 R1 E3 + i3 R1 i1 R1= 0,= 0,= 0.The junction rule (looks at point a) gives us i1 + i2 + i3 = 0. Use this to eliminate i2 from the secondloop equation,E2 + i1 R2 + i3 R2 + 2i3 R1 E3 = 0,and then combine this with the the third equation to eliminate i3 ,2E1 R2 E3 R2 + 2i3 R1 R2 + 2E2 R1 + 2i3 R1 R2 + 4i3 R1 2E3 R1 = 0,ori3 =2E3 R1 + E3 R2 E1 R2 2E2 R1= 0.582 A.24R1 R2 + 4R1Then we can nd i1 fromi1 =E3 E2 i3 R2 2i3 R1= 0.668 A,R2where the negative sign indicates the current is down.Finally, we can nd i2 = (i1 + i3 ) = 0.0854 A.(b) Start at a and go to b (nal minus initial!),+i2 R2 E2 = 3.60 V.E31-14 (a) The current through the circuit is i = E/(r + R). The power delivered to R is thenP = iV = i2 R = E 2 R/(r + R)2 . Evaluate dP/dR and set it equal to zero to nd the maximum.ThendPrR0== E 2R,dR(r + R)3which has the solution r = R.(b) When r = R the power isP = E 2R1E2=.(R + R)24rE31-15 (a) We rst use P = F v to nd the power output by the electric motor. Then P =(2.0 N)(0.50 m/s) = 1.0 W.The potential dierence across the motor is V m = E ir. The power output from the motor isthe rate of energy dissipation, so P m = V m i. Combining these two expressions,= (E ir) i,= Ei i2 r,0 = i2 r + Ei P m ,0 = (0.50 )i2 (2.0 V)i + (1.0 W).PmRearrange and solve for i,i=(2.0 V) (2.0 V)2 4(0.50 )(1.0 W),2(0.50 )84which has solutions i = 3.4 A and i = 0.59 A.(b) The potential dierence across the terminals of the motor is V m = E ir which if i = 3.4 Ayields V m = 0.3 V, but if i = 0.59 A yields V m = 1.7 V. The battery provides an emf of 2.0 V; itisnt possible for the potential dierence across the motor to be larger than this, but both solutionsseem to satisfy this constraint, so we will move to the next part and see what happens.(c) So what is the signicance of the two possible solutions? It is a consequence of the fact thatpower is related to the current squared, and with any quadratics we expect two solutions. Bothare possible, but it might be that only one is stable, or even that neither is stable, and a smallperturbation to the friction involved in turning the motor will cause the system to break down. Wewill learn in a later chapter that the eective resistance of an electric motor depends on the speedat which it is spinning, and although that wont aect the problem here as worded, it will aect thephysical problem that provided the numbers in this problem!E31-16 req = 4r = 4(18 ) = 72 . The current is i = (27 V)/(72 ) = 0.375 A.E31-17 In parallel connections of two resistors the eective resistance is less than the smallerresistance but larger than half the smaller resistance. In series connections of two resistors theeective resistance is greater than the larger resistance but less than twice the larger resistance.Since the eective resistance of the parallel combination is less than either single resistanceand the eective resistance of the series combinations is larger than either single resistance we canconclude that 3.0 must have been the parallel combination and 16 must have been the seriescombination.The resistors are then 4.0 and 12 resistors.E31-18 Points B and C are eectively the same point!(a) The three resistors are in parallel. Then req = R/3.(b) See (a).(c) 0, since there is no resistance between B and C.E31-19 Focus on the loop through the battery, the 3.0 , and the 5.0 resistors. The loop ruleyields(12.0 V) = i[(3.0 ) + (5.0 )] = i(8.0 ).The potential dierence across the 5.0 resistor is thenV = i(5.0 ) = (5.0 )(12.0 V)/(8.0 ) = 7.5 V.E31-20 Each lamp draws a current of (500 W)/(120 V) = 4.17 A. Furthermore, the fuse cansupport (15 A)/(4.17 A) = 3.60 lamps. That is a maximum of 3.E31-21 The current in the series combination is is = E/(R1 + R2 ). The power dissipated isP s = iE = E 2 /(R1 + R2 ).In a parallel arrangement R1 dissipates P1 = i1 E = E 2 /R1 . A similar expression exists for R2 ,so the total power dissipated is P p = E 2 (1/R1 + 1/R2 ).The ratio is 5, so 5 = P p /P s = (1/R1 + 1/R2 )(R1 + R2 ), or 5R1 R2 = (R1 + R2 )2 . Solving forR2 yields 2.618R1 or 0.382R1 . Then R2 = 262 or R2 = 38.2 .85E31-22 Combining n identical resistors in series results in an equivalent resistance of req = nR.Combining n identical resistors in parallel results in an equivalent resistance of req = R/n. If theresistors are arranged in a square array consisting of n parallel branches of n series resistors, thenthe eective resistance is R. Each will dissipate a power P , together they will dissipate n2 P .So we want nine resistors, since four would be too small.E31-23parallel:(a) Work through the circuit one step at a time. We rst add R2 , R3 , and R4 in11111=++=Re42.0 61.6 75.0 18.7 We then add this resistance in series with R1 ,Re = (112 ) + (18.7 ) = 131 .(b) The current through the battery is i = E/R = (6.22 V)/(131 ) = 47.5 mA. This is also thecurrent through R1 , since all the current through the battery must also go through R1 .The potential dierence across R1 is V1 = (47.5 mA)(112 ) = 5.32 V. The potential dierenceacross each of the three remaining resistors is 6.22 V 5.32 V = 0.90 V.The current through each resistor is theni2i3i4= (0.90 V)/(42.0 ) = 21.4 mA,= (0.90 V)/(61.6 ) = 14.6 mA,= (0.90 V)/(75.0 ) = 12.0 mA.E31-24 The equivalent resistance of the parallel part is r = R2 R/(R2 + R). The equivalentresistance for the circuit is r = R1 + R2 R/(R2 + R). The current through the circuit is i = E/r.The potential dierence across R is V = E i R1 , orV= E(1 R1 /r),R2 + RR1 R2 + R1 R + RR2RR2= E.R1 R2 + R1 R + RR2= E 1 R1,Since P = iV = (V )2 /R,P = E22RR2.(R1 R2 + R1 R + RR2 )2Set dP/dR = 0, the solution is R = R1 R2 /(R1 + R2 ).E31-25 (a) First add the left two resistors in series; the eective resistance of that branch is2R. Then add the right two resistors in series; the eective resistance of that branch is also 2R.Now we combine the three parallel branches and nd the eective resistance to be11114=+ +=,Re2R R 2R2Ror Re = R/2.(b) First we add the right two resistors in series; the eective resistance of that branch is 2R.We then combine this branch with the resistor which connects points F and H. This is a parallelconnection, so the eective resistance is1113=+=,Re2R R2R86or 2R/3.This value is eectively in series with the resistor which connects G and H, so the total is5R/3.Finally, we can combine this value in parallel with the resistor that directly connects F and Gaccording to1138=+=,ReR 5R5Ror Re = 5R/8.E31-26 The resistance of the second resistor is r2 = (2.4 V)/(0.001 A) = 2400 . The potentialdierence across the rst resistor is (12 V) (2.4 V) = 9.6 V. The resistance of the rst resistor is(9.6 V)/(0.001 A) = 9600 .E31-27 See Exercise 31-26. The resistance ratio is(0.95 0.1 V)r1=,r 1 + r2(1.50 V)orr2(1.50 V)= 1.r1(0.95 0.1 V)The allowed range for the ratio r2 /r1 is between 0.5625 and 0.5957.We can choose any standard resistors we want, and we could use any tolerance, but then wewill need to check our results. 22 and 39 would work; as would 27 and 47. There are otherchoices.E31-28 Consider any junction other than A or B. Call this junction point 0; label the fournearest junctions to this as points 1, 2, 3, and 4. The current through the resistor that linkspoint 0 to point 1 is i1 = V01 /R, where V01 is the potential dierence across the resistor, soV01 = V0 V1 , where V0 is the potential at the junction 0, and V1 is the potential at the junction1. Similar expressions exist for the other three resistor.For the junction 0 the net current must be zero; there is no way for charge to accumulate on thejunction. Then i1 + i2 + i3 + i4 = 0, and this meansV01 /R + V02 /R + V03 /R + V04 /R = 0orV01 + V02 + V03 + V04 = 0.Let V0i = V0 Vi , and then rearrange,4V0 = V1 + V2 + V3 + V4 ,orV0 =1(V1 + V2 + V3 + V4 ) .4E31-29 The current through the radio is i = P/V = (7.5 W)/(9.0 V) = 0.83 A. The radiowas left one for 6 hours, or 2.16104 s. The total charge to ow through the radio in that time is(0.83 A)(2.16104 s) = 1.8104 C.E31-30 The power dissipated by the headlights is (9.7 A)(12.0 V) = 116 W. The power requiredby the engine is (116 W)/(0.82) = 142 W, which is equivalent to 0.190 hp.87E31-31 (a) P = (120 V)(120 V)/(14.0 ) = 1030 W.(b) W = (1030 W)(6.42 h) = 6.61 kW h. The cost is $0.345.E31-32E31-33We want to apply either Eq. 31-21,PR = i2 R,or Eq. 31-22,PR = (VR )2 /R,depending on whether we are in series (the current is the same through each bulb), or in parallel(the potential dierence across each bulb is the same. The brightness of a bulb will be measured byP , even though P is not necessarily a measure of the rate radiant energy is emitted from the bulb.(b) If the bulbs are in parallel then PR = (VR )2 /R is how we want to compare the brightness.The potential dierence across each bulb is the same, so the bulb with the smaller resistance isbrighter.(b) If the bulbs are in series then PR = i2 R is how we want to compare the brightness. Bothbulbs have the same current, so the larger value of R results in the brighter bulb.One direct consequence of this can be tried at home. Wire up a 60 W, 120 V bulb and a 100 W,120 V bulb in series. Which is brighter? You should observe that the 60 W bulb will be brighter.2E31-34 (a) j = i/A = (25 A)/(0.05 in) = 3180 A/in = 4.93106 A/m2 .(b) E = j = (1.69108 m)(4.93106 A/m2 ) = 8.33102 V/m.(c) V = Ed = (8.33102 V/m)(305 m) = 25 V.(d) P = iV = (25 A)(25 V) = 625 W.E31-35 (a) The bulb is on for 744 hours. The energy consumed is (100 W)(744 h) = 74.4 kW h,at a cost of (74.4)(0.06) = $4.46.(b) r = V 2 /P = (120 V)2 /(100 W) = 144 .(c) i = P/V = (100 W)/(120 V) = 0.83 A.E31-36 P = (V )2 /r and r = r0 (1 + T ). ThenP =(500 W)P0== 660 W1 + T1 + (4.0104 /C )(600C )E31-37 (a) n = q/e = it/e, son = (485103 A)(95109 s)/(1.61019 C) = 2.881011 .(b) iav = (520/s)(485103 A)(95109 s) = 2.4105 A.(c) P p = ip V = (485103 A)(47.7106 V) = 2.3106 W; while P a = ia V = (2.4105 A)(47.7610 V) = 1.14103 W.E31-38 r = L/A = (3.5105 m)(1.96102 m)/(5.12103 m)2 = 8.33103 .(a) i = P/r = (1.55 W)/(8.33103 ) = 13.6 A, soj = i/A = (13.6 A)/(5.12103 m)2 = 1.66105 A/m2 .(b) V =Pr =(1.55 W)(8.33103 ) = 0.114 V.88E31-39(a) The current through the wire isi = P/V = (4800 W)/(75 V) = 64 A,The resistance of the wire isR = V /i = (75 V)/(64 A) = 1.17 .The length of the wire is then found fromL=RA(1.17 )(2.6106 m2 )== 6.1 m.(5.0107 m)One could easily wind this much nichrome to make a toaster oven. Of course allowing 64 Amps tobe drawn through household wiring will likely blow a fuse.(b) We want to combine the above calculations into one formula, soL=thenL=RAAV /iA(V )2==,P(2.6106 m2 )(110 V)2= 13 m.(4800 W)(5.0107 m)Hmm. We need more wire if the potential dierence is increased? Does this make sense? Yes, itdoes. We need more wire because we need more resistance to decrease the current so that the samepower output occurs.E31-40 (a) The energy required to bring the water to boiling is Q = mCT . The time requiredisQ(2.1 kg)(4200 J/kg)(100 C 18.5 C)t=== 2.22103 s0.77P0.77(420 W)(b) The additional time required to boil half of the water away ist=E31-41mL/2(2.1 kg)(2.26106 J/kg)/2== 7340 s.0.77P0.77(420 W)(a) Integrate both sides of Eq. 31-26;q0dqq ECt= 0dt,RCtqln(q EC)|0lnq ECECq ECECqt,RC 0t= ,RC= = et/RC ,= EC 1 et/RC .That wasnt so bad, was it?89(b) Rearrange Eq. 31-26 in order to get q terms on the left and t terms on the right, thenintegrate;qq0dqqt= 0dt,RCtqln q|q0lnqq0qq0qt,RC 0t= ,RC= = et/RC ,= q0 et/RC .That wasnt so bad either, was it?E31-42 (a) C = RC = (1.42106 )(1.80106 F) = 2.56 s.(b) q0 = CV = (1.80106 F)(11.0 V) = 1.98105 C.(c) t = C ln(1 q/q0 ), sot = (2.56 s) ln(1 15.5106 C/1.98105 C) = 3.91 s.E31-43 Solve n = t/C = ln(1 0.99) = 4.61.E31-44 (a) V = E(1 et/C ), soC = (1.28106 s)/ ln(1 5.00 V/13.0 V) = 2.64106 s(b) C = C /R = (2.64106 s)/(15.2103 ) = 1.731010 FE31-45 (a) V = Eet/C , soC = (10.0 s)/ ln(1.06 V/100 V) = 2.20 s(b) V = (100 V)e17 s/2.20 s = 4.4102 V.E31-46 V = Eet/C and C = RC, sotttR===.C ln(V /V0 )(220109 F) ln(0.8 V/5 V)4.03107 FIf t is between 10.0 s and 6.0 ms, then R is betweenR = (10106 s)/(4.03107 F) = 24.8,andR = (6103 s)/(4.03107 F) = 14.9103 .E31-47 The charge on the capacitor needs to build up to a point where the potential across thecapacitor is VL = 72 V, and this needs to happen within 0.5 seconds. This means that we want tosolveCVL = CE 1 eT /RCfor R knowing that T = 0.5 s. This expression can be written asR=T(0.5 s)== 2.35106 .C ln(1 VL /E)(0.15 C) ln(1 (72 V)/(95 V))90E31-48 (a) q0 = 2U C = 2(0.50 J)(1.0106 F) = 1103 C.(b) i0 = V0 /R = q0 /RC = (1103 C)/(1.0106 )(1.0106 F) = 1103 A.(c) VC = V0 et/C , soVC =(1103 C) t/(1.0106 )(1.0106 F)e= (1000 V)et/(1.0 s)(1.0106 F)Note that VR = VC .(d) PR = (VR )2 /R, soPR = (1000 V)2 e2t/(1.0 s) /(1106 ) = (1 W)e2t/(1.0 s) .E31-49 (a) i = dq/dt = Eet/C /R, soi=66(4.0 V)e(1.0 s)/(3.010 )(1.010 F) = 9.55107 A.(3.0106 )(b) PC = iV = (E 2 /R)et/C (1 et/C ), soPC =66(4.0 V)2 (1.0 s)/(3.0106 )(1.0106 F)e1 e(1.0 s)/(3.010 )(1.010 F) = 1.08106 W.(3.0106 )(c) PR = i2 R = (E 2 /R)e2t/C , soPR =(4.0 V)2 2(1.0 s)/(3.0106 )(1.0106 F)e= 2.74106 W.(3.0106 )(d) P = PR + PC , orP = 2.74106 W + 1.08106 W = 3.82106 WE31-50 The rate of energy dissipation in the resistor isPR = i2 R = (E 2 /R)e2t/C .EvaluatingE 2 2t/RCE2edt =C,R 020but that is the original energy stored in the capacitor.PR dt =P31-1The terminal voltage of the battery is given by V = E ir, so the internal resistance isr=E V(12.0 V) (11.4 V)== 0.012 ,i(50 A)so the battery appears within specs.The resistance of the wire is given byR=V(3.0 V)== 0.06 ,i(50 A)so the cable appears to be bad.What about the motor? Trying it,R=V(11.4 V) (3.0 V)== 0.168 ,i(50 A)so it appears to be within spec.91P31-2Traversing the circuit we haveE ir1 + E ir2 iR = 0,so i = 2E/(r1 + r2 + R). The potential dierence across the rst battery is thenV1 = E ir1 = E 1 2r1r 1 + r2 + R=Er2 r 1 + Rr1 + r 2 + RThis quantity will only vanish if r2 r1 + R = 0, or r1 = R + r2 . Since r1 > r2 this is actuallypossible; R = r1 r2 .P31-3V = E iri and i = E/(ri + R), soV = ER,ri + RThere are then two simultaneous equations:(0.10 V)(500 ) + (0.10 V)ri = E(500 )and(0.16 V)(1000 ) + (0.16 V)ri = E(1000 ),with solution(a) ri = 1.5103 and(b) E = 0.400 V.2(c) The cell receives energy from the sun at a rate (2.0 mW/cm )(5.0 cm2 ) = 0.010 W. The cell22converts energy at a rate of V /R = (0.16 V) /(1000 ) = 0.26 %P31-4(a) The emf of the battery can be found fromE = iri + V l = (10 A)(0.05 ) + (12 V) = 12.5 V(b) Assume that resistance is not a function of temperature. The resistance of the headlights isthenrl = (12.0 V)/(10.0 A) = 1.2 .The potential dierence across the lights when the starter motor is on isV l = (8.0 A)(1.2 ) = 9.6 V,and this is also the potential dierence across the terminals of the battery. The current through thebattery is thenE V(12.5 V) (9.6 V)i=== 58 A,ri(0.05 )so the current through the motor is 50 Amps.P31-5(a) The resistivities areA = rA A/L = (76.2106 )(91.0104 m2 )/(42.6 m) = 1.63108 m,andB = rB A/L = (35.0106 )(91.0104 m2 )/(42.6 m) = 7.48109 m.(b) The current is i = V /(rA + rB ) = (630 V)/(111.2 ) = 5.67106 A. The current densityis thenj = (5.67106 A)/(91.0104 m2 ) = 6.23108 A/m2 .(c) EA = A j = (1.63108 m)(6.23108 A/m2 ) = 10.2 V/m and EB = B j = (7.48109 m)(6.23108 A/m2 ) = 4.66 V/m.(d) VA = EA L = (10.2 V/m)(42.6 m) = 435 V and VB = EB L = (4.66 V/m)(42.6 m) = 198 V.92P31-6 Set up the problem with the traditional presentation of the Wheatstone bridge problem.Then the symmetry of the problem (ip it over on the line between x and y) implies that there is nocurrent through r. As such, the problem is equivalent to two identical parallel branches each withtwo identical series resistances.Each branch has resistance R + R = 2R, so the overall circuit has resistance1111=+= ,Req2R 2RRso Req = R.P31-7P31-8 (a) The loop through R1 is trivial: i1 = E2 /R1 = (5.0 V)/(100 ) = 0.05 A. The loopthrough R2 is only slightly harder: i2 = (E2 + E3 E1 )/R2 = 0.06 A.(b) Vab = E3 + E2 = (5.0 V) + (4.0 V) = 9.0 V.P31-9 (a) The three way light-bulb has two laments (or so we are told in the question). Thereare four ways for these two laments to be wired: either one alone, both in series, or both inparallel. Wiring the laments in series will have the largest total resistance, and since P = V 2 /Rthis arrangement would result in the dimmest light. But we are told the light still operates at thelowest setting, and if a lament burned out in a series arrangement the light would go out.We then conclude that the lowest setting is one lament, the middle setting is another lament,and the brightest setting is both laments in parallel.(b) The beauty of parallel settings is that then power is additive (it is also addictive, but thatsa dierent eld.) One lament dissipates 100 W at 120 V; the other lament (the one that burnsout) dissipates 200 W at 120 V, and both together dissipate 300 W at 120 V.The resistance of one lament is thenR=(120 V)2(V )2== 144 .P(100 W)The resistance of the other lament isR=(V )2(120 V)2== 72 .P(200 W)P31-10 We can assume that R contains all of the resistance of the resistor, the battery and theammeter, thenR = (1.50 V)/(1.0 m/A) = 1500 .For each of the following parts we apply R + r = V /i, so(a) r = (1.5 V)/(0.1 mA) (1500 ) = 1.35104 ,(b) r = (1.5 V)/(0.5 mA) (1500 ) = 1.5103 ,(c) r = (1.5 V)/(0.9 mA) (1500 ) = 167.(d) R = (1500 ) (18.5 ) = 1482 P31-11(a) The eective resistance of the parallel branches on the middle and the right isR2 R 3.R2 + R393The eective resistance of the circuit as seen by the battery is thenR1 +R2 R3R1 R 2 + R1 R3 + R2 R3=,R2 + R3R 2 + R3The current through the battery isi=ER 2 + R3,R1 R 2 + R1 R3 + R2 R3The potential dierence across R1 is thenV1 = ER2 + R 3R1 ,R 1 R2 + R 1 R 3 + R 2 R 3while V3 = E V1 , orV3 = ER 2 R3,R1 R2 + R 1 R3 + R 2 R 3so the current through the ammeter isi3 =ori3 = (5.0 V)V3R2=E,R3R1 R2 + R 1 R3 + R 2 R 3(4 )= 0.45 A.(2 )(4 ) + (2 )(6 ) + (4 )(6 )(b) Changing the locations of the battery and the ammeter is equivalent to swapping R1 andR3 . But since the expression for the current doesnt change, then the current is the same.P31-12V1 + V2 = VS + VX ; if Va = Vb , then V1 = VS . Using the rst expression,ia (R1 + R2 ) = ib (RS + RX ),using the second,i a R1 = i b R2 .Dividing the rst by the second,1 + R2 /R1 = 1 + RX /RS ,or RX = RS (R2 /R1 ).P31-13P31-14Lv = Q/m and Q/t = P = iV , soLv =iV(5.2 A)(12 V)== 2.97106 J/kg.m/t(21106 kg/s)P31-15 P = i2 R. W = pV , where V is volume. p = mg/A and V = Ay, where y is the heightof the piston. Then P = dW/dt = mgv. Combining all of this,v=i2 R(0.240 A)2 (550 )== 0.274 m/s.mg(11.8 kg)(9.8 m/s2 )94P31-16(a) Since q = CV , thenq = (32106 F) (6 V) + (4 V/s)(0.5 s) (2 V/s2 )(0.5 s)2 = 2.4104 C.(b) Since i = dq/dt = C dV /dt, theni = (32106 F) (4 V/s) 2(2 V/s2 )(0.5 s) = 6.4105 A.(c) Since P = iV ,P = [ (4 V/s) 2(2 V/s2 )(0.5 s) (6 V) + (4 V/s)(0.5 s) (2 V/s2 )(0.5 s)2 = 4.8104 W.P31-17(a) We have P = 30P0 and i = 4i0 . ThenR=P30P030==R0 .i2(4i0 )216We dont really care what happened with the potential dierence, since knowing the change inresistance of the wire should give all the information we need.The volume of the wire is a constant, even upon drawing the wire out, so LA = L0 A0 ; theproduct of the length and the cross sectional area must be a constant.Resistance is given by R = L/A, but A = L0 A0 /L, so the length of the wire isL=A0 L0 R=30 A0 L0 R0= 1.37L0 .16(b) We know that A = L0 A0 /L, soA=P31-18A0LA0 == 0.73A0 .L01.37(a) The capacitor charge as a function of time is given by Eq. 31-27,q = CE 1 et/RC ,while the current through the circuit (and the resistor) is given by Eq. 31-28,i=E t/RCe.RThe energy supplied by the emf isU=Ei dt = Edq = Eq;but the energy in the capacitor is UC = qV /2 = Eq/2.(b) Integrating,E2E2EqUR = i2 Rdt =e2t/RC dt ==.R2C295P31-19The capacitor charge as a function of time is given by Eq. 31-27,q = CE 1 et/RC ,while the current through the circuit (and the resistor) is given by Eq. 31-28,i=E t/RCe.RThe energy stored in the capacitor is given byU=q2,2Cso the rate that energy is being stored in the capacitor isPC =dUq dqq== i.dtC dtCThe rate of energy dissipation in the resistor isPR = i2 R,so the time at which the rate of energy dissipation in the resistor is equal to the rate of energystorage in the capacitor can be found by solvingPC2i RiRCECet/RCet/RCt= PR ,q=i,C= q,= CE 1 et/RC ,= 1/2,= RC ln 2.96E32-1 Apply Eq. 32-3, F = qv B.All of the paths which involve left hand turns are positive particles (path 1); those paths whichinvolve right hand turns are negative particle (path 2 and path 4); and those paths which dont turninvolve neutral particles (path 3).E32-2 (a) The greatest magnitude of force is F = qvB = (1.61019 C)(7.2106 m/s)(83103 T) =9.61014 N. The least magnitude of force is 0.(b) The force on the electron is F = ma; the angle between the velocity and the magnetic eldis , given by ma = qvB sin . Then(9.11031 kg)(4.91016 m/s2 )(1.61019 C)(7.2106 m/s)(83103 T) = arcsin= 28 .E32-3 (a) v = E/B = (1.5103 V/m)/(0.44 T) = 3.4103 m/s.E32-4 (a) v = F/qB sin = (6.481017 N/(1.601019 C)(2.63103 T) sin(23.0 ) = 3.94105 m/s.(b) K = mv 2 /2 = (938 MeV/c2 )(3.94105 m/s)2 /2 = 809 eV.E32-5The magnetic force on the proton isFB = qvB = (1.61019 C)(2.8107 m/s)(30eex6 T) = 1.31016 N.The gravitational force on the proton ismg = (1.71027 kg)(9.8 m/s2 ) = 1.71026 N.The ratio is then 7.6109 . If, however, you carry the number of signicant digits for the intermediateanswers farther you will get the answer which is in the back of the book.E32-6 The speed of the electron is given by v =v=2qV /m, or2(1000 eV)/(5.1105 eV/c2 ) = 0.063c.The electric eld between the plates is E = (100 V)/(0.020 m) = 5000 V/m. The required magneticeld is thenB = E/v = (5000 V/m)/(0.063c) = 2.6104 T.E32-7 Both have the same velocity. Then K p /K e = mp v 2 /me v 2 = mp /me =.E32-8 The speed of the ion is given by v =v=2qV /m, or2(10.8 keV)/(6.01)(932 MeV/c2 ) = 1.96103 c.The required electric eld is E = vB = (1.96103 c)(1.22 T) = 7.17105 V/m.E32-9(a) For a charged particle moving in a circle in a magnetic eld we apply Eq. 32-10;r=mv(9.111031 kg)(0.1)(3.00108 m/s)== 3.4104 m.|q|B(1.61019 C)(0.50 T)(b) The (non-relativistic) kinetic energy of the electron isK=11mv 2 = (0.511 MeV)(0.10c)2 = 2.6103 MeV.2297E32-10 (a) v = 2K/m = 2(1.22 keV)/(511 keV/c2 ) = 0.0691c.(b) B = mv/qr = (9.111031 kg)(0.0691c)/(1.601019 C)(0.247 m) = 4.78104 T.(c) f = qB/2m = (1.601019 C)(4.78104 T)/2(9.111031 kg) = 1.33107 Hz.(d) T = 1/f = 1/(1.33107 Hz) = 7.48108 s.E32-11 (a) v = 2K/m = 2(350 eV)/(511 keV/c2 ) = 0.037c.(b) r = mv/qB = (9.111031 kg)(0.037c)/(1.601019 C)(0.20T) = 3.16104 m.E32-12 The frequency is f = (7.00)/(1.29103 s) = 5.43103 Hz. The mass is given by m =qB/2f , or(1.601019 C)(45.0103 T)m== 2.111025 kg = 127 u.2(5.43103 Hz)E32-13(a) Apply Eq. 32-10, but rearrange it asv=|q|rB2(1.61019 C)(0.045 m)(1.2 T)== 2.6106 m/s.m4.0(1.661027 kg)(b) The speed is equal to the circumference divided by the period, soT =2r2m24.0(1.661027 kg)=== 1.1107 s.v|q|B2(1.6 1019 C)(1.2 T)(c) The (non-relativistic) kinetic energy isK=|q|2 r2 B(21.61019 C)2 (0.045 m)2 (1.2 T)2== 2.241014 J.2m2(4.01.661027 kg))To change to electron volts we need merely divide this answer by the charge on one electron, soK=(d) V =Kq(2.241014 J)= 140 keV.(1.61019 C)= (140 keV)/(2e) = 70 V.E32-14 (a) R = mv/qB = (938 MeV/c2 )(0.100c)/e(1.40 T) = 0.223 m.(b) f = qB/2m = e(1.40 T)/2(938 MeV/c2 ) = 2.13107 Hz.2E32-15 (a) K /K p = (q /m )/(q p 2 /mp ) = 22 /4 = 1.2(b) K d /K p = (q d /md )/(q p 2 /mp ) = 12 /2 = 1/2.E32-16 (a) K = qV . Then K p = eV , K d = eV , and K = 2eV .(b) r = sqrt2mK/qB. Then rd /rp = (2/1)(1/1)/(1/1) = 2.(c) r = sqrt2mK/qB. Then r /rp = (4/1)(2/1)/(2/1) = 2.E32-17 r = 2mK/|q|B = ( m/|q|)( 2K/B). All three particles are traveling with the samekinetic energy in the same magnetic eld. The relevant factors are in front; we just need to comparethe mass and charge of each of the three particles.(a) The radius of the deuteron path is 12 rp .(b) The radius of the alpha particle path is 24 rp = rp .98E32-18 The neutron, being neutral, is unaected by the magnetic eld and moves o in a linetangent to the original path. The proton moves at the same original speed as the deuteron and hasthe same charge, but since it has half the mass it moves in a circle with half the radius.E32-19 (a) The proton momentum would be pc = qcBR = e(3.0108 m/s)(41106 T)(6.4106 m) =7.9104 MeV. Since 79000 MeV is much, much greater than 938 MeV the proton is ultra-relativistic.Then E pc, and since = E/mc2 we have = p/mc. Inverting,v=c1E32-20 (a) Classically, R =R=1=21m2 c2m2 c21 0.99993.2p2p22mK/qB, or2(0.511 MeV/c2 )(10.0 MeV)/e(2.20 T) = 4.84103 m.(b) This would be an ultra-relativistic electron, so K E pc, then R = p/qB = K/qBc, orR = (10.0 MeV)/e(2.2 T)(3.00108 m/s) = 1.52102 m.(c) The electron is eectively traveling at the speed of light, so T = 2R/c, orT = 2(1.52102 m)/(3.00108 m/s) = 3.181010 s.This result does depend on the speed!E32-21Use Eq. 32-10, except we rearrange for the mass,m=|q|rB2(1.601019 C)(4.72 m)(1.33 T)== 9.431027 kgv0.710(3.00108 m/s)However, if it is moving at this velocity then the mass which we have here is not the true mass,but a relativistic correction. For a particle moving at 0.710c we have=11v 2 /c2=11 (0.710)2= 1.42,so the true mass of the particle is (9.431027 kg)/(1.42) = 6.641027 kg. The number of nucleonspresent in this particle is then (6.641027 kg)/(1.671027 kg) = 3.97 4. The charge was +2,which implies two protons, the other two nucleons would be neutrons, so this must be an alphaparticle.E32-22 (a) Since 950 GeV is much, much greater than 938 MeV the proton is ultra-relativistic. = E/mc2 , sov1m2 c4m2 c4= 1 2 = 11 0.9999995.cE22E 2(b) Ultra-relativistic motion requires pc E, soB = pc/qRc = (950 GeV)/e(750 m)(3.00108 m/s) = 4.44 T.99E32-23 First use 2f = qB/m. The use K = q 2 B 2 R2 /2m = mR2 (2f )2 /2. The number ofturns is n = K/2qV , on average the particle is located at a distance R/ 2 from the center, so thedistance traveled is x = n2R/ 2 = n 2R. Combining, 3 3 32 R mf 22 (0.53 m)3 (2 932103 keV/c2 )(12106 /s)2x=== 240 m.qVe(80 kV)E32-24 The particle moves in a circle. x = R sin t and y = R cos t.E32-25 We will use Eq. 32-20, E H = v d B, except we will not take the derivation through to Eq.32-21. Instead, we will set the drift velocity equal to the speed of the strip. We will, however, setE H = V H /w. Thenv=EHV H /w(3.9106 V)/(0.88102 m)=== 3.7101 m/s.BB(1.2103 T)E32-26 (a) v = E/B = (40106 V)/(1.2102 m)/(1.4 T) = 2.4103 m/s.(b) n = (3.2 A)(1.4 T)/(1.61019 C)(9.5106 m)(40106 V) = 7.41028 /m3 .; Silver.E32-27 E H = v d B and v d = j/ne. Combine and rearrange.E32-28 (a) Use the result of the previous exercise and E c = j.(b) (0.65 T)/(8.491028 /m3 )(1.601019 C)(1.69108 m) = 0.0028.E32-29Since L is perpendicular to B can useFB = iLB.Equating the two forces,iLBi= mg,(0.0130 kg)(9.81 m/s2 )mg=== 0.467 A.LB(0.620 m)(0.440 T)Use of an appropriate right hand rule will indicate that the current must be directed to the rightin order to have a magnetic force directed upward.E32-30 F = iLB sin = (5.12 103 A)(100 m)(58 106 T) sin(70 ) = 27.9 N. The direction ishorizontally west.E32-31 (a) We use Eq. 32-26 again, and since the (horizontal) axle is perpendicular to thevertical component of the magnetic eld,i=F(10, 000 N)== 3.3108 A.BL(10 T)(3.0 m)(b) The power lost per ohm of resistance in the rails is given byP/r = i2 = (3.3108 A)2 = 1.11017 W.(c) If such a train were to be developed the rails would melt well before the train left the station.100E32-32 F = idB, so a = F/m = idB/m. Since a is constant, v = at = idBt/m. The direction isto the left.E32-33 Only the component of B is of interest. Then F =jdF = iBy dx, or3.2F = (5.0 A)(8103 T/m2 )x2 dx = 0.414 N.1.2The direction is k.E32-34 The magnetic force will have two components: one will lift vertically (Fy = F sin ), theother push horizontally (Fx = F cos ). The rod will move when Fx > (W Fy ). We are interestedin the minimum value for F as a function of . This occurs whenddF=ddWcos + sin = 0.This happens when = tan . Then = arctan(0.58) = 30 , andF =(0.58)(1.15 kg)(9.81 m/s2 )= 5.66 Ncos(30 ) + (0.58) sin(30 )is the minimum force. Then B = (5.66 N)/(53.2 A)(0.95 m) = 0.112 T.E32-35 We choose that the eld points from the shorter side to the longer side.(a) The magnetic eld is parallel to the 130 cm side so there is no magnetic force on that side.The magnetic force on the 50 cm side has magnitudeFB = iLB sin ,where is the angle between the 50 cm side and the magnetic eld. This angle is larger than 90 ,but the sine can be found directly from the triangle,sin =(120 cm)= 0.923,(130 cm)and then the force on the 50 cm side can be found byFB = (4.00 A)(0.50 m)(75.0103 T)(120 cm)= 0.138 N,(130 cm)and is directed out of the plane of the triangle.The magnetic force on the 120 cm side has magnitudeFB = iLB sin ,where is the angle between the 1200 cm side and the magnetic eld. This angle is larger than180 , but the sine can be found directly from the triangle,sin =(50 cm)= 0.385,(130 cm)and then the force on the 50 cm side can be found byFB = (4.00 A)(1.20 m)(75.0103 T)and is directed into the plane of the triangle.(b) Look at the three numbers above.101(50 cm)= 0.138 N,(130 cm)E32-36 = N iAB sin , so = (20)(0.1 A)(0.12 m)(0.05 m)(0.5 T) sin(90 33 ) = 5.0103 N m.E32-37 The external magnetic eld must be in the plane of the clock/wire loop. The clockwisecurrent produces a magnetic dipole moment directed into the plane of the clock.(a) Since the magnetic eld points along the 1 pm line and the torque is perpendicular to boththe external eld and the dipole, then the torque must point along either the 4 pm or the 10 pm line.Applying Eq. 32-35, the direction is along the 4 pm line. It will take the minute hand 20 minutesto get there.(b) = (6)(2.0 A)(0.15 m)2 (0.07 T) = 0.059 N m.P32-1 Since F must be perpendicular to B then B must be along k. The magnitude of v is2 + (35)2 km/s = 53.1 km/s; the magnitude of F is2 + (4.8)2 fN = 6.38 fN. Then(40)(4.2)B = F/qv = (6.381015 N)/(1.61019 C)(53.1103 m/s) = 0.75 T.or B = 0.75 T k.P32-2a = (q/m)(E + v B). For the initial velocity given,v B = (15.0103 m/s)(400106 T) (12.0103 m/s)(400106 T)k.jBut since there is no acceleration in the or k direction this must be oset by the electric eld.jConsequently, two of the electric eld components are Ey = 6.00 V/m and Ez = 4.80 V/m. Thethird component of the electric eld is the source of the acceleration, soEx = max /q = (9.111031 kg)(2.001012 m/s2 )/(1.601019 C) = 11.4 V/m.P32-3 (a) Consider rst the cross product, v B. The electron moves horizontally, there is acomponent of the B which is down, so the cross product results in a vector which points to the leftof the electrons path.But the force on the electron is given by F = qv B, and since the electron has a negative chargethe force on the electron would be directed to the right of the electrons path.(b) The kinetic energy of the electrons is much less than the rest mass energy, so this is nonrelativistic motion. The speed of the electron is then v = 2K/m, and the magnetic force on theelectron is FB = qvB, where we are assuming sin = 1 because the electron moves horizontallythrough a magnetic eld with a vertical component. We can ignore the eect of the magnetic eldshorizontal component because the electron is moving parallel to this component.The acceleration of the electron because of the magnetic force is thena ==qvBqB=mm2K,m(1.601019 C)(55.0106 T)(9.111031 kg)2(1.921015 J)= 6.271014 m/s2 .(9.111031 kg)(c) The electron travels a horizontal distance of 20.0 cm in a time oft=(20.0 cm)2K/m=(20.0 cm)2(1.921015 J)/(9.111031 kg)= 3.08109 s.In this time the electron is accelerated to the side through a distance ofd=11 2at = (6.271014 m/s2 )(3.08109 s)2 = 2.98 mm.22102P32-4 (a) d needs to be larger than the turn radius, so R d; but 2mK/q 2 B 2 = R2 d2 , orB 2mK/q 2 d2 .(b) Out of the page.P32-5 Only undeected ions emerge from the velocity selector, so v = E/B. The ions are thendeected by B with a radius of curvature of r = mv/qB; combining and rearranging, q/m =E/rBB .P32-6 The ions are given a kinetic energy K = qV ; they are then deected with a radius ofcurvature given by R2 = 2mK/q 2 B 2 . But x = 2R. Combine all of the above, and m = B 2 qx2 /8V.P32-7(a) Start with the equation in Problem 6, and take the square root of both sides to get12B2q8Vm=x,and then take the derivative of x with respect to m,1 dm =2 m12B2q8Vdx,and then consider nite dierences instead of dierential quantities,m =mB 2 q2V12x,(b) Invert the above expression,x =2VmB 2 q12m,and then put in the given values,x ==2(7.33103 V)27 kg)(0.520 T)2 (1.601019 C)(35.0)(1.66108.02 mm.12(2.0)(1.661027 kg),Note that we used 35.0 u for the mass; if we had used 37.0 u the result would have been closer tothe answer in the back of the book.P32-8 (a) B = 2V m/qr2 = 2(0.105 MV)(238)(932 MeV/c2 )/2e(0.973 m)2 = 5.23107 T.(b) The number of atoms in a gram is 6.021023 /238 = 2.531021 . The current is then(0.090)(2.531021 )(2)(1.61019 C)/(3600 s) = 20.2 mA.P32-9 (a) q.(b) Regardless of speed, the orbital period is T = 2m/qB. But they collide halfway around acomplete orbit, so t = m/qB.P32-10103P32-11(a) The period of motion can be found from the reciprocal of Eq. 32-12,T =2m2(9.111031 kg)== 7.86108 s.|q|B(1.601019 C)(455106 T)(b) We need to nd the velocity of the electron from the kinetic energy,v=2K/m =2(22.5 eV)(1.601019 J/eV)/(9.111031 kg) = 2.81106 m/s.The velocity can written in terms of components which are parallel and perpendicular to the magneticeld. Thenv|| = v cos and v = v sin .The pitch is the parallel distance traveled by the electron in one revolution, sop = v|| T = (2.81106 m/s) cos(65.5 )(7.86108 s) = 9.16 cm.(c) The radius of the helical path is given by Eq. 32-10, except that we use the perpendicularvelocity component, soR=(9.111031 kg)(2.81106 m/s) sin(65.5 )mv== 3.20 cm|q|B(1.601019 C)(455106 T)bP32-12 F = i a dl B. dl has two components, those parallel to the path, say dx and thoseperpendicular, say dy. Then the integral can be written asbbdx B +F=ady B.abBut B is constant, and can be removed from the integral. a dx = l, a vector that points from a tobb. a dy = 0, because there is no net motion perpendicular to l.P32-13 qvy B = Fx = m dvx /dt; qvx B = Fy = m dvy /dt. Taking the time derivative of thesecond expression and inserting into the rst we getqvy B = m mqBd2 vy,dt2which has solution vy = v sin(mt/qB), where v is a constant. Using the second equation we ndthat there is a similar solution for vx , except that it is out of phase, and so vx = v cos(mt/qB).Integrating,qBvx = vx dt = v cos(mt/qB) =sin(mt/qB).mSimilarly,qBvy = vy dt = v sin(mt/qB) =cos(mt/qB).mThis is the equation of a circle.P32-14dL = + + kdz. B is uniform, so that the integral can be written asidx jdy F=ibut sincedx =( + + kdz) B = i Bidx jdy idy =dx + i Bjdz = 0, the entire expression vanishes.104dy + ik Bdz,P32-15The current pulse provides an impulse which is equal toF dt =BiL dt = BLi dt = BLq.This gives an initial velocity of v0 = BLq/m, which will cause the rod to hop to a height of2h = v0 /2g = B 2 L2 q 2 /2m2 g.Solving for q,q=mBL2gh =(0.013 kg)(0.12 T)(0.20 m)2(9.8 m/s2 )(3.1 m) = 4.2 C.P32-16P32-17 The torque on a current carrying loop depends on the orientation of the loop; themaximum torque occurs when the plane of the loop is parallel to the magnetic eld. In this case themagnitude of the torque is from Eq. 32-34 with sin = 1 = N iAB.The area of a circular loop is A = r2 where r is the radius, but since the circumference is C = 2r,we can writeC2A=.4The circumference is not the length of the wire, because there may be more than one turn. Instead,C = L/N , where N is the number of turns.Finally, we can write the torque as = NiL2iL2 BB=,4N 24Nwhich is a maximum when N is a minimum, or N = 1.P32-18 dF = i dL B; the direction of dF will be upward and somewhat toward the center. Land B are a right angles, but only the upward component of dF will survive the integration as thecentral components will cancel out by symmetry. HenceF = iB sin P32-19dL = 2riB sin .The torque on the cylinder from gravity is g = mgr sin ,where r is the radius of the cylinder. The torque from magnetism needs to balance this, somgr sin = N iAB sin = N i2rLB sin ,ori=mg(0.262 kg)(9.8 m/s2 )== 1.63 A.2N LB2(13)(0.127 m)(0.477 T)105E33-1 (a) The magnetic eld from a moving charge is given by Eq. 33-5. If the protons aremoving side by side then the angle is = /2, soB=0 qv4 r2and we are interested is a distance r = d. The electric eld at that distance isE=1 q,4 0 r2where in both of the above expressions q is the charge of the source proton.On the receiving end is the other proton, and the force on that proton is given byF = q(E + v B).The velocity is the same as that of the rst proton (otherwise they wouldnt be moving side by side.)This velocity is then perpendicular to the magnetic eld, and the resulting direction for the crossproduct will be opposite to the direction of E. Then for balance,E1 q4 0 r210 0= vB,0 qv= v,4 r2= v2 .We can solve this easily enough, and we nd v 3 108 m/s.(b) This is clearly a relativistic speed!E33-2 B = 0 i/2d = (4 107 T m/A)(120 A)/2(6.3 m) = 3.8106 T. This will deect thecompass needle by as much as one degree. However, there is unlikely to be a place on the Earthssurface where the magnetic eld is 210 T. This was likely a typo, and should probably have been21.0 T. The deection would then be some ten degrees, and that is signicant.E33-3 B = 0 i/2d = (4107 T m/A)(50 A)/2(1.3103 m) = 37.7103 T.E33-4 (a) i = 2dB/0 = 2(8.13102 m)(39.0106 T)/(4107 T m/A) = 15.9 A.(b) Due East.E33-5UseB=0 i(4107 N/A2 )(1.6 1019 C)(5.6 1014 s1 )== 1.2108 T.2d2(0.0015 m)E33-6 Zero, by symmetry. Any contributions from the top wire are exactly canceled by contributions from the bottom wire.E33-7 B = 0 i/2d = (4107 T m/A)(48.8 A)/2(5.2102 m) = 1.88104 T.F = qv B. All cases are either parallel or perpendicular, so either F = 0 or F = qvB.(a) F = qvB = (1.601019 C)(1.08107 m/s)(1.88104 T) = 3.241016 N. The direction ofF is parallel to the current.(b) F = qvB = (1.601019 C)(1.08107 m/s)(1.88104 T) = 3.241016 N. The direction ofF is radially outward from the current.(c) F = 0.106E33-8 We want B1 = B2 , but with opposite directions. Then i1 /d1 = i2 /d2 , since all constantscancel out. Then i2 = (6.6 A)(1.5 cm)/(2.25 cm) = 4.4 A, directed out of the page.E33-9 For a single long straight wire, B = 0 i/2d but we need a factor of 2 since there aretwo wires, then i = dB/0 . FinallydB(0.0405 m)(296, T)== 30 A0(4107 N/A2 )i=E33-10 (a) The semi-circle part contributes half of Eq. 33-21, or 0 i/4R. Each long straight wirecontributes half of Eq. 33-13, or 0 i/4R. Add the three contributions and getBa =0 i4R2+1=(4107 N/A2 )(11.5 A)4(5.20103 m)2+1= 1.14103 T.The direction is out of the page.(b) Each long straight wire contributes Eq. 33-13, or 0 i/2R. Add the two contributions andget0 i(4107 N/A2 )(11.5 A)== 8.85104 T.Ba =R(5.20103 m)The direction is out of the page.E33-11 z 3 = 0 iR2 /2B = (4 107 N/A2 )(320)(4.20 A)(2.40102 m)2 /2(5.0106 T) = 9.73102 m3 . Then z = 0.46 m.E33-12 The circular part contributes a fraction of Eq. 33-21, or 0 i/4R. Each long straightwire contributes half of Eq. 33-13, or 0 i/4R. Add the three contributions and get0 i( 2).4RB=The goal is to get B = 0 that will happen if = 2 radians.E33-13 There are four current segments that could contribute to the magnetic eld. The straightsegments, however, contribute nothing because the straight segments carry currents either directlytoward or directly away from the point P .That leaves the two rounded segments. Each contribution to B can be found by starting withEq. 33-21, or 0 i/4b. The direction is out of the page.There is also a contribution from the top arc; the calculations are almost identical except thatthis is pointing into the page and r = a, so 0 i/4a. The net magnetic eld at P is thenB = B1 + B2 =0 i41 1b a.E33-14 For each straight wire segment use Eq. 33-12. When the length of wire is L, the distanceto the center is W/2; when the length of wire is W the distance to the center is L/2. There are fourterms, but they are equal in pairs, soB=0 i4=20 i L2 + W 24LWL2 /4+W 2 /4+L2W2+WL WL1074W,+ W 2 /420 i L2 + W 2=.WLLL2 /4E33-15 We imagine the ribbon conductor to be a collection of thin wires, each of thickness dxand carrying a current di. di and dx are related by di/dx = i/w. The contribution of one of thesethin wires to the magnetic eld at P is dB = 0 di/2x, where x is the distance from this thin wireto the point P . We want to change variables to x and integrate, soB=dB =0 i dx0 i=2wx2wdx.xThe limits of integration are from d to d + w, soB=0 iln2wd+wd.E33-16 The elds from each wire are perpendicular at P . Each contributes an amount B =0 i/2d, but since they are perpendicular there is a net eld of magnitude B = 2B 2 = 20 i/2d.Note that a = 2d, so B = 0 i/a.(a) B = (4107 T m/A)(115 A)/(0.122 m) = 3.77104 T. The direction is to the left.(b) Same numerical result, except the direction is up.E33-17 Follow along with Sample Problem 33-4.Reversing the direction of the second wire (so that now both currents are directed out of thepage) will also reverse the direction of B2 . ThenB0 i112 b + x b x(b x) (b + x),b2 x2x.2 b2x= B1 B2 ===0 i20 i,E33-18 (b) By symmetry, only the horizontal component of B survives, and must point to theright.(a) The horizontal component of the eld contributed by the top wire is given byB=0 i b/20 ib0 isin ==,2h2h h(4R2 + b2 )since h is the hypotenuse, or h = R2 + b2 /4. But there are two such components, one from thetop wire, and an identical component from the bottom wire.E33-19loop,(a) We can use Eq. 33-21 to nd the magnetic eld strength at the center of the large0 i(4107 T m/A)(13 A)== 6.8105 T.2R2(0.12 m)(b) The torque on the smaller loop in the center is given by Eq. 32-34,B= = N iA B,but since the magnetic eld from the large loop is perpendicular to the plane of the large loop, andthe plane of the small loop is also perpendicular to the plane of the large loop, the magnetic eld isin the plane of the small loop. This means that |A B| = AB. Consequently, the magnitude of thetorque on the small loop is = N iAB = (50)(1.3 A)()(8.2103 m)2 (6.8105 T) = 9.3107 N m.108E33-20 (a) There are two contributions to the eld. One is from the circular loop, and is givenby 0 i/2R. The other is from the long straight wire, and is given by 0 i/2R. The two elds areout of the page and parallel, so0 iB=(1 + 1/).2R(b) The two components are now at right angles, soB=0 i2R1 + 1/ 2 .The direction is given by tan = 1/ or = 18 .E33-21 The force per meter for any pair of parallel currents is given by Eq. 33-25, F/L = 0 i2 /2d,where d is the separation. The direction of the force is along the line connecting the intersection ofthe currents with the perpendicular plane. Each current experiences three forces; two are at right22angles and equal in magnitude, so |F12 + F14 |/L = F12 + F14 /L = 20 i2 /2a. The third forcepoints parallel to this sum, but d = a, so the resultant force is20 i20 i24 107 N/A2 (18.7 A)2 F=+ =( 2 + 1/ 2) = 6.06104 N/m.L2a2(0.245 m)2 2aIt is directed toward the center of the square.E33-22 By symmetry we expect the middle wire to have a net force of zero; the two on the outsidewill each be attracted toward the center, but the answers will be symmetrically distributed.For the wire which is the farthest left,F0 i2=L21111+++a 2a 3a 4a=4 107 N/A2 (3.22 A)22(0.083 m)1+1 1 1+ +2 3 4= 5.21105 N/m.For the second wire over, the contributions from the two adjacent wires should cancel. ThisleavesF0 i2=L2E33-2311+ +2a 3a=4 107 N/A2 (3.22 A)22(0.083 m)1 1+2 3= 2.08105 N/m.(a) The force on the projectile is given by the integral ofdF = i dl Bover the length of the projectile (which is w). The magnetic eld strength can be found from addingtogether the contributions from each rail. If the rails are circular and the distance between them issmall compared to the length of the wire we can use Eq. 33-13,B=0 i,2xwhere x is the distance from the center of the rail. There is one problem, however, because theseare not wires of innite length. Since the current stops traveling along the rail when it reaches theprojectile we have a rod that is only half of an innite rod, so we need to multiply by a factor of1/2. But there are two rails, and each will contribute to the eld, so the net magnetic eld strengthbetween the rails is0 i0 iB=+.4x 4(2r + w x)109In that last term we have an expression that is a measure of the distance from the center of thelower rail in terms of the distance x from the center of the upper rail.The magnitude of the force on the projectile is thenr+wF= iB dx,r==0 i2 r+w 11+4 rx 2r + w x0 i2r+w2 ln4rdx,The current through the projectile is down the page; the magnetic eld through the projectile isinto the page; so the force on the projectile, according to F = il B, is to the right.(b) Numerically the magnitude of the force on the rail isF =(450103 A)2 (4107 N/A2 )ln2(0.067 m) + (0.012 m)(0.067 m)= 6.65103 NThe speed of the rail can be found from either energy conservation so we rst nd the work doneon the projectile,W = F d = (6.65103 N)(4.0 m) = 2.66104 J.This work results in a change in the kinetic energy, so the nal speed isv=2K/m =2(2.66104 J)/(0.010 kg) = 2.31103 m/s.E33-24 The contributions from the left end and the right end of the square cancel out. This leavesthe top and the bottom. The net force is the dierence, or=(4107 N/A2 )(28.6 A)(21.8 A)(0.323 m)2=F3.27103 N.112 m)(1.1010(10.30102 m),E33-25 The magnetic force on the upper wire near the point d isFB =0 ia ib L0 ia ib L 0 ia ib Lx,2(d + x)2d2d2where x is the distance from the equilibrium point d. The equilibrium magnetic force is equal to theforce of gravity mg, so near the equilibrium point we can writexFB = mg mg .dThere is then a restoring force against small perturbations of magnitude mgx/d which correspondsto a spring constant of k = mg/d. This would give a frequency of oscillation off=12k/m =12g/d,which is identical to the pendulum.E33-26 B = (4107 N/A2 )(3.58 A)(1230)/(0.956m) = 5.79103 T.110E33-27 The magnetic eld inside an ideal solenoid is given by Eq. 33-28 B = 0 in, where n isthe turns per unit length. Solving for n,n=B(0.0224 T)== 1.00103 /m1 .0 i(4107 N/A2 )(17.8 A)The solenoid has a length of 1.33 m, so the total number of turns isN = nL = (1.00103 /m1 )(1.33 m) = 1330,and since each turn has a length of one circumference, then the total length of the wire which makesup the solenoid is (1330)(0.026 m) = 109 m.E33-28 From the solenoid we haveB s = 0 nis = 0 (11500/m)(1.94 mA) = 0 (22.3A/m).From the wire we have0 iw0 (6.3 A)0== (1.002 A)2r2rrThese elds are at right angles, so we are interested in when tan(40 ) = B w /B s , orBw =r=(1.002 A)= 5.35102 m.tan(40 )(22.3 A/m)E33-29 Let u = z d. ThenB=0 niR22d+L/2dL/2du,[R2 + u2 ]3/22==u0 niR2R 2 R 2 + u20 ni2d+L/2,dL/2d + L/2R2+ (d +L/2)2d L/2R2+ (d L/2)2.If L is much, much greater than R and d then |L/2 d| >> R, and R can be ignored in thedenominator of the above expressions, which then simplify to=0 ni2=B0 ni2d + L/2R2+ (d +L/2)2d + L/2(d +L/2)2d L/2R2+ (d L/2)2d L/2(d L/2)2..= 0 in.It is important that we consider the relative size of L/2 and d!E33-30 The net current in the loop is 1i0 + 3i0 + 7i0 6i0 = 5i0 . ThenB ds = 50 i0 .E33-31 (a) The path is clockwise, so a positive current is into page. The net current is 2.0 A out,so B ds = 0 i0 = 2.5106 T m.(b) The net current is zero, so B ds = 0.111E33-32 Let R0 be the radius of the wire. On the surface of the wire B0 = 0 i/2R0 .Outside the wire we have B = 0 i/2R, this is half B0 when R = 2R0 .2Inside the wire we have B = 0 iR/2R0 , this is half B0 when R = R0 /2.E33-33looks like(a) We dont want to reinvent the wheel. The answer is found from Eq. 33-34, except it0 ir.2c2(b) In the region between the wires the magnetic eld looks like Eq. 33-13,B=B=0 i.2rThis is derived on the right hand side of page 761.(c) Amperes law (Eq. 33-29) is B ds = 0 i, where i is the current enclosed. Our Amperianloop will still be a circle centered on the axis of the problem, so the left hand side of the aboveequation will reduce to 2rB, just like in Eq. 33-32. The right hand side, however, depends onthe net current enclosed which is the current i in the center wire minus the fraction of the currentenclosed in the outer conductor. The cross sectional area of the outer conductor is (a2 b2 ), sothe fraction of the outer current enclosed in the Amperian loop isi(r2 b2 )r 2 b2=i 2.2 b2 )(aa b2The net current in the loop is theniir 2 b2a2 r2=i 2,a2 b2a b2so the magnetic eld in this region isB=0 i a2 r2.2r a2 b2(d) This part is easy since the net current is zero; consequently B = 0.E33-34 (a) Amperes law (Eq. 33-29) is B ds = 0 i, where i is the current enclosed. OurAmperian loop will still be a circle centered on the axis of the problem, so the left hand side of theabove equation will reduce to 2rB, just like in Eq. 33-32. The right hand side, however, dependson the net current enclosed which is the fraction of the current enclosed in the conductor. The crosssectional area of the conductor is (a2 b2 ), so the fraction of the current enclosed in the Amperianloop is(r2 b2 )r 2 b2i=i 2.(a2 b2 )a b2The magnetic eld in this region isB=0 i r2 b2.2r a2 b2(b) If r = a, thenB=0 i a2 b20 i=,2a a2 b22awhich is what we expect.112If r = b, then0 i b2 b2= 0,2b a2 b2B=which is what we expect.If b = 0, thenB=0 i r2 020 ir=2r a2 022a2which is what I expected.E33-35 The magnitude of the magnetic eld due to the cylinder will be zero at the center ofthe cylinder and 0 i0 /2(2R) at point P . The magnitude of the magnetic eld eld due to thewire will be 0 i/2(3R) at the center of the cylinder but 0 i/2R at P . In order for the neteld to have dierent directions in the two locations the currents in the wire and pipe must be indierent direction. The net eld at the center of the pipe is 0 i/2(3R), while that at P is then0 i0 /2(2R) 0 i/2R. Set these equal and solve for i;i/3 = i0 /2 i,or i = 3i0 /8.E33-36 (a) B = (4107 N/A2 )(0.813 A)(535)/2(0.162 m) = 5.37104 T.(b) B = (4107 N/A2 )(0.813 A)(535)/2(0.162 m + 0.052 m) = 4.07104 T.E33-37 (a) A positive particle would experience a magnetic force directed to the right for amagnetic eld out of the page. This particle is going the other way, so it must be negative.(b) The magnetic eld of a toroid is given by Eq. 33-36,0 iN,2rwhile the radius of curvature of a charged particle in a magnetic eld is given by Eq. 32-10mvR=.|q|BB=We use the R to distinguish it from r. Combining,R=2mvr,0 iN |q|so the two radii are directly proportional. This meansR/(11 cm) = (110 cm)/(125 cm),so R = 9.7 cm.P33-1The eld from one coil is given by Eq. 33-19B=0 iR2.2(R2 + z 2 )3/2There are N turns in the coil, so we need a factor of N . There are two coils and we are interestedin the magnetic eld at P , a distance R/2 from each coil. The magnetic eld strength will be twicethe above expression but with z = R/2, soB=20 N iR280 N i=.2 + (R/2)2 )3/22(R(5)3/2 R113P33-2(a) Change the limits of integration that lead to Eq. 33-12:B=0 id4L0(z 2dz,+ d2 )3/2L==z0 id,4 (z 2 + d2 )1/2 00 idL.4 (L2 + d2 )1/2(b) The angle in Eq. 33-11 would always be 0, so sin = 0, and therefore B = 0.P33-3 This problem is the all important derivation of the Helmholtz coil properties.(a) The magnetic eld from one coil isB1 =0 N iR2.2(R2 + z 2 )3/2The magnetic eld from the other coil, located a distance s away, but for points measured from therst coil, is0 N iR2B2 =.2(R2 + (z s)2 )3/2The magnetic eld on the axis between the coils is the sum,B=0 N iR20 N iR2+.2(R2 + z 2 )3/22(R2 + (z s)2 )3/2Take the derivative with respect to z and get30 N iR230 N iR2dB=z(z s).2 + z 2 )5/22 + (z s)2 )5/2dz2(R2(RAt z = s/2 this expression vanishes! We expect this by symmetry, because the magnetic eld willbe strongest in the plane of either coil, so the mid-point should be a local minimum.(b) Take the derivative again andd2 Bdz 230 N iR2150 N iR2 2+z2 + z 2 )5/22(R2(R2 + z 2 )5/230 N iR2150 N iR2+(z s)2 .2 + (z s)2 )5/22 + (z s)2 )5/22(R2(R= We could try and simplify this, but we dont really want to; we instead want to set it equal to zero,then let z = s/2, and then solve for s. The second derivative will equal zero when3(R2 + z 2 ) + 15z 2 3(R2 + (z s)2 ) + 15(z s)2 = 0,and is z = s/2 this expression will simplify to30(s/2)2 = 6(R2 + (s/2)2 ),4(s/2)2 = R2 ,s = R.114P33-4 (a) Each of the side of the square is a straight wire segment of length a which contributesa eld strength ofa0 i,B=4r a2 /4 + r2where r is the distance to the point on the axis of the loop, soa2 /4 + z 2 .r=This eld is not parallel to the z axis; the z component is Bz = B(a/2)/r. There are four of thesecontributions. The o axis components cancel. Consequently, the eld for the square isB=40 i4raa2 /4+r22=0 i2r2=0 i2 /4 + z 2 )2(a=a/2,raa2 /4 + r2,a2a2 /2 + z 2,40 ia2.2)+ 4z2a2 + 4z 2(a2(b) When z = 0 this reduces toB=40 i a240 i=.(a2 ) 2a22 aP33-5 (a) The polygon has n sides. A perpendicular bisector of each side can be drawn to thecenter and has length x where x/a = cos(/n). Each side has a length L = 2a sin(/n). Each of theside of the polygon is a straight wire segment which contributes a eld strength ofB=0 i4xLL2 /4+ x2,This eld is parallel to the z axis. There are n of these contributions. The o axis componentscancel. Consequently, the eld for the polygon= n0 i4xL2 /4= n0 i4L2 /4= nBL0 i 1tan(/n),2 a+ x22+ x2,tan(/n),since (L/2)2 + x2 = a2 .(b) Evaluate:lim n tan(/n) = lim n sin(/n) n/n = .nnThen the answer to part (a) simplies toB=0 i.2a115P33-6 For a square loop of wire we have four nite length segments each contributing a termwhich looks like Eq. 33-12, except that L is replaced by L/4 and d is replaced by L/8. Then at thecenter,L/4160 i0 i.=B=42 /64 + L2 /644L/8 L2LFor a circular loop R = L/2 so0 i0=.2RLSince 16/ 2 > , the square wins. But only by some 7%!B=P33-7 We want to use the dierential expression in Eq. 33-11, except that the limits of integration are going to be dierent. We have four wire segments. From the top segment,B10 id4 z 2 + d2=0 i4d=3L/4,L/43L/4(3L/4)2+d2L/4(L/4)2 + d2.For the top segment d = L/4, so this simplies even further to0 i B1 =2(3 5 + 5) .10LThe bottom segment has the same integral, but d = 3L/4, so0 i B3 =2( 5 + 5) .30LBy symmetry, the contribution from the right hand side is the same as the bottom, so B2 = B3 ,and the contribution from the left hand side is the same as that from the top, so B4 = B1 . Addingall four terms, 20 i3 2(3 5 + 5) + 2( 5 + 5) ,B =30L20 i =(2 2 + 10).3LP33-8 Assume a current ring has a radius r and a width dr, the charge on the ring is dq = 2r dr,where = q/R2 . The current in the ring is di = dq/2 = r dr. The ring contributes a elddB = 0 di/2r. Integrate over all the rings:RB=0 r dr/2r = 0 R/2 = q/2R.0P33-9B = 0 in and mv = qBr. Combine, andi=P33-10mv(5.11105 eV/c2 )(0.046c)== 0.271 A.0 qrn(4107 N/A2 )e(0.023 m)(10000/m)This shape is a triangle with area A = (4d)(3d)/2 = 6d2 . The enclosed current is theni = jA = (15 A/m2 )6(0.23 m)2 = 4.76 AThe line integral is then0 i = 6.0106 T m.116P33-11 Assume that B does vary as the picture implies. Then the line integral along the pathshown must be nonzero, since B l on the right is not zero, while it is along the three other sides.Hence B dl is non zero, implying some current passes through the dotted path. But it doesnt,so B cannot have an abrupt change.P33-12 (a) Sketch an Amperian loop which is a rectangle which enclosed N wires, has a verticalsides with height h, and horizontal sides with length L. Then B dl = 0 N i. Evaluate the integralalong the four sides. The vertical side contribute nothing, since B is perpendicular to h, and thenB h = 0. If the integral is performed in a counterclockwise direction (it must, since the senseof integration was determined by assuming the current is positive), we get BL for each horizontalsection. Then10 iN= 0 in.B=2L2(b) As a then tan1 (a/2R) /2. Then B 0 i/2a. If we assume that i is made up ofseveral wires, each with current i0 , then i/a = i0 n.P33-13 Apply Amperes law with an Amperian loop that is a circle centered on the center ofthe wire. ThenB ds = B ds = B ds = 2rB,because B is tangent to the path and B is uniform along the path by symmetry. The currentenclosed isienc = j dA.This integral is best done in polar coordinates, so dA = (dr)(r d), and thenrienc2=(j0 r/a) rdr d,00r=r2 dr,2j0 /a0=2j0 3r .3aWhen r = a the current enclosed is i, soi=2j0 a23ior j0 =.32a2The magnetic eld strength inside the wire is found by gluing together the two parts of Ampereslaw,2rBB2j0 3r ,3a0 j0 r2,3a20 ir.2a3= 0==117P33-14 (a) According to Eq. 33-34, the magnetic eld inside the wire without a hole has magnitudeB = 0 ir/2R2 = 0 jr/2 and is directed radially. If we superimpose a second current to create thehole, the additional eld at the center of the hole is zero, so B = 0 jb/2. But the current in theremaining wire isi = jA = j(R2 a2 ),soB=0 ib.2(R2 a2 )118E34-1 B = B A = (42106 T)(2.5 m2 ) cos(57 ) = 5.7105 Wb.E34-2 |E| = |dB /dt| = A dB/dt = (/4)(0.112 m)2 (0.157 T/s) = 1.55 mV.E34-3(a) The magnitude of the emf induced in a loop is given by Eq. 34-4,|E| = NdB,dt= N (12 mWb/s2 )t + (7 mWb/s)There is only one loop, and we want to evaluate this expression for t = 2.0 s, so|E| = (1) (12 mWb/s2 )(2.0 s) + (7 mWb/s) = 31 mV.(b) This part isnt harder. The magnetic ux through the loop is increasing when t = 2.0 s. Theinduced current needs to ow in such a direction to create a second magnetic eld to oppose thisincrease. The original magnetic eld is out of the page and we oppose the increase by pointing theother way, so the second eld will point into the page (inside the loop).By the right hand rule this means the induced current is clockwise through the loop, or to theleft through the resistor.E34-4 E = dB /dt = A dB/dt.(a) E = (0.16 m)2 (0.5 T)/(2 s) = 2.0102 V.(b) E = (0.16 m)2 (0.0 T)/(2 s) = 0.0102 V.(c) E = (0.16 m)2 (0.5 T)/(4 s) = 1.0102 V.E34-5 (a) R = L/A = (1.69108 m)[()(0.104 m)]/[(/4)(2.50103 m)2 ] = 1.12103 .(b) E = iR = (9.66 A)(1.12103 ) = 1.08102 V. The required dB/dt is then given bydBE== (1.08102 V)/(/4)(0.104 m)2 = 1.27 T/s.dtAE34-6 E = A B/t = AB/t. The power is P = iE = E 2 /R. The energy dissipated isE = P t =E 2 tA2 B 2=.RRtE34-7 (a) We could re-derive the steps in the sample problem, or we could start with the endresult. Well start with the result,diE = N A0 n,dtexcept that we have gone ahead and used the derivative instead of the .The rate of change in the current isdi= (3.0 A/s) + (1.0 A/s2 )t,dtso the induced emf is=(130)(3.46104 m2 )(4107 Tm/A)(2.2104 /m) (3.0A/s) + (2.0A/s2 )t ,=E(3.73103 V) + (2.48103 V/s)t.(b) When t = 2.0 s the induced emf is 8.69103 V, so the induced current isi = (8.69103 V)/(0.15 ) = 5.8102 A.119E34-8 (a) i = E/R = N A dB/dt. Note that A refers to the area enclosed by the outer solenoidwhere B is non-zero. This A is then the cross sectional area of the inner solenoid! Theni=1di(120)(/4)(0.032 m)2 (4107 N/A2 )(220102 /m) (1.5 A)N A0 n == 4.7103 A.Rdt(5.3 )(0.16 s)E34-9 P = Ei = E 2 /R = (A dB/dt)2 /(L/a), where A is the area of the loop and a is the crosssectional area of the wire. But a = d2 /4 and A = L2 /4, soP =L3 d264dBdt2=(0.525 m)3 (1.1103 m)2(9.82103 T/s)2 = 4.97106 W.64(1.69108 m)E34-10 B = BA = B(2.3 m)2 /2. EB = dB /dt = AdB/dt, orEB = (2.3 m)2[(0.87 T/s)] = 2.30 V,2so E = (2.0 V) + (2.3 V) = 4.3 V.E34-11 (a) The induced emf, as a function of time, is given by Eq. 34-5, E(t) = dB (t)/dtThis emf drives a current through the loop which obeys E(t) = i(t)R Combining,i(t) = 1 dB (t).R dtSince the current is dened by i = dq/dt we can writedq(t)1 dB (t)=.dtR dtFactor out the dt from both sides, and then integrate:1dB (t),R1dq(t) = dB (t),R1q(t) q(0) =(B (0) B (t))Rdq(t)= (b) No. The induced current could have increased from zero to some positive value, then decreasedto zero and became negative, so that the net charge to ow through the resistor was zero. Thiswould be like sloshing the charge back and forth through the loop.E34-12 P hiB = 2B = 2N BA. Then the charge to ow through isq = 2(125)(1.57 T)(12.2104 m2 )/(13.3 ) = 3.60102 C.E34-13 The part above the long straight wire (a distance b a above it) cancels out contributionsbelow the wire (a distance b a beneath it). The ux through the loop is thenaB =2ab0 i0 ibb dr =ln2r2120a2a b.The emf in the loop is thenE =dB0 b=lndt2a2a b[2(4.5 A/s2 )t (10 A/s)].Evaluating,E=4107 N/A2 (0.16 m)ln2(0.12 m)2(0.12 m) (0.16 m)[2(4.5 A/s2 )(3.0 s)(10 A/s)] = 2.20107 V.E34-14 Use Eq. 34-6: E = BDv = (55106 T)(1.10 m)(25 m/s) = 1.5103 V.E34-15 If the angle doesnt vary then the ux, given by=BAis constant, so there is no emf.E34-16 (a) Use Eq. 34-6: E = BDv = (1.18T)(0.108 m)(4.86 m/s) = 0.619 V.(b) i = (0.619 V)/(0.415 ) = 1.49 A.(c) P = (0.619 V)(1.49 A) = 0.922 W.(d) F = iLB = (1.49 A)(0.108 m)(1.18T) = 0.190 N.(e) P = F v = (0.190 V)(4.86 m/s) = 0.923 W.E34-17 The magnetic eld is out of the page, and the current through the rod is down. Then Eq.32-26 F = iL B shows that the direction of the magnetic force is to the right; furthermore, sinceeverything is perpendicular to everything else, we can get rid of the vector nature of the problemand write F = iLB. Newtons second law gives F = ma, and the acceleration of an object from restresults in a velocity given by v = at. Combining,v(t) =iLBt.mE34-18 (b) The rod will accelerate as long as there is a net force on it. This net force comes fromF = iLB. The current is given by iR = E BLv, so as v increases i decreases. When i = 0 the rodstops accelerating and assumes a terminal velocity.(a) E = BLv will give the terminal velocity. In this case, v = E/BL.E34-19E34-20 The acceleration is a = R 2 ; since E = BR2 /2, we can nda = 4E 2 /B 2 R3 = 4(1.4 V)2 /(1.2 T)2 (5.3102 m)3 = 3.7104 m/s2 .E34-21 We will use the results of Exercise 11 that were worked out above. All we need to do isnd the initial ux; ipping the coil up-side-down will simply change the sign of the ux.SoB (0) = B A = (59 T)()(0.13 m)2 sin(20 ) = 1.1106 Wb.121Then using the results of Exercise 11 we haveqN(B (0) B (t)),R950=((1.1106 Wb) (1.1106 Wb)),85 = 2.5105 C.=E34-22 (a) The ux through the loop isvtB =a+Ldx0draThe emf is thenE =0 i0 ivt a + L=ln.2r2a0 iv a + LdB=ln.dt2aPutting in the numbers,E=(4107 N/A2 )(110 A)(4.86 m/s) (0.0102 m) + (0.0983 m)ln= 2.53104 V.2(0.0102 m)(b) i = E/R = (2.53104 V)/(0.415 ) = 6.10104 A.(c) P = i2 R = (6.10104 A)2 (0.415 ) = 1.54107 W.(d) F = Bil dl, ora+L0 i0 i a + L= illn.F = ildr2r2aaPutting in the numbers,F = (6.10104 A)(4107 N/A2 )(110 A) (0.0102 m) + (0.0983 m)ln= 3.17108 N.2(0.0102 m)(e) P = F v = (3.17108 N)(4.86 m/s) = 1.54107 W.E34-23(a) Starting from the beginning, Eq. 33-13 givesB=0 i.2yThe ux through the loop is given byB dA,B =but since the magnetic eld from the long straight wire goes through the loop perpendicular to theplane of the loop this expression simplies to a scalar integral. The loop is a rectangular, so usedA = dx dy, and let x be parallel to the long straight wire.Combining the above,D+bBa=D==0D+b0 i2y0 idya,2yD0 iD+ba ln2D122dx dy,(b) The ux through the loop is a function of the distance D from the wire. If the loop movesaway from the wire at a constant speed v, then the distance D varies as vt. The induced emf is thenEdB,dt0 iba.2 t(vt + b)= =The current will be this emf divided by the resistance R. The back-of-the-book answer is somewhatdierent; the answer is expressed in terms of D instead if t. The two answers are otherwise identical.E34-24 (a) The area of the triangle is A = x2 tan /2. In this case x = vt, soB = B(vt)2 tan /2,and thenE = 2Bv 2 t tan /2,(b) t = E/2Bv 2 tan /2, sot=(56.8 V)= 2.08 s.2(0.352 T)(5.21 m/s)2 tan(55 )E34-25 E = N BA, so=(24 V)= 39.4 rad/s.(97)(0.33 T)(0.0190 m2 )Thats 6.3 rev/second.E34-26 (a) The frequency of the emf is the same as the frequency of rotation, f .(b) The ux changes by BA = Ba2 during a half a revolution. This is a sinusoidal change, sothe amplitude of the sinusoidal variation in the emf is E = B /2. Then E = B 2 a2 f .E34-27 We can use Eq. 34-10; the emf is E = BA sin t, This will be a maximum whensin t = 1. The angular frequency, is equal to = (1000)(2)/(60) rad/s = 105 rad/s Themaximum emf is thenE = (3.5 T) [(100)(0.5 m)(0.3 m)] (105 rad/s) = 5.5 kV.E34-28 (a) The amplitude of the emf is E = BA, soA = E/2f B = (150 V)/2(60/s)(0.50 T) = 0.798m2 .(b) Divide the previous result by 100. A = 79.8 cm2 .E34-29 dB /dt = A dB/dt = A(8.50 mT/s).(a) For this pathE ds = dB /dt = (0.212 m)2 (8.50 mT/s) = 1.20 mV.(b) For this pathE ds = dB /dt = (0.323 m)2 (8.50 mT/s) = 2.79 mV.123(c) For this pathE ds = dB /dt = (0.323 m)2 (8.50 mT/s) (0.323 m)2 (8.50 mT/s) = 1.59 mV.E34-30 dB /dt = A dB/dt = A(6.51 mT/s), while E ds = 2rE.(a) The path of integration is inside the solenoid, soE=(0.022 m)(6.51 mT/s)r2 (6.51 mT/s)== 7.16105 V/m.2r2(b) The path of integration is outside the solenoid, soE=E34-31r2 (6.51 mT/s)(0.063 m)2 (6.51 mT/s)== 1.58104 V/m2R2(0.082 m)The induced electric eld can be found from applying Eq. 34-13,E ds = dB.dtWe start with the left hand side of this expression. The problem has cylindrical symmetry, so theinduced electric eld lines should be circles centered on the axis of the cylindrical volume. If wechoose the path of integration to lie along an electric eld line, then the electric eld E will beparallel to ds, and E will be uniform along this path, soE ds =E ds = Eds = 2rE,where r is the radius of the circular path.Now for the right hand side. The ux is contained in the path of integration, so B = Br2 .All of the time dependence of the ux is contained in B, so we can immediately write2rE = r2dBr dBor E = .dt2 dtWhat does the negative sign mean? The path of integration is chosen so that if our right hand ngerscurl around the path our thumb gives the direction of the magnetic eld which cuts through thepath. Since the eld points into the page a positive electric eld would have a clockwise orientation.Since B is decreasing the derivative is negative, but we get another negative from the equationabove, so the electric eld has a positive direction.Now for the magnitude.E = (4.82102 m)(10.7103 T /s)/2 = 2.58104 N/C.The acceleration of the electron at either a or c then has magnitudea = Eq/m = (2.58104 N/C)(1.601019 C)/(9.111031 kg) = 4.53107 m/s2 .P34-1 The induced current is given by i = E/R. The resistance of the loop is given by R = L/A,where A is the cross sectional area. Combining, and writing in terms of the radius of the wire, wehaver2 Ei=.L124The length of the wire is related to the radius of the wire because we have a xed mass. The totalvolume of the wire is r2 L, and this is related to the mass and density by m = r2 L. Eliminatingr we havemEi=.L2The length of the wire loop is the same as the circumference, which is related to the radius R of theloop by L = 2R. The emf is related to the changing ux by E = dB /dt, but if the shape of theloop is xed this becomes E = A dB/dt. Combining all of this,i=mA dB.(2R)2 dtWe dropped the negative sign because we are only interested in absolute values here.Now A = R2 , so this expression can also be written asi=mR2 dBm dB=.(2R)2 dt4 dtP34-2 For the lower surface B A = (76103 T)(/2)(0.037 m)2 cos(62 ) = 7.67105 Wb. Forthe upper surface B A = (76103 T)(/2)(0.037 m)2 cos(28 ) = 1.44104 Wb.. The induced emfis thenE = (7.67105 Wb + 1.44104 Wb)/(4.5103 s) = 4.9102 V.P34-3 (a) We are only interested in the portion of the ring in the yz plane. Then E = (3.32103 T/s)(/4)(0.104 m)2 = 2.82105 V.(b) From c to b. Point your right thumb along x to oppose the increasing B eld. Your rightngers will curl from c to b.P34-4 E N A, but A = r2 and N 2r = L, so E 1/N . This means use only one loop tomaximize the emf.P34-5 This is a integral best performed in rectangular coordinates, then dA = (dx)(dy). Themagnetic eld is perpendicular to the surface area, so B dA = B dA. The ux is thenBB dA ==aa(4 T/m s2 )t2 y dy dx,=00= (4 T/m s2 )t2=B dA,1 2a a,2(2 T/m s2 )a3 t2 .But a = 2.0 cm, so this becomesB = (2 T/m s2 )(0.02 m)3 t2 = (1.6105 Wb/s2 )t2 .The emf around the square is given bydB= (3.2105 Wb/s2 )t,dtand at t = 2.5 s this is 8.0105 V. Since the magnetic eld is directed out of the page, a positiveemf would be counterclockwise (hold your right thumb in the direction of the magnetic eld andyour ngers will give a counter clockwise sense around the loop). But the answer was negative, sothe emf must be clockwise.E =125P34-6then(a) Far from the plane of the large loop we can approximate the large loop as a dipole, andB=0 iR2.2x3The ux through the small loop is thenB = r2 B =0 i 2 r2 R2.2x3(b) E = dB /dt, so30 i 2 r2 R2v.2x4(c) Anti-clockwise when viewed from above.E=P34-7thenThe magnetic eld is perpendicular to the surface area, so B dA = B dA. The ux isB dA =B =B dA = BA,since the magnetic eld is uniform. The area is A = r2 , where r is the radius of the loop. Theinduced emf isdBdrE == 2rB .dtdtIt is given that B = 0.785 T, r = 1.23 m, and dr/dt = 7.50102 m/s. The negative sign indicatea decreasing radius. ThenE = 2(1.23 m)(0.785 T)(7.50102 m/s) = 0.455 V.P34-8 (a) dB /dt = B dA/dt, but dA/dt is A/t, where A is the area swept out during onerotation and t = 1/f . But the area swept out is R2 , so|E| =dB= f BR2 .dt(b) If the output current is i then the power is P = Ei. But P = = 2f , so=P34-9P= iBR2 /2.2f(a) E = dB /dt, and B = B A,soE = BLv cos .The component of the force of gravity on the rod which pulls it down the incline is FG = mg sin .The component of the magnetic force on the rod which pulls it up the incline is FB = BiL cos .Equating,BiL cos = mg sin ,and since E = iR,v=EmgR sin = 2 2.BL cos B L cos2 (b) P = iE = E 2 /R = B 2 L2 v 2 cos2 /R = mgv sin . This is identical to the rate of change ofgravitational potential energy.126P34-10 Let the cross section of the wire be a.(a) R = L/a = (r + 2r)/a; with numbers,R = (3.4103 )(2 + ).(b) B = Br2 /2; with numbers,B = (4.32103 Wb).(c) i = E/R = Br2 /2R = Bar/2( + 2), ori=Batr.(t2 + 4)Take the derivative and set it equal to zero,0=4 at2,(t2 + 4)2so at2 = 4, or = 1 at2 = 2 rad.2(d) = 2, so2(0.15 T)(1.2106 m2 ) 2(12 rad/s )(2 rad)(0.24 m)i=(1.7108 m)(6 rad)= 2.2 A.P34-11 It does say approximate, so we will be making some rather bold assumptions here. Firstwe will nd an expression for the emf. Since B is constant, the emf must be caused by a change inthe area; in this case a shift in position. The small square where B = 0 has a width a and sweepsaround the disk with a speed r. An approximation for the emf is then E = Bar. This emf causesa current. We dont know exactly where the current ows, but we can reasonably assume that itoccurs near the location of the magnetic eld. Let us assume that it is constrained to that region ofthe disk. The resistance of this portion of the disk is the approximatelyR=1L1 a1== ,A attwhere we have assumed that the current is owing radially when dening the cross sectional area ofthe resistor. The induced current is then (on the order of)EBar== Bart.R1/(t)This current experiences a breaking force according to F = BIl, soF = B 2 a2 rt,where l is the length through which the current ows, which is a.Finally we can nd the torque from = rF , and = B 2 a2 r2 t.127P34-12 The induced electric eld in the ring is given by Eq. 34-11: 2RE = |dB /dt|. Thiselectric eld will result in a force on the free charge carries (electrons?), given by F = Ee. Theacceleration of the electrons is then a = Ee/me . Thena=dBe.2Rme dtIntegrate both sides with respect to time to nd the speed of the electrons.a dtvedBdt,2Rme dtdBe2Rme,eB .2Rme===The current density is given by j = nev, and the current by iA = ia2 . Combining,i=ne2 a2P hiB .2RmeActually, it should be pointed out that P hiB refers to the change in ux from external sources. Thecurrent induced in the wire will produce a ux which will exactly oset P hiB so that the net uxthrough the superconducting ring is xed at the value present when the ring became superconducting.P34-13 Assume that E does vary as the picture implies. Then the line integral along the pathshown must be nonzero, since E l on the right is not zero, while it is along the three other sides.Hence E dl is non zero, implying a change in the magnetic ux through the dotted path. But itdoesnt, so E cannot have an abrupt change.P34-14The electric eld a distance r from the center is given byE=r2 dB/dTr dB=.2r2 dtThis eld is directed perpendicular to the radial lines.Dene h to be the distance from the center of the circle to the center of the rod, and evaluateE = E ds,E==dBrhdx,dt2rdB Lh.dt 2But h2 = R2 (L/2)2 , soE=P34-15dB Ldt 2R2 (L/2)2 .(a) B = r2 B av , soE=E(0.32 m)=2(0.28 T)(120 ) = 34 V/m.2r2(b) a = F/m = Eq/m = (33.8 V/m)(1.61019 C)/(9.11031 kg) = 6.01012 m/s2 .128E35-1 If the Earths magnetic dipole moment were produced by a single current around the core,then that current would bei=(8.0 1022 J/T)== 2.1109 AA(3.5 106 m)2E35-2 (a) i = /A = (2.33 A m2 )/(160)(0.0193 m)2 = 12.4 A.(b) = B = (2.33 A m2 )(0.0346 T) = 8.06102 N m.E35-3 (a) Using the right hand rule a clockwise current would generate a magnetic moment whichwould be into the page. Both currents are clockwise, so add the moments: = (7.00 A)(0.20 m)2 + (7.00 A)(0.30 m)2 = 2.86 A m2 .(b) Reversing the current reverses the moment, so = (7.00 A)(0.20 m)2 (7.00 A)(0.30 m)2 = 1.10 A m2 .E35-4 (a) = iA = (2.58 A)(0.16 m)2 = 0.207 A m2 .(b) = B sin = (0.207 A m2 )(1.20 T) sin(41 ) = 0.163 N m.E35-5(a) The result from Problem 33-4 for a square loop of wire wasB(z) =40 ia2.(4z 2 + a2 )(4z 2 + 2a2 )1/2For z much, much larger than a we can ignore any a terms which are added to or subtracted fromz terms. This means that4z 2 + a2 4z 2 and (4z 2 + 2a2 )1/2 2z,but we cant ignore the a2 in the numerator.The expression for B then simplies toB(z) =0 ia2,2z 3which certainly looks like Eq. 35-4.(b) We can rearrange this expression and getB(z) =0ia2 ,2z 3where it is rather evident that ia2 must correspond to , the dipole moment, in Eq. 35-4. So thatmust be the answer.E35-6 = iA = (0.2 A)(0.08 m)2 = 4.02103 A m2 ; = .n(a) For the torque, = B = (9.65104 N m) + (7.24104 N m) + (8.08104 N m)k.ij(b) For the magnetic potential energy,U = B = [(0.60)(0.25 T)] = 0.603103 J.129E35-7 = iA = i(a2 + b2 /2) = i(a2 + b2 )/2.E35-8 If the distance to P is very large compared to a or b we can write the Law of Biot andSavart as0 i s rB=.4 r3s is perpendicular to r for the left and right sides, so the left side contributesB1 =0 ib,4 (x + a/2)2and the right side contributesB3 = 0 ib.4 (x a/2)2The top and bottom sides each contribute an equal amountB2 = B4 =0 i a sin 0 i a(b/2).4 x2 + b2 /44 x3Add the four terms, expand the denominators, and keep only terms in x3 ,B=0 i ab0 =.34 x4 x3The negative sign indicates that it is into the page.E35-9(a) The electric eld at this distance from the proton isE=14(8.851012 C2 /Nm2 )(1.601019 C)= 5.141011 N/C.(5.291011 m)2(b) The magnetic eld at this from the proton is given by the dipole approximation,B(z)===0 ,2z 3(4107 N/A2 )(1.411026 A/m2 ),2(5.291011 m)31.90102 TE35-10 1.50 g of water has (2)(6.02 1023 )(1.5)/(18) = 1.00 1023 hydrogen nuclei. If all arealigned the net magnetic moment would be = (1.001023 )(1.411026 J/T) = 1.41103 J/T.The eld strength is thenB=0 (1.41103 J/T)= (1.00107 N/A2 )= 9.31013 T.4 x3(5.33 m)3E35-11 (a) There is eectively a current of i = f q = q/2. The dipole moment is then = iA =(q/2)(r2 ) = 1 qr2 .2(b) The rotational inertia of the ring is mr2 so L = I = mr2 . Then(1/2)qr2q==.Lmr2 2m130E35-12 The mass of the bar is3m = V = (7.87 g/cm )(4.86 cm)(1.31 cm2 ) = 50.1 g.The number of atoms in the bar isN = (6.021023 )(50.1 g)/(55.8 g) = 5.411023 .The dipole moment of the bar is then = (5.411023 )(2.22)(9.271024 J/T) = 11.6 J/T.(b) The torque on the magnet is = (11.6 J/T)(1.53 T) = 17.7 N m.E35-13The magnetic dipole moment is given by = M V , Eq. 35-13. Then = (5, 300 A/m)(0.048 m)(0.0055 m)2 = 0.024 A m2 .E35-14 (a) The original eld is B0 = 0 in. The eld will increase to B = m B0 , so the increase isB = (1 1)0 in = (3.3104 )(4107 N/A2 )(1.3 A)(1600/m) = 8.6107 T.(b) M = (1 1)B0 /0 = (1 1)in = (3.3104 )(1.3 A)(1600/m) = 0.69 A/m.E35-15 The energy to ip the dipoles is given by U = 2B. The temperature is thenT =2B4(1.21023 J/T)(0.5 T)== 0.58 K.3k/23(1.381023 J/K)E35-16 The Curie temperature of iron is 770 C, which is 750 C higher than the surface temperature. This occurs at a depth of (750 C)/(30 C /km) = 25 km.E35-17 (a) Look at the gure. At 50% (which is 0.5 on the vertical axis), the curve is atB0 /T 0.55 T/K. Since T = 300 K, we have B0 165 T.(b) Same gure, but now look at the 90% mark. B0 /T 1.60 T/K, so B0 480 T.(c) Good question. I think both elds are far beyond our current abilities.E35-18 (a) Look at the gure. At 50% (which is 0.5 on the vertical axis), the curve is at B0 /T 0.55 T/K. Since B0 = 1.8 T, we have T (1.8 T)/(0.55 T/K) = 3.3 K.(b) Same gure, but now look at the 90% mark. B0 /T 1.60 T/K, so T (1.8 T)/(1.60 T/K) =1.1 K.E35-19 Since (0.5 T)/(10 K) = 0.05 T/K, and all higher temperatures have lower values of theratio, and this puts all points in the region near where Curies Law (the straight line) is valid, thenthe answer is yes.E35-20 Using Eq. 35-19,n =Mr M(108g/mol)(511103 A/m)VM=== 8.741021 A/m2NA(10490 kg/m3 )(6.021023 /mol)131E35-21 (a) B = 0 /2z 3 , soB=(4107 N/A2 )(1.51023 J/T)= 9.4106 T.2(10109 m)3(b) U = 2B = 2(1.51023 J/T)(9.4106 T) = 2.821028 J.E35-22 B = (43106 T)(295, 000106 m2 ) = 1.3107 Wb.E35-23 (a) Well assume, however, that all of the iron atoms are perfectly aligned. Then thedipole moment of the earth will be related to the dipole moment of one atom byEarth = N Fe ,where N is the number of iron atoms in the magnetized sphere. If mA is the relative atomic massof iron, then the total mass isN mAmA Earthm==,AA Fewhere A is Avogadros number. Next, the volume of a sphere of mass m ismmA EarthV ==,A Fewhere is the density.And nally, the radius of a sphere with this volume would ber=3V41/3=3Earth mA4Fe A1/3.Now we nd the radius by substituting in the known values,r=3(8.01022 J/T)(56 g/mol)1/3= 1.8105 m.34(14106 g/m )(2.11023 J/T)(6.01023 /mol)(b) The fractional volume is the cube of the fractional radius, so the answer is(1.8105 m/6.4106 )3 = 2.2105 .E35-24 (a) At magnetic equator Lm = 0, so0 (1.00107 N/A2 )(8.01022 J/T)== 31T.4r3(6.37106 m)3B=There is no vertical component, so the inclination is zero.(b) Here Lm = 60 , soB=0 4r31 + 3 sin2 Lm =(1.00107 N/A2 )(8.01022 J/T)(6.37106 m)31 + 3 sin2 (60 ) = 56T.The inclination is given byarctan(B v /B h ) = arctan(2 tan Lm ) = 74 .(c) At magnetic north pole Lm = 90 , soB=0 2(1.00107 N/A2 )(8.01022 J/T)== 62T.32r(6.37106 m)3There is no horizontal component, so the inclination is 90 .132E35-25 This problem is eectively solving 1/r3 = 1/2 for r measured in Earth radii. Thenr = 1.26rE , and the altitude above the Earth is (0.26)(6.37106 m) = 1.66106 m.E35-26 The radial distance from the center is r = (6.37106 m) (2900103 m) = 3.47106 m.The eld strength isB=0 2(1.00107 N/A2 )(8.01022 J/T)== 380T.32r(3.47106 m)3E35-27 Here Lm = 90 11.5 = 78.5 , soB=0 4r31 + 3 sin2 Lm =(1.00107 N/A2 )(8.01022 J/T)(6.37106 m)31 + 3 sin2 (78.5 ) = 61T.The inclination is given byarctan(B v /B h ) = arctan(2 tan Lm ) = 84 .E35-28 The ux out the other end is (1.6103 T)(0.13 m)2 = 85Wb. The net ux throughthe surface is zero, so the ux through the curved surface is 0 (85Wb) (25Wb) = 60Wb..The negative indicates inward.E35-29 The total magnetic ux through a closed surface is zero. There is inward ux on facesone, three, and ve for a total of -9 Wb. There is outward ux on faces two and four for a total of+6 Wb. The dierence is +3 Wb; consequently the outward ux on the sixth face must be +3 Wb.E35-30 The stable arrangements are (a) and (c). The torque in each case is zero.E35-31 The eld on the x axis between the wires isB=0 i211+2r + x 2r x.Since B dA = 0, we can assume the ux through the curved surface is equal to the ux throughthe xz plane within the cylinder. This ux isrB= Lr0 i11+2 2r + x 2r x3rrln ln,r3rdx,0 i20 i= Lln 3.= LP35-1 We can imagine the rotating disk as being composed of a number of rotating rings ofradius r, width dr, and circumference 2r. The surface charge density on the disk is = q/R2 ,and consequently the (dierential) charge on any ring isdq = (2r)(dr) =2qrdrR2The rings rotate with angular frequency , or period T = 2/. The eective (dierential) currentfor each ring is thendqqrdi ==dr.TR2133Each ring contributes to the magnetic moment, and we can glue all of this together as ==d,r2 di,R==qr3 dr,R20qR2 .4P35-2 (a) The sphere can be sliced into disks. The disks can be sliced into rings. Each ring hassome charge qi , radius ri , and mass mi ; the period of rotation for a ring is T = 2/, so the currentin the ring is qi /T = qi /2. The magnetic moment is22i = (qi /2)ri = qi ri /2.2Note that this is closely related to the expression for angular momentum of a ring: li = mi ri .Equating,i = qi li /2mi .If both mass density and charge density are uniform then we can write qi /mi = q/m,=d = (q/2m)dl = qL/2mFor a solid sphere L = I = 2mR2 /5, so = qR2 /5.(b) See part (a)P35-3 (a) The orbital speed is given by K = mv 2 /2. The orbital radius is given by mv = qBr, orr = mv/qB. The frequency of revolution is f = v/2r. The eective current is i = qf . Combiningall of the above to nd the dipole moment, = iA = qvvrmv 2Kr2 = q=q= .2r22qBB(b) Since q and m cancel out of the above expression the answer is the same!(c) Work it out:M=(5.281021 /m3 )(6.211020 J) (5.281021 /m3 )(7.581021 J)=+= 312 A/m.V(1.18 T)(1.18 T)P35-4 (b) Point the thumb or your right hand to the right. Your ngers curl in the direction ofthe current in the wire loop.(c) In the vicinity of the wire of the loop B has a component which is directed radially outward.Then B ds has a component directed to the left. Hence, the net force is directed to the left.P35-5 (b) Point the thumb or your right hand to the left. Your ngers curl in the direction of thecurrent in the wire loop.(c) In the vicinity of the wire of the loop B has a component which is directed radially outward.Then B ds has a component directed to the right. Hence, the net force is directed to the right.134P35-6 (a) Let x = B/kT . Adopt the convention that N+ refers to the atoms which have parallelalignment and N those which are anti-parallel. Then N+ + N = N , soN+ = N ex /(ex + ex ),andN = N ex /(ex + ex ),Note that the denominators are necessary so that N+ + N = N . Finally,M = (N+ N ) = N(b) If BkT then x is very small and ex 1 x. The above expression reduces toM = N(c) If Bex ex.ex + ex(1 + x) (1 x)2 B= N x =.(1 + x) + (1 x)kTkT then x is very large and ex while ex 0. The above expression reducestoN = N.P35-7(a) Centripetal acceleration is given by a = r 2 . Thena a02= r(0 + )2 r0 ,= 2r0 + r(0 )2 , 2r0 .(b) The change in centripetal acceleration is caused by the additional magnetic force, which hasmagnitude FB = qvB = erB. Then =a a0eB=.2r02mNote that we boldly canceled against 0 in this last expression; we are assuming that is small,and for these problems it is.P35-8 (a) i = /A = (8.01022 J/T)/(6.37106 m)2 = 6.3108 A.(b) Far enough away both elds act like perfect dipoles, and can then cancel.(c) Close enough neither eld acts like a perfect dipole and the elds will not cancel.P35-9(a) B =B h 2 + B v 2 , soB=0 4r3cos2 Lm + 4 sin2 Lm =(b) tan i = B v /B h = 2 sin Lm / cos Lm = 2 tan Lm .1350 4r31 + 3 sin2 Lm .E36-1The important relationship is Eq. 36-4, written asB =(5.0 mA)(8.0 mH)iL== 1.0107 WbN(400)E36-2 (a) = (34)(2.62103 T)(0.103 m)2 = 2.97103 Wb.(b) L = /i = (2.97103 Wb)/(3.77 A) = 7.88104 H.E36-3 n = 1/d, where d is the diameter of the wire. Then0 A(4107 H/m)(/4)(4.10102 m)2L= 0 n2 A = 2 == 2.61104 H/m.ld(2.52103 m)2E36-4 (a) The emf supports the current, so the current must be decreasing.(b) L = E/(di/dt) = (17 V)/(25103 A/s) = 6.8104 H.E36-5(a) Eq. 36-1 can be used to nd the inductance of the coil.L=EL(3.0 mV)== 6.0104 H.di/dt(5.0 A/s)(b) Eq. 36-4 can then be used to nd the number of turns in the coil.N=iL(8.0 A)(6.0104 H)== 120B(40Wb)E36-6 Use the equation between Eqs. 36-9 and 36-10.=(4107 H/m)(0.81 A)(536)(5.2102 m) (5.2102 m) + (15.3102 m)ln,2(15.3102 m)=B1.32106 Wb.E36-7 L = m 0 n2 Al = m 0 N 2 A/l, orL = (968)(4107 H/m)(1870)2 (/4)(5.45102 m)2 /(1.26 m) = 7.88 H.E36-8 In each case apply E = Li/t.(a) E = (4.6 H)(7 A)/(2103 s) = 1.6104 V.(b) E = (4.6 H)(2 A)/(3103 s) = 3.1103 V.(c) E = (4.6 H)(5 A)/(1103 s) = 2.3104 V.E36-9 (a) If two inductors are connected in parallel then the current through each inductor willadd to the total current through the circuit, i = i1 + i2 , Take the derivative of the current withrespect to time and then di/dt = di1 /dt + di2 /dt,The potential dierence across each inductor is the same, so if we divide by E and apply we getdi/dtdi1 /dt di2 /dt=+,EEEButdi/dt1= ,EL136so the previous expression can also be written as111=+.LeqL1L2(b) If the inductors are close enough together then the magnetic eld from one coil will inducecurrents in the other coil. Then we will need to consider mutual induction eects, but that is a topicnot covered in this text.E36-10 (a) If two inductors are connected in series then the emf across each inductor will add tothe total emf across both, E = E1 + E2 ,Then the current through each inductor is the same, so if we divide by di/dt and apply we getE1E2E=+,di/dtdi/dt di/dtButE= L,di/dtso the previous expression can also be written asLeq = L1 + L2 .(b) If the inductors are close enough together then the magnetic eld from one coil will inducecurrents in the other coil. Then we will need to consider mutual induction eects, but that is a topicnot covered in this text.E36-11 Use Eq. 36-17, but rearrange:L =t(1.50 s)== 0.317 s.ln[i0 /i]ln[(1.16 A)/(10.2103 A)]Then R = L/L = (9.44 H)/(0.317 s) = 29.8 .E36-12 (a) There is no current through the resistor, so ER = 0 and then EL = E.(b) EL = Ee2 = (0.135)E.(c) n = ln(EL /E) = ln(1/2) = 0.693.E36-13(a) From Eq. 36-4 we nd the inductance to beL=N B(26.2103 Wb)== 4.78103 H.i(5.48 A)Note that B is the ux, while the quantity N B is the number of ux linkages.(b) We can nd the time constant from Eq. 36-14,L = L/R = (4.78103 H)/(0.745 ) = 6.42103 s.The we can invert Eq. 36-13 to gett= L ln 1 Ri(t)E,= (6.42103 s) ln 1 (0.745 A)(2.53 A)(6.00 V)137= 2.42103 s.E36-14 (a) Rearrange:EEiRRdtL= iR + L==di,dtL di,R dtdi.E/R i(b) Integrate:t0RdtLR tLE t/LeRE1 et/LRi==di,i E/R0i + E/Rln,E/R= i + E/R,= i.E36-15 di/dt = (5.0 A/s). ThenE = iR + Ldi= (3.0 A)(4.0 ) + (5.0 A/s)t(4.0 ) + (6.0 H)(5.0 A/s) = 42 V + (20 V/s)t.dtE36-16 (1/3) = (1 et/L ), soL = E36-17(5.22 s)t== 12.9 s.ln(2/3)ln(2/3)We want to take the derivative of the current in Eq. 36-13 with respect to time,diE 1 t/LE=e= et/L .dtR LLThen L = (5.0102 H)/(180 ) = 2.78104 s. Using this we nd the rate of change in the currentwhen t = 1.2 ms to be34di(45 V)=e(1.210 s)/(2.7810 s) = 12 A/s.2 H)dt((5.010E36-18 (b) Consider some time ti :EL (ti ) = Eeti /L .Taking a ratio for two dierent times,EL (t1 )= e(t2 t1 )/L ,EL (t2 )orL =t2 t1(2 ms) (1 ms)== 3.58 msln[EL (t1 )/EL (t2 )]ln[(18.24 V)/(13.8 V)](a) Choose any time, andE = EL et/L = (18.24 V)e(1 ms)/(3.58 ms) = 24 V.138E36-19 (a) When the switch is just closed there is no current through the inductor. So i1 = i2 isgiven byE(100 V)i1 === 3.33 A.R1 + R2(10 ) + (20 )(b) A long time later there is current through the inductor, but it is as if the inductor has noeect on the circuit. Then the eective resistance of the circuit is found by rst nding the equivalentresistance of the parallel part1/(30 ) + 1/(20 ) = 1/(12 ),and then nding the equivalent resistance of the circuit(10 ) + (12 ) = 22 .Finally, i1 = (100 V)/(22 ) = 4.55 A andV2 = (100 V) (4.55 A)(10 ) = 54.5 V;consequently, i2 = (54.5 V)/(20 ) = 2.73 A. It didnt ask, but i2 = (4.55 A) (2.73 A) = 1.82 A.(c) After the switch is just opened the current through the battery stops, while that through theinductor continues on. Then i2 = i3 = 1.82 A.(d) All go to zero.E36-20 (a) For toroids L = 0 N 2 h ln(b/a)/2. The number of turns is limited by the inner radius:N d = 2a. In this case,N = 2(0.10 m)/(0.00096 m) = 654.The inductance is thenL=(4107 H/m)(654)2 (0.02 m) (0.12 m)ln= 3.1104 H.2(0.10 m)(b) Each turn has a length of 4(0.02 m) = 0.08 m. The resistance is thenR = N (0.08 m)(0.021 /m) = 1.10 The time constant isL = L/R = (3.1104 H)/(1.10 ) = 2.8104 s.E36-21 (I) When the switch is just closed there is no current through the inductor or R2 , so thepotential dierence across the inductor must be 10 V. The potential dierence across R1 is always10 V when the switch is closed, regardless of the amount of time elapsed since closing.(a) i1 = (10 V)/(5.0 ) = 2.0 A.(b) Zero; read the above paragraph.(c) The current through the switch is the sum of the above two currents, or 2.0 A.(d) Zero, since the current through R2 is zero.(e) 10 V, since the potential across R2 is zero.(f) Look at the results of Exercise 36-17. When t = 0 the rate of change of the current isdi/dt = E/L. Thendi/dt = (10 V)/(5.0 H) = 2.0 A/s.(II) After the switch has been closed for a long period of time the currents are stable and theinductor no longer has an eect on the circuit. Then the circuit is a simple two resistor parallelnetwork, each resistor has a potential dierence of 10 V across it.139(a) Still 2.0 A; nothing has changed.(b) i2 = (10 V)/(10 ) = 1.0 A.(c) Add the two currents and the current through the switch will be 3.0 A.(d) 10 V; see the above discussion.(e) Zero, since the current is no longer changing.(f) Zero, since the current is no longer changing.E36-22 U = (71 J/m3 )(0.022 m3 ) = 1.56 J. Then using U = i2 L/2 we geti=2U/L =2(1.56 J)/(0.092 H) = 5.8 A.E36-23 (a) L = 2U/i2 = 2(0.0253 J)/(0.062 A)2 = 13.2 H.(b) Since the current is squared in the energy expression, doubling the current would quadruplethe energy. Then i = 2i0 = 2(0.062 A) = 0.124 A.E36-24 (a) B = 0 in and u = B 2 /20 , oru = 0 i2 n2 /2 = (4107 N/A2 )(6.57 A)2 (950/0.853 m)2 /2 = 33.6 J/m3 .(b) U = uAL = (33.6 J/m3 )(17.2104 m2 )(0.853 m) = 4.93102 J.E36-25 uB = B 2 /20 , and from Sample Problem 33-2 we know B, henceuB =(12.6 T)2= 6.32107 J/m3 .2(4107 N/A2 )E36-26 (a) uB = B 2 /20 , souB =(1001012 T)213= 2.5102 eV/cm .2(4107 N/A2 ) (1.61019 J/eV)(b) x = (10)(9.461015 m) = 9.461016 m. Using the results from part (a) expressed in J/m3 wend the energy contained isU = (3.981015 J/m3 )(9.461016 m)3 = 3.41036 JE36-27 The energy density of an electric eld is given by Eq. 36-23; that of a magnetic eld isgiven by Eq. 36-22. Equating,0E2=E2=1 2B ,20B.0 0The answer is thenE = (0.50 T)/ (8.851012 C2 /N m2 )(4107 N/A2 ) = 1.5108 V/m.E36-28 The rate of internal energy increase in the resistor is given by P = iVR . The rate ofenergy storage in the inductor is dU/dt = Li di/dt = iVL . Since the current is the same throughboth we want to nd the time when VR = VL . Using Eq. 36-15 we nd1 et/L = et/L ,ln 2 = t/L ,so t = (37.5 ms) ln 2 = 26.0 ms.140E36-29 (a) Start with Eq. 36-13:= E(1 et/L )/R,iiR1E= et/L ,=t,ln(1 iR/E)=(5.20103 s),ln[1 (1.96103 A)(10.4103 )/(55.0 V)]=L1.12102 s.Then L = L R = (1.12102 s)(10.4103 ) = 116 H.(b) U = (1/2)(116 H)(1.96103 A)2 = 2.23104 J.E36-30 (a) U = Eq; q =idt.U= E==E1 et/L dt,R2E2t + L et/L ,R0E2E 2 L t/Lt + 2 (e 1).RRUsing the numbers provided,L = (5.48 H)/(7.34 ) = 0.7466 s.ThenU=(12.2 V)2(2 s) + (0.7466 s)(e(2 s)/0.7466 s) 1) = 26.4 J(7.34 )(b) The energy stored in the inductor is UL = Li2 /2, orUL==LE 22R26.57 J.1 et/L2dt,(c) UR = U UL = 19.8 J.E36-31This shell has a volume ofV =4(RE + a)3 RE 3 .3Since a << RE we can expand the polynomials but keep only the terms which are linear in a. ThenV 4RE 2 a = 4(6.37106 m)2 (1.6104 m) = 8.21018 m3 .The magnetic energy density is found from Eq. 36-22,uB =1 2(60106 T)2B == 1.43103 J/m3 .202(4107 N/A2 )The total energy is then (1.43103 J/m3 )(8.2eex18m3 ) = 1.21016 J.141E36-32 (a) B = 0 i/2r and uB = B 2 /20 = 0 i2 /8 2 r2 , oruB = (4107 H/m)(10 A)2 /8 2 (1.25103 m)2 = 1.0 J/m3 .(b) E = V /l = iR/l and uE =0E2/2 =0i2(R/l)2 /2. ThenuE = (8.851012 F/m)(10 A)2 (3.3103 /m)2 /2 = 4.81015 J/m3 .E36-33 i =2U/L =2(11.2106 J)/(1.48103 H) = 0.123 A.E36-34 C = q 2 /2U = (1.63106 C)2 /2(142106 J) = 9.36109 F.E36-35 1/2f =LC so L = 1/4 2 f 2 C, orL = 1/4 2 (10103 Hz)2 (6.7106 F) = 3.8105 H.E36-36 qmax 2 /2C = Limax 2 /2, orimax = qmax / LC = (2.94106 C)/ (1.13103 H)(3.88106 F) = 4.44102 A.E36-37 Closing a switch has the eect of shorting out the relevant circuit element, whicheectively removes it from the circuit. If S1 is closed we have C = RC or C = C /R, if instead S2is closed we have L = L/R or L = RL , but if instead S3 is closed we have a LC circuit which willoscillate with period2T == 2 LC.Substituting from the expressions above,T =2= 2 L C .E36-38 The capacitors can be used individually, or in series, or in parallel. The four possiblecapacitances are then 2.00F, 5.00F, 2.00F + 5.00F = 7.00F, and (2.00F )(5.00F)(2.00F +5.00F) = 1.43F. The possible resonant frequencies are then121LC= f,121(10.0 mH)(1.43F )= 1330 Hz,121(10.0 mH)(2.00F )= 1130 Hz,121(10.0 mH)(5.00F )= 712 Hz,121(10.0 mH)(7.00F )= 602 Hz.142E36-39 (a) k = (8.13 N)/(0.0021 m) = 3.87103 N/m. = k/m =89.3 rad/s.(b) T = 2/ = 2/(89.3 rad/s) = 7.03102 s.(c) LC = 1/ 2 , soC = 1/(89.3 rad/s)2 (5.20 H) = 2.41105 F.(3870 N/m)/(0.485 kg) =E36-40 The period of oscillation is T = 2 LC = 2 (52.2mH)(4.21F) = 2.95 ms. It requiresone-quarter period for the capacitor to charge, or 0.736 ms.E36-41 (a) An LC circuit oscillates so that the energy is converted from all magnetic to allelectrical twice each cycle. It occurs twice because once the energy is magnetic with the currentowing in one direction through the inductor, and later the energy is magnetic with the currentowing the other direction through the inductor.The period is then four times 1.52s, or 6.08s.(b) The frequency is the reciprocal of the period, or 164000 Hz.(c) Since it occurs twice during each oscillation it is equal to half a period, or 3.04s.E36-42 (a) q = CV = (1.13109 F)(2.87 V) = 3.24109 C.(c) U = q 2 /2C = (3.24109 C)2 /2(1.13109 F) = 4.64109 J.(b) i = 2U/L = 2(4.64109 J)/(3.17103 H) = 1.71103 A.E36-43 (a) im = q m and q m = CV m . Multiplying the second expression by L we get Lq m =V m / 2 . Combining, Lim = V m . Thenf=(50 V)== 6.1103 /s.22(0.042 H)(0.031 A)(b) See (a) above.(c) C = 1/ 2 L = 1/(26.1103 /s)2 (0.042 H) = 1.6108 F.E36-44 (a) f = 1/2 LC = 1/2 (6.2106 F)(54103 H) = 275 Hz.(b) Note that from Eq. 36-32 we can deduce imax = qmax . The capacitor starts with a chargeq = CV = (6.2106 F)(34 V) = 2.11104 C. Then the current amplitude isimax = qmax / LC = (2.11104 C)/ (6.2106 F)(54103 H) = 0.365 A.E36-45 (a) = 1/ LC = 1/ (10106 F)(3.0103 H) = 5800 rad/s.(b) T = 2/ = 2/(5800 rad/s) = 1.1103 s.E36-46 f = (2105 Hz)(1 + /180 ). C = 4 2 /f 2 L, soC=4 2(9.9107 F)=.(2105 Hz)2 (1 + /180 )2 (1 mH)(1 + /180 )22E36-47 (a) UE = UB /2 and UE + UB = U , so 3UE = U , or 3(q 2 /2C) = qmax /2C, so q = qmax / 3.(b) Solve q = qmax cos t, ort=Tarccos 1/ 3 = 0.152T.2143E36-48 (a) Add the contribution from the inductor and the capacitor,U=(b) q m =(c) im =(3.83106 C)2(24.8103 H)(9.16103 A)2+= 1.99106 J.22(7.73106 F)2(7.73106 F)(1.99106 J) = 5.55106 C.2(1.99106 J)/(24.8103 H) = 1.27102 A.E36-49 (a) The frequency of such a system is given by Eq. 36-26, f = 1/2 LC. Note thatmaximum frequency occurs with minimum capacitance. Thenf1=f2C2=C1(365 pF)= 6.04.(10 pF)(b) The desired ratio is 1.60/0.54 = 2.96 Adding a capacitor in parallel will result in an eectivecapacitance given byC 1,e = C1 + C add ,with a similar expression for C2 . We want to choose C add so thatf1=f2C 2,e= 2.96.C 1,eSolving,C 2,eC2 + C addC add= C 1,e (2.96)2 ,= (C1 + C add )8.76,C2 8.76C1=,7.76(365 pF) 8.76(10 pF)== 36 pF.7.76The necessary inductance is thenL=14 2 f 2 C=14 2 (0.54106 Hz)2 (4011012 F)= 2.2104 H.E36-50 The key here is that UE = C(V )2 /2. We want to charge a capacitor with one-ninth thecapacitance to have three times the potential dierence. Since 32 = 9, it is reasonable to assumethat we want to take all of the energy from the rst capacitor and give it to the second. ClosingS1 and S2 will not work, because the energy will be shared. Instead, close S2 until the capacitorhas completely discharged into the inductor, then simultaneously open S2 while closing S1 . Theinductor will then discharge into the second capacitor. Open S1 when it is full.E36-51 (a) = 1/ LC.qm =im= (2.0 A) (3.0103 H)(2.7106 F) = 1.80104 C(b) dUC /dt = qi/C. Since q = q m sin t and i = im cos t then dUC /dt is a maximum whensin t cos t is a maximum, or when sin 2t is a maximum. This happens when 2t = /2, ort = T /8.(c) dUC /dt = q m im /2C, ordUC /dt = (1.80104 C)(2.0 A)/2(2.7106 F) = 67 W.144E36-52 After only complete cycles q = qmax eRt/2L . Not only that, but t = N , where = 2/ .Finally, = (1/LC) (R/2L)2 . Since the rst term under the square root is so much larger thanthe second, we can ignore the eect of damping on the frequency, and simply use = 1/ LC.Thenq = qmax eN R /2L = qmax eN R LC/L = qmax eN R C/L .Finally, RC/L = (7.22 ) (3.18 F)/(12.3 H) = 1.15102 . ThenN =5N =5N =5:::q = (6.31C)e5(0.0115) = 5.96C,q = (6.31C)e10(0.0115) = 5.62C,q = (6.31C)e100(0.0115) = 1.99C.E36-53 Use Eq. 36-40, and since U q 2 , we want to solve eRt/L = 1/2, thenLln 2.Rt=E36-54 Start by assuming that the presence of the resistance does not signicantly change thefrequency. Then = 1/ LC, q = qmax eRt/2L , t = N , and = 2/. Combining,q = qmax eN R /2L = qmax eN R LC/L = qmax eN R C/L .ThenR=L/Cln(q/qmax ) = NIt remains to be veried that 1/LCE36-55be(220mH)/(12F)ln(0.99) = 8700 .(50)(R/2L)2 .The damped (angular) frequency is given by Eq. 36-41; the fractional change would then= 1 1 (R/2L)2 = 1 Setting this equal to 0.01% and then solving for R,R=P36-14L(1 (1 0.0001)2 ) =C1 (R2 C/4L).4(12.6103 H)(1.9999104 ) = 2.96 .(1.15106 F)The inductance of a toroid isL=0 N 2 h bln .2aIf the toroid is very large and thin then we can write b = a + , where << a. The natural log thencan be approximated asbln = ln 1 + .aaaThe product of and h is the cross sectional area of the toroid, while 2a is the circumference,which we will call l. The inductance of the toroid then reduces toL0 N 2 A0 N 2 =.2 alBut N is the number of turns, which can also be written as N = nl, where n is the turns per unitlength. Substitute this in and we arrive at Eq. 36-7.145P36-2 (a) Since ni is the net current per unit length and is this case i/W , we can simply writeB = 0 i/W .(b) There is only one loop of wire, soL = B /i = BA/i = 0 iR2 /W i = 0 R2 /W.P36-3 Choose the y axis so that it is parallel to the wires and directly between them. The eldin the plane between the wires isB=11+d/2 + x d/2 x0 i2.The ux per length l of the wires isd/2aB= lB dx = ld/2+a==d/2a0 i2d/2+a0 i d/2a12 d/2+a d/2 + x0 i d a2lln.2a2lThe inductance is thenL=11+d/2 + x d/2 xdx,dx,B0 l d a=ln.iaP36-4 (a) Choose the y axis so that it is parallel to the wires and directly between them. Theeld in the plane between the wires isB=0 i211+d/2 + x d/2 x.The ux per length l between the wires isd/2a1=B dx =d/2+a==0 i2d/2ad/2+a0 i d/2a12 d/2+a d/2 + x0 i d a2ln.2a211+d/2 + x d/2 xdx,dx,The eld in the plane inside one of the wires, but still between the centers isB=0 i21d/2 x+d/2 + xa2.The additional ux is thend/22=2B dx = 2d/2a=20 i2lnd/20 i2d1+da 2d/2a.1461d/2 x+d/2 + xa2dx,The ux per meter between the axes of the wire is the sum, orB0 id 1ln +,a 2(4107 H/m)(11.3 A)==ln(21.8, mm) 1+(1.3 mm)2,1.5105 Wb/m.=(b) The fraction f inside the wires isf===d1d 1+/ ln +,da 2a 2(21.8, mm)1+/(21.8, mm) (1.3 mm) 20.09.ln(21.8, mm) 1+(1.3 mm)2,(c) The net ux is zero for parallel currents.P36-5The magnetic eld in the region between the conductors of a coaxial cable is given byB=0 i,2rso the ux through an area of length l, width b a, and perpendicular to B isB=B dA =b==B dA,l0 idz dr,a0 2r0 il bln .2aWe evaluated this integral is cylindrical coordinates: dA = (dr)(dz). As you have been warned somany times before, learn these dierentials!The inductance is then0 l bB=ln .L=i2aP36-6 (a) So long as the fuse is not blown there is eectively no resistance in the circuit. Thenthe equation for the current is E = L di/dt, but since E is constant, this has a solution i = Et/L.The fuse blows when t = imax L/E = (3.0 A)(5.0 H)/(10 V) = 1.5 s.(b) Note that once the fuse blows the maximum steady state current is 2/3 A, so there must bean exponential approach to that current.P36-7 The initial rate of increase is di/dt = E/L. Since the steady state current is E/R, thecurrent will reach the steady state value in a time given by E/R = i = Et/L, or t = L/R. But thatsL .1P36-8 (a) U = 2 Li2 = (152 H)(32 A)2 /2 = 7.8104 J.(b) If the coil develops at nite resistance then all of the energy in the eld will be dissipated asheat. The mass of Helium that will boil o ism = Q/Lv = (7.8104 J)/(85 J/mol)/(4.00g/mol) = 3.7 kg.147P36-9(a) B = (0 N i)/(2r), soB20 N 2 i2=.208 2 r2u=(b) U =u dV =urdr d dz. The eld inside the toroid is uniform in z and , sobU0 N 2 i2r dr,8 2 r2ah0 N 2 i2bln .4a= 2h=(c) The answers are the same!P36-10 The energy in the inductor is originally U = Li2 /2. The internal energy in the resistor0increases at a rate P = i2 R. ThenP dt = R0P36-11density isi2 e2t/L dt =00Li2Ri2 L0= 0.22(a) In Chapter 33 we found the magnetic eld inside a wire carrying a uniform current0 ir.2R2B=The magnetic energy density in this wire isuB =1 20 i2 r2B =.208 2 R4We want to integrate in cylindrical coordinates over the volume of the wire. Then the volumeelement is dV = (dr)(r d)(dz), soUB=uB dV,Rl2=0==00 i2 l4R40 i2 l.160R0 i2 r2d dz rdr,8 2 R4r3 dr,0(b) SolveUB =L 2i2for L, andL=P36-122UB0 l=.2i81/C = 1/C1 + 1/C2 and L = L1 + L2 . Then1= LC =(L1 + L2 )C1 C2=C1 + C214822C2 /0 + C1 /01=.C1 + C20P36-13(a) There is no current in the middle inductor; the loop equation becomesLd2 qqd2 qq+ +L 2 += 0.dt2CdtCTry q = q m cos t as a solution:L 2 +11 L 2 += 0;CCwhich requires = 1/ LC.(b) Look at only the left hand loop; the loop equation becomesLqd2 qd2 q+ + 2L 2 = 0.2dtCdtTry q = q m cos t as a solution:L 2 +which requires = 1/ 3LC.P36-141 2L 2 = 0;C(b) ( )/ is the fractional shift; this can also be written as11 (LC)(R/2L)2 1,=1 R2 C/4L 1,=P36-15=1(100 )2 (7.3106 F) 1 = 2.1103 .4(4.4 H)We start by focusing on the charge on the capacitor, given by Eq. 36-40 asq = q m eRt/2L cos( t + ).After one oscillation the cosine term has returned to the original value but the exponential term hasattenuated the charge on the capacitor according toq = q m eRT /2L ,where T is the period. The fractional energy loss on the capacitor is thenU0 Uq2= 1 2 = 1 eRT /L .U0qmFor small enough damping we can expand the exponent. Not only that, but T = 2/, soU 2R/L.U149P36-16We are given 1/2 = et/2L when t = 2n/ . Then =2nnR2n==.t2(L/R) ln 2L ln 2From Eq. 36-41,2 2( )( + )( )2= (R/2L)2 ,= (R/2L)2 , (R/2L)2 ,(R/2L)2= ,2 22(ln 2)=,8 2 n20.0061=.n2150E37-1The frequency, f , is related to the angular frequency by = 2f = 2(60 Hz) = 377 rad/sThe current is alternating because that is what the generator is designed to produce. It does thisthrough the conguration of the magnets and coils of wire. One complete turn of the generator will(could?) produce one cycle; hence, the generator is turning 60 times per second. Not only doesthis set the frequency, it also sets the emf, since the emf is proportional to the speed at which thecoils move through the magnetic eld.E37-2 (a) XL = L, sof = XL /2L = (1.28103 )/2(0.0452 H) = 4.51103 /s.(b) XC = 1/C, soC = 1/2f XC = 1/2(4.51103 /s)(1.28103 ) = 2.76108 F.(c) The inductive reactance doubles while the capacitive reactance is cut in half.E37-3 (a) XL = XC implies L = 1/C or = 1/ LC, so = 1/ (6.23103 H)(11.4106 F) = 3750 rad/s.(b) XL = L = (3750 rad/s)(6.23103 H) = 23.4 (c) See (a) above.E37-4 (a) im = E/XL = E/L, soim = (25.0 V)/(377 rad/s)(12.7 H) = 5.22103 A.(b) The current and emf are 90 out of phase. When the current is a maximum, E = 0.(c) t = arcsin[E(t)/E m ], sot = arcsin(13.8 V)= 0.585 rad.(25.0 V)andi = (5.22103 A) cos(0.585) = 4.35103 A.(d) Taking energy.E37-5 (a) The reactance of the capacitor is from Eq. 37-11, XC = 1/C. The AC generatorfrom Exercise 4 has E = (25.0 V) sin(377 rad/s)t. So the reactance isXC =11== 647 .C(377 rad/s)(4.1F)The maximum value of the current is found from Eq. 37-13,im =(VC )max )(25.0 V)== 3.86102 A.XC(647 )(b) The generator emf is 90 out of phase with the current, so when the current is a maximumthe emf is zero.151(c) The emf is -13.8 V whent = arcsin(13.8 V)= 0.585 rad.(25.0 V)The current leads the voltage by 90 = /2, soi = im sin(t ) = (3.86102 A) sin(0.585 /2) = 3.22102 A.(d) Since both the current and the emf are negative the product is positive and the generator issupplying energy to the circuit.E37-6 R = (L 1/omegaC)/ tan and = 2f = 2(941/s) = 5910 rad/s , soR=(5910 rad/s)(88.3103 H) 1/(5910 rad/s)(937109 F)= 91.5 .tan(75 )E37-7E37-8 (a) XL doesnt change, so XL = 87 .(b) XC = 1/C = 1/2(60/s)(70106 F) = 37.9.(c) Z = (160 )2 + (87 37.9 )2 = 167 .(d) im = (36 V)/(167 ) = 0.216 A.(e) tan = (87 37.9 )/(160 ) = 0.3069, so = arctan(0.3069) = 17 .E37-9 A circuit is considered inductive if XL > XC , this happens when im lags E m . If, on theother hand, XL < XC , and im leads E m , we refer to the circuit as capacitive. This is discussed onpage 850, although it is slightly hidden in the text of column one.(a) At resonance, XL = XC . Since XL = L and XC = 1/C we expect that XL grows withincreasing frequency, while XC decreases with increasing frequency.Consequently, at frequencies above the resonant frequency XL > XC and the circuit is predominantly inductive. But what does this really mean? It means that the inductor plays a major role inthe current through the circuit while the capacitor plays a minor role. The more inductive a circuitis, the less signicant any capacitance is on the behavior of the circuit.For frequencies below the resonant frequency the reverse is true.(b) Right at the resonant frequency the inductive eects are exactly canceled by the capacitiveeects. The impedance is equal to the resistance, and it is (almost) as if neither the capacitor orinductor are even in the circuit.E37-10 The net y component is XC XL . The net x component is R. The magnitude of theresultant isZ = R2 + (XC XL )2 ,while the phase angle istan =(XC XL ).RE37-11 Yes.At resonance = 1/ (1.2 H)(1.3106 F) = 800 rad/s and Z = R. Then im = E/Z =(10 V)/(9.6 ) = 1.04 A, so[VL ]m = im XL = (1.08 A)(800 rad/s)(1.2 H) = 1000 V.152E37-12 (a) Let O = XL XC and A = R, then H 2 = A2 + O2 = Z 2 , sosin = (XL XC )/Zandcos = R/Z.E37-13 (a) The voltage across the generator is the generator emf, so when it is a maximum fromSample Problem 37-3, it is 36 V. This corresponds to t = /2.(b) The current through the circuit is given by i = im sin(t ). We found in Sample Problem37-3 that im = 0.196 A and = 29.4 = 0.513 rad.For a resistive load we apply Eq. 37-3,VR = im R sin(t ) = (0.196 A)(160) sin((/2) (0.513)) = 27.3 V.(c) For a capacitive load we apply Eq. 37-12,VC = im XC sin(t /2) = (0.196 A)(177) sin((0.513)) = 17.0 V.(d) For an inductive load we apply Eq. 37-7,VL = im XL sin(t + /2) = (0.196 A)(87) sin( (0.513)) = 8.4 V.(e) (27.3 V) + (17.0 V) + (8.4 V) = 35.9 V.E37-14 If circuit 1 and 2 have the same resonant frequency then L1 C1 = L2 C2 . The seriescombination for the inductors isL = L1 + L2 ,The series combination for the capacitors is1/C = 1/C1 + 1/C2 ,soC1 C2L1 C1 C2 + L2 C2 C1== L1 C1 ,C1 + C2C1 + C2which is the same as both circuit 1 and 2.LC = (L1 + L2 )E37-15 (a) Z = (125 V)/(3.20 A) = 39.1 .(b) Let O = XL XC and A = R, then H 2 = A2 + O2 = Z 2 , socos = R/Z.Using this relation,R = (39.1 ) cos(56.3 ) = 21.7 .(c) If the current leads the emf then the circuit is capacitive.E37-16 (a) Integrating over a single cycle,1TTsin2 t dt=0=1 T 1(1 cos 2t) dt,T 0 211T = .2T2(b) Integrating over a single cycle,1TTsin t cos t dt01T= 0.T=15301sin 2tdt,2E37-17The resistance would be given by Eq. 37-32,R=P av(0.10)(746 W)== 177 .2irms(0.650 A)2This would not be the same as the direct current resistance of the coils of a stopped motor, becausethere would be no inductive eects.E37-18 Since irms = E rms /Z, thenP av = i2 rms R =E37-19 (a) Z =(160 )2 + (177 )2 = 239 ; thenP av =(b) Z =E 2 rms R.Z21 (36 V)2 (160 )= 1.82 W.2(239 )2(160 )2 + (87 )2 = 182 ; thenP av =1 (36 V)2 (160 )= 3.13 W.2(182 )2E37-20 (a) Z = (12.2 )2 + (2.30 )2 = 12.4 (b) P av = (120 V)2 (12.2 )/(12.4 )2 = 1140 W.(c) irms = (120 V)/(12.4 ) = 9.67 A.E37-21 The rms value of any sinusoidal quantity is related to the maximum value by 2 v rms =vmax . Since this factor of 2 appears in all of the expressions, we can conclude that if the rms valuesare equal then so are the maximum values. This means that(VR )max = (VC )max = (VL )maxor im R = im XC = im XL or, with one last simplication, R = XL = XC . Focus on the right handside of the last equality. If XC = XL then we have a resonance condition, and the impedance (seeEq. 37-20) is a minimum, and is equal to R. Then, according to Eq. 37-21,im =Em,Rwhich has the immediate consequence that the rms voltage across the resistor is the same as therms voltage across the generator. So everything is 100 V.E37-22 (a) The antenna is in-tune when the impedance is a minimum, or = 1/ LC. Sof = /2 = 1/2(8.22106 H)(0.2701012 F) = 1.07108 Hz.(b) irms = (9.13 V)/(74.7 ) = 1.22107 A.(c) XC = 1/2f C, soVC = iXC = (1.22107 A)/2(1.07108 Hz)(0.270 1012 F) = 6.72104 V.154E37-23 Assuming no inductors or capacitors in the circuit, then the circuit eectively behaves asa DC circuit. The current through the circuit is i = E/(r + R). The power delivered to R is thenP = iV = i2 R = E 2 R/(r + R)2 . Evaluate dP/dR and set it equal to zero to nd the maximum.ThenrRdP= E 2R,0=dR(r + R)3which has the solution r = R.E37-24 (a) Since P av = im 2 R/2 = E m 2 R/2Z 2 , then P av is a maximum when Z is a minimum, andvise-versa. Z is a minimum at resonance, when Z = R and f = 1/2 LC. When Z is a minimumC = 1/4 2 f 2 L = 1/4 2 (60 Hz)2 (60 mH) = 1.2107 F.(b) Z is a maximum when XC is a maximum, which occurs when C is very small, like zero.(c) When XC is a maximum P = 0. When P is a maximum Z = R soP = (30 V)2 /2(5.0 ) = 90 W.(d) The phase angle is zero for resonance; it is 90 for innite XC or XL .(e) The power factor is zero for a system which has no power. The power factor is one for asystem in resonance.E37-25(a) The resistance is R = 15.0 . The inductive reactance isXC =11== 61.3 .1 )(4.72F)C2(550 sThe inductive reactance is given byXL = L = 2(550 s1 )(25.3 mH) = 87.4 .The impedance is thenZ=2(15.0 )2 + ((87.4 ) (61.3 )) = 30.1 .Finally, the rms current isE rms(75.0 V)== 2.49 A.Z(30.1 )(b) The rms voltages between any two points is given byirms =(V )rms = irms Z,where Z is not the impedance of the circuit but instead the impedance between the two points inquestion. When only one device is between the two points the impedance is equal to the reactance(or resistance) of that device.Were not going to show all of the work here, but we will put together a nice table for youPointsabbccdbdacImpedance ExpressionZ=RZ = XCZ = XLZ = |XL XC |2Z = R 2 + XCImpedance ValueZ = 15.0 Z = 61.3 Z = 87.4 Z = 26.1 Z = 63.1 (V )rms ,37.4 V,153 V,218 V,65 V,157 V,Note that this last one was Vac , and not Vad , because it is more entertaining. You probablyshould use Vad for your homework.(c) The average power dissipated from a capacitor or inductor is zero; that of the resistor isPR = [(VR )rms ]2 /R = (37.4 V)2 /(15.0) = 93.3 W.155E37-26 (a) The energy stored in the capacitor is a function of the charge on the capacitor; althoughthe charge does vary with time it varies periodically and at the end of the cycle has returned to theoriginal values. As such, the energy stored in the capacitor doesnt change from one period to thenext.(b) The energy stored in the inductor is a function of the current in the inductor; although thecurrent does vary with time it varies periodically and at the end of the cycle has returned to theoriginal values. As such, the energy stored in the inductor doesnt change from one period to thenext.(c) P = Ei = E m im sin(t) sin(t ), so the energy generated in one cycle isTUTP dt = E m im=0sin(t) sin(t )dt,0T= E m imsin(t) sin(t )dt,0=TE m im cos .2(d) P = im 2 R sin2 (t ), so the energy dissipated in one cycle isTUTsin2 (t )dt,P dt = im 2 R=00Tsin2 (t )dt,= im 2 R0T 2im R.2=(e) Since cos = R/Z and E m /Z = im we can equate the answers for (c) and (d).E37-27Apply Eq. 37-41,V s = V pNs(780)= (150 V)= 1.8103 V.Np(65)E37-28 (a) Apply Eq. 37-41,V s = V pNs(10)= (120 V)= 2.4 V.Np(500)(b) is = (2.4 V)/(15 ) = 0.16 A;ip = isNs(10)= (0.16 A)= 3.2103 A.Np(500)E37-29 The autotransformer could have a primary connected between taps T1 and T2 (200 turns),T1 and T3 (1000 turns), and T2 and T3 (800 turns).The same possibilities are true for the secondary connections. Ignoring the one-to-one connectionsthere are 6 choices three are step up, and three are step down. The step up ratios are 1000/200 = 5,800/200 = 4, and 1000/800 = 1.25. The step down ratios are the reciprocals of these three values.156E37-30 = (1.69108 m)[1 (4.3103 /C )(14.6 C)] = 1.58108 m. The resistance ofthe two wires isL(1.58108 m)2(1.2103 m)R=== 14.9 .A(0.9103 m)2P = i2 R = (3.8 A)2 (14.9 ) = 220 W.E37-31 The supply current isip = (0.270 A)(74103 V/ 2)/(220 V) = 64.2 A.The potential drop across the supply lines isV = (64.2 A)(0.62 ) = 40 V.This is the amount by which the supply voltage must be increased.E37-32 Use Eq. 37-46:N p /N s =(1000 )/(10 ) = 10.P37-1 (a) The emf is a maximum when t /4 = /2, so t = 3/4 = 3/4(350 rad/s) =6.73103 s.(b) The current is a maximum when t 3/4 = /2, so t = 5/4 = 5/4(350 rad/s) =1.12102 s.(c) The current lags the emf, so the circuit contains an inductor.(d) XL = E m /im and XL = L, soL=Em(31.4 V)== 0.144 H.im (0.622 A)(350 rad/s)P37-2 (a) The emf is a maximum when t /4 = /2, so t = 3/4 = 3/4(350 rad/s) =6.73103 s.(b) The current is a maximum when t+/4 = /2, so t = /4 = /4(350 rad/s) = 2.24103 s.(c) The current leads the emf, so the circuit contains an capacitor.(d) XC = E m /im and XC = 1/C, soC=im(0.622 A)== 5.66105 F.E m(31.4 V)(350 rad/s)P37-3 (a) Since the maximum values for the voltages across the individual devices are proportional to the reactances (or resistances) for devices in series (the constant of proportionality is themaximum current), we have XL = 2R and XC = R.From Eq. 37-18,XL XC2R Rtan === 1,RRor = 45 .(b) The impedance of the circuit, in terms of the resistive element, isZ = R2 + (XL XC )2 = R2 + (2R R)2 = 2 R.But E m = im Z, so Z = (34.4 V)/(0.320 A) = 108. Then we can use our previous work to nds thatR = 76.157P37-4 When the switch is open the circuit is an LRC circuit. In position 1 the circuit is an RLCcircuit, but the capacitance is equal to the two capacitors of C in parallel, or 2C. In position 2 thecircuit is a simple LC circuit with no resistance.The impedance when the switch is in position 2 is Z2 = |L 1/C|. ButZ2 = (170 V)/(2.82 A) = 60.3 .The phase angle when the switch is open is 0 = 20 . Buttan 0 =Z2L 1/C=,RRso R = (60.3 )/ tan(20 ) = 166 .The phase angle when the switch is in position 1 istan 1 =L 1/2C,Rso L 1/2C = (166 ) tan(10 ) = 29.2 . Equating the L part,(29.2 ) + 1/2CC==(60.3 ) + 1/C,1/2(377 rad/s)[(60.3 ) + (29.2 )] = 1.48105 F.Finally,L=(60.3) + 1/(377 rad/s)(1.48105 F)= 0.315 H.(377 rad/s)P37-5 All three wires have emfs which vary sinusoidally in time; if we choose any two wires thephase dierence will have an absolute value of 120 . We can then choose any two wires and expect(by symmetry) to get the same result. We choose 1 and 2. The potential dierence is thenV1 V2 = V m (sin t sin(t 120 )) .We need to add these two sine functions to get just one. We use11sin sin = 2 sin ( ) cos ( + ).22ThenV1 V 2112V m sin (120 ) cos (2t 120 ),223= 2V m () cos(t 60 ),2=3V m sin(t + 30 ).=P37-6 (a) cos = cos(42 ) = 0.74.(b) The current leads.(c) The circuit is capacitive.(d) No. Resonance circuits have a power factor of one.(e) There must be at least a capacitor and a resistor.(f) P = (75 V)(1.2 A)(0.74)/2 = 33 W.158P37-7 (a) = 1/ LC = 1/ (0.988 H)(19.3106 F) = 229 rad/s.(b) im = (31.3 V)/(5.12 ) = 6.11 A.(c) The current amplitude will be halved when the impedance is doubled, or when Z = 2R. Thisoccurs when 3R2 = (L 1/C)2 , or3R2 2 = 4 L2 2 2 L/C + 1/C 2 .The solution to this quadratic is2 =2L + 3CR2 9C 2 R4 + 12CR2 L,2L2 Cso 1 = 224.6 rad/s and 2 = 233.5 rad/s.(d) / = (8.9 rad/s)/(229 rad/s) = 0.039.P37-8 (a) The current amplitude will be halved when the impedance is doubled, or when Z = 2R.This occurs when 3R2 = (L 1/C)2 , or3R2 2 = 4 L2 2 2 L/C + 1/C 2 .The solution to this quadratic is2 =2L + 3CR2 9C 2 R4 + 12CR2 L,2L2 CNote that = + ; with a wee bit of algebra,22(+ + ) = + .Also, + + 2. Hence,assuming that CR29C 2 R4 + 12CR2 L,2 2L C 2 R 9C 2 R2 + 12LC,2LR 9 2 C 2 R2 + 12,2LR 9CR2 /L + 12,2L3R,L4L/3.P37-9P37-10P37-11Use Eq. 37-46.(a) The resistance of this bulb isR=(V )2(120 V)2== 14.4 .P(1000 W)159The power is directly related to the brightness; if the bulb is to be varied in brightness by a factor of5 then it would have a minimum power of 200 W. The rms current through the bulb at this powerwould beirms = P/R = (200 W)/(14.4 ) = 3.73 A.The impedance of the circuit must have beenZ=E rms(120 V)== 32.2 .irms(3.73 A)The inductive reactance would then beXL =Z 2 R2 =(32.2 )2 (14.4 )2 = 28.8 .Finally, the inductance would beL = XL / = (28.8 )/(2(60.0 s1 )) = 7.64 H.(b) One could use a variable resistor, and since it would be in series with the lamp a value of32.2 14.4 = 17.8 would work. But the resistor would get hot, while on average there is no power radiated from a pureinductor.160E38-1 The maximum value occurs where r = R; there Bmax = 1 0 0 R dE/dt. For r < R B is2half of Bmax when r = R/2. For r > R B is half of Bmax when r = 2R. Then the two values of rare 2.5 cm and 10.0 cm.E38-2 For a parallel plate capacitor E = /id =00and the ux is then E = A/0= q/ 0 . ThendEdqddV== CV = C.dtdtdtdtE38-3 Use the results of Exercise 2, and change the potential dierence across the plates of thecapacitor at a rateid(1.0 mA)dV=== 1.0 kV/s.dtC(1.0F)Provide a constant current to the capacitori=ddVdQ= CV = C= id .dtdtdtE38-4 Since E is uniform between the plates E = EA, regardless of the size of the region ofinterest. Since j d = id /A,id1 dEdEjd === 0.0AAdtdtE38-5 (a) In this case id = i = 1.84 A.(b) Since E = q/ 0 A, dE/dt = i/ 0 A, ordE/dt = (1.84 A)/(8.851012 F/m)(1.22 m)2 = 1.401011 V/m.(c) id = 0 dE /dt = 0 adE/dt. a here refers to the area of the smaller square. Combine thiswith the results of part (b) andid = ia/A = (1.84 A)(0.61 m/1.22 m)2 = 0.46 A.(d)B ds = 0 id = (4107 H/m)(0.46 A) = 5.78107 T m.E38-6 Substitute Eq. 38-8 into the results obtained in Sample Problem 38-1. Outside the capacitorE = R2 E, so0 0 R2 dE0B==id .2rdt2rInside the capacitor the enclosed ux is E = r2 E; but we want instead to dene id in terms ofthe total E inside the capacitor as was done above. Consequently, inside the conductorB=E38-7have0 r2R20 R2dEdt=0 rid .2R2Since the electric eld is uniform in the area and perpendicular to the surface area weE =E dA =E dA = EdA = EA.The displacement current is thenid =0AdEdE= (8.851012 F/m)(1.9 m2 ).dtdt161(a) In the rst region the electric eld decreases by 0.2 MV/m in 4s, soid = (8.851012 F/m)(1.9 m2 )(0.2106 V/m)= 0.84 A.(4106 s)(b) The electric eld is constant so there is no change in the electric ux, and hence there is nodisplacement current.(c) In the last region the electric eld decreases by 0.4 MV/m in 5s, soid = (8.851012 F/m)(1.9 m2 )(0.4106 V/m)= 1.3 A.(5106 s)E38-8 (a) Because of the circular symmetry B ds = 2rB, where r is the distance from thecenter of the circular plates. Not only that, but id = j d A = r2 j d . Equate these two expressions,andB = 0 rj d /2 = (4107 H/m)(0.053 m)(1.87101 A/m)/2 = 6.23107 T.(b) dE/dt = id / 0 A = j d /0= (1.87101 A/m)/(8.851012 F/m) = 2.111012 V/m.E38-9 The magnitude of E is given byE=(162 V)sin 2(60/s)t;(4.8103 m)Using the results from Sample Problem 38-1,Bm=0 0 R dE2dt,t=0=(4107 H/m)(8.851012 F/m)(0.0321 m)(162 V)2(60/s),2(4.8103 m)=2.271012 T.E38-10 (a) Eq. 33-13 from page 764 and Eq. 33-34 from page 762.(b) Eq. 27-11 from page 618 and the equation from Ex. 27-25 on page 630.(c) The equations from Ex. 38-6 on page 876.(d) Eqs. 34-16 and 34-17 from page 785.E38-11 (a) Consider the path abef a. The closed line integral consists of two parts: b e ande f a b. ThendE ds = dtcan be written asE ds +beE ds = ef abdabef .dtNow consider the path bcdeb. The closed line integral consists of two parts: b c d e ande b. ThendE ds = dtcan be written asE ds +bcdeE ds = eb162dbcde .dtThese two expressions can be added together, and sinceE ds = ebE dsbewe getE ds +E ds = ef abbcded(abef + bcde ) .dtThe left hand side of this is just the line integral over the closed path ef adcde; the right hand sideis the net change in ux through the two surfaces. Then we can simplify this expression asE ds = d.dt(b) Do everything above again, except substitute B for E.(c) If the equations were not self consistent we would arrive at dierent values of E and Bdepending on how we dened our surfaces. This multi-valued result would be quite unphysical.E38-12 (a) Consider the part on the left. It has a shared surface s, and the other surfaces l.Applying Eq. I,ql /0=E dA =E dA +sE dA.lNote that dA is directed to the right on the shared surface.Consider the part on the right. It has a shared surface s, and the other surfaces r. Applying Eq.I,qr /0=E dA =E dA +sE dA.rNote that dA is directed to the left on the shared surface.Adding these two expressions will result in a canceling out of the partE dAssince one is oriented opposite the other. We are left withqr + ql0E dA +=rE dA =E dA.lE38-13E38-14 (a) Electric dipole is because the charges are separating like an electric dipole. Magneticdipole because the current loop acts like a magnetic dipole.E38-15A series LC circuit will oscillate naturally at a frequencyf=1=22 LCWe will need to combine this with v = f , where v = c is the speed of EM waves.We want to know the inductance required to produce an EM wave of wavelength = 550109 m,so2(550 109 m)2L=== 5.01 1021 H.4 2 c2 C4 2 (3.00 108 m/s)2 (17 1012 F)This is a small inductance!163E38-16 (a) B = E/c, and B must be pointing in the negative y direction in order that the wavebe propagating in the positive x direction. Then Bx = Bz = 0, andBy = Ez /c = (2.34104 V/m)/(3.00108 m/s) = (7.801013 T) sin k(x ct).(b) = 2/k = 2/(9.72106 /m) = 6.46107 m.E38-17 The electric and magnetic eld of an electromagnetic wave are related by Eqs. 38-15and 38-16,(321V/m)EB=== 1.07 pT.c(3.00 108 m/s)E38-18 Take the partial of Eq. 38-14 with respect to x, Ex x2Ex2 B,x t2B= .xt= Take the partial of Eq. 38-17 with respect to t, Bt x2BtxEquate, and let 00 E,t t2 E= 0 0 2 .t= 00= 1/c2 , then2E1 2E= 2 2.x2c tRepeat, except now take the partial of Eq. 38-14 with respect to t, and then take the partial of Eq.38-17 with respect to x.E38-19 (a) Since sin(kx t) is of the form f (kx t), then we only need do part (b).(b) The constant E m drops out of the wave equation, so we need only concern ourselves withf (kx t). Letting g = kx t,2fg 22ft22gtgt2f,x2 2 f g= c2 2gxg= c ,x= ck.= c22,E38-20 Use the right hand rule.E38-21 U = P t = (1001012 W)(1.0109 s) = 1.0105 J.E38-22 E = Bc = (28109 T)(3.0108 m/s) = 8.4 V/m. It is in the positive x direction.164E38-23 Intensity is given by Eq. 38-28, which is simply an expression of power divided by surfacearea. To nd the intensity of the TV signal at -Centauri we need to nd the distance in meters;r = (4.30 light-years)(3.00108 m/s)(3.15107 s/year) = 4.06 1016 m.The intensity of the signal when it has arrived at out nearest neighbor is thenI=(960 kW)P2== 4.63 1029 W/m4r24(4.06 1016 m)2E38-24 (a) From Eq. 38-22, S = cB 2 /0 . B = B m sin t. The time average is dened as1TTS dt =0TcB m 20 Tcos2 t dt =0cB m 2.20(b) S av = (3.0108 m/s)(1.0104 T)2 /2(4107 H/m) = 1.2106 W/m2 .E38-25 I = P/4r2 , sor=E38-26 uE =0E2/2 =(1.0103 W)/4(130 W/m2 ) = 0.78 m.P/4I =20 (cB) /2= B 2 /20 = uB .E38-27 (a) Intensity is related to distance by Eq. 38-28. If r1 is the original distance from thestreet lamp and I1 the intensity at that distance, thenI1 =P2.4r1There is a similar expression for the closer distance r2 = r1 162 m and the intensity at that distanceI2 = 1.50I1 . We can combine the two expression for intensity,I2P24r2=2r1=r11.50I1 ,P= 1.502,4r11.50r2 , 2=1.50 (r1 162 m).The last line is easy enough to solve and we nd r1 = 883 m.(b) No, we cant nd the power output from the lamp, because we were never provided with anabsolute intensity reference.E38-28 (a) E m =Em =20 cI, so2(4107 H/m)(3.00108 m/s)(1.38103 W/m2 ) = 1.02103 V/m.(b) B m = E m /c = (1.02103 V/m)/(3.00108 m/s) = 3.40106 T.E38-29 (a) B m = E m /c = (1.96 V/m)/(3.00108 m/s) = 6.53109 T.(b) I = E m 2 /20 c = (1.96 V)2 /2(4107 H/m)(3.00108 m/s) = 5.10103 W/m2 .(c) P = 4r2 I = 4(11.2 m)2 (5.10103 W/m2 ) = 8.04 W.165E38-30 (a) The intensity isI=(11012 W)P== 1.961027 W/m2 .A4(6.37106 m)2The power received by the Arecibo antenna isP = IA = (1.961027 W/m2 )(305 m)2 /4 = 1.41022 W.(b) The power of the transmitter at the center of the galaxy would beP = IA = (1.961027 W)(2.3104 ly)2 (9.461015 m/ly)2 = 2.91014 W.E38-31(a) The electric eld amplitude is related to the intensity by Eq. 38-26,I=E2m,20 corEm =20 cI =2(4107 H/m)(3.00108 m/s)(7.83W/m2 ) = 7.68102 V/m.(b) The magnetic eld amplitude is given byBm =Em(7.68 102 V/m)== 2.56 1010 Tc(3.00 108 m/s)(c) The power radiated by the transmitter can be found from Eq. 38-28,P = 4r2 I = 4(11.3 km)2 (7.83W/m2 ) = 12.6 kW.E38-32 (a) The power incident on (and then reected by) the target craft is P1 = I1 A = P0 A/2r2 .The intensity of the reected beam is I2 = P1 /2r2 = P0 A/4 2 r4 . ThenI2 = (183103 W)(0.222 m2 )/4 2 (88.2103 m)4 = 1.701017 W/m2 .(b) Use Eq. 38-26:Em =(c) B rms20 cI = 2(4107 H/m)(3.00108 m/s)(1.701017 W/m2 ) = 1.13107 V/m.= E m / 2c = (1.13107 V/m)/ 2(3.00108 m/s) = 2.661016 T.E38-33 Radiation pressure for absorption is given by Eq. 38-34, but we need to nd the energyabsorbed before we can apply that. We are given an intensity, a surface area, and a time, soU = (1.1103 W/m2 )(1.3 m2 )(9.0103 s) = 1.3107 J.The momentum delivered isp = (U )/c = (1.3107 J)/(3.00108 m/s) = 4.3102 kg m/s.E38-34 (a) F/A = I/c = (1.38103 W/m2 )/(3.00108 m/s) = 4.60106 Pa.(b) (4.60106 Pa)/(101105 Pa) = 4.551011 .E38-35 F/A = 2P/Ac = 2(1.5109 W)/(1.3106 m2 )(3.0108 m/s) = 7.7106 Pa.166E38-36 F/A = P/4r2 c, soF/A = (500 W)/4(1.50 m)2 (3.00108 m/s) = 5.89108 Pa.E38-37 (a) F = IA/c, soF =(1.38103 W/m2 )(6.37106 m)2= 5.86108 N.(3.00108 m/s)E38-38 (a) Assuming MKSA, the units arem C V sNNsmFV N== 2 .s m m Ams Vm m Cmm s(b) Assuming MKSA, the units areA2 V NA2 J N1 JJ=== 2 .N m AmN Cm Amsm mm sE38-39 We can treat the object as having two surfaces, one completely reecting and the othercompletely absorbing. If the entire surface has an area A then the absorbing part has an area f Awhile the reecting part has area (1 f )A. The average force is then the sum of the force on eachpart,I2IF av = f A + (1 f )A,ccwhich can be written in terms of pressure asF avI= (2 f ).AcE38-40 We can treat the object as having two surfaces, one completely reecting and the othercompletely absorbing. If the entire surface has an area A then the absorbing part has an area f Awhile the reecting part has area (1 f )A. The average force is then the sum of the force on eachpart,I2IF av = f A + (1 f )A,ccwhich can be written in terms of pressure asF avI= (2 f ).AcThe intensity I is that of the incident beam; the reected beam will have an intensity (1 f )I.Each beam will contribute to the energy density I/c and (1 f )I/c, respectively. Add these twoenergy densities to get the net energy density outside the surface. The result is (2 f )I/c, which isthe left hand side of the pressure relation above.E38-41 The bullet density is = N m/V . Let V = Ah; the kinetic energy density is K/V =122 N mv /Ah. h/v, however, is the time taken for N balls to strike the surface, so thatP =FN mvN mv 22K===.AAtAhV167E38-42 F = IA/c; P = IA; a = F/m; and v = at. Combine:v = P t/mc = (10103 W)(86400 s)/(1500 kg)(3108 m/s) = 1.9103 m/s.E38-43 The force of radiation on the bottom of the cylinder is F = 2IA/c. The force of gravityon the cylinder isW = mg = HAg.Equating, 2I/c = Hg. The intensity of the beam is given by I = 4P/d2 . Solving for H,H=8(4.6 W)8P= 4.9107 m.=28 m/s)(1200 kg/m3 )(9.8 m/s2 )(2.6103 m)2cgd(3.010E38-44 F = 2IA/c. The value for I is in Ex. 38-37, among other places. ThenF = (1.38103 W/m2 )(3.1106 m2 )/(3.00108 m/s) = 29 N.P38-1 For the two outer circles use Eq. 33-13. For the inner circle use E = V /d, Q = CV ,C = 0 A/d, and i = dQ/dt. Theni=dQ0 A dV==dtd dt0AdE.dtThe change in ux is dE /dt = A dE/dt. ThenB dl = 00dE= 0 i,dtso B = 0 i/2r.P38-2 (a) id = i. Assuming V = (174103 V) sin t, then q = CV and i = dq/dt = Cd(V )/dt.Combine, and use = 2(50.0/s),id = (1001012 F)(174103 V)2(50.0/s) = 5.47103 A.P38-3 (a) i = id = 7.63A.(b) dE /dt = id / 0 = (7.63A)/(8.851012 F/m) = 8.62105 V/m.(c) i = dq/dt = Cd(V )/dt; C = 0 A/d; [d(V )/dt]m = E m . Combine, andd=0AC=0 AE m i=(8.851012 F/m)(0.182 m)2 (225 V)(128 rad/s)= 3.48103 m.(7.63A)P38-4 (a) q = i dt = t dt = t2 /2.(b) E = / 0 = q/ 0 A = t2 /2R2 0 .(d) 2rB = 0 0 r2 dE/dt, soB = 0 r(dE/dt)/2 = 0 rt/2R2 .(e) Check Exercise 38-10!P38-5(a) E = E and B = B k. Then S = E B/0 , orjS = EB/mu0 i.Energy only passes through the yz faces; it goes in one face and out the other. The rate is P =SA = EBa2 /mu0 .(b) The net change is zero.168P38-6(a) For a sinusoidal time dependence |dE/dt|m = E m = 2f E m . Then|dE/dt|m = 2(2.4109 /s)(13103 V/m) = 1.961014 V/m s.(b) Using the result of part (b) of Sample Problem 38-1,B=11(4107 H/m)(8.91012 F/m)(2.4102 m) (1.961014 V/m s) = 1.3105 T.22P38-7 Look back to Chapter 14 for a discussion on the elliptic orbit. On page 312 it is pointedout that the closest distance to the sun is Rp = a(1 e) while the farthest distance is Ra = a(1 + e),where a is the semi-major axis and e the eccentricity.The fractional variation in intensity isIIIp Ia,IaIp 1,IaRa 2 1,Rp 2==(1 + e)2 1.(1 e)2=We need to expand this expression for small e using (1 + e)2 1 + 2e, and (1 e)2 1 + 2e, andnally (1 + 2e)2 1 + 4e. Combining,I (1 + 2e)2 1 4e.IP38-8 The beam radius grows as r = (0.440 rad)R, where R is the distance from the origin. Thebeam intensity isI=P38-9P(3850 W)== 4.3102 W.2r(0.440 rad)2 (3.82108 m)2Eq. 38-14 requiresExE m k cos kx sin tEmkB,t= B m cos kx sin t,= B m .= Eq. 38-17 requiresEt0 0 E m sin kx cos t0 0 E m 00B,x= B m k sin kx cos t,= B m k.= Dividing one expression by the other,0 0 k 2 = 2 ,169or1=c= k00.Not only that, but E m = cB m . Youve seen an expression similar to this before, and youll seeexpressions similar to it again.(b) Well assume that Eq. 38-21 is applicable here. ThenS==1EmBm=sin kx sin t cos kx cos t,00E2msin 2kx sin 2t40 cis the magnitude of the instantaneous Poynting vector.(c) The time averaged power ow across any surface is the value of1TTS dA dt,0where T is the period of the oscillation. Well just gloss over any concerns about direction, andassume that the S will be constant in direction so that we will, at most, need to concern ourselvesabout a constant factor cos . We can then deal with a scalar, instead of vector, integral, and wecan integrate it in any order we want. We want to do the t integration rst, because an integral oversin t for a period T = 2/ is zero. Then we are done!(d) There is no energy ow; the energy remains inside the container.P38-10(a) The electric eld is parallel to the wire and given byE = V /d = iR/d = (25.0 A)(1.00 /300 m) = 8.33102 V/m(b) The magnetic eld is in rings around the wire. Using Eq. 33-13,B=0 i(4107 H/m)(25 A)== 4.03103 T.2r2(1.24103 m)(c) S = EB/0 , soS = (8.33102 V/m)(4.03103 T)/(4107 H/m) = 267 W/m2 .P38-11(a) Weve already calculated B previously. It isB=0 iEwhere i = .2rRThe electric eld of a long straight wire has the form E = k/r, where k is some constant. ButbV = E ds = E dr = k ln(b/a).aIn this problem the inner conductor is at the higher potential, sok=VE=,ln(b/a)ln(b/a)170and then the electric eld isE.r ln(b/a)This is also a vector eld, and if E is positive the electric eld points radially out from the centralconductor.(b) The Poynting vector is1S=E B;0E=E is radial while B is circular, so they are perpendicular. Assuming that E is positive the directionof S is away from the battery. Switching the sign of E (connecting the battery in reverse) will ipthe direction of both E and B, so S will pick up two negative signs and therefore still point awayfrom the battery.The magnitude isE2EBS==02R ln(b/a)r2(c) We want to evaluate a surface integral in polar coordinates and so dA = (dr)(rd). We havealready established that S is pointing away from the battery parallel to the central axis. Then wecan integratePS dA ==b2=a0bE2d r dr,2R ln(b/a)r2E2dr,R ln(b/a)r=a=S dA,E2.R(d) Read part (b) above.P38-12 (a) B is oriented as rings around the cylinder. If the thumb is in the direction of currentthen the ngers of the right hand grip ion the direction of the magnetic eld lines. E is directedparallel to the wire in the direction of the current. S is found from the cross product of these two,and must be pointing radially inward.(b) The magnetic eld on the surface is given by Eq. 33-13:B = 0 i/2a.The electric eld on the surface is given byE = V /l = iR/lThen S has magnitudei iRi2 R=.2a l2alS dA is only evaluated on the surface of the cylinder, not the end caps. S is everywhere parallelto dA, so the dot product reduces to S dA; S is uniform, so it can be brought out of the integral;dA = 2al on the surface.Hence,S = EB/0 =S dA = i2 R,as it should.171P38-13 (a) f = vlambda = (3.00108 m/s)/(3.18 m) = 9.43107 Hz.(b) Bmust be directed along the z axis. The magnitude isB = E/c = (288 V/m)/(3.00108 m/s) = 9.6107 T.(c) k = 2/ = 2/(3.18 m) = 1.98/m while = 2f , so = 2(9.43107 Hz) = 5.93108 rad/s.(d) I = E m B m /20 , soI=(288 V)(9.6107 T)= 110 W.2(4107 H/m)(e) P = I/c = (110 W)/(3.00108 m/s) = 3.67107 Pa.P38-14 (a) B is oriented as rings around the cylinder. If the thumb is in the direction of currentthen the ngers of the right hand grip ion the direction of the magnetic eld lines. E is directedparallel to the wire in the direction of the current. S is found from the cross product of these two,and must be pointing radially inward.(b) The magnitude of the electric eld isE=VQQit===.dCd0A0AThe magnitude of the magnetic eld on the outside of the plates is given by Sample Problem 38-1,B=0 0 R dE0 0 iR0 0 R==E.2dt2 0A2tS has magnitudeS=EB0R 2=E .02tIntegrating,S dA =0R2tE 2 2Rd = Ad0Et2.But E is linear in t, so d(E 2 )/dt = 2E 2 /t; and thenS dA = Adddt12.0E2P38-15 (a) I = P/A = (5.00103 W)/(1.05)2 (633109 m)2 = 3.6109 W/m2 .(b) p = I/c = (3.6109 W/m2 )/(3.00108 m/s) = 12 Pa(c) F = pA = P/c = (5.00103 W)/(3.00108 m/s) = 1.671011 N.(d) a = F/m = F/V , soa=(1.671011 N)= 2.9103 m/s2 .4(4880 kg/m3 )(1.05)3 (633109 )3 /3172P38-16The force from the sun is F = GM m/r2 . The force from radiation pressure isF =2P A2IA=.c4r2 cEquating,A=soA=4GM m,2P/c4(6.671011 N m2 /kg2 )(1.991030 kg)(1650 kg)= 1.06106 m2 .2(3.91026 W)/(3.0108 m/s)Thats about one square kilometer.173E39-1 Both scales are logarithmic; choose any data point from the right hand side such asc = f (1 Hz)(3108 m) = 3108 m/s,and another from the left hand side such asc = f (11021 Hz)(31013 m) = 3108 m/s.E39-2 (a) f = v/ = (3.0108 m/s)/(1.0104 )(6.37106 m) = 4.7103 Hz. If we assume thatthis is the data transmission rate in bits per second (a generous assumption), then it would take 140days to download a web-page which would take only 1 second on a 56K modem!(b) T = 1/f = 212 s = 3.5 min.E39-3(a) Apply v = f . Thenf = (3.0108 m/s)/(0.0671015 m) = 4.51024 Hz.(b) = (3.0108 m/s)/(30 Hz) = 1.0107 m.E39-4 Dont simply take reciprocal of linewidth! f = c/, so f = (c/2 ). Ignore the negative,andf = (3.00108 m/s)(0.010109 m)/(632.8109 m)2 = 7.5109 Hz.E39-5 (a) We refer to Fig. 39-6 to answer this question. The limits are approximately 520 nmand 620 nm.(b) The wavelength for which the eye is most sensitive is 550 nm. This corresponds to to afrequency off = c/ = (3.00 108 m/s)/(550 109 m) = 5.45 1014 Hz.This frequency corresponds to a period of T = 1/f = 1.83 1015 s.E39-6 f = c/. The number of complete pulses is f t, orf t = ct/ = (3.00108 m/s)(4301012 s)/(520109 m) = 2.48105 .E39-7 (a) 2(4.34 y) = 8.68 y.(b) 2(2.2106 y) = 4.4106 y.E39-8 (a) t = (150103 m)/(3108 m/s) = 5104 s.(b) The distance traveled by the light is (1.51011 m) + 2(3.8108 m), sot = (1.511011 m)/(3108 m/s) = 503 s.(c) t = 2(1.31012 m)/(3108 m/s) = 8670 s.(d) 1054 6500 5400 BC.E39-9 This is a question of how much time it takes light to travel 4 cm, because the light traveledfrom the Earth to the moon, bounced o of the reector, and then traveled back. The time to travel4 cm is t = (0.04 m)/(3 108 m/s) = 0.13 ns. Note that I interpreted the question dierently thanthe answer in the back of the book.174E39-10 Consider any incoming ray. The path of the ray can be projected onto the xy plane, thexz plane, or the yz plane. If the projected rays is exactly reected in all three cases then the threedimensional incoming ray will be reected exactly reversed. But the problem is symmetric, so it issucient to show that any plane works.Now the problem has been reduced to Sample Problem 39-2, so we are done.E39-11 We will choose the mirror to lie in the xy plane at z = 0. There is no loss of generalityin doing so; we had to dene our coordinate system somehow. The choice is convenient in thatany normal is then parallel to the z axis. Furthermore, we can arbitrarily dene the incident rayto originate at (0, 0, z1 ). Lastly, we can rotate the coordinate system about the z axis so that thereected ray passes through the point (0, y3 , z3 ).The point of reection for this ray is somewhere on the surface of the mirror, say (x2 , y2 , 0). Thisdistance traveled from the point 1 to the reection point 2 isd12 =(0 x2 )2 + (0 y2 )2 + (z1 0)2 =22x2 + y2 + z12and the distance traveled from the reection point 2 to the nal point 3 isd23 =(x2 0)2 + (y2 y3 )2 + (0 z3 )2 =2x2 + (y2 y3 )2 + z3 .2The only point which is free to move is the reection point, (x2 , y2 , 0), and that point can onlymove in the xy plane. Fermats principle states that the reection point will be such to minimizethe total distance,d12 + d23 =22x2 + y2 + z1 +22x2 + (y2 y3 )2 + z3 .2We do this minimization by taking the partial derivative with respect to both x2 and y2 . Butwe can do part by inspection alone. Any non-zero value of x2 can only add to the total distance,regardless of the value of any of the other quantities. Consequently, x2 = 0 is one of the conditionsfor minimization.We are done! Although you are invited to nish the minimization process, once we know thatx2 = 0 we have that point 1, point 2, and point 3 all lie in the yz plane. The normal is parallel tothe z axis, so it also lies in the yz plane. Everything is then in the yz plane.E39-12 Refer to Page 442 of Volume 1.E39-13 (a) 1 = 38 .(b) (1.58) sin(38 ) = (1.22) sin 2 . Then 2 = arcsin(0.797) = 52.9 .E39-14 ng = nv sin 1 / sin 2 = (1.00) sin(32.5 )/ sin(21.0 ) = 1.50.E39-15 n = c/v = (3.00108 m/s)/(1.92108 m/s) = 1.56.E39-16 v = c/n = (3.00108 m/s)/(1.46) = 2.05108 m/s.E39-17 The speed of light in a substance with index of refraction n is given by v = c/n. Anelectron will then emit Cerenkov radiation in this particular liquid if the speed exceedsv = c/n = (3.00 108 m/s)/(1.54) = 1.95108 m/s.175E39-18 Since t = d/v = nd/c, t = n d/c. Thent = (1.00029 1.00000)(1.61103 m)/(3.00108 m/s) = 1.56109 s.E39-19 The angle of the refracted ray is 2 = 90 , the angle of the incident ray can be found bytrigonometry,(1.14 m)tan 1 == 1.34,(0.85 m)or 1 = 53.3 .We can use these two angles, along with the index of refraction of air, to nd that the index ofrefraction of the liquid from Eq. 39-4,n1 = n2(sin 90 )sin 2= 1.25.= (1.00)sin 1(sin 53.3 )There are no units attached to this quantity.E39-20 For an equilateral prism = 60 . Thenn=sin[ + ]/2sin[(37 ) + (60 )]/2== 1.5.sin[/2]sin[(60 )/2]E39-21E39-22 t = d/v; but L/d = cos 2 =t=nL2c 1 sin 21 sin2 2 and v = c/n. Combining,n2 L=cn22 sin 1(1.63)2 (0.547 m)=(3108 m/s)= 3.07109 s.2(1.632 ) sin (24 )E39-23 The ray of light from the top of the smokestack to the life ring is 1 , where tan 1 = L/hwith h the height and L the distance of the smokestack.Snells law gives n1 sin 1 = n2 sin 2 , so1 = arcsin[(1.33) sin(27 )/(1.00)] = 37.1 .Then L = h tan 1 = (98 m) tan(37.1 ) = 74 m.E39-24 The length of the shadow on the surface of the water isx1 = (0.64 m)/ tan(55 ) = 0.448 m.The ray of light which forms the end of the shadow has an angle of incidence of 35 , so the raytravels into the water at an angle of2 = arcsin(1.00)sin(35 )(1.33)= 25.5 .The ray travels an additional distancex2 = (2.00 m 0.64 m)/ tan(90 25.5 ) = 0.649 mThe total length of the shadow is(0.448 m) + (0.649 m) = 1.10 m.176E39-25 Well rely heavily on the gure for our arguments. Let x be the distance between thepoints on the surface where the vertical ray crosses and the bent ray crosses.2xd app1dIn this exercise we will take advantage of the fact that, for small angles , sin tan Inthis approximation Snells law takes on the particularly simple form n1 1 = n2 2 The two angleshere are conveniently found from the gure,1 tan 1 =and2 tan 2 =x,dxdapp.Inserting these two angles into the simplied Snells law, as well as substituting n1 = n and n2 = 1.0,n1 1xnddapp= n2 2 ,x=,dappd=.nE39-26 (a) You need to address the issue of total internal reection to answer this question.(b) Rearrangesin[ + ]/2n=sin[/2]/and = ( + )/2 to get = arcsin (n sin[/2]) = arcsin ((1.60) sin[(60 )/2]) = 53.1 .E39-27 Use the results of Ex. 39-35. The apparent thickness of the carbon tetrachloride layer, asviewed by an observer in the water, isdc,w = nw dc /nc = (1.33)(41 mm)/(1.46) = 37.5 mm.The total thickness from the water perspective is then (37.5 mm) + (20 mm) = 57.5 mm. Theapparent thickness of the entire system as view from the air is thendapp = (57.5 mm)/(1.33) = 43.2 mm.177E39-28 (a) Use the results of Ex. 39-35. dapp = (2.16 m)/(1.33) = 1.62 m.(b) Need a diagram here!E39-29 (a) n = /n = (612 nm)/(1.51) = 405 nm.(b) L = nLn = (1.51)(1.57 pm) = 2.37 pm. There is actually a typo: the p in pm wassupposed to be a . This makes a huge dierence for part (c)!E39-30 (a) f = c/ = (3.00108 m/s)/(589 nm) = 5.091014 Hz.(b) n = /n = (589 nm)/(1.53) = 385 nm.(c) v = f = (5.091014 Hz)(385 nm) = 1.96108 m/s.E39-31 (a) The second derivative ofL=isa2 + x2 +b2 + (d x)2a2 (b2 + (d 2)2 )3/2 + b2 (a2 + x2 )3/2.(b2 + (d 2)2 )3/2 (a2 + x2 )3/2This is always a positive number, so dL/dx = 0 is a minimum.(a) The second derivative ofL = n1isa2 + x2 + n2b2 + (d x)2n1 a2 (b2 + (d 2)2 )3/2 + n2 b2 (a2 + x2 )3/2.(b2 + (d 2)2 )3/2 (a2 + x2 )3/2This is always a positive number, so dL/dx = 0 is a minimum.E39-32 (a) The angle of incidence on the face ac will be 90 . Total internal reection occurswhen sin(90 ) > 1/n, or < 90 arcsin[1/(1.52)] = 48.9 .(b) Total internal reection occurs when sin(90 ) > nw /n, or < 90 arcsin[(1.33)/(1.52)] = 29.0 .E39-33(a) The critical angle is given by Eq. 39-17,c = sin1n2(1.586)= sin1= 72.07 .n1(1.667)(b) Critical angles only exist when attempting to travel from a medium of higher index ofrefraction to a medium of lower index of refraction; in this case from A to B.E39-34 If the re is at the waters edge then the light travels along the surface, entering thewater near the sh with an angle of incidence of eectively 90 . Then the angle of refraction inthe water is numerically equivalent to a critical angle, so the sh needs to look up at an angle of = arcsin(1/1.33) = 49 with the vertical. Thats the same as 41 with the horizontal.178E39-35 Light can only emerge from the water if it has an angle of incidence less than the criticalangle, or < c = arcsin 1/n = arcsin 1/(1.33) = 48.8 .The radius of the circle of light is given by r/d = tan c , where d is the depth. The diameter is twicethis radius, or2(0.82 m) tan(48.8 ) = 1.87 m.E39-36 The refracted angle is given by n sin 1 = sin(39 ). This ray strikes the left surface withan angle of incidence of 90 1 . Total internal reection occurs whensin(90 1 ) = 1/n;but sin(90 1 ) = cos 1 , so we can combine and get tan = sin(39 ) with solution 1 = 32.2 . Theindex of refraction of the glass is thenn = sin(39 )/ sin(32.2) = 1.18.E39-37 The light strikes the quartz-air interface from the inside; it is originally white, so ifthe reected ray is to appear bluish (reddish) then the refracted ray should have been reddish(bluish). Since part of the light undergoes total internal reection while the other part does not,then the angle of incidence must be approximately equal to the critical angle.(a) Look at Fig. 39-11, the index of refraction of fused quartz is given as a function of thewavelength. As the wavelength increases the index of refraction decreases. The critical angle is afunction of the index of refraction; for a substance in air the critical angle is given by sin c = 1/n.As n decreases 1/n increases so c increases. For fused quartz, then, as wavelength increases c alsoincreases.In short, red light has a larger critical angle than blue light. If the angle of incidence is midwaybetween the critical angle of red and the critical angle of blue, then the blue component of the lightwill experience total internal reection while the red component will pass through as a refracted ray.So yes, the light can be made to appear bluish.(b) No, the light cant be made to appear reddish. See above.(c) Choose an angle of incidence between the two critical angles as described in part (a). Usinga value of n = 1.46 from Fig. 39-11,c = sin1 (1/1.46) = 43.2 .Getting the eect to work will require considerable sensitivity.E39-38 (a) There needs to be an opaque spot in the center of each face so that no refracted rayemerges. The radius of the spot will be large enough to cover rays which meet the surface at lessthan the critical angle. This means tan c = r/d, where d is the distance from the surface to thespot, or 6.3 mm. Sincec = arcsin 1/(1.52) = 41.1 ,then r = (6.3 mm) tan(41.1 ) = 5.50 mm.(b) The circles have an area of a = (5.50 mm)2 = 95.0 mm2 . Each side has an area of (12.6 mm)2 ;the fraction covered is then (95.0 mm2 )/(12.6 mm)2 = 0.598.E39-39 For uc the relativistic Doppler shift simplies tof = f0 u/c = u/0 ,sou = 0 f = (0.211 m)f.179E39-40 c = f , so 0 = f + f . Then / = f /f . Furthermore, f0 f , from Eq. 39-21,is f0 u/c for small enough u. Thenf f0u== .f0cE39-41The Doppler theory for light givesf = f01 u/c1u2 /c21 (0.2)= f01 (0.2)2= 0.82 f0 .The frequency is shifted down to about 80%, which means the wavelength is shifted up by anadditional 25%. Blue light (480 nm) would appear yellow/orange (585 nm).E39-42 Use Eq. 39-20:f = f01 u/c1u2 /c2= (100 Mhz)1 (0.892)1 (0.892)2= 23.9 MHz.E39-43 (a) If the wavelength is three times longer then the frequency is one-third, so for theclassical Doppler shiftf0 /3 = f0 (1 u/c),or u = 2c.(b) For the relativistic shift,f0 /3= f01 u/c1 u2 /c2,1 u2 /c2 = 3(1 u/c),c2 u2 = 9(c u)2 ,0 = 10u2 18uc + 8c2 .The solution is u = 4c/5.E39-44 (a) f0 /f = /0 . This shift is small, so we apply the approximation:u=c01= (3108 m/s)(462 nm)1(434 nm)= 1.9107 m/s.(b) A red shift corresponds to objects moving away from us.E39-45 The sun rotates once every 26 days at the equator, while the radius is 7.0108 m. Thespeed of a point on the equator is thenv=2R2(7.0108 m)== 2.0103 m/s.T(2.2106 s)This corresponds to a velocity parameter of = u/c = (2.0103 m/s)/(3.0108 m/s) = 6.7106 .This is a case of small numbers, so well use the formula that you derived in Exercise 39-40: = = (553 nm)(6.7106 ) = 3.7103 nm.180E39-46 Use Eq. 39-23 written as(1 u/c)2 = 2 (1 + u/c),0which can be rearranged asu/c =2 2(540 nm)2 (620 nm)20== 0.137.2 + 2(540 nm)2 + (620 nm)20The negative sign means that you should be going toward the red light.E39-47 (a) f1 = cf /(c + v) and f2 = cf /(c v).f = (f2 f ) (f f1 ) = f2 + f1 2f,soffcc+ 2,c+v cv2v 2,c2 v 22(8.65105 m/s)2,(3.00108 m/s)2 (8.65105 m/s)2===1.66105 .=(b) f1 = f (c u)/sqrtc2 u2 and f2 = f (c + u)/ c2 u2 .f = (f2 f ) (f f1 ) = f2 + f1 2f,soff===2c 2, u22(3.00108 m/s)c2(3.00108 m/s)2 (8.65105 m/s)2 2,8.3106 .E39-48 (a) No relative motion, so every 6 minutes.(b) The Doppler eect at this speed is1 u/c1u2 /c2=1 (0.6)1 (0.6)2= 0.5;this means the frequency is one half, so the period is doubled to 12 minutes.(c) If C send the signal at the instant the signal from A passes, then the two signals travel togetherto C, so C would get Bs signals at the same rate that it gets As signals: every six minutes.E39-49E39-50 The transverse Doppler eect is = 0 / 1 u2 /c2 . Then = (589.00 nm)/ 1 (0.122)2 = 593.43 nm.The shift is (593.43 nm) (589.00 nm) = 4.43 nm.181E39-51The frequency observed by the detector from the rst source is (Eq. 39-31)f = f11 (0.717)2 = 0.697f1 .The frequency observed by the detector from the second source is (Eq. 39-30)f = f21 (0.717)20.697f2=.1 + (0.717) cos 1 + (0.717) cos We need to equate these and solve for . Then0.697f11 + 0.717 cos cos 0.697f2,1 + 0.717 cos = f2 /f1 ,= (f2 /f1 1) /0.717,= 101.1 .=Subtract from 180 to nd the angle with the line of sight.E39-52P39-1 Consider the triangle in Fig. 39-45. The true position corresponds to the speed of light,the opposite side corresponds to the velocity of earth in the orbit. Then = arctan(29.8103 m/s)/(3.00108 m/s) = 20.5 .P39-2isThe distance to Jupiter from point x is dx = rj re . The distance to Jupiter from point yd2 =22re + rj .The dierence in distance is related to the time according to(d2 d1 )/t = c,so(778109 m)2 + (150109 m)2 (778109 m) + (150109 m)= 2.7108 m/s.(600 s)P39-3 sin(30 )/(4.0 m/s) = sin /(3.0 m/s). Then = 22 . Water waves travel more slowly inshallower water, which means they always bend toward the normal as they approach land.P39-4 (a) If the ray is normal to the waters surface then it passes into the water undeected.Once in the water the problem is identical to Sample Problem 39-2. The reected ray in the wateris parallel to the incident ray in the water, so it also strikes the water normal, and is transmittednormal.(b) Assume the ray strikes the water at an angle 1 . It then passes into the water at an angle2 , wherenw sin 2 = na sin 1 .Once the ray is in the water then the problem is identical to Sample Problem 39-2. The reectedray in the water is parallel to the incident ray in the water, so it also strikes the water at an angle2 . When the ray travels back into the air it travels with an angle 3 , wherenw sin 2 = na sin 3 .Comparing the two equations yields 1 = 3 , so the outgoing ray in the air is parallel to the incomingray.182P39-5(a) As was done in Ex. 39-25 above we use the small angle approximation ofsin tan The incident angle is ; if the light were to go in a straight line we would expect it to strike adistance y1 beneath the normal on the right hand side. The various distances are related to theangle by tan y1 /t.The light, however, does not go in a straight line, it is refracted according to (the small angleapproximation to) Snells law, n1 1 = n2 2 , which we will simplify further by letting 1 = , n2 = n,and n1 = 1, = n2 . The point where the refracted ray does strike is related to the angle by2 tan 2 = y2 /t. Combining the three expressions,y1 = ny2 .The dierence, y1 y2 is the vertical distance between the displaced ray and the original ray asmeasured on the plate glass. A little algebra yieldsy1 y2= y1 y1 /n,= y1 (1 1/n) ,n1= t.nThe perpendicular distance x is related to this dierence bycos = x/(y1 y2 ).In the small angle approximation cos 1 2 /2. If is suciently small we can ignore the squareterm, and x y2 y1 .(b) Remember to use radians and not degrees whenever the small angle approximation is applied.Then(1.52) 1x = (1.0 cm)(0.175 rad)= 0.060 cm.(1.52)P39-6(a) At the top layer,n1 sin 1 = sin ;at the next layer,n2 sin 2 = n1 sin 1 ;at the next layer,n3 sin 3 = n2 sin 2 .Combining all three expressions,n3 sin 3 = sin .(b) 3 = arcsin[sin(50 )/(1.00029)] = 49.98 . Then shift is (50 ) (49.98 ) = 0.02 .P39-7 The big idea of Problem 6 is that when light travels through layers the angle that itmakes in any layer depends only on the incident angle, the index of refraction where that incidentangle occurs, and the index of refraction at the current point.That means that light which leaves the surface of the runway at 90 to the normal will make ananglen0 sin 90 = n0 (1 + ay) sin 183at some height y above the runway. It is mildly entertaining to note that the value of n0 is unimportant, only the value of a!The expression1sin = 1 ay1 + aycan be used to nd the angle made by the curved path against the normal as a function of y. Theslope of the curve at any point is given bydycos = tan(90 ) = cot =.dxsin Now we need to know cos . It is1 sin2 cos =Combining2ay.2aydy,dx1 ayand now we integrate. We will ignore the ay term in the denominator because it will always besmall compared to 1. Thendhdx =00d=dy,2ay2h=a2(1.7 m)= 1500 m.(1.5106 m1 )P39-8 The energy of a particle is given by E 2 = p2 c2 + m2 c4 . This energy is related to the massby E = mc2 . is related to the speed by = 1/ 1 u2 /c2 . Rearranging,uc1=2=1=1m2 c2,+ m2 c2p2.p2 + m2 c2p2Since n = c/u we can write this asm2 c2=p2mc2pcn=1+1+1+(135 MeV)(145 MeV)2n=1+(106 MeV)(145 MeV)2n=2.For the pion,= 1.37.For the muon,184= 1.24.P39-9 (a) Before adding the drop of liquid project the light ray along the angle so that = 0.Increase slowly until total internal reection occurs at angle 1 . Thenng sin 1 = 1is the equation which can be solved to nd ng .Now put the liquid on the glass and repeat the above process until total internal reection occursat angle 2 . Thenng sin 2 = nl .Note that ng < ng for this method to work.(b) This is not terribly practical.P39-10 Let the internal angle at Q be Q . Then n sin Q = 1, because it is a critical angle. Letthe internal angle at P be P . Then P + Q = 90 . Combine this with the other formula and1 = n sin(90 P ) = n cos Q = n1 sin2 P .Not only that, but sin 1 = n sin P , or1 (sin 1 )2 /n2 ,1=nwhich can be solved for n to yieldn=1 + sin2 1 .(b) The largest value of the sine function is one, so nmax =2.P39-11 (a) The fraction of light energy which escapes from the water is dependent on the criticalangle. Light radiates in all directions from the source, but only that which strikes the surface at anangle less than the critical angle will escape. This critical angle issin c = 1/n.We want to nd the solid angle of the light which escapes; this is found by integrating2c=sin d d.00This is not a hard integral to do. The result is = 2(1 cos c ).There are 4 steradians in a spherical surface, so the fraction which escapes isf=11(1 cos c ) = (1 221 sin2 c ).The last substitution is easy enough. We never needed to know the depth h.(b) f = 1 (1 1 (1/(1.3))2 ) = 0.18.2185P39-12to(a) The beam of light strikes the face of the ber at an angle and is refracted accordingn1 sin 1 = sin .The beam then travels inside the ber until it hits the cladding interface; it does so at an angle of90 1 to the normal. It will be reected if it exceeds the critical angle ofn1 sin c = n2 ,or ifsin(90 1 ) n2 /n1 ,which can be written ascos 1 n2 /n1 .but if this is the cosine, then we can use sin2 + cos2 = 1 to nd the sine, and1 n2 /n2 .21sin 1 Combine this with the rst equation andn2 n2 .12 arcsin(b) = arcsin(1.58)2 (1.53)2 = 23.2 .P39-13 Consider the two possible extremes: a ray of light can propagate in a straight linedirectly down the axis of the ber, or it can reect o of the sides with the minimum possible angleof incidence. Start with the harder option.The minimum angle of incidence that will still involve reection is the critical angle, sosin c =n2.n1This light ray has farther to travel than the ray down the ber axis because it is traveling at anangle. The distance traveled by this ray isL = L/ sin c = Ln1,n2The time taken for this bouncing ray to travel a length L down the ber is thent =LL n1L n21==.vcc n2Now for the easier ray. It travels straight down the ber in a timet=The dierence ist t = t =LcLn1 .cn21 n1n2=Ln1(n1 n2 ).cn2(b) For the numbers in Problem 12 we havet =(350103 m)(1.58)((1.58) (1.53)) = 6.02105 s.(3.00108 m/s)(1.53)186P39-14P39-15 We can assume the airplane speed is small compared to the speed of light, and use Eq.39-21. f = 990 Hz; so|f | = f0 u/c = u/0 ,hence u = (990/s)(0.12 m) = 119 m/s. The actual answer for the speed of the airplane is half thisbecause there were two Doppler shifts: once when the microwaves struck the plane, and one when thereected beam was received by the station. Hence, the plane approaches with a speed of 59.4 m/s.187E40-1 (b) Since i = o, vi = di/dt = do/dt = vo .(a) In order to change from the frame of reference of the mirror to your own frame of referenceyou need to subtract vo from all velocities. Then your velocity is vo v0 = 0, the mirror is movingwith velocity 0 vo = vo and your image is moving with velocity vo vo = 2vo .E40-2 You are 30 cm from the mirror, the image is 10 cm behind the mirror. You need to focus40 cm away.E40-3 If the mirror rotates through an angle then the angle of incidence will increase by anangle , and so will the angle of reection. But that means that the angle between the incidentangle and the reected angle has increased by twice.E40-4 Sketch a line from Sarah through the right edge of the mirror and then beyond. Sarah cansee any image which is located between that line and the mirror. By similar triangles, the imageof Bernie will be d/2 = (3.0 m)/2 = 1/5 m from the mirror when it becomes visible. Since i = o,Bernie will also be 1.5 m from the mirror.E40-5 The images are fainter than the object. Several sample rays are shown.E40-6 The image is displaced. The eye would need to look up to see it.E40-7 The apparent depth of the swimming pool is given by the work done for Exercise 3925, dapp = d/n The water then appears to be only 186 cm/1.33 = 140 cm deep. The apparentdistance between the light and the mirror is then 250 cm + 140 cm = 390 cm; consequently theimage of the light is 390 cm beneath the surface of the mirror.E40-8 Three. There is a single direct image in each mirror and one more image of an image inone of the mirrors.188E40-9 We want to know over what surface area of the mirror are rays of light reected from theobject into the eye. By similar triangles the diameter of the pupil and the diameter of the part ofthe mirror (d) which reects light into the eye are related by(5.0 mm)d=,(10 cm)(24 cm) + (10 cm)which has solution d = 1.47 mm The area of the circle on the mirror isA = (1.47 mm)2 /4 = 1.7 mm2 .E40-10 (a) Seven; (b) Five; and (c) Two. This is a problem of symmetry.E40-11 Seven. Three images are the ones from Exercise 8. But each image has an image in theceiling mirror. That would make a total of six, except that you also have an image in the ceilingmirror (look up, eh?). So the total is seven!E40-12 A point focus is not formed. The envelope of rays is called the caustic. You can see asimilar eect when you allow light to reect o of a spoon onto a table.E40-13 The image is magnied by a factor of 2.7, so the image distance is 2.7 times farther fromthe mirror than the object. An important question to ask is whether or not the image is real orvirtual. If it is a virtual image it is behind the mirror and someone looking at the mirror could seeit. If it were a real image it would be in front of the mirror, and the man, who serves as the objectand is therefore closer to the mirror than the image, would not be able to see it.So we shall assume that the image is virtual. The image distance is then a negative number.The focal length is half of the radius of curvature, so we want to solve Eq. 40-6, with f = 17.5 cmand i = 2.7o1110.63= +=,(17.5 cm)o 2.7oowhich has solution o = 11 cm.E40-14 The image will be located at a point given by111111= ==.ifo(10 cm) (15 cm)(30 cm)The vertical scale is three times the horizontal scale in the gure below.189E40-15 This problem requires repeated application of 1/f = 1/o + 1/i, r = 2f , m = i/o, orthe properties of plane, convex, or concave mirrors. All dimensioned variables below (f, r, i, o) aremeasured in centimeters.(a) Concave mirrors have positive focal lengths, so f = +20; r = 2f = +40;1/i = 1/f 1/o = 1/(20) 1/(10) = 1/(20);m = i/o = (20)/(10) = 2; the image is virtual and upright.(b) m = +1 for plane mirrors only; r = for at surface; f = /2 = ; i = o = 10; theimage is virtual and upright.(c) If f is positive the mirror is concave; r = 2f = +40;1/i = 1/f 1/o = 1/(20) 1/(30) = 1/(60);m = i/o = (60)/(30) = 2; the image is real and inverted.(d) If m is negative then the image is real and inverted; only Concave mirrors produce real images(from real objects); i = mo = (0.5)(60) = 30;1/f = 1/o + 1/i = 1/(30) + 1/(60) = 1/(20);r = 2f = +40.(e) If r is negative the mirror is convex; f = r/2 = (40)/2 = 20;1/o = 1/f 1/i = 1/(20) 1/(10) = 1/(20);m = (10)/(20) = 0.5; the image is virtual and upright.(f) If m is positive the image is virtual and upright; if m is less than one the image is reduced,but only convex mirrors produce reduced virtual images (from real objects); f = 20 for convexmirrors; r = 2f = 40; let i = mo = o/10, then1/f = 1/o + 1/i = 1/o 10/o = 9/o,so o = 9f = 9(20) = 180; i = o/10 = (180)/10 = 18.(g) r is negative for convex mirrors, so r = 40; f = r/2 = 20; convex mirrors produce onlyvirtual upright images (from real objects); so i is negative; and1/o = 1/f 1/i = 1/(20) 1/(4) = 1/(5);m = i/o = (4)/(5) = 0.8.(h) Inverted images are real; only concave mirrors produce real images (from real objects);inverted images have negative m; i = mo = (0.5)(24) = 12;1/f = 1/o + 1/i = 1/(24) + 1/(12) = 1/(8);r = 2f = 16.190E40-16 Use the angle denitions provided by Eq. 40-8. From triangle OaI we have + = 2,while from triangle IaC we have + = .Combining to eliminate we get = 2.Substitute Eq. 40-8 and eliminate s,21 1 = ,oiror112+=,o irwhich is the same as Eq. 40-4 if i i and r r.E40-17 (a) Consider the point A. Light from this point travels along the line ABC and will beparallel to the horizontal center line from the center of the cylinder. Since the tangent to a circledenes the outer limit of the intersection with a line, this line must describe the apparent size.(b) The angle of incidence of ray AB is given bysin 1 = r/R.The angle of refraction of ray BC is given bysin 2 = r /R.Snells law, and a little algebra, yieldsn1 sin 1rn1Rnr= n2 sin 2 ,r= n2 ,R= r .In the last line we used the fact that n2 = 1, because it is in the air, and n1 = n, the index ofrefraction of the glass.E40-18 This problem requires repeated application of (n2 n1 )/r = n1 /o + n2 /i. All dimensionedvariables below (r, i, o) are measured in centimeters.(a)(1.5) (1.0) (1.0)= 0.08333,(30)(10)so i = (1.5)/(0.08333) = 18, and the image is virtual.(b)(1.0)(1.5)+= 0.015385,(10)(13)so r = (1.5 1.0)/(0.015385) = 32.5, and the image is virtual.(c)(1.5) (1.0)(1.5)= 0.014167,(30)(600)191so o = (1.0)/(0.014167) = 71. The image was real since i > 0.(d) Rearrange the formula to solve for n2 , thenn21 1ir= n11 1+ro.Substituting the numbers,n211(20) (20)= (1.0)11+(20) (20),which has any solution for n2 ! Since i < 0 the image is virtual.(e)(1.5) (1.0)+= 0.016667,(10)(6)so r = (1.0 1.5)/(0.016667) = 30, and the image is virtual.(f)(1.0)(1.0) (1.5)= 0.15,(30)(7.5)so o = (1.5)/(0.15) = 10. The image was virtual since i < 0.(g)(1.0) (1.5) (1.5)= 3.81102 ,(30)(70)so i = (1.0)/(3.81102 ) = 26, and the image is virtual.(h) Solving Eq. 40-10 for n2 yieldsn2 = n1son2 = (1.5)1/o + 1/r,1/r 1/i1/(100) + 1/(30)= 1.01/(30) 1/(600)and the image is real.E40-19 (b) If the beam is small we can use Eq. 40-10. Parallel incoming rays correspond to anobject at innity. Solving for n2 yieldsn2 = n11/o + 1/r,1/r 1/iso if o and i = 2r, thenn2 = (1.0)1/ + 1/r= 2.01/r 1/2r(c) There is no solution if i = r!E40-20 The image will be located at a point given by111111= ==.ifo(10 cm) (6 cm)(15 cm)192E40-21The image location can be found from Eq. 40-15,111111= ==,ifo(30 cm) (20 cm)12 cmso the image is located 12 cm from the thin lens, on the same side as the object.E40-22 For a double convex lens r1 > 0 and r2 < 0 (see Fig. 40-21 and the accompanying text).Then the problem states that r2 = r1 /2. The lens makers equation can be applied to get1= (n 1)f11r1r2=3(n 1),r1so r1 = 3(n 1)f = 3(1.5 1)(60 mm) = 90 mm, and r2 = 45 mm.E40-23 The object distance is essentially o = , so 1/f = 1/o + 1/i implies f = i, and the imageforms at the focal point. In reality, however, the object distance is not innite, so the magnicationis given by m = i/o f /o, where o is the Earth/Sun distance. The size of the image is thenhi = ho f /o = 2(6.96108 m)(0.27 m)/(1.501011 m) = 2.5 mm.The factor of two is because the suns radius is given, and we need the diameter!E40-24 (a) The at side has r2 = , so 1/f = (n 1)/r, where r is the curved side. Thenf = (0.20 m)/(1.5 1) = 0.40 m.(b) 1/i = 1/f 1/o = 1/(0.40 m) 1/(0.40 m) = 0. Then i is .E40-25 (a) 1/f = (1.5 1)[1/(0.4 m) 1/(0.4 m)] = 1/(0.40 m).(b) 1/f = (1.5 1)[1/() 1/(0.4 m)] = 1/(0.80 m).(c) 1/f = (1.5 1)[1/(0.4 m) 1/(0.6 m)] = 1/(2.40 m).(d) 1/f = (1.5 1)[1/(0.4 m) 1/(0.4 m)] = 1/(0.40 m).(e) 1/f = (1.5 1)[1/() 1/(0.8 m)] = 1/(0.80 m).(f) 1/f = (1.5 1)[1/(0.6 m) 1/(0.4 m)] = 1/(2.40 m).E40-26 (a) 1/f = (n 1)[1/(r) 1/r], so 1/f = 2(1 n)/r. 1/i = 1/f 1/o so if o = r, then1/i = 2(1 n)/r 1/r = (1 2n)/r,so i = r/(1 2n). For n > 0.5 the image is virtual.(b) For n > 0.5 the image is virtual; the magnication ism = i/o = r/(1 2n)/r = 1/(2n 1).E40-27According to the denitions, o = f + x and i = f + x . Starting with Eq. 40-15,1 1+oii+ooi2f + x + x(f + x)(f + x )2f 2 + f x + f xf2===1,f1,f1,f= f 2 + f x + f x + xx ,= xx .193E40-28 (a) You cant determine r1 , r2 , or n. i is found from1111==,i+10 +20+20the image is real and inverted. m = (20)/(20) = 1.(b) You cant determine r1 , r2 , or n. The lens is converging since f is positive. i is found from1111==,i+10 +510the image is virtual and upright. m = (10)/(+5) = 2.(c) You cant determine r1 , r2 , or n. Since m is positive and greater than one the lens isconverging. Then f is positive. i is found from1111==,i+10 +510the image is virtual and upright. m = (10)/(+5) = 2.(d) You cant determine r1 , r2 , or n. Since m is positive and less than one the lens is diverging.Then f is negative. i is found from1111==,i10 +53.3the image is virtual and upright. m = (3.3)/(+5) = 0.66.(e) f is found from1111= (1.5 1)=.f+30 30+30The lens is converging. i is found from1111==,i+30 +1015the image is virtual and upright. m = (15)/(+10) = 1.5.(f) f is found from1111= (1.5 1)=.f30 +3030The lens is diverging. i is found from1111==,i30 +107.5the image is virtual and upright. m = (7.5)/(+10) = 0.75.(g) f is found from1111= (1.5 1)=.f30 60120The lens is diverging. i is found from1111==,i120 +109.2the image is virtual and upright. m = (9.2)/(+10) = 0.92.(h) You cant determine r1 , r2 , or n. Upright images have positive magnication. i is found fromi = (0.5)(10) = 5;194f is found from1111=+=,f+10 510so the lens is diverging.(h) You cant determine r1 , r2 , or n. Real images have negative magnication. i is found fromi = (0.5)(10) = 5;f is found from1111=+ =,f+10 5+3.33so the lens is converging.E40-29 o + i = 0.44 m = L, so11 111L= + = +=,foio Loo(L o)which can also be written as o2 oL + f L = 0. This has solutiono=L(0.44 m) L2 4f L=2(0.44 m) 4(0.11 m)(0.44 m)= 0.22 m.2There is only one solution to this problem, but sometimes there are two, and other times there arenone!E40-30 (a) Real images (from real objects) are only produced by converging lenses.(b) Since hi = h0 /2, then i = o/2. But d = i+o = o+o/2 = 3o/2, so o = 2(0.40 m)/3 = 0.267 m,and i = 0.133 m.(c) 1/f = 1/o + 1/i = 1/(0.267 m) + 1/(0.133 m) = 1/(0.0889 m).E40-31 Step through the exercise one lens at a time. The object is 40 cm to the left of aconverging lens with a focal length of +20 cm. The image from this rst lens will be located bysolving111111= ==,ifo(20 cm) (40 cm)40 cmso i = 40 cm. Since i is positive it is a real image, and it is located to the right of the converginglens. This image becomes the object for the diverging lens.The image from the converging lens is located 40 cm - 10 cm from the diverging lens, but it islocated on the wrong side: the diverging lens is in the way so the rays which would form the imagehit the diverging lens before they have a chance to form the image. That means that the real imagefrom the converging lens is a virtual object in the diverging lens, so that the object distance for thediverging lens is o = 30 cm.The image formed by the diverging lens is located by solving111111= ==,ifo(15 cm) (30 cm)30 cmor i = 30 cm. This would mean the image formed by the diverging lens would be a virtual image,and would be located to the left of the diverging lens.The image is virtual, so it is upright. The magnication from the rst lens ism1 = i/o = (40 cm)/(40 cm)) = 1;195the magnication from the second lens ism2 = i/o = (30 cm)/(30 cm)) = 1;which implies an overall magnication of m1 m2 = 1.E40-32 (a) The parallel rays of light which strike the lens of focal length f will converge on thefocal point. This point will act like an object for the second lens. If the second lens is located adistance L from the rst then the object distance for the second lens will be L f . Note that thiswill be a negative value for L < f , which means the object is virtual. The image will form at a point1/i = 1/(f ) 1/(L f ) = L/f (f L).Note that i will be positive if L < f , so the rays really do converge on a point.(b) The same equation applies, except switch the sign of f . Then1/i = 1/(f ) 1/(L f ) = L/f (L f ).This is negative for L < f , so there is no real image, and no converging of the light rays.(c) If L = 0 then i = , which means the rays coming from the second lens are parallel.E40-33 The image from the converging lens is found from1111==i1(0.58 m) (1.12 m)1.20 mso i1 = 1.20 m, and the image is real and inverted.This real image is 1.97 m 1.20 m = 0.77 m in front of the plane mirror. It acts as an objectfor the mirror. The mirror produces a virtual image 0.77 m behind the plane mirror. This image isupright relative to the object which formed it, which was inverted relative to the original object.This second image is 1.97 m + 0.77 m = 2.74 m away from the lens. This second image acts as anobject for the lens, the image of which is found from1111==i3(0.58 m) (2.74 m)0.736 mso i3 = 0.736 m, and the image is real and inverted relative to the object which formed it, which wasinverted relative to the original object. So this image is actually upright.E40-34 (a) The rst lens forms a real image at a location given by1/i = 1/f 1/o = 1/(0.1 m) 1/(0.2 m) = 1/(0.2 m).The image and object distance are the same, so the image has a magnication of 1. This image is0.3 m 0.2 m = 0.1 m from the second lens. The second lens forms an image at a location given by1/i = 1/f 1/o = 1/(0.125 m) 1/(0.1 m) = 1/(0.5 m).Note that this puts the nal image at the location of the original object! The image is magnied bya factor of (0.5 m)/(0.1 m) = 5.(c) The image is virtual, but inverted.196E40-35 If the two lenses pass the same amount of light then the solid angle subtended by eachlens as seen from the respective focal points must be the same. If we assume the lenses have thesame round shape then we can write this as do /f o = de /f e . Thendefo== m ,dofeor de = (72 mm)/36 = 2 mm.E40-36 (a) f = (0.25 m)/(200) 1.3 mm. Then 1/f = (n 1)(2/r) can be used to nd r;r = 2(n 1)f = 2(1.5 1)(1.3 mm) = 1.3 mm.(b) The diameter would be twice the radius. In eect, these were tiny glass balls.E40-37 (a) In Fig. 40-46(a) the image is at the focal point. This means that in Fig. 40-46(b)i = f = 2.5 cm, even though f = f . Solving,1111=+=.f(36 cm) (2.5 cm)2.34 cm(b) The eective radii of curvature must have decreased.E40-38 (a) s = (25 cm) (4.2 cm) (7.7 cm) = 13.1 cm.(b) i = (25 cm) (7.7 cm) = 17.3 cm. Then1111==o(4.2 cm) (17.3 cm)5.54 cm.The object should be placed 5.5 4.2 = 1.34 cm beyond F1 .(c) m = (17.3)/(5.5) = 3.1.(d) m = (25 cm)/(7.7 cm) = 3.2.(e) M = mm = 10.E40-39 Microscope magnication is given by Eq. 40-33. We need to rst nd the focal lengthof the objective lens before we can use this formula. We are told in the text, however, that themicroscope is constructed so the at the object is placed just beyond the focal point of the objectivelens, then f ob 12.0 mm. Similarly, the intermediate image is formed at the focal point of theeyepiece, so f ey 48.0 mm. The magnication is thenm=s(250 mm)(285 mm)(250 mm)== 124.f ob f ey(12.0 mm)(48.0 mm)A more accurate answer can be found by calculating the real focal length of the objective lens, whichis 11.4 mm, but since there is a huge uncertainty in the near point of the eye, I see no point in tryingto be more accurate than this.P40-1 The old intensity is Io = P/4d2 , where P is the power of the point source. With themirror in place there is an additional amount of light which needs to travel a total distance of 3din order to get to the screen, so it contributes an additional P/4(3d)2 to the intensity. The newintensity is thenIn = P/4d2 + P/4(3d)2 = (10/9)P/4d2 = (10/9)Io .197P40-2(a) vi = di/dt; but i = f o/(o f ) and f = r/2 sovi =ddtro2o r2r2o r=do=dtr2o r2vo .(b) Put in the numbers!vi = (15 cm)2(75 cm) (15 cm)2(5.0 cm/s) = 6.2102 cm/s.(c) Put in the numbers!vi = (15 cm)2(7.7 cm) (15 cm)2(5.0 cm/s) = 70 m/s(d) Put in the numbers!vi = (15 cm)2(0.15 cm) (15 cm)2(5.0 cm/s) = 5.2 cm/s.P40-3 (b) There are two ends to the object of length L, one of these ends is a distance o1 fromthe mirror, and the other is a distance o2 from the mirror. The images of the two ends will belocated at i1 and i2 .Since we are told that the object has a short length L we will assume that a dierential approachto the problem is in order. ThenL = o = o1 o2 and L = i = i1 i2 ,Finding the ratio of L /L is then reduced toLidi=.LodoWe can take the derivative of Eq. 40-15 with respect to changes in o and i,di do+ 2 = 0,i2oorLdii2= 2 = m2 ,Ldoowhere m is the lateral magnication.(a) Since i is given by111of= =,ifoofthe fraction i/o can also be writtenioff==.oo(o f )ofThenLi2=o2198fof2P40-4 The left surface produces an image which is found from n/i = (n 1)/R 1/o, but sincethe incoming rays are parallel we take o = and the expression simplies to i = nR/(n 1). Thisimage is located a distance o = 2R i = (n 2)R/(n 1) from the right surface, and the imageproduced by this surface can be found from1/i = (1 n)/(R) n/o = (n 1)/R n(n 1)/(n 2)R = 2(1 n)/(n 2)R.Then i = (n 2)R/2(n 1).P40-5 The 1 in Eq. 40-18 is actually nair ; the assumption is that the thin lens is in the air. Ifthat isnt so, then we need to replace 1 with n , so Eq. 40-18 becomesnnnn=.o|i |r1A similar correction happens to Eq. 40-21:nnnn+=.|i |ir2Adding these two equations,nn+= (n n )oi11r1r2This yields a focal length given by1nn=fnP40-611r1r2Start with Eq. 40-41 1+oi|f | |f |+oi11+y y==1,f|f |,f= 1,where + is when f is positive and is when f is negative.The plot on the right is for +, that on the left for .Real image and objects occur when y or y is positive.199..P40-7 (a) The image (which will appear on the screen) and object are a distance D = o + iapart. We can use this information to eliminate one variable from Eq. 40-15,1 1+oi11+o DoDo(D o)o2 oD + f D1,f1,f1,f====0.This last expression is a quadratic, and we would expect to get two solutions for o. These solutionswill be of the form something plus/minus something else; the distance between the two locationsfor o will evidently be twice the something else, which is thend = o+ o =(D)2 4(f D) =D(D 4f ).(b) The ratio of the image sizes is m+ /m , or i+ o /i o+ . Now it seems we must nd the actualvalues of o+ and o . From the quadratic in part (a) we haveo =so the ratio isDD(D 4f )Dd=,22o=o+DdD+d.But i = o+ , and vice-versa, so the ratio of the image sizes is this quantity squared.P40-8 1/i = 1/f 1/o implies i = f o/(o f ). i is only real if o f . The distance between theimage and object isofo2y =i+o=+o=.ofofThis quantity is a minimum when dy/do = 0, which occurs when o = 2f . Then i = 2f , and y = 4f .P40-9 (a) The angular size of each lens is the same when viewed from the shared focal point. Thismeans W1 /f1 = W2 /f2 , orW2 = (f2 /f1 )W1 .(b) Pass the light through the diverging lens rst; choose the separation of the lenses so thatthe focal point of the converging lens is at the same location as the focal point of the diverging lenswhich is on the opposite side of the diverging lens.(c) Since I 1/A, where A is the area of the beam, we have I 1/W 2 . Consequently,I2 /I1 = (W1 /W2 )2 = (f1 /f2 )2P40-10The location of the image in the mirror is given by111= .ifa+b200The location of the image in the plate is given by i = a, which is located at b a relative to themirror. Equating,11+ba b+a2b2 a2bb2 a2===a ==1,f1,f2bf,b2 2bf ,(7.5 cm)2 2(7.5 cm)(28.2 cm) = 21.9 cm.P40-11 Well solve the problem by nding out what happens if you put an object in front of thecombination of lenses.Let the object distance be o1 . The rst lens will create an image at i1 , where111=i1f1o1This image will act as an object for the second lens.If the rst image is real (i1 positive) then the image will be on the wrong side of the secondlens, and as such the real image will act like a virtual object. In short, o2 = i1 will give the correctsign to the object distance when the image from the rst lens acts like an object for the second lens.The image formed by the second lens will then be at1i2===11 ,f2o211+ ,f2i2111+ .f2f1o1In this case it appears as if the combination11+f2f1is equivalent to the reciprocal of a focal length. We will go ahead and make this connection, and111f1 + f2=+=.ff2f1f1 f2The rest is straightforward enough.P40-12(a) The image formed by the rst lens can be found from1111==.i1f12f12f1This is a distance o2 = 2(f1 + f2 ) = 2f2 from the mirror. The image formed by the mirror is at animage distance given by1111==.i2f22f22f2Which is at the same point as i1 !. This means it will act as an object o3 in the lens, and, reversingthe rst step, produce a nal image at O, the location of the original object. There are then threeimages formed; each is real, same size, and inverted. Three inversions nets an inverted image. Thenal image at O is therefore inverted.201P40-13(a) Place an object at o. The image will be at a point i given by111= ,ifoor i = f o/(o f ).(b) The lens must be shifted a distance i i, ori i=fo 1.of(c) The range of motion isi =(0.05 m)(1.2 m) 1 = 5.2 cm.(1.2 m) (0.05 m)P40-14 (a) Because magnication is proportional to 1/f .(b) Using the results of Problem 40-11,111=+ ,ff2f1so P = P1 + P2 .P40-15 We want the maximum linear motion of the train to move no more than 0.75 mm on thelm; this means we want to nd the size of an object on the train that will form a 0.75 mm image.The object distance is much larger than the focal length, so the image distance is approximatelyequal to the focal length. The magnication is then m = i/o = (3.6 cm)/(44.5 m) = 0.00081.The size of an object on the train that would produce a 0.75 mm image on the lm is then0.75 mm/0.00081 = 0.93 m.How much time does it take the train to move that far?t=(0.93 m)= 25 ms.(135 km/hr)(1/3600 hr/s)P40-16 (a) The derivation leading to Eq. 40-34 depends only on the fact that two convergingoptical devices are used. Replacing the objective lens with an objective mirror doesnt changeanything except the ray diagram.(b) The image will be located very close to the focal point, so |m| f /o, andhi = (1.0 m)(16.8 m)= 8.4103 m(2000 m)(c) f e = (5 m)/(200) = 0.025 m. Note that we were given the radius of curvature, not the focallength, of the mirror!202E41-1In this problem we look for the location of the third-order bright fringe, so = sin1m(3)(554 109 m)= sin1= 12.5 = 0.22 rad.d(7.7 106 m)E41-2 d1 sin = gives the rst maximum; d2 sin = 2 puts the second maximum at the locationof the rst. Divide the second expression by the rst and d2 = 2d1 . This is a 100% increase in d.E41-3 y = D/d = (512109 m)(5.4 m)/(1.2103 m) = 2.3103 m.E41-4 d = / sin = (592109 m)/ sin(1.00 ) = 3.39105 m.E41-5 Since the angles are very small, we can assume sin for angles measured in radians.If the interference fringes are 0.23 apart, then the angular position of the rst bright fringeis 0.23 away from the central maximum. Eq. 41-1, written with the small angle approximationin mind, is d = for this rst (m = 1) bright fringe. The goal is to nd the wavelength whichincreases by 10%. To do this we must increase the right hand side of the equation by 10%, whichmeans increasing by 10%. The new wavelength will be = 1.1 = 1.1(589 nm) = 650 nmE41-6 Immersing the apparatus in water will shorten the wavelengths to /n. Start with d sin 0 =; and then nd from d sin = /n. Combining the two expressions, = arcsin[sin 0 /n] = arcsin[sin(0.20 )/(1.33)] = 0.15 .E41-7 The third-order fringe for a wavelength will be located at y = 3D/d, where y is measuredfrom the central maximum. Then y isy1 y2 = 3(1 2 )D/d = 3(612109 m 480109 m)(1.36 m)/(5.22103 m) = 1.03104 m.E41-8 = arctan(y/D); = d sin = (0.120 m) sin[arctan(0.180 m/2.0 m)] = 1.08102 m.Then f = v/ = (0.25 m/s)/(1.08102 m) = 23 Hz.E41-9A variation of Eq. 41-3 is in order:ym =m+12DdWe are given the distance (on the screen) between the rst minima (m = 0) and the tenth minima(m = 9). Then(50 cm)18 mm = y9 y0 = 9,(0.15 mm)or = 6104 mm = 600 nm.E41-10 The maximum maxima is given by the integer part ofm = d sin(90 )/ = (2.0 m)/(0.50 m) = 4.Since there is no integer part, the maximum maxima occurs at 90 . These are point sourcesradiating in both directions, so there are two central maxima, and four maxima each with m = 1,m = 2, and m = 3. But the m = 4 values overlap at 90 , so there are only two. The total is 16.203E41-11This gure should explain it well enough.E41-12 y = D/d = (589109 m)(1.13 m)/(0.18103 m) = 3.70103 m.E41-13 Consider Fig. 41-5, and solve it exactly for the information given. For the tenth brightfringe r1 = 10 + r2 . There are two important triangles:2r2 = D2 + (y d/2)2and2r1 = D2 + (y + d/2)2Solving to eliminate r2 ,D2 + (y + d/2)2 =D2 + (y d/2)2 + 10.This has solutiony = 54D2 + d2 1002.d2 1002The solution predicted by Eq. 41-1 isy =10dD2 + y 2 ,ory = 5d24D2. 1002The fractional error is y /y 1, or4D24D2 1,+ d2 1002or4(40 mm)2 1 = 3.1104 .4(40 mm)2 + (2 mm)2 100(589106 mm)2E41-14 (a) x = c/t = (3.00108 m/s)/(1108 s) = 3 m.(b) No.204E41-15 Leading by 90 is the same as leading by a quarter wavelength, since there are 360 in acircle. The distance from A to the detector is 100 m longer than the distance from B to the detector.Since the wavelength is 400 m, 100 m corresponds to a quarter wavelength.So a wave peak starts out from source A and travels to the detector. When it has traveled aquarter wavelength a wave peak leaves source B. But when the wave peak from A has traveleda quarter wavelength it is now located at the same distance from the detector as source B, whichmeans the two wave peaks arrive at the detector at the same time.They are in phase.E41-16 The rst dark fringe involves waves radians out of phase. Each dark fringe after thatinvolves an additional 2 radians of phase dierence. So the mth dark fringe has a phase dierenceof (2m + 1) radians.E41-17 I = 4I0 cos22dsin , so for this problem we want to plotI/I0 = cos22(0.60 mm)sin (600109 m)= cos2 (6280 sin ) .E41-18 The resultant quantity will be of the form A sin(t + ). Solve the problem by looking att = 0; then y1 = 0, but x1 = 10, and y2 = 8 sin 30 = 4 and x2 = 8 cos 30 = 6.93. Then the resultantis of lengthA = (4)2 + (10 + 6.93)2 = 17.4,and has an angle given by = arctan(4/16.93) = 13.3 .E41-19 (a) We want to know the path length dierence of the two sources to the detector.Assume the detector is at x and the second source is at y = d. The distance S1 D is x; thedistance S2 D is x2 + d2 . The dierence is x2 + d2 x. If this dierence is an integral numberof wavelengths then we have a maximum; if instead it is a half integral number of wavelengths wehave a minimum. For part (a) we are looking for the maxima, so we set the path length dierenceequal to m and solve for xm .x2 + d2 xmmx2 + d2mx2 + d2mxm= m,= (m + xm )2 ,= m2 2 + 2mxm + x2 ,md 2 m2 2=2mThe rst question we need to ask is what happens when m = 0. The right hand side becomesindeterminate, so we need to go back to the rst line in the above derivation. If m = 0 then d2 = 0;since this is not true in this problem, there is no m = 0 solution.In fact, we may have even more troubles. xm needs to be a positive value, so the maximumallowed value for m will be given bym2 2 < d2 ,m < d/ = (4.17 m)/(1.06 m) = 3.93;but since m is an integer, m = 3 is the maximum value.205The rst three maxima occur at m = 3, m = 2, and m = 1. These maxima are located atx3x2=x1(4.17 m)2 (3)2 (1.06 m)2= 1.14 m,2(3)(1.06 m)(4.17 m)2 (2)2 (1.06 m)2= 3.04 m,2(2)(1.06 m)(4.17 m)2 (1)2 (1.06 m)2= 7.67 m.2(1)(1.06 m)==Interestingly enough, as m decreases the maxima get farther away!(b) The closest maxima to the origin occurs at x = 6.94 cm. What then is x = 0? It is a localminimum, but the intensity isnt zero. It corresponds to a point where the path length dierence is3.93 wavelengths. It should be half an integer to be a complete minimum.E41-20 The resultant can be written in the form A sin(t + ). Consider t = 0. The threecomponents can be written as10 sin 0 = 0,14 sin 26 = 6.14,4.7 sin(41 ) = 3.08,0 + 6.14 3.08 = 3.06.y1y2y3y====x1x2x3x= 10 cos 0 = 10,= 14 cos 26 = 12.6,= 4.7 cos(41 ) = 3.55,= 10 + 12.6 + 3.55 = 26.2.andThen A =(3.06)2 + (26.2)2 = 26.4 and = arctan(3.06/26.2) = 6.66 .E41-21 The order of the indices of refraction is the same as in Sample Problem 41-4, sod = /4n = (620 nm)/4(1.25) = 124 nm.E41-22 Follow the example in Sample Problem 41-3.=2dn2(410 nm)(1.50)1230 nm==.m 1/2m 1/2m 1/2The result is only in the visible range when m = 3, so = 492 nm.E41-23 (a) Light from above the oil slick can be reected back up from the top of the oil layeror from the bottom of the oil layer. For both reections the light is reecting o a substance witha higher index of refraction so both reected rays pick up a phase change of . Since both waveshave this phase the equation for a maxima is112d + n + n = mn .22Remember that n = /n, where n is the index of refraction of the thin lm. Then 2nd = (m 1)is the condition for a maxima. We know n = 1.20 and d = 460 nm. We dont know m or . It might206seem as if there isnt enough information to solve the problem, but we can. We need to nd thewavelength in the visible range (400 nm to 700 nm) which has an integer m. Trial and error mightwork. If = 700 nm, then m ism=2nd2(1.20)(460 nm)+1=+ 1 = 2.58(700 nm)But m needs to be an integer. If we increase m to 3, then=2(1.20)(460 nm)= 552 nm(3 1)which is in the visible range. So the oil slick will appear green.(b) One of the most profound aspects of thin lm interference is that wavelengths which aremaximally reected are minimally transmitted, and vice versa. Finding the maximally transmittedwavelengths is the same as nding the minimally reected wavelengths, or looking for values of mthat are half integer.The most obvious choice is m = 3.5, and then=2(1.20)(460 nm)= 442 nm.(3.5 1)E41-24 The condition for constructive interference is 2nd = (m 1/2). Assuming a minimumvalue of m = 1 one ndsd = /4n = (560 nm)/4(2.0) = 70 nm.E41-25 The top surface contributes a phase dierence of , so the phase dierence because of thethickness is 2, or one complete wavelength. Then 2d = /n, or d = (572 nm)/2(1.33) = 215 nm.E41-26 The wave reected from the rst surface picks up a phase shift of . The wave which isreected o of the second surface travels an additional path dierence of 2d. The interference willbe bright if 2d + n /2 = mn results in m being an integer.m = 2nd/ + 1/2 = 2(1.33)(1.21106 m)/(585109 m) + 1/2 = 6.00,so the interference is bright.E41-27 As with the oil on the water in Ex. 41-23, both the light which reects o of the acetoneand the light which reects o of the glass undergoes a phase shift of . Then the maxima forreection are given by 2nd = (m 1). We dont know m, but at some integer value of m we have = 700 nm. If m is increased by exactly 1 then we are at a minimum of = 600 nm. Consequently,22(1.25)d = (m 1)(700 nm) and 2(1.25)d = (m 1/2)(600 nm),we can set these two expressions equal to each other to nd m,(m 1)(700 nm) = (m 1/2)(600 nm),so m = 4. Then we can nd the thickness,d = (4 1)(700 nm)/2(1.25) = 840 nm.207E41-28 The wave reected from the rst surface picks up a phase shift of . The wave which isreected o of the second surface travels an additional path dierence of 2d. The interference willbe bright if 2d + n /2 = mn results in m being an integer. Then 2nd = (m 1/2)1 is bright, and2nd = m2 is dark. Divide one by the other and (m 1/2)1 = m2 , som = 1 /2(1 2 ) = (600 nm)/2(600 nm 450 nm) = 2,then d = m2 /2n = (2)(450 nm)/2(1.33) = 338 nm.E41-29 Constructive interference happens when 2d = (m 1/2). The minimum value for m ism = 1; the maximum value is the integer portion of 2d/+1/2 = 2(4.8105 m)/(680109 m)+1/2 =141.67, so mmax = 141. There are then 141 bright bands.E41-30 (a) A half wavelength phase shift occurs for both the air/water interface and the water/oilinterface, so if d = 0 the two reected waves are in phase. It will be bright!(b) 2nd = 3, or d = 3(475 nm)/2(1.20) = 594 nm.E41-31 There is a phase shift on one surface only, so the bright bands are given by 2nd = (m 1/2). Let the rst band be given by 2nd1 = (m1 1/2). The last bright band is then given by2nd2 = (m1 + 9 1/2). Subtract the two equations to get the change in thickness:d = 9/2n = 9(630 nm)/2(1.50) = 1.89 m.E41-32 Apply Eq. 41-21: 2nd = m. In one case we have2nair = (4001),in the other,2nvac = (4000).Equating, nair = (4001)/(4000) = 1.00025.E41-33(a) We can start with the last equation from Sample Problem 41-5,r=1(m )R,2and solve for m,r21+R 2In this exercise R = 5.0 m, r = 0.01 m, and = 589 nm. Thenm=m=(0.01 m)2= 34(589 nm)(5.0 m)is the number of rings observed.(b) Putting the apparatus in water eectively changes the wavelength to(589 nm)/(1.33) = 443 nm,so the number of rings will now bem=(0.01 m)2= 45.(443 nm)(5.0 m)208(10 1 )R, while (1.27 cm) =2the other, and (1.42 cm)/(1.27 cm) = n, or n = 1.25.E41-34 (1.42 cm) =(10 1 )R/n. Divide one expression by2E41-35 (0.162 cm) = (n 1 )R, while (0.368 cm) =2sions, the divide one by the other, and nd(n + 20 1 )R. Square both expres2(n + 19.5)/(n 0.5) = (0.368 cm/0.162 cm)2 = 5.16which can be rearranged to yieldn=19.5 + 5.16 0.5= 5.308.5.16 1Oops! That should be an integer, shouldnt it? The above work is correct, which means that therereally arent bright bands at the specied locations. Im just going to gloss over that fact and solvefor R using the value of m = 5.308. ThenR = r2 /(m 1/2) = (0.162 cm)2 /(5.308 0.5)(546 nm) = 1.00 m.Well, at least we got the answer which is in the back of the book...E41-36 Pretend the ship is a two point source emitter, one h above the water, and one h belowthe water. The one below the water is out of phase by half a wavelength. Then d sin = , whered = 2h, gives the angle for theta for the rst minimum./2h = (3.43 m)/2(23 m) = 7.46102 = sin H/D,so D = (160 m)/(7.46102 ) = 2.14 km.E41-37 The phase dierence is 2/n times the path dierence which is 2d, so = 4d/n = 4nd/.We are given that d = 100109 m and n = 1.38.(a) = 4(1.38)(100109 m)/(450109 m) = 3.85. Then(3.85)I= cos2= 0.12.I02The reected ray is diminished by 1 0.12 = 88%.(b) = 4(1.38)(100109 m)/(650109 m) = 2.67. ThenI(2.67)= cos2= 0.055.I02The reected ray is diminished by 1 0.055 = 95%.E41-38 The change in the optical path length is 2(d d/n), so 7/n = 2d(1 1/n), ord=7(589109 m)= 4.9106 m.2(1.42) 2209E41-39 When M2 moves through a distance of /2 a fringe has will be produced, destroyed, andthen produced again. This is because the light travels twice through any change in distance. Thewavelength of light is then2(0.233 mm)= 588 nm.=792E41-40 The change in the optical path length is 2(d d/n), so 60 = 2d(1 1/n), orn=11== 1.00030.9 m)/2(5102 m)1 60/2d1 60(50010P41-1 (a) This is a small angle problem, so we use Eq. 41-4. The distance to the screen is2 20 m, because the light travels to the mirror and back again. Thend=D(632.8 nm)(40.0 m)== 0.253 mm.y(0.1 m)(b) Placing the cellophane over one slit will cause the interference pattern to shift to the left orright, but not disappear or change size. How does it shift? Since we are picking up 2.5 waves thenwe are, in eect, swapping bright fringes for dark fringes.P41-2The change in the optical path length is d d/n, so 7/n = d(1 1/n), ord=7(550109 m)= 6.64106 m.(1.58) 1P41-3 The distance from S1 to P is r1 =(x + d/2)2 + y 2 . The distance from S2 to P is2 + y 2 . The dierence in distances is xed at some value, say c, so thatr2 = (x d/2)r 1 r22 2r1 r2 + r222(r1 + r2 c2 )222 22 22(r1 r2 ) 2c (r1 + r2 ) + c4(2xd)2 2c2 (2x2 + d2 /2 + 2y 2 ) + c44x2 d2 4c2 x2 c2 d2 4c2 y 2 + c44(d2 c2 )x2 4c2 y 22r1=======c,c2 ,2 24r1 r2 ,0,0,0,c2 (d2 c2 ).Yes, that is the equation of a hyperbola.P41-4 The change in the optical path length for each slit is nt t, where n is the correspondingindex of refraction. The net change in the path dierence is then n2 t n1 t. Consequently, m =t(n2 n1 ), so(5)(480109 m)t== 8.0106 m.(1.7) (1.4)P41-5 The intensity is given by Eq. 41-17, which, in the small angle approximation, can bewritten asdI = 4I0 cos2.210The intensity will be half of the maximum when1= cos22ord/2d=,42which will happen if = /2d.P41-6 Follow the construction in Fig. 41-10, except that one of the electric eld amplitudes istwice the other. The resultant eld will have a length given byE(2E0 + E0 cos )2 + (E0 sin )2 ,== E05 + 4 cos ,so squaring this yieldsIP41-7mum is2d sin ,d sin = I0 1 + 8 cos2,Imd sin =1 + 8 cos2.9= I0 5 + 4 cosWe actually did this problem in Exercise 41-27, although slightly dierently. One maxi2(1.32)d = (m 1/2)(679 nm),the other is2(1.32)d = (m + 1/2)(485 nm).Set these equations equal to each other,(m 1/2)(679 nm) = (m + 1/2)(485 nm),and nd m = 3. Then the thickness isd = (3 1/2)(679 nm)/2(1.32) = 643 nm.P41-8(a) Since we are concerned with transmission there is a phase shift for two rays, so2d = mnThe minimum thickness occurs when m = 1; solving for d yieldsd=(525109 m)== 169109 m.2n2(1.55)(b) The wavelengths are dierent, so the other parts have diering phase dierences.(c) The nearest destructive interference wavelength occurs when m = 1.5, or = 2nd = 2(1.55)1.5(169109 m) = 393109 m.This is blue-violet.211P41-9 It doesnt matter if we are looking at bright are dark bands. It doesnt even matter if weconcern ourselves with phase shifts. All that cancels out. Consider 2d = m; thend = (10)(480 nm)/2 = 2.4 m.P41-10(a) Apply 2d = m. Thend = (7)(600109 m)/2 = 2100109 m.(b) When water seeps in it introduces an extra phase shift. Point A becomes then a bright fringe,and the equation for the number of bright fringes is 2nd = m. Solving for m,m = 2(1.33)(2100109 m)/(600109 m) = 9.3;this means that point B is almost, but not quite, a dark fringe, and there are nine of them.P41-11(a) Look back at the work for Sample Problem 41-5 where it was found1(m )R,2rm =We can write this as1rm =12mmRand expand the part in parentheses in a binomial expansion,1 1 rm 1 mR.2 2mWe will do the same with1(m + 1 )R,2rm+1 =expandingrm+1 =1+12mto getrm+1 1+Thenr or1 12 2mmRmR.1 mR,2m1R/m.2(b) The area between adjacent rings is found from the dierence,r 22A = rm+1 rm ,and into this expression we will substitute the exact values for rm and rm+1 ,11A = (m + 1 )R (m )R ,22= R.Unlike part (a), we did not need to assume mfor all m.1 in order to arrive at this expression; it is exact212P41-12 The path length shift that occurs when moving the mirror as distance x is 2x. This means = 22x/ = 4x/. The intensity is thenI = 4I0 cos22132xE42-1 = a sin = (0.022 mm) sin(1.8 ) = 6.91107 m.E42-2 a = / sin = (0.10109 m)/ sin(0.12103 rad/2) = 1.7 m.E42-3 (a) This is a valid small angle approximation problem: the distance between the pointson the screen is much less than the distance to the screen. Then(0.0162 m)= 7.5 103 rad.(2.16 m)(b) The diraction minima are described by Eq. 42-3,a sin = m,a sin(7.5 10 rad) = (2)(441 109 m),a = 1.18 104 m.3E42-4 a = / sin = (633109 m)/ sin(1.97 /2) = 36.8 m.E42-5 (a) We again use Eq. 42-3, but we will need to throw in a few extra subscripts todistinguish between which wavelength we are dealing with. If the angles match, then so will the sineof the angles. We then have sin a,1 = sin b,2 or, using Eq. 42-3,(1)a(2)b=,aafrom which we can deduce a = 2b .(b) Will any other minima coincide? We want to solve for the values of ma and mb that will beintegers and have the same angle. Using Eq. 42-3 one more time,mb bma a=,aaand then substituting into this the relationship between the wavelengths, ma = mb /2. whenever mbis an even integer ma is an integer. Then all of the diraction minima from a are overlapped by aminima from b .E42-6 The angle is given by sin = 2/a. This is a small angle, so we can use the small angleapproximation of sin = y/D. Theny = 2D/a = 2(0.714 m)(593109 m)/(420106 m) = 2.02 mm.E42-7 Small angles, so y/D = sin = /a. Thena = D/y = (0.823 m)(546109 m)/(5.20103 m/2) = 1.73104 m.E42-8 (b) Small angles, so y/D = m/a. Thena = mD/y = (5 1)(0.413 m)(546109 m)/(0.350103 m) = 2.58 mm.(a) = arcsin(/a) = arcsin[(546109 m)/(2.58 mm)] = 1.21102 .E42-9 Small angles, so y/D = m/a. Theny = mD/a = (2 1)(2.94 m)(589109 m)/(1.16103 m) = 1.49103 m.214E42-10 Doubling the width of the slit results in a narrowing of the diraction pattern. Since thewidth of the central maximum is eectively cut in half, then there is twice the energy in half thespace, producing four times the intensity.E42-11(a) This is a small angle approximation problem, so = (1.13 cm)/(3.48 m) = 3.25 103 rad.(b) A convenient measure of the phase dierence, is related to through Eq. 42-7,=(25.2 106 m)asin(3.25 103 rad) = 0.478 radsin =(538 109 m)(c) The intensity at a point is related to the intensity at the central maximum by Eq. 42-8,I=Imsin 2=sin(0.478 rad)(0.478 rad)2= 0.926E42-12 Consider Fig. 42-11; the angle with the vertical is given by ( )/2. For Fig. 42-10(d)the circle has wrapped once around onto itself so the angle with the vertical is (3 )/2. Substitute into this expression and the angel against the vertical is 3/2 .Use the result from Problem 42-3 that tan = for the maxima. The lowest non-zero solutionis = 4.49341 rad. The angle against the vertical is then 0.21898 rad, or 12.5 .E42-13 Drawing heavily from Sample Problem 42-4,x = arcsinx a= arcsin1.3910= 2.54 .Finally, = 2x = 5.1 .E42-14 (a) Rayleighs criterion for resolving images (Eq. 42-11) requires that two objects have anangular separation of at leastR = sin11.22d= sin11.22(540 109 )(4.90 103 m)= 1.34 104 rad(b) The linear separation is y = D = (1.34 104 rad)(163103 m) = 21.9 m.E42-15 (a) Rayleighs criterion for resolving images (Eq. 42-11) requires that two objects havean angular separation of at leastR = sin11.22d= sin11.22(562 109 )(5.00 103 m)= 1.37 104 rad.(b) Once again, this is a small angle, so we can use the small angle approximation to nd thedistance to the car. In that case R = y/D, where y is the headlight separation and D the distanceto the car. Solving,D = y/R = (1.42 m)/(1.37 104 rad) = 1.04 104 m,or about six or seven miles.215E42-16 y/D = 1.22/a; orD = (5.20103 m)(4.60103 /m)/1.22(542109 m) = 36.2 m.E42-17The smallest resolvable angular separation will be given by Eq. 42-11,R = sin11.22d= sin11.22(565 109 m)(5.08 m)= 1.36 107 rad,The smallest objects resolvable on the Moons surface by this telescope have a size y wherey = DR = (3.84 108 m)(1.36 107 rad) = 52.2 mE42-18 y/D = 1.22/a; ory = 1.22(1.57102 m)(6.25103 m)/(2.33 m) = 51.4 mE42-19 y/D = 1.22/a; orD = (4.8102 m)(4.3103 /m)/1.22(0.12109 m) = 1.4106 m.E42-20 y/D = 1.22/a; ord = 1.22(550109 m)(160103 m)/(0.30 m) = 0.36 m.E42-21Using Eq. 42-11, we nd the minimum resolvable angular separation is given byR = sin11.22d= sin11.22(475 109 m)(4.4 103 m)= 1.32 104 radThe dots are 2 mm apart, so we want to stand a distance D away such thatD > y/R = (2 103 m)/(1.32 104 rad) = 15 m.E42-22 y/D = 1.22/a; ory = 1.22(500109 m)(354103 m)/(9.14 m/2) = 4.73102 m.E42-23 (a) = v/f . Now use Eq. 42-11: = arcsin 1.22(1450 m/s)(25103 Hz)(0.60 m)= 6.77 .(b) Following the same approach, = arcsin 1.22(1450 m/s)(1103 Hz)(0.60 m)has no real solution, so there is no minimum.216E42-24 (a) = v/f . Now use Eq. 42-11: = arcsin 1.22(3108 m/s)(220109 Hz)(0.55 m)= 0.173 .This is the angle from the central maximum; the angular width is twice this, or 0.35 .(b) use Eq. 42-11:(0.0157 m) = arcsin 1.22= 0.471 .(2.33 m)This is the angle from the central maximum; the angular width is twice this, or 0.94 .E42-25The linear separation of the fringes is given byyD= = or y =Dddfor suciently small d compared to .E42-26 (a) d sin = 4 gives the location of the fourth interference maximum, while a sin = gives the location of the rst diraction minimum. Hence, if d = 4a there will be no fourthinterference maximum!(b) Since d sin mi = mi gives the interference maxima and a sin md = md gives the diractionminima, and d = 4a, then whenever mi = 4md there will be a missing maximum.E42-27 (a) The central diraction envelope is contained in the range = arcsinaThis angle corresponds to the mth maxima of the interference pattern, wheresin = m/d = m/2a.Equating, m = 2, so there are three interference bands, since the m = 2 band is washed out bythe diraction minimum.(b) If d = a then = and the expression reduces toIsin2 ,2sin2 (2)= Im,222sin = 2I m,= I m cos2 where = 2 , which is the same as replacing a by 2a.E42-28 Remember that the central peak has an envelope width twice that of any other peak.Ignoring the central maximum there are (11 1)/2 = 5 fringes in any other peak envelope.217E42-29 (a) The rst diraction minimum is given at an angle such that a sin = ; the orderof the interference maximum at that point is given by d sin = m. Dividing one expression by theother we get d/a = m, with solution m = (0.150)/(0.030) = 5. The fact that the answer is exactly 5implies that the fth interference maximum is squelched by the diraction minimum. Then there areonly four complete fringes on either side of the central maximum. Add this to the central maximumand we get nine as the answer.(b) For the third fringe m = 3, so d sin = 3. Then is Eq. 42-14 is 3, while in Eq. 42-16isaa 3== 3 , ddso the relative intensity of the third fringe is, from Eq. 42-17,(cos 3)2P42-1sin(3a/d)(3a/d)2= 0.255.y = mD/a. Theny = (10)(632.8109 m)(2.65 m)/(1.37103 m) = 1.224102 m.The separation is twice this, or 2.45 cm.P42-2 If a then the diraction pattern is extremely tight, and there is eectively no light atP . In the event that either shape produces an interference pattern at P then the other shape mustproduce an equal but opposite electric eld vector at that point so that when both patterns fromboth shapes are superimposed the eld cancel.But the intensity is the eld vector squared; hence the two patterns look identical.P42-3(a) We want to take the derivative of Eq. 42-8 with respect to , sodId2sin ,sin cos sin = I m2 2sin = I m 2 3 ( cos sin ) .=dImd,This equals zero whenever sin = 0 or cos = sin ; the former is the case for a minima while thelatter is the case for the maxima. The maxima case can also be written astan = .(b) Note that as the order of the maxima increases the solutions get closer and closer to oddintegers times /2. The solutions are = 0, 1.43, 2.46, etc.(c) The m values are m = / 1/2, and correspond tom = 0.5, 0.93, 1.96, etc.These values will get closer and closer to integers as the values are increased.218P42-4The outgoing beam strikes the moon with a circular spot of radiusr = 1.22D/a = 1.22(0.69106 m)(3.82108 m)/(2 1.3 m) = 123 m.The light is not evenly distributed over this circle.If P0 is the power in the light, thenRP0 =I r dr d = 2I r dr,0where R is the radius of the central peak and I is the angular intensity. For a ar/D, thenP0 = 2ImDa2/20sin2 d 2ImDa we can write2(0.82).Then the intensity at the center falls o with distance D as2Im = 1.9 (a/D) P0The fraction of light collected by the mirror on the moon is thenP1 /P0 = 1.9(2 1.3 m)(0.69106 m)(3.82108 m)2(0.10 m)2 = 5.6106 .The fraction of light collected by the mirror on the Earth is thenP2 /P1 = 1.9(2 0.10 m)(0.69106 m)(3.82108 m)2(1.3 m)2 = 5.6106 .Finally, P2 /P0 = 31011 .P42-5 (a) The ring is reddish because it occurs at the blue minimum.(b) Apply Eq. 42-11 for blue light:d = 1.22/ sin = 1.22(400 nm)/ sin(0.375 ) = 70 m.(c) Apply Eq. 42-11 for red light: = arcsin (1.22(700 nm)/(70 m)) 0.7 ,which occurs 3 lunar radii from the moon.P42-6 The diraction pattern is a property of the speaker, not the interference between the speakers. The diraction pattern should be unaected by the phase shift. The interference pattern,however, should shift up or down as the phase of the second speaker is varied.P42-7 (a) The missing fringe at = 5 is a good hint as to what is going on. There should besome sort of interference fringe, unless the diraction pattern has a minimum at that point. Thiswould be the rst minimum, soa sin(5 ) = (440 109 m)would be a good measure of the width of each slit. Then a = 5.05 106 m.219(b) If the diraction pattern envelope were not present we could expect that the fourth interference maxima beyond the central maximum would occur at this point, and thend sin(5 ) = 4(440 109 m)yieldingd = 2.02 105 m.(c) Apply Eq. 42-17, where = m and=a maasin ==m= m/4. ddThen for m = 1 we haveI1 = (7)sin(/4)(/4)I2 = (7)sin(2/4)(2/4)while for m = 2 we haveThese are in good agreement with the gure.2202= 5.7;2= 2.8.E43-1 (a) d = (21.5103 m)/(6140) = 3.50106 m.(b) There are a number of angles allowed:=====arcsin[(1)(589109 m)/(3.50106 m)] = 9.7 ,arcsin[(2)(589109 m)/(3.50106 m)] = 19.5 ,arcsin[(3)(589109 m)/(3.50106 m)] = 30.3 ,arcsin[(4)(589109 m)/(3.50106 m)] = 42.3 ,arcsin[(5)(589109 m)/(3.50106 m)] = 57.3 .E43-2 The distance between adjacent rulings isd=(2)(612109 m)= 2.235106 m.sin(33.2 )The number of lines is thenN = D/d = (2.86102 m)/(2.235106 m) = 12, 800.E43-3 We want to nd a relationship between the angle and the order number which is linear.Well plot the data in this representation, and then use a least squares t to nd the wavelength.The data to be plotted ism12317.637.365.2sin 0.3020.6060.908m-1-2-3-17.6-37.1-65.0sin -0.302-0.603-0.906On my calculator I get the best straight line t as0.302m + 8.33 104 = sin m ,which means that = (0.302)(1.73 m) = 522 nm.E43-4 Although an approach like the solution to Exercise 3 should be used, well assume that eachmeasurement is perfect and error free. Then randomly choosing the third maximum,=E43-5d sin (5040109 m) sin(20.33 )== 586109 m.m(3)(a) The principle maxima occur at points given by Eq. 43-1,sin m = m .dThe dierence of the sine of the angle between any two adjacent orders issin m+1 sin m = (m + 1)m = .dddUsing the information provided we can nd d fromd=(600 109 )== 6 m.sin m+1 sin m(0.30) (0.20)221It doesnt take much imagination to recognize that the second and third order maxima were given.(b) If the fourth order maxima is missing it must be because the diraction pattern envelope hasa minimum at that point. Any fourth order maxima should have occurred at sin 4 = 0.4. If it is adiraction minima thena sin m = m where sin m = 0.4We can solve this expression and nda=m(600 109 m)=m= m1.5 m.sin m(0.4)The minimum width is when m = 1, or a = 1.5 m.(c) The visible orders would be integer values of m except for when m is a multiple of four.E43-6 (a) Find the maximum integer value of m = d/ = (930 nm)/(615 nm) = 1.5, hencem = 1, 0, +1; there are three diraction maxima.(b) The rst order maximum occurs at = arcsin(615 nm)/(930 nm) = 41.4 .The width of the maximum is =(615 nm)= 7.87104 rad,(1120)(930 nm) cos(41.4 )or 0.0451 .E43-7 The fth order maxima will be visible if d/ 5; this meansd(1103 m)== 635109 m.5(315 rulings)(5)E43-8 (a) The maximum could be the rst, and then=(1103 m) sin(28 )d sin == 2367109 m.m(200)(1)Thats not visible. The rst visible wavelength is at m = 4, then=d sin (1103 m) sin(28 )== 589109 m.m(200)(4)The next is at m = 5, then=d sin (1103 m) sin(28 )== 469109 m.m(200)(5)Trying m = 6 results in an ultraviolet wavelength.(b) Yellow-orange and blue.222E43-9A grating with 400 rulings/mm has a slit separation ofd=1= 2.5 103 mm.400 mm1To nd the number of orders of the entire visible spectrum that will be present we need only considerthe wavelength which will be on the outside of the maxima. That will be the longer wavelengths, sowe only need to look at the 700 nm behavior. Using Eq. 43-1,d sin = m,and using the maximum angle 90 , we ndm<(2.5 106 m)d== 3.57,(700 109 m)so there can be at most three orders of the entire spectrum.E43-10 In this case d = 2a. Since interference maxima are given by sin = m/d while diractionminima are given at sin = m /a = 2m /d then diraction minima overlap with interferencemaxima whenever m = 2m . Consequently, all even m are at diraction minima and thereforevanish.E43-11 If the second-order spectra overlaps the third-order, it is because the 700 nm second-orderline is at a larger angle than the 400 nm third-order line.Start with the wavelengths multiplied by the appropriate order parameter, then divide both sideby d, and nally apply Eq. 43-1.2(700 nm) > 3(400 nm),2(700 nm)3(400 nm)>,ddsin 2,=700 > sin 3,=400 ,regardless of the value of d.E43-12 Fig. 32-2 shows the path length dierence for the right hand side of the grating as d sin .If the beam strikes the grating at ang angle then there will be an additional path length dierenceof d sin on the right hand side of the gure. The diraction pattern then has two contributions tothe path length dierence, these add to gived(sin + sin psi) = m.E43-13E43-14 Let d sin i = i and 1 + 20 = 2 . Thensin 2 = sin 1 cos(20 ) + cos 1 sin(20 ).Rearranging,sin 2 = sin 1 cos(20 ) +1 sin2 1 sin(20 ).Substituting the equations together yields a rather nasty expression,21=cos(20 ) +dd1 (1 /d)2 sin(20 ).223Rearranging,2(2 1 cos(20 )) = d2 2 sin2 (20 ).1Use 1 = 430 nm and 2 = 680 nm, then solve for d to nd d = 914 nm. This corresponds to 1090rulings/mm.E43-15 The shortest wavelength passes through at an angle of1 = arctan(50 mm)/(300 mm) = 9.46 .This corresponds to a wavelength of1 =(1103 m) sin(9.46 )= 470109 m.(350)The longest wavelength passes through at an angle of2 = arctan(60 mm)/(300 mm) = 11.3 .This corresponds to a wavelength of2 =(1103 m) sin(11.3 )= 560109 m.(350)E43-16 (a) = /R = /N m, so = (481 nm)/(620 rulings/mm)(5.05 mm)(3) = 0.0512 nm.(b) mm is the largest integer smaller than d/, ormm 1/(481109 m)(620 rulings/mm) = 3.35,so m = 3 is highest order seen.E43-17The required resolving power of the grating is given by Eq. 43-10R=(589.0 nm)== 982.(589.6 nm) (589.0 nm)Our resolving power is then R = 1000.Using Eq. 43-11 we can nd the number of grating lines required. We are looking at the secondorder maxima, soR(1000)N=== 500.m(2)E43-18 (a) N = R/m = /m, soN=(415.5 nm)= 23100.(2)(415.496 nm 415.487 nm)(b) d = w/N , where w is the width of the grating. Then = arcsinm(23100)(2)(415.5109 m)= arcsin= 27.6 .d(4.15102 m)224E43-19 N = R/m = /m, soN=(656.3 nm)= 3650(1)(0.180 nm)E43-20 Start with Eq. 43-9:D=E43-21d sin /tan m==.d cos d cos (a) We nd the ruling spacing by Eq. 43-1,d=m(3)(589 nm)== 9.98 m.sin msin(10.2 )(b) The resolving power of the grating needs to be at least R = 1000 for the third-order line; seethe work for Ex. 43-17 above. The number of lines required is given by Eq. 43-11,N=R(1000)== 333,m(3)so the width of the grating (or at least the part that is being used) is 333(9.98 m) = 3.3 mm.E43-22 (a) Condition (1) is satised ifd 2(600 nm)/ sin(30 ) = 2400 nm.The dispersion is maximal for the smallest d, so d = 2400 nm.(b) To remove the third order requires d = 3a, or a = 800 nm.E43-23 (a) The angles of the rst three orders are1=2=3=(1)(589109 m)(40000)= 18.1 ,(76103 m)(2)(589109 m)(40000)= 38.3 ,arcsin(76103 m)(3)(589109 m)(40000)arcsin= 68.4 .(76103 m)arcsinThe dispersion for each order isD1=D2=D3=(1)(40000)360= 3.2102 /nm,(76106 nm) cos(18.1 ) 2(2)(40000)360= 7.7102 /nm,(76106 nm) cos(38.3 ) 2(3)(40000)360= 2.5101 /nm.(76106 nm) cos(68.4 ) 2(b) R = N m, soR1R2R3===(40000)(1) = 40000,(40000)(2) = 80000,(40000)(3) = 120000.225E43-24 d = m/2 sin , sod=E43-25(2)(0.122 nm)= 0.259 nm.2 sin(28.1 )Bragg reection is given by Eq. 43-122d sin = m,where the angles are measured not against the normal, but against the plane. The value of d dependson the family of planes under consideration, but it is at never larger than a0 , the unit cell dimension.We are looking for the smallest angle; this will correspond to the largest d and the smallest m.That means m = 1 and d = 0.313 nm. Then the minimum angle is = sin1(1)(29.3 1012 m)= 2.68 .2(0.313 109 m)E43-26 2d/ = sin 1 and 2d/2 = sin 2 . Then2 = arcsin[2 sin(3.40 )] = 6.81 .E43-27We apply Eq. 43-12 to each of the peaks and nd the productm = 2d sin .The four values are 26 pm, 39 pm, 52 pm, and 78 pm. The last two values are twice the rst two,so the wavelengths are 26 pm and 39 pm.E43-28 (a) 2d sin = m, sod=(3)(96.7 pm)= 171 pm.2 sin(58.0 )(b) = 2(171 pm) sin(23.2 )/(1) = 135 pm.E43-29 The angle against the face of the crystal is 90 51.3 = 38.7 . The wavelength is = 2(39.8 pm) sin(38.7 )/(1) = 49.8 pm.E43-30 If > 2d then /2d > 1. But/2d = sin /m.This means that sin > m, but the sine function can never be greater than one.E43-31 There are too many unknowns. It is only possible to determine the ratio d/.E43-32 A wavelength will be diracted if m = 2d sin . The possible solutions are34==2(275 pm) sin(47.8)/(3) = 136 pm,2(275 pm) sin(47.8)/(4) = 102 pm.226E43-33We use Eq. 43-12 to rst nd d;d=m(1)(0.261 109 m)== 1.45 1010 m.2 sin 2 sin(63.8 )d is the spacing between the planes in Fig. 43-28; it correspond to half of the diagonal distancebetween two cell centers. Then(2d)2 = a2 + a2 ,00ora0 =2d =2(1.45 1010 m) = 0.205 nm.E43-34 Diraction occurs when 2d sin = m. The angles in this case are then given bysin = m(0.125109 m)= (0.248)m.2(0.252109 m)There are four solutions to this equation. They are 14.4 , 29.7 , 48.1 , and 82.7 . They involverotating the crystal from the original orientation (90 42.4 = 47.6 ) by amounts47.6 14.447.6 29.747.6 48.147.6 82.7====33.2 ,17.9 ,0.5 ,35.1 .P43-1 Since the slits are so narrow we only need to consider interference eects, not diractioneects. There are three waves which contribute at any point. The phase angle between adjacentwaves is = 2d sin /.We can add the electric eld vectors as was done in the previous chapters, or we can do it in adierent order as is shown in the gure below.Then the vectors sum toE(1 + 2 cos ).We need to square this quantity, and then normalize it so that the central maximum is the maximum.Then(1 + 4 cos + 4 cos2 )I = Im.9227P43-2(a) Solve for I = I m /2, this occurs when3 = 1 + 2 cos ,2or = 0.976 rad. The corresponding angle x isx (0.976)==.2d2d6.44dBut = 2x , so.3.2d(b) For the two slit pattern the half width was found to be = /2d. The half width in thethree slit case is smaller. P43-3 (a) and (b) A plot of the intensity quickly reveals that there is an alternation of largemaximum, then a smaller maximum, etc. The large maxima are at = 2n, the smaller maximaare half way between those values.(c) The intensity at these secondary maxima is thenI = Im(1 + 4 cos + 4 cos2 )Im=.99Note that the minima are not located half-way between the maxima!P43-4 Covering up the middle slit will result in a two slit apparatus with a slit separation of 2d.The half width, as found in Problem 41-5, is then = /2(2d), = /4d,which is narrower than before covering up the middle slit by a factor of 3.2/4 = 0.8.P43-5 (a) If N is large we can treat the phasors as summing to form a exible line of lengthN E. We then assume (incorrectly) that the secondary maxima occur when the loop wraps aroundon itself as shown in the gures below. Note that the resultant phasor always points straight up.This isnt right, but it is close to reality.228The length of the resultant depends on how many loops there are. For k = 0 there are none.For k = 1 there are one and a half loops. The circumference of the resulting circle is 2N E/3, thediameter is N E/3. For k = 2 there are two and a half loops. The circumference of the resultingcircle is 2N E/5, the diameter is N E/5. The pattern for higher k is similar: the circumferenceis 2N E/(2k + 1), the diameter is N E/(k + 1/2).The intensity at this approximate maxima is proportional to the resultant squared, or(N E)2.(k + 1/2)2 2Ik but I m is proportional to (N E)2 , soIk = I m1.(k + 1/2)2 2(b) Near the middle the vectors simply fold back on one another, leaving a resultant of E. ThenIk (E)2 =so(N E)2,N2Im,N2(c) Let have the values which result in sin = 1, and then the two expressions are identical!Ik =P43-6 (a) v = f , so v = f + f . Assuming v = 0, we have f /f = /. Ignore thenegative sign (we dont need it here). ThenR=fc==,ffand thencc=.RN m(b) The ray on the top gets there rst, the ray on the bottom must travel an additional distanceof N d sin . It takes a timet = N d sin /cf =to do this.(c) Since m = d sin , the two resulting expression can be multiplied together to yield(f )(t) =c N d sin = 1.N mcThis is almost, but not quite, one of Heisenbergs uncertainty relations!P43-7 (b) We sketch parallel lines which connect centers to form almost any right triangle similarto the one shown in the Fig. 43-18. The triangle will have two sides which have integer multiplelengths of the lattice spacing a0 . The hypotenuse of the triangle will then have length h2 + k 2 a0 ,where h and k are the integers. In Fig. 43-18 h = 2 while k = 1. The number of planes which cutthe diagonal is equal to h2 + k 2 if, and only if, h and k are relatively prime. The inter-planar spacingis thenh2 + k 2 a0a0.d==2 + k22 + k2hh229(a) The next ve spacings are thenh = 1,k = 1,dh = 1,k = 2,dh = 1,k = 3,dh = 2,k = 3,dh = 1,k = 4,d= a0 / 2,= a0 / 5,= a0 / 10,= a0 / 13,= a0 / 17.P43-8 The middle layer cells will also diract a beam, but this beam will be exactly out of phasewith the top layer. The two beams will then cancel out exactly because of destructive interference.230E44-1 (a) The direction of propagation is determined by considering the argument of the sinefunction. As t increases y must decrease to keep the sine function looking the same, so the waveis propagating in the negative y direction.(b) The electric eld is orthogonal (perpendicular) to the magnetic eld (so Ex = 0) and thedirection of motion (so Ey = 0); Consequently, the only non-zero term is Ez . The magnitude of Ewill be equal to the magnitude of B times c. Since S = E B/0 , when B points in the positivex direction then E must point in the negative z direction in order that S point in the negative ydirection. ThenEz = cB sin(ky + t).(c) The polarization is given by the direction of the electric eld, so the wave is linearly polarizedin the z direction.E44-2 Let one wave be polarized in the x direction and the other in the y direction. Then the net22electric eld is given by E 2 = Ex + Ey , or2E 2 = E0 sin2 (kz t) + sin2 (kz t + ) ,where is the phase dierence. We can consider any point in space, including z = 0, and thenaverage the result over a full cycle. Since merely shifts the integration limits, then the result isindependent of . Consequently, there are no interference eects.E44-3 (a) The transmitted intensity is I0 /2 = 6.1103 W/m2 . The maximum value of the electriceld isEm =20 cI =2(1.26106 H/m)(3.00108 m/s)(6.1103 W/m2 ) = 2.15 V/m.(b) The radiation pressure is caused by the absorbed half of the incident light, sop = I/c = (6.1103 W/m2 )/(3.00108 m/s) = 2.031011 Pa.E44-4 The rst sheet transmits half the original intensity, the second transmits an amount proportional to cos2 . Then I = (I0 /2) cos2 , or = arccos2I/I0 = arccos2(I0 /3)/I0 35.3 .E44-5 The rst sheet polarizes the un-polarized light, half of the intensity is transmitted, soI1 = 1 I0 .2The second sheet transmits according to Eq. 44-1,I2 = I1 cos2 =11I0 cos2 (45 ) = I0 ,24and the transmitted light is polarized in the direction of the second sheet.The third sheet is 45 to the second sheet, so the intensity of the light which is transmittedthrough the third sheet is11I3 = I2 cos2 = I0 cos2 (45 ) = I0 .48231E44-6 The transmitted intensity through the rst sheet is proportional to cos2 , the transmittedintensity through the second sheet is proportional to cos2 (90 ) = sin2 . ThenI = I0 cos2 sin2 = (I0 /4) sin2 2,or11arcsin 4I/I0 = arcsin22Note that 70.4 is also a valid solution!=4(0.100I0 )/I0 = 19.6 .E44-7 The rst sheet transmits half of the original intensity; each of the remaining sheets transmitsan amount proportional to cos2 , where = 30 . Then1I=cos2 I023=16(cos(30 )) = 0.2112E44-8 The rst sheet transmits an amount proportional to cos2 , where = 58.8 . The secondsheet transmits an amount proportional to cos2 (90 ) = sin2 . ThenI = I0 cos2 sin2 = (43.3 W/m2 ) cos2 (58.8 ) sin2 (58.8 ) = 8.50 W/m2 .E44-9 Since the incident beam is unpolarized the rst sheet transmits 1/2 of the original intensity.The transmitted beam then has a polarization set by the rst sheet: 58.8 to the vertical. The secondsheet is horizontal, which puts it 31.2 to the rst sheet. Then the second sheet transmits cos2 (31.2 )of the intensity incident on the second sheet. The nal intensity transmitted by the second sheetcan be found from the product of these terms,I = (43.3 W/m2 )12cos2 (31.2 ) = 15.8 W/m2 .E44-10 p = arctan(1.53/1.33) = 49.0 .E44-11 (a) The angle for complete polarization of the reected ray is Brewsters angle, and isgiven by Eq. 44-3 (since the rst medium is air)p = tan1 n = tan1 (1.33) = 53.1 .(b) Since the index of refraction depends (slightly) on frequency, then so does Brewsters angle.E44-12 (b) Since r + p = 90 , p = 90 (31.8 ) = 58.2 .(a) n = tan p = tan(58.2 ) = 1.61.E44-13 The angles are betweenp = tan1 n = tan1 (1.472) = 55.81 .andp = tan1 n = tan1 (1.456) = 55.52 .E44-14 The smallest possible thickness t will allow for one half a wavelength phase dierence forthe o and e waves. Then nt = /2, ort = (525109 m)/2(0.022) = 1.2105 m.232E44-15 (a) The incident wave is at 45 to the optical axis. This means that there are twocomponents; assume they originally point in the +y and +z direction. When they travel throughthe half wave plate they are now out of phase by 180 ; this means that when one component is inthe +y direction the other is in the z direction. In eect the polarization has been rotated by 90 .(b) Since the half wave plate will delay one component so that it emerges 180 later than itshould, it will in eect reverse the handedness of the circular polarization.(c) Pretend that an unpolarized beam can be broken into two orthogonal linearly polarizedcomponents. Both are then rotated through 90 ; but when recombined it looks like the originalbeam. As such, there is no apparent change.E44-16 The quarter wave plate has a thickness of x = /4n, so the number of plates that canbe cut is given byN = (0.250103 m)4(0.181)/(488109 m) = 371.P44-1 Intensity is proportional to the electric eld squared, so the original intensity reaching theeye is I0 , with components I h = (2.3)2 I v , and thenI0 = I h + I v = 6.3I v or I v = 0.16I0 .Similarly, I h = (2.3)2 I v = 0.84I0 .(a) When the sun-bather is standing only the vertical component passes, while(b) when the sun-bather is lying down only the horizontal component passes.P44-2 The intensity of the transmitted light which was originally unpolarized is reduced to I u /2,regardless of the orientation of the polarizing sheet. The intensity of the transmitted light whichwas originally polarized is between 0 and I p , depending on the orientation of the polarizing sheet.Then the maximum transmitted intensity is I u /2 + I p , while the minimum transmitted intensity isI u /2. The ratio is 5, soI u /2 + I pIp5==1+2 ,I u /2Iuor I p /I u = 2. Then the beam is 1/3 unpolarized and 2/3 polarized.P44-3Each sheet transmits a fractioncos2 = cos2N.There are N sheets, so the fraction transmitted through the stack iscos2NN.We want to evaluate this in the limit as N .As N gets larger we can use a small angle approximation to the cosine function,1cos x 1 x2 for x2The the transmitted intensity is11 22 N22332N.1This expression can also be expanded in a binomial expansion to get1 2N1 2,2 N2which in the limit as N approaches 1.The stack then transmits all of the light which makes it past the rst lter. Assuming the lightis originally unpolarized, then the stack transmits half the original intensity.P44-4 (a) Stack several polarizing sheets so that the angle between any two sheets is sucientlysmall, but the total angle is 90 .(b) The transmitted intensity fraction needs to be 0.95. Each sheet will transmit a fractioncos2 , where = 90 /N , with N the number of sheets. Then we want to solve0.95 = cos2 (90 /N )Nfor N . For large enough N , will be small, so we can expand the cosine function ascos2 = 1 sin2 1 2 ,so0.95 1 (/2N )2N 1 N (/2N )2 ,which has solution N = 2 /4(0.05) = 49.P44-5 Since passing through a quarter wave plate twice can rotate the polarization of a linearlypolarized wave by 90 , then if the light passes through a polarizer, through the plate, reects o thecoin, then through the plate, and through the polarizer, it would be possible that when it passesthrough the polarizer the second time it is 90 to the polarizer and no light will pass. You wont seethe coin.On the other hand if the light passes rst through the plate, then through the polarizer, then isreected, the passes again through the polarizer, all the reected light will pass through he polarizerand eventually work its way out through the plate. So the coin will be visible.Hence, side A must be the polarizing sheet, and that sheet must be at 45 to the optical axis.P44-6(a) The displacement of a ray is given bytan k = yk /t,so the shift isy = t(tan e tan o ).Solving for each angle,e=arcsino=arcsin1sin(38.8 )(1.486)1sin(38.8 )(1.658)= 24.94 ,= 22.21 .The shift is theny = (1.12102 m) (tan(24.94) tan(22.21)) = 6.35104 m.(b) The e-ray bends less than the o-ray.(c) The rays have polarizations which are perpendicular to each other; the o-wave being polarizedalong the direction of the optic axis.(d) One ray, then the other, would disappear.234P44-7 The method is outline in Sample Problem 44-24; use a polarizing sheet to pick out the o-rayor the e-ray.235E45-1(a) The energy of a photon is given by Eq. 45-1, E = hf , soE = hf =hc.Putting in best numbershc =(6.626068761034 J s)(2.99792458108 m/s) = 1.23984106 eV m.(1.6021764621019 C)This means that hc = 1240 eV nm is accurate to almost one part in 8000!(b) E = (1240 eV nm)/(589 nm) = 2.11 eV.E45-2 Using the results of Exercise 45-1,=(1240 eV nm)= 2100 nm,(0.60 eV)which is in the infrared.E45-3 Using the results of Exercise 45-1,E1 =(1240 eV nm)= 3.31 eV,(375 nm)E2 =(1240 eV nm)= 2.14 eV,(580 nm)andThe dierence is E = (3.307 eV) (2.138 eV) = 1.17 eV.E45-4 P = E/t, so, using the result of Exercise 45-1,P = (100/s)(1240 eV nm)= 230 eV/s.(540 nm)Thats a small 3.681017 W.E45-5 When talking about the regions in the suns spectrum it is more common to refer towavelengths than frequencies. So we will use the results of Exercise 45-1(a), and solve = hc/E = (1240 eV nm)/E.The energies are between E = (1.01018 J)/(1.61019 C) = 6.25 eV and E = (1.01016 J)/(1.61019 C) = 625 eV. These energies correspond to wavelengths between 198 nm and 1.98 nm; this isthe ultraviolet range.E45-6 The energy per photon is E = hf = hc/. The intensity is power per area, which is energyper time per area, soPEnhchc nI====.AAtAtA tBut R = n/t is the rate of photons per unit time. Since h and c are constants and I and A are equalfor the two beams, we have R1 /1 = R2 /2 , orR1 /R2 = 1 /2 .236E45-7 (a) Since the power is the same, the bulb with the larger energy per photon will emitsfewer photons per second. Since longer wavelengths have lower energies, the bulb emitting 700 nmmust be giving o more photons per second.(b) How many more photons per second? If E1 is the energy per photon for one of the bulbs,then N1 = P/E1 is the number of photons per second emitted. The dierence is thenN1 N2 =PPP=(1 2 ),E1E2hcorN1 N2 =(130 W)((700109 m) (400109 m)) = 1.961020 .(6.631034 Js)(3.00108 m/s)E45-8 Using the results of Exercise 45-1, the energy of one photon isE=(1240 eV nm)= 1.968 eV,(630 nm)The total light energy given o by the bulb isE t = P t = (0.932)(70 W)(730 hr)(3600 s/hr) = 1.71108 J.The number of photons isn=Et(1.71108 J)== 5.431026 .E0(1.968 eV)(1.61019 J/eV)E45-9 Apply Wiens law, Eq. 45-4, max T = 2898 m K; soT =(2898 m K)= 91106 K.(321012 m)Actually, the wavelength was supposed to be 32 m. Then the temperature would be 91 K.E45-10 Apply Wiens law, Eq. 45-4, max T = 2898 m K; so=(2898 m K)= 1.45 m.(0.0020 K)This is in the radio region, near 207 on the FM dial.E45-11 The wavelength of the maximum spectral radiancy is given by Wiens law, Eq. 45-4,max T = 2898 m K.Applying to each temperature in turn,(a) = 1.06103 m, which is in the microwave;(b) = 9.4106 m, which is in the infrared;(c) = 1.6106 m, which is in the infrared;(d) = 5.0107 m, which is in the visible;(e) = 2.91010 m, which is in the x-ray;(f) = 2.91041 m, which is in a hard gamma ray.237E45-12 (a) Apply Wiens law, Eq. 45-4, max T = 2898 m K.; so=(2898 m K)= 5.00107 m.(5800 K)Thats blue-green.(b) Apply Wiens law, Eq. 45-4, max T = 2898 m K.; soT =(2898 m K)= 5270 K.(550109 m)E45-13 I = T 4 and P = IA. ThenP = (5.67108 W/m2 K4 )(1900 K)4 (0.5103 m)2 = 0.58 W.E45-14 Since I T 4 , doubling T results in a 24 = 16 times increase in I. Then the new powerlevel is(16)(12.0 mW) = 192 mW.E45-15(a) We want to apply Eq. 45-6,R(, T ) =12c2 h.5hc/kT 1 eWe know the ratio of the spectral radiancies at two dierent wavelengths. Dividing the aboveequation at the rst wavelength by the same equation at the second wavelength,3.5 =5 ehc/1 kT 11,5 ehc/2 kT 12where 1 = 200 nm and 2 = 400 nm. We can considerably simplify this expression if we letx = ehc/2 kT ,because since 2 = 21 we would haveehc/1 kT = e2hc/2 kT = x2 .Then we get3.5 =125x2 11=(x + 1).x132We will use the results of Exercise 45-1 for the exponents and then rearrange to getT =hc(3.10 eV)== 7640 K.1 k ln(111)(8.62105 eV/K) ln(111)(b) The method is the same, except that instead of 3.5 we have 1/3.5; this means the equationfor x is11=(x + 1),3.532with solution x = 8.14, so thenT =hc(3.10 eV)== 17200 K.1 k ln(8.14)(8.62105 eV/K) ln(8.14)238E45-16 hf = , sof=(5.32 eV)== 1.281015 Hz.h(4.141015 eV s)E45-17 Well use the results of Exercise 45-1. Visible red light has an energy ofE=(1240 eV nm)= 1.9 eV.(650 nm)The substance must have a work function less than this to work with red light. This means thatonly cesium will work with red light. Visible blue light has an energy ofE=(1240 eV nm)= 2.75 eV.(450 nm)This means that barium, lithium, and cesium will work with blue light.E45-18 Since K m = hf ,K m = (4.141015 eV s)(3.191015 Hz) (2.33 eV) = 10.9 eV.E45-19(a) Use the results of Exercise 45-1 to nd the energy of the corresponding photon,E=hc(1240 eV nm)== 1.83 eV.(678 nm)Since this energy is less than than the minimum energy required to remove an electron then thephoto-electric eect will not occur.(b) The cut-o wavelength is the longest possible wavelength of a photon that will still result inthe photo-electric eect occurring. That wavelength is=(1240 eV nm)(1240 eV nm)== 544 nm.E(2.28 eV)This would be visible as green.E45-20 (a) Since K m = hc/ ,Km =(1240 eV nm)(4.2 eV) = 2.0 eV.(200 nm)(b) The minimum kinetic energy is zero; the electron just barely makes it o the surface.(c) V s = K m /q = 2.0 V.(d) The cut-o wavelength is the longest possible wavelength of a photon that will still result inthe photo-electric eect occurring. That wavelength is=(1240 eV nm)(1240 eV nm)== 295 nm.E(4.2 eV)E45-21 K m = qV s = 4.92 eV. But K m = hc/ , so=(1240 eV nm)= 172 nm.(4.92 eV + 2.28 eV)239E45-22 (a) K m = qV s and K m = hc/ . We have two dierent values for qV s and , sosubtracting this equation from itself yieldsq(V s,1 V s,2 ) = hc/1 hc/2 .Solving for 2 ,2===hc,hc/1 q(V s,1 V s,2 )(1240 eV nm),(1240 eV nm)/(491 nm) (0.710 eV) + (1.43 eV)382 nm.(b) K m = qV s and K m = hc/ , so = (1240 eV nm)/(491 nm) (0.710 eV) = 1.82 eV.E45-23(a) The stopping potential is given by Eq. 45-11,hf ,eeV0 =soV0 =(1240 eV nm) (1.85 eV= 1.17 V.e(410 nme(b) These are not relativistic electrons, sov=2K/m = c 2K/mc2 = c 2(1.17 eV)/(0.511106 eV) = 2.14103 c,or v = 64200 m/s.E45-24 It will have become the stopping potential, orhf ,eeV0 =soV0 =(2.49 eV)(4.141015 eV m)(6.331014 /s) = 0.131 V.(1.0e)(1.0e)E45-25E45-26 (a) Using the results of Exercise 45-1,=(1240 eV nm)= 62 pm.(20103 eV)(b) This is in the x-ray region.E45-27 (a) Using the results of Exercise 45-1,E=(1240 eV nm)= 29, 800 eV.(41.6103 nm)(b) f = c/ = (3108 m/s)/(41.6 pm) = 7.211018 /s.(c) p = E/c = 29, 800 eV/c = 2.98104 eV/c.240E45-28 (a) E = hf , sof=(0.511106 eV)= 1.231020 /s.(4.141015 eV s)(b) = c/f = (3108 m/s)/(1.231020 /s) = 2.43 pm.(c) p = E/c = (0.511106 eV)/c.E45-29 The initial momentum of the system is the momentum of the photon, p = h/. Thismomentum is imparted to the sodium atom, so the nal speed of the sodium is v = p/m, where mis the mass of the sodium. Thenv=h(6.631034 J s)== 2.9 cm/s.m(589109 m)(23)(1.71027 kg)E45-30 (a) C = h/mc = hc/mc2 , soC =(1240 eV nm)= 2.43 pm.(0.511106 eV)(c) Since E = hf = hc/, and = h/mc = hc/mc2 , thenE = hc/ = mc2 .(b) See part (c).E45-31 The change in the wavelength of a photon during Compton scattering is given by Eq.45-17,h =+(1 cos ).mcWell use the results of Exercise 45-30 to save some time, and let h/mc = C , which is 2.43 pm.(a) For = 35 , = (2.17 pm) + (2.43 pm)(1 cos 35 ) = 2.61 pm.(b) For = 115 , = (2.17 pm) + (2.43 pm)(1 cos 115 ) = 5.63 pm.E45-32 (a) Well use the results of Exercise 45-1:(1240 eV nm)= 2.43 pm.(0.511106 eV)=(b) The change in the wavelength of a photon during Compton scattering is given by Eq. 45-17, =+h(1 cos ).mcWell use the results of Exercise 45-30 to save some time, and let h/mc = C , which is 2.43 pm. = (2.43 pm) + (2.43 pm)(1 cos 72 ) = 4.11 pm.(c) Well use the results of Exercise 45-1:E=(1240 eV nm)= 302 keV.(4.11 pm)241E45-3345-17,The change in the wavelength of a photon during Compton scattering is given by Eq.h(1 cos ).mcWe are not using the expression with the form because and E are not simply related.The wavelength is related to frequency by c = f , while the frequency is related to the energyby Eq. 45-1, E = hf . Then =E= E E = hf hf ,11= hc, = hc.Into this last expression we substitute the Compton formula. ThenE =h2 (1 cos ).mNow E = hf = hc/, and we can divide this on both sides of the above equation. Also, = c/f ,and we can substitute this into the right hand side of the above equation. Both of these steps resultinEhf=(1 cos ).Emc2Note that mc2 is the rest energy of the scattering particle (usually an electron), while hf is theenergy of the scattered photon.E45-34 The wavelength is related to frequency by c = f , while the frequency is related to theenergy by Eq. 45-1, E = hf . ThenEEE= E E = hf hf ,11= hc, ,= hc=, + But E/E = 3/4, so3 + 3 = 4,or = 3.E45-35 The maximum shift occurs when = 180 , som = 2h(1240 eV nm)=2= 2.641015 m.mc938 MeV)E45-36 Since E = hf frequency shifts are identical to energy shifts. Then we can use the resultsof Exercise 45-33 to get(0.9999)(6.2 keV)(0.0001) =(1 cos ),(511 keV)which has solution = 7.4 .(b) (0.0001)(6.2 keV) = 0.62 eV.242E45-3745-17,(a) The change in wavelength is independent of the wavelength and is given by Eq. =(1240 eV nm)hc(1 cos ) = 2= 4.85103 nm.2mc(0.511106 eV)(b) The change in energy is given byEhc hc ,fi11= hci + i==(1240 eV nm),11(9.77 pm) + (4.85 pm) (9.77 pm)= 42.1 keV(c) This energy went to the electron, so the nal kinetic energy of the electron is 42.1 keV.E45-38 For = 90 = h/mc. ThenEE==hf=1,hf + h/mc. + h/mc1But h/mc = 2.43 pm for the electron (see Exercise 45-30).(a) E/E = (2.43 pm)/(3.00 cm + 2.43 pm) = 8.11011 .(b) E/E = (2.43 pm)/(500 nm + 2.43 pm) = 4.86106 .(c) E/E = (2.43 pm)/(0.100 nm + 2.43 pm) = 0.0237.(d) E/E = (2.43 pm)/(1.30 pm + 2.43 pm) = 0.651.E45-39 We can use the results of Exercise 45-33 to get(0.10) =(0.90)(215 keV)(1 cos ),(511 keV)which has solution = 42/6 .E45-40 (a) A crude estimate is that the photons cant arrive more frequently than once every10 8s. That would provide an emission rate of 108 /s.(b) The power output would beP = (108 )(1240 eV nm)= 2.3108 eV/s,(550 nm)which is 3.61011 W!E45-41 We can follow the example of Sample Problem 45-6, and apply = 0 (1 v/c).(a) Solving for 0 ,0 =(588.995 nm)= 588.9944 nm.(1 (300 m/s)(3108 m/s)243(b) Applying Eq. 45-18,v = h(6.61034 J s)== 3102 m/s.m(22)(1.71027 kg)(590109 m)(c) Emitting another photon will slow the sodium by about the same amount.E45-42 (a) (430 m/s)/(0.15 m/s) 2900 interactions.(b) If the argon averages a speed of 220 m/s, then it requires interactions at the rate of(2900)(220 m/s)/(1.0 m) = 6.4105 /sif it is going to slow down in time.P45-1 The radiant intensity is given by Eq. 45-3, I = T 4 . The power that is radiated throughthe opening is P = IA, where A is the area of the opening. But energy goes both ways through theopening; it is the dierence that will give the net power transfer. Then44P net = (I0 I1 )A = A T0 T1 .Put in the numbers, andP net = (5.67108 W/m2 K4 )(5.20 104 m2 ) (488 K)4 (299 K)4 = 1.44 W.P45-2(a) I = T 4 and P = IA. Then T 4 = P/A, orT =4(100 W)= 3248 K.(5.67108 W/m2 K4 )(0.28103 m)(1.8102 m)Thats 2980 C.(b) The rate that energy is radiated o is given by dQ/dt = mC dT /dt. The mass is found fromm = V , where V is the volume. This can be combined with the power expression to yieldAT 4 = V CdT /dt,which can be integrated to yieldt =V C33(1/T2 1/T1 ).3APutting in numbers,t==(19300kg/m3 )(0.28103 m)(132J/kgC)[1/(2748 K)3 1/(3248 K)3 ],3(5.67108 W/m2 K4 )(4)20 ms.P45-3 Light from the sun will heat-up the thin black screen. As the temperature of the screenincreases it will begin to radiate energy. When the rate of energy radiation from the screen is equalto the rate at which the energy from the sun strikes the screen we will have equilibrium. We needrst to nd an expression for the rate at which energy from the sun strikes the screen.The temperature of the sun is T S . The radiant intensity is given by Eq. 45-3, I S = T S 4 . Thetotal power radiated by the sun is the product of this radiant intensity and the surface area of thesun, soP S = 4rS 2 T S 4 ,244where rS is the radius of the sun.Assuming that the lens is on the surface of the Earth (a reasonable assumption), then we cannd the power incident on the lens if we know the intensity of sunlight at the distance of the Earthfrom the sun. That intensity isPSPSIE ==,A4RE 2where RE is the radius of the Earths orbit. Combining,I E = T S 42rSREThe total power incident on the lens is thenP lens = I E Alens = T S 4rSRE2rl 2 ,where rl is the radius of the lens. All of the energy that strikes the lens is focused on the image, sothe power incident on the lens is also incident on the image.The screen radiates as the temperature increases. The radiant intensity is I = T 4 , where T isthe temperature of the screen. The power radiated is this intensity times the surface area, soP = IA = 2ri 2 T 4 .The factor of 2 is because the screen has two sides, while ri is the radius of the image. Set thisequal to P lens ,2rS2ri 2 T 4 = T S 4rl 2 ,REor21rS rlT 4 = T S4.2RE r iThe radius of the image of the sun divided by the radius of the sun is the magnication of thelens. But magnication is also related to image distance divided by object distance, sorii= |m| = ,rSoDistances should be measured from the lens, but since the sun is so far from the Earth, we wont befar o in stating o RE . Since the object is so far from the lens, the image will be very, very closeto the focal point, so we can also state i f . Thenrif=,rSREso the expression for the temperature of the thin black screen is considerably simplied toT4 =1 4TS2rlf2.Now we can put in some of the numbers.T =1(5800 K)21/4(1.9 cm)= 1300 K.(26 cm)245P45-4The derivative of R with respect to is10hc c2 hhc6 (e( k T ) 1)+2 c3 h2 e( k T )hc7 (e( k T ) 1)2 k T.Ohh, thats ugly. Setting it equal to zero allows considerable simplication, and we are left with(5 x)ex = 5,where x = hc/kT . The solution is found numerically to be x = 4.965114232. Then=(1240 eV nm)2.898103 m K=.5 eV/K)T(4.965)(8.6210TP45-5 (a) If the planet has a temperature T , then the radiant intensity of the planet will beIT 4 , and the rate of energy radiation from the planet will beP = 4R2 T 4 ,where R is the radius of the planet.A steady state planet temperature requires that the energy from the sun arrive at the same rateas the energy is radiated from the planet. The intensity of the energy from the sun a distance rfrom the sun isP sun /4r2 ,and the total power incident on the planet is thenP = R2P sun.4r2Equating,4R2 T 4T4= R2=P sun,4r2P sun.16r2(b) Using the last equation and the numbers from Problem 3,1T = (5800 K)2(6.96108 m)= 279 K.(1.51011 m)Thats about 43 F.P45-6 (a) Change variables as suggested, then = hc/xkT and d = (hc/x2 kT )dx. Integrate(note the swapping of the variables of integration picks up a minus sign):I===2c2 h (hc/x2 kT )dx,(hc/xkT )5ex 12k 4 T 4x3 dx,h3 c2ex 12 5 k 4 4T .15h3 c2246P45-7(a) P = E/t = nhf /t = (hc/)(n/t), where n/t is the rate of photon emission. Then(100 W)(589109 m)= 2.961020 /s.(6.631034 J s)(3.00108 m/s)n/t =(b) The ux at a distance r is the rate divided by the area of the sphere of radius r, sor=(2.961020 /s)= 4.8107 m.4(1104 /m2 s)(c) The photon density is the ux divided by the speed of light; the distance is then(2.961020 /s)= 280 m.4(1106 /m3 )(3108 m/s)r=(d) The ux is given by(2.961020 /s)= 5.891018 /m2 s.4(2.0 m)2The photon density is(5.891018 m2 s)/(3.00108 m/s) = 1.961010 /m3 .P45-8Momentum conservation requiresp = pe ,while energy conservation requiresE + mc2 = Ee .Square both sides of the energy expression and2E + 2E mc2 + m2 c42E + 2E mc22 2p c + 2E mc22= Ee = p2 c2 + m2 c4 ,e2 2= pe c ,= p2 c2 .eBut the momentum expression can be used here, and the result is2E mc2 = 0.Not likely.P45-9be(a) Since qvB = mv 2 /r, v = (q/m)rB The kinetic energy of (non-relativistic) electrons willK=orK=11 q 2 (rB)2mv 2 =,22m1 (1.61019 C)(188106 T m)2 = 3.1103 eV.2 (9.11031 kg)(b) Use the results of Exercise 45-1,=(1240 eV nm) 3.1103 eV = 1.44104 eV.(71103 nm)247P45-10P45-11then(a) The maximum value of is 2h/mc. The maximum energy lost by the photon isEhc hc ,fi11= hci + i2h/mc= hc,( + 2h/mc)=,where in the last line we wrote for i . The energy given to the electron is the negative of this, soKmax =2h2.m( + 2h/mc)Multiplying through by 12 = (E/hc)2 we getKmax =2E 2.mc2 (1 + 2hc/mc2 )orKmax =E2.mc2 /2 + E(b) The answer isKmax =(17.5 keV)2= 1.12 keV.(511 eV)/2 + (17.5 keV)248E46-1(a) Apply Eq. 46-1, = h/p. The momentum of the bullet isp = mv = (0.041 kg)(960 m/s) = 39kg m/s,so the corresponding wavelength is = h/p = (6.631034 J s)/(39kg m/s) = 1.71035 m.(b) This length is much too small to be signicant. How much too small? If the radius of thegalaxy were one meter, this distance would correspond to the diameter of a proton.E46-2 (a) = h/p and p2 /2m = K, then= hc2mc2 K=1.226 nm(1240 eV nm) =.K2(511 keV) K(b) K = eV , so=1.226 nm=eV1.5 Vnm.VE46-3 For non-relativistic particles = h/p and p2 /2m = K, so = hc/ 2mc2 K.(a) For the electron,(1240 eV nm)== 0.0388 nm.2(511 keV)(1.0 keV)(c) For the neutron,=(1240 MeV fm)2(940 MeV)(0.001 MeV)= 904 fm.(b) For ultra-relativistic particles K E pc, so=hc(1240 eV nm)== 1.24 nm.E(1000 eV)E46-4 For non-relativistic particles p = h/ and p2 /2m = K, so K = (hc)2 /2mc2 2 . ThenK=E46-5(1240 eV nm)2= 4.34106 eV.2(5.11106 eV)(589 nm)2(a) Apply Eq. 46-1, p = h/. The proton speed would then bev=hhc(1240 MeV fm)=c 2 =c= 0.0117c.mmc (938 MeV)(113 fm)This is good, because it means we were justied in using the non-relativistic equations. Thenv = 3.51106 m/s.(b) The kinetic energy of this electron would beK=11mv 2 = (938 MeV)(0.0117)2 = 64.2 keV.22The potential through which it would need to be accelerated is 64.2 kV.249E46-6 (a) K = qV and p =2mK. Then2(22)(932 MeV/c2 )(325 eV) = 3.65106 eV/c.p=(b) = h/p, so=(1240 eV nm)hc== 3.39104 nm.pc(3.65106 eV/c)cE46-7 (a) For non-relativistic particles = h/p and p2 /2m = K, so = hc/ 2mc2 K. For thealpha particle,(1240 MeV fm)== 5.2 fm.2(4)(932 MeV)(7.5 MeV)(b) Since the wavelength of the alpha is considerably smaller than the distance to the nucleuswe can ignore the wave nature of the alpha particle.E46-8 (a) For non-relativistic particles p = h/ and p2 /2m = K, so K = (hc)2 /2mc2 2 . ThenK=(1240 keV pm)2= 15 keV.2(511keV)(10 pm)2(b) For ultra-relativistic particles K E pc, soE=E46-9(1240 keV pm)hc== 124 keV.(10 pm)The relativistic relationship between energy and momentum isE 2 = p2 c2 + m2 c4 ,and if the energy is very large (compared to mc2 ), then the contribution of the mass to the aboveexpression is small, andE 2 p2 c2 .Then from Eq. 46-1,=hhchc(1240 MeV f m)==== 2.5102 fm.ppcE(50103 MeV)E46-10 (a) K = 3kT /2, p ===2mK, and = h/p, sohch=,3mkT3mc2 kT(1240 MeV f m)3(4)(932MeV)(8.621011 MeV/K)(291 K)= 74 pm.(b) pV = N kT ; assuming that each particle occupies a cube of volume d3 = V0 then the interparticle spacing is d, sod=3V /N =3(1.381023 J/K)(291 K)= 3.4 nm.(1.01105 Pa)250E46-11 p = mv and p = h/, so m = h/v. Taking the ratio,vme== (1.813104 )(3) = 5.439104 .me veThe mass of the unknown particle is thenm=(0.511 MeV/c2 )= 939.5 MeV.(5.439104 )That would make it a neutron.E46-12 (a) For non-relativistic particles = h/p and p2 /2m = K, so = hc/ 2mc2 K.For the electron,(1240 eV nm)== 1.0 nm.2(5.11105 eV)(1.5 eV)For ultra-relativistic particles K E pc, so for the photon=hc(1240 eV nm)== 830 nm.E(1.5 eV)(b) Electrons with energies that high are ultra-relativistic. Both the photon and the electron willthen have the same wavelength;=E46-13hc(1240 MeV fm)== 0.83 fm.E(1.5 GeV)(a) The classical expression for kinetic energy isp = 2mK,so=hhc==p2mc2 K(1240 keV pm)2(511 keV)(25.0 keV)= 7.76 pm.(a) The relativistic expression for momentum ispc = sqrtE 2 m2 c4 =Then=hc=pc(mc2 + K)2 m2 c4 =K 2 + 2mc2 K.(1240 keV pm)(25.0 keV)2 + 2(511 keV)(25.0 keV)= 7.66 pm.E46-14 We want to match the wavelength of the gamma to that of the electron. For the gamma, = hc/E . For the electron, K = p2 /2m = h2 /2m2 . Combining,K=2Eh22E =.2mh2 c22mc2With numbers,K=(136keV)2= 18.1 keV.2(511 keV)That would require an accelerating potential of 18.1 kV.251E46-15 First nd thewavelength of the neutrons. For non-relativistic particles = h/p andp2 /2m = K, so = hc/ 2mc2 K. Then(1240 keV pm)=2(940103 keV)(4.2103 keV)= 14 pm.Bragg reection occurs when 2d sin = , so = arcsin(14 pm)/2(73.2 pm) = 5.5 .E46-16 This is merely a Bragg reection problem. Then 2d sin = m, or= arcsin(1)(11 pm)/2(54.64 pm) = 5.78 ,= arcsin(2)(11 pm)/2(54.64 pm) = 11.6 ,= arcsin(3)(11 pm)/2(54.64 pm) = 17.6 .E46-17 (a) Since sin 52 = 0.78, then 2(/d) = 1.57 > 1, so there is no diraction order otherthan the rst.(b) For an accelerating potential of 54 volts we have /d = 0.78. Increasing the potential willincrease the kinetic energy, increase the momentum, and decrease the wavelength. d wont change.The kinetic energy is increased by a factor of 60/54 = 1.11, the momentum increases by a factor of1.11 = 1.05, so the wavelength changes by a factor of 1/1.05 = 0.952. The new angle is then = arcsin(0.952 0.78) = 48 .E46-18 First nd thewavelength of the electrons. For non-relativistic particles = h/p andp2 /2m = K, so = hc/ 2mc2 K. Then=(1240 keV pm)2(511 keV)(0.380 keV)= 62.9 pm.This is now a Bragg reection problem. Then 2d sin = m, or=========arcsin(1)(62.9arcsin(2)(62.9arcsin(3)(62.9arcsin(4)(62.9arcsin(5)(62.9arcsin(6)(62.9arcsin(7)(62.9arcsin(8)(62.9arcsin(9)(62.9pm)/2(314pm)/2(314pm)/2(314pm)/2(314pm)/2(314pm)/2(314pm)/2(314pm)/2(314pm)/2(314pm) = 5.74 ,pm) = 11.6 ,pm) = 17.5 ,pm) = 23.6 ,pm) = 30.1 ,pm) = 36.9 ,pm) = 44.5 ,pm) = 53.3 ,pm) = 64.3 .But the odd orders vanish (see Chapter 43 for a discussion on this).E46-19 Since f t 1/2, we havef = 1/2(0.23 s) = 0.69/s.252E46-20 Since f t 1/2, we havef = 1/2(0.10109 s) = 1.61010 /s.The bandwidth wouldnt t in the frequency allocation!E46-21Apply Eq. 46-9,E h4.141015 eV s)== 7.6105 eV.2t2(8.71012 s)This is much smaller than the photon energy.E46-22 Apply Heisenberg twice:E1 =4.141015 eV s)= 5.49108 eV.2(12109 s)E2 =4.141015 eV s)= 2.86108 eV.2(23109 s)andThe sum is E transition = 8.4108 eV.E46-23 Apply Heisenberg:p =6.631034 J s)= 8.81024 kg m/s.2(121012 m)E46-24 p = (0.5 kg)(1.2 s) = 0.6 kg m/s. The position uncertainty would then bex =E46-25(0.6 J/s)= 0.16 m.2(0.6 kg m/s)We want v v, which means p p. Apply Eq. 46-8, andx hh.2p2pAccording to Eq. 46-1, the de Broglie wavelength is related to the momentum by = h/p,sox .2E46-26 (a) A particle conned in a (one dimensional) box of size L will have a position uncertaintyof no more than x L. The momentum uncertainty will then be no less thanp sop hh.2x2L(6.631034 J s)= 11024 kg m/s.2(1010 m)253(b) Assuming that p p, we haveh,2Land then the electron will have a (minimum) kinetic energy ofpEorEh2p2.2 mL22m8(hc)2(1240 keV pm)2== 0.004 keV.8 2 mc2 L28 2 (511 keV)(100 pm)2E46-27 (a) A particle conned in a (one dimensional) box of size L will have a position uncertainty of no more than x L. The momentum uncertainty will then be no less thanp sop hh.2x2L(6.631034 J s)= 11020 kg m/s.2(1014 m)(b) Assuming that p p, we haveh,2Land then the electron will have a (minimum) kinetic energy ofpEorEp2h2.2 mL22m8(1240 MeV fm)2(hc)2== 381 MeV.8 2 mc2 L28 2 (0.511 MeV)(10 fm)2This is so large compared to the mass energy of the electron that we must consider relativistic eects.It will be very relativistic (3810.5!), so we can use E = pc as was derived in Exercise 9. ThenE=hc(1240 MeV fm)== 19.7 MeV.2L2(10 fm)This is the total energy; so we subtract 0.511 MeV to get K = 19 MeV.E46-28 We want to nd L when T = 0.01. This means solvingT(0.01)EE1e2kL ,U0U0(5.0 eV)(5.0 eV)= 161(6.0 eV)(6.0 eV)=16= 2.22e2k L ,ln(4.5103 ) = 2(5.12109 /m)L,5.31010 m = L.254e2k L ,E46-29The wave number, k, is given byk=2hc2mc2 (U0 E).(a) For the proton mc2 = 938 MeV, sok=2(1240 MeV fm)2(938 MeV)(10 MeV 3.0 MeV) = 0.581 fm1 .The transmission coecient is thenT = 16(3.0 MeV)(10 MeV)1(3.0 MeV)(10 MeV)1e2(0.581 fm )(10 fm) = 3.0105 .(b) For the deuteron mc2 = 2 938 MeV, sok=2(1240 MeV fm)2(2)(938 MeV)(10 MeV 3.0 MeV) = 0.821 fm1 .The transmission coecient is thenT = 16(3.0 MeV)(10 MeV)1(3.0 MeV)(10 MeV)1e2(0.821 fm )(10 fm) = 2.5107 .E46-30 The wave number, k, is given byk=2hc2mc2 (U0 E).(a) For the proton mc2 = 938 MeV, sok=2(1240 keV pm)2(938 MeV)(6.0 eV 5.0 eV) = 0.219 pm1 .We want to nd T . This means solvingT===EE1e2kL ,U0U0(5.0 eV)(5.0 eV)161(6.0 eV)(6.0 eV)16e2(0.2191012)(0.7109 )1.610133 .A current of 1 kA corresponds toN = (1103 C/s)/(1.61019 C) = 6.31021 /sprotons per seconds. The time required for one proton to pass is thent = 1/(6.31021 /s)(1.610133 ) = 9.910110 s.Thats 10104 years!255,P46-1We will interpret low energy to mean non-relativistic. Then=hh.=p2mn KThe diraction pattern is then given byd sin = m = mh/2mn K,where m is diraction order while mn is the neutron mass. We want to investigate the spread bytaking the derivative of with respect to K,mhd cos d = dK.2 2mn K 3Divide this by the original equation, and thencos dKd = .sin 2KRearrange, change the dierential to a dierence, and then = tan K.2KWe dropped the negative sign out of laziness; but the angles are in radians, so we need to multiplyby 180/ to convert to degrees.P46-2P46-3We want to solveT = 16EU01EU0e2kL ,for E. Unfortunately, E is contained in k sincek=2hc2mc2 (U0 E).We can do this by iteration. The maximum possible value forEU01EU0is 1/4; using this value we can get an estimate for k:(0.001) = 16(0.25)e2kL ,ln(2.5104 ) = 2k(0.7 nm),5.92/ nm = k.Now solve for E:E= U0 (hc)2 k 2 8mc2 2 ,(1240 eV nm)2 (5.92/nm)2= (6.0 eV) ,8 2 (5.11105 eV)= 4.67 eV.256Put this value for E back into the transmission equation to nd a new k:TEE1e2kL ,U0U0(4.7 eV)(4.7 eV)161(6.0 eV)(6.0 eV)= 16(0.001)=e2kL ,ln(3.68104 ) = 2k(0.7 nm),5.65/ nm = k.Now solve for E using this new, improved, value for k:= U0 (hc)2 k 2 8mc2 2 ,(1240 eV nm)2 (5.65/nm)2= (6.0 eV) ,8 2 (5.11105 eV)= 4.78 eV.EKeep at it. Youll eventually stop around E = 5.07 eV.P46-4 (a) A one percent increase in the barrier height means U0 = 6.06 eV.For the electron mc2 = 5.11105 eV, sok=2(1240 eV nm)2(5.11105 eV)(6.06 eV 5.0 eV) = 5.27 nm1 .We want to nd T . This means solvingT===EE1e2kL ,U0U0(5.0 eV)(5.0 eV)161(6.06 eV)(6.06 eV)16e2(5.27)(0.7) ,1.44103 .Thats a 16% decrease.(b) A one percent increase in the barrier thickness means L = 0.707 nm.For the electron mc2 = 5.11105 eV, sok=2(1240 eV nm)2(5.11105 eV)(6.0 eV 5.0 eV) = 5.12 nm1 .We want to nd T . This means solvingT===EE1e2kL ,U0U0(5.0 eV)(5.0 eV)161(6.0 eV)(6.0 eV)16e2(5.12)(0.707) ,1.59103 .Thats a 8.1% decrease.(c) A one percent increase in the incident energy means E = 5.05 eV.For the electron mc2 = 5.11105 eV, sok=2(1240 eV nm)2(5.11105 eV)(6.0 eV 5.05 eV) = 4.99 nm1 .257We want to nd T . This means solvingT===EE1e2kL ,U0U0(5.05 eV)(5.05 eV)161(6.0 eV)(6.0 eV)16e2(4.99)(0.7) ,1.97103 .Thats a 14% increase.P46-5First, the rule for exponentsei(a+b) = eia eib .Then apply Eq. 46-12, ei = cos + i sin ,cos(a + b) + i sin(a + b) = (cos a + i sin a)(sin b + i sin b).Expand the right hand side, remembering that i2 = 1,cos(a + b) + i sin(a + b) = cos a cos b + i cos a sin b + i sin a cos b sin a sin b.Since the real part of the left hand side must equal the real part of the right and the imaginary partof the left hand side must equal the imaginary part of the right, we actually have two equations.They arecos(a + b) = cos a cos b sin a sin bandsin(a + b) = cos a sin b + sin a cos b.P46-6258E47-1(a) The ground state energy level will be given byE1 =(6.63 1034 J s)2h2== 3.1 1010 J.28mL8(9.11 1031 kg)(1.4 1014 m)2The answer is correct, but the units make it almost useless. We can divide by the electron charge toexpress this in electron volts, and then E = 1900 MeV. Note that this is an extremely relativisticquantity, so the energy expression loses validity.(b) We can repeat what we did above, or we can apply a trick that is often used in solvingthese problems. Multiplying the top and the bottom of the energy expression by c2 we getE1 =(hc)28(mc2 )L2Then(1240 MeV fm)2= 1.0 MeV.8(940 MeV)(14 fm)2(c) Finding an neutron inside the nucleus seems reasonable; but nding the electron would not.The energy of such an electron is considerably larger than binding energies of the particles in thenucleus.E1 =E47-2 SolveEn =n2 (hc)28(mc2 )L2for L, thenL ===nhc,8mc2 En(3)(1240 eV nm)8(5.11105 eV)(4.7 eV)0.85 nm.,E47-3 Solve for E4 E1 :E4 E1===42 (hc)212 (hc)2,8(mc2 )L28(mc2 )L2(16 1)(1240 eV nm)2,8(5.11105 )(0.253 nm)288.1 eV.E47-4 Since E 1/L2 , doubling the width of the well will lower the ground state energy to(1/2)2 = 1/4, or 0.65 eV.E47-5 (a) Solve for E2 E1 :E2 E1===22 h212 h2,8mL28mL2(3)(6.631034 J s)2,8(40)(1.671027 kg)(0.2 m)26.21041 J.(b) K = 3kT /2 = 3(1.381023 J/K)(300 K)/2 = 6.211021 . The ratio is 11020 .(c) T = 2(6.21041 J)/3(1.381023 J/K) = 3.01018 K.259E47-6 (a) The fractional dierence is (En+1 En )/En , orEnEn===h2h2h2 n2/ n2,228mL8mL8mL2(n + 1)2 n2,n22n + 1.n2(n + 1)2(b) As n the fractional dierence goes to zero; the system behaves as if it is continuous.E47-7Then(a) We will take advantage of the trick that was developed in part (b) of Exercise 47-1.En = n 2(1240 eV nm)2(hc)2= (15)2= 8.72 keV.8mc2 L8(0.511 106 eV)(0.0985 nm)2(b) The magnitude of the momentum is exactly known because E = p2 /2m. This momentum isgiven bypc = 2mc2 E = 2(511 keV)(8.72 keV) = 94.4 keV.What we dont know is in which direction the particle is moving. It is bouncing back and forthbetween the walls of the box, so the momentum could be directed toward the right or toward theleft. The uncertainty in the momentum is thenp = pwhich can be expressed in terms of the box size L byp = p =2mE =n 2 h2nh=.24L2L(c) The uncertainty in the position is 98.5 pm; the electron could be anywhere inside the well.E47-8 The probability distribution function isP2 =22xsin2.LLWe want to integrate over the central third, orL/6P=L/6==22xsin2dx,LL1 /3sin2 d, /30.196.E47-9 (a) Maximum probability occurs when the argument of the cosine (sine) function is k([k + 1/2]). This occurs whenx = N L/2nfor odd N .(b) Minimum probability occurs when the argument of the cosine (sine) function is [k + 1/2](k). This occurs whenx = N L/2nfor even N .260E47-10 In Exercise 47-21 we show that the hydrogen levels can be written asEn = (13.6 eV)/n2 .(a) The Lyman series is the series which ends on E1 . The least energetic state starts on E2 . Thetransition energy isE2 E1 = (13.6 eV)(1/12 1/22 ) = 10.2 eV.The wavelength is=(1240 eV nm)hc== 121.6 nm.E(10.2 eV)(b) The series limit is0 E1 = (13.6 eV)(1/12 ) = 13.6 eV.The wavelength is=E47-11hc(1240 eV nm)== 91.2 nm.E(13.6 eV)The ground state of hydrogen, as given by Eq. 47-21, isE1 = me4(9.109 1031 kg)(1.602 1019 C)4== 2.179 1018 J.8 2 h28(8.854 1012 F/m)2 (6.626 1034 J s)20In terms of electron volts the ground state energy isE1 = (2.179 1018 J)/(1.602 1019 C) = 13.60 eV.E47-12 In Exercise 47-21 we show that the hydrogen levels can be written asEn = (13.6 eV)/n2 .(c) The transition energy isE = E3 E1 = (13.6 eV)(1/12 1/32 ) = 12.1 eV.(a) The wavelength is=hc(1240 eV nm)== 102.5 nm.E(12.1 eV)(b) The momentum isp = E/c = 12.1 eV/c.E47-13 In Exercise 47-21 we show that the hydrogen levels can be written asEn = (13.6 eV)/n2 .(a) The Balmer series is the series which ends on E2 . The least energetic state starts on E3 . Thetransition energy isE3 E2 = (13.6 eV)(1/22 1/32 ) = 1.89 eV.The wavelength is=hc(1240 eV nm)== 656 nm.E(1.89 eV)261(b) The next energetic state starts on E4 . The transition energy isE4 E2 = (13.6 eV)(1/22 1/42 ) = 2.55 eV.The wavelength is=hc(1240 eV nm)== 486 nm.E(2.55 eV)(c) The next energetic state starts on E5 . The transition energy isE5 E2 = (13.6 eV)(1/22 1/52 ) = 2.86 eV.The wavelength is=hc(1240 eV nm)== 434 nm.E(2.86 eV)(d) The next energetic state starts on E6 . The transition energy isE6 E2 = (13.6 eV)(1/22 1/62 ) = 3.02 eV.The wavelength is=hc(1240 eV nm)== 411 nm.E(3.02 eV)(e) The next energetic state starts on E7 . The transition energy isE7 E2 = (13.6 eV)(1/22 1/72 ) = 3.12 eV.The wavelength is=hc(1240 eV nm)== 397 nm.E(3.12 eV)E47-14 In Exercise 47-21 we show that the hydrogen levels can be written asEn = (13.6 eV)/n2 .The transition energy isE =hc(1240 eV nm)== 10.20 eV.(121.6 nm)This must be part of the Lyman series, so the higher state must beEn = (10.20 eV) (13.6 eV) = 3.4 eV.That would correspond to n = 2.E47-15 The binding energy is the energy required to remove the electron. If the energy of theelectron is negative, then that negative energy is a measure of the energy required to set the electronfree.The rst excited state is when n = 2 in Eq. 47-21. It is not necessary to re-evaluate the constantsin this equation every time, instead, we start fromEn =E1where E1 = 13.60 eV.n2Then the rst excited state has energyE2 =(13.6 eV)= 3.4 eV.(2)2The binding energy is then 3.4 eV.262E47-16 rn = a0 n2 , son=E47-17(847 pm)/(52.9 pm) = 4.(a) The energy of this photon isE=hc(1240 eV nm)== 0.96739 eV.(1281.8 nm)The nal state of the hydrogen must have an energy of no more than 0.96739, so the largestpossible n of the nal state isn<13.60 eV/0.96739 eV = 3.75,so the nal n is 1, 2, or 3. The initial state is only slightly higher than the nal state. The jumpfrom n = 2 to n = 1 is too large (see Exercise 15), any other initial state would have a larger energydierence, so n = 1 is not the nal state.So what level might be above n = 2? Well tryn=13.6 eV/(3.4 eV 0.97 eV) = 2.36,which is so far from being an integer that we dont need to look farther. The n = 3 state has energy13.6 eV/9 = 1.51 eV. Then the initial state could ben=13.6 eV/(1.51 eV 0.97 eV) = 5.01,which is close enough to 5 that we can assume the transition was n = 5 to n = 3.(b) This belongs to the Paschen series.E47-18 In Exercise 47-21 we show that the hydrogen levels can be written asEn = (13.6 eV)/n2 .(a) The transition energy isE = E4 E1 = (13.6 eV)(1/12 1/42 ) = 12.8 eV.(b) All transitions n m are allowed for n 4 and m < n. The transition energy will be of theformEn Em = (13.6 eV)(1/m2 1/n2 ).The six possible results are 12.8 eV, 12.1 eV, 10.2 eV, 2.55 eV, 1.89 eV, and 0.66 eV.E47-19 E = h/2t, soE = (4.141015 eV s)/2(1108 s) = 6.6108 eV.E47-20 (a) According to electrostatics and uniform circular motion,mv 2 /r = e2 /4 0 r2 ,orv=e2==4 0 mr263e4e2=.4 2 h2 n 22 0 hn0Putting in the numbers,v=(1.61019 C)22.18106 m/s=.2(8.851012 F/m)(6.631034 J s)nnIn this case n = 1.(b) = h/mv, = (6.631034 J s)/(9.111031 kg)(2.18106 m/s) = 3.341010 m.(c) /a0 = (3.341010 m)/(5.291011 ) = 6.31 2. Actually, it is exactly 2.E47-21 In order to have an inelastic collision with the 6.0 eV neutron there must exist a transitionwith an energy dierence of less than 6.0 eV. For a hydrogen atom in the ground state E1 = 13.6 eVthe nearest state isE2 = (13.6 eV)/(2)2 = 3.4 eV.Since the dierence is 10.2 eV, it will not be possible for the 6.0 eV neutron to have an inelasticcollision with a ground state hydrogen atom.E47-22 (a) The atom is originally in the state n given byn=(13.6 eV)/(0.85 eV) = 4.The state with an excitation energy of 10.2 eV, isn=(13.6 eV)/(13.6 eV 10.2 eV) = 2.The transition energy is thenE = (13.6 eV)(1/22 1/42 ) = 2.55 eV.E47-23 According to electrostatics and uniform circular motion,mv 2 /r = e2 /4 0 r2 ,orv=e2==4 0 mre442 h2 n 20The de Broglie wavelength is then=h2 0 hn=.mvme2The ratio of /r is2 0 hn== kn,rme2 a0 n2where k is a constant. As n the ratio goes to zero.264=e2.2 0 hnE47-24 In Exercise 47-21 we show that the hydrogen levels can be written asEn = (13.6 eV)/n2 .The transition energy isE = E4 E1 = (13.6 eV)(1/12 1/42 ) = 12.8 eV.The momentum of the emitted photon isp = E/c = (12.8 eV)/c.This is the momentum of the recoiling hydrogen atom, which then has velocityv=ppc(12.8 eV)=c=(3.00108 m/s) = 4.1 m/s.mmc2(932 MeV)E47-25 The rst Lyman line is the n = 1 to n = 2 transition. The second Lyman line is then = 1 to n = 3 transition. The rst Balmer line is the n = 2 to n = 3 transition. Since the photonfrequency is proportional to the photon energy (E = hf ) and the photon energy is the energydierence between the two levels, we havefnm =Em E nhwhere the En is the hydrogen atom energy level. Thenf13E 3 E1,hE 3 E2 + E2 E1E3 E2E2 E 1==+,hhh= f23 + f12 .=E47-26 UseEn = Z 2 (13.6 eV)/n2 .(a) The ionization energy of the ground state of He+ isEn = (2)2 (13.6 eV)/(1)2 = 54.4 eV.(b) The ionization energy of the n = 3 state of Li2+ isEn = (3)2 (13.6 eV)/(3)2 = 13.6 eV.E47-27(a) The energy levels in the He+ spectrum are given byEn = Z 2 (13.6 eV)/n2 ,where Z = 2, as is discussed in Sample Problem 47-6. The photon wavelengths for the n = 4 seriesare thenhchc/E4==,En E 41 En /E4265which can also be written as===16hc/(54.4 eV),1 16/n216hcn2 /(54.4 eV),n2 16Cn2,n2 16where C = hc/(3.4 eV) = 365 nm.(b) The wavelength of the rst line is the transition from n = 5,=(365 nm)(5)2= 1014 nm.(5)2 (4)2The series limit is the transition from n = , so = 365 nm.(c) The series starts in the infrared (1014 nm), and ends in the ultraviolet (365 nm). So it mustalso include some visible lines.E47-28 We answer these questions out of order!(a) n = 1.(b) r = a0 = 5.291011 m.(f) According to electrostatics and uniform circular motion,mv 2 /r = e2 /4 0 r2 ,orv=e2==4 0 mre442 h2 n 20=e2.2 0 hnPutting in the numbers,v=(1.61019 C)2= 2.18106 m/s.2(8.851012 F/m)(6.631034 J s)(1)(d) p = (9.111031 kg)(2.18106 m/s) = 1.991024 kg m/s.(e) = v/r = (2.18106 m/s)/(5.291011 m) = 4.121016 rad/s.(c) l = pr = (1.991024 kg m/s)(5.291011 m) = 1.051034 J s.(g) F = mv 2 /r, soF = (9.111031 kg)(2.18106 m/s)2 /(5.291011 m) = 8.18108 N.(h) a = (8.18108 N)/(9.111031 kg) = 8.981022 m/s2 .(i) K = mv 2 /r, orK=(9.111031 kg)(2.18106 m/s)2= 2.161018 J = 13.6 eV.2(k) E = 13.6 eV.(j) P = E K = (13.6 eV) (13.6 eV) = 27.2 eV.266E47-29 For each r in the quantity we have a factor of n2 .(a) n is proportional to n.(b) r is proportional to n2 .(f) v is proportional to 1/r, or 1/n.(d) p is proportional to v, or 1/n.(e) is proportional to v/r, or 1/n3 .(c) l is proportional to pr, or n.(g) f is proportional to v 2 /r, or 1/n4 .(h) a is proportional to F , or 1/n4 .(i) K is proportional to v 2 , or 1/n2 .(j) E is proportional to 1/n2 .(k) P is proportional to 1/n2 .E47-30 (a) Using the results of Exercise 45-1,E1 =(1240 eV nm)= 1.24105 eV.(0.010 nm)(b) Using the results of Problem 45-11,E2(1.24105 eV)2== 40.5104 eV.mc2 /2 + E(5.11105 eV)/2 + (1.24105 eV)Kmax =(c) This would likely knock the electron way out of the atom.E47-31 The energy of the photon in the series limit is given byE limit = (13.6 eV)/n2 ,where n = 1 for Lyman, n = 2 for Balmer, and n = 3 for Paschen. The wavelength of the photon islimit =(1240 eV nm) 2n = (91.17 nm)n2 .(13.6 eV)The energy of the longest wavelength comes from the transition from the nearest level, orE long =(13.6 eV) (13.6 eV)2n + 1= (13.6 eV).(n + 1)2n2[n(n + 1)]2The wavelength of the photon islong =(1240 eV nm)[n(n + 1)]2[n(n + 1)]2= (91.17 nm).2(13.6 eV)n2n + 1(a) The wavelength interval long limit , or = (91.17 nm)n2 (n + 1)2 n2 (2n + 1)n4= (91.17 nm).2n + 12n + 1For n = 1, = 30.4 nm. For n = 2, = 292 nm. For n = 3, = 1055 nm.(b) The frequency interval is found fromf =E limit E long(13.6 eV)1(3.291015 /s)==.15 eV s) (n + 1)2h(4.1410(n + 1)2For n = 1, f = 8.231014 Hz. For n = 2, f = 3.661014 Hz. For n = 3, f = 2.051014 Hz.267E47-32E47-33(a) Well use Eqs. 47-25 and 47-26. At r = 0 2 (0) =1 2(0)/a01== 2150 nm3 ,3ea0a30whileP (0) = 4(0)2 2 (0) = 0.(b) At r = a0 we have 2 (a0 ) = f rac1a3 e2(a0 )/a0 =0e2= 291 nm3 ,a30andP (a0 ) = 4(a0 )2 2 (a0 ) = 10.2 nm1 .E47-34 Assume that (a0 ) is a reasonable estimate for (r) everywhere inside the small sphere.Thene20.1353=.2 =a3a300The probability of nding it in a sphere of radius 0.05a0 is0.05a04(0.1353)4r2 dr= (0.1353)(0.05)3 = 2.26105 .a3300E47-35 Using Eq. 47-26 the ratio of the probabilities ise2P (a0 )(a0 )2 e2(a0 )/a0== 4 = 1.85.2 e2(2a0 )/a0P (2a0 )4e(2a0 )E47-36 The probability is1.016a0P=a04r2 e2r/a0dr,a301 2.032 2 uu e du,2 2= 0.00866.=E47-37 If l = 3 then ml can be 0, 1, 2, or 3.(a) From Eq. 47-30, Lz = ml h/2.. So Lz can equal 0, h/2, h/, or 3h/2.(b) From Eq. 47-31, = arccos(ml / l(l + 1)), so can equal 90 , 73.2 , 54.7 , or 30.0 .(c) The magnitude of L is given by Eq. 47-28,L=l(l + 1)h= 3h/.2E47-38 The maximum possible value of ml is 5. Apply Eq. 47-31: = arccos(5)(5)(5 + 1)268= 24.1 .E47-39 Use the hint.p x =rp x =rxp r=rL =h,2h,2h,2h.2E47-40 Note that there is a typo in the formula; P (r) must have dimensions of one over length.The probability isP=0r4 er/a0dr,24a501=u4 eu du,24 0= 1.00What does it mean? It means that if we look for the electron, we will nd it somewhere.E47-41 (a) Find the maxima by taking the derivative and setting it equal to zero.r(2a r)(4a2 6ra + r2 ) rdP=e = 0.dr8a60The solutions are r = 0, r = 2a, and 4a2 6ra + r2 = 0. The rst two correspond to minima (seeFig. 47-14). The other two are the solutions to the quadratic, or r = 0.764a0 and r = 5.236a0 .(b) Substitute these two values into Eq. 47-36. The results areP (0.764a0 ) = 0.981 nm1 .andP (5.236a0 ) = 3.61 nm1 .E47-42 The probability is5.01a0P=5.00a0=r2 (2 r/a0 )2 er/a0dr,8a300.01896.E47-43 n = 4 and l = 3, while ml can be any of3, 2, 1, 0, 1, 2, 3,while ms can be either 1/2 or 1/2. There are 14 possible states.E47-44 n must be greater than l, so n 4. |ml | must be less than or equal to l, so |ml | 3. msis 1/2 or 1/2.E47-45If ml = 4 then l 4. But n l + 1, so n > 4. We only know that ms = 1/2.269E47-46 There are 2n2 states in a shell n, so if n = 5 there are 50 states.E47-47 Each is in the n = 1 shell, the l = 0 angular momentum state, and the ml = 0 state. Butone is in the state ms = +1/2 while the other is in the state ms = 1/2.E47-48 Apply Eq. 47-31: = arccosand = arccos(+1/2)(1/2)(1/2 + 1)(1/2)(1/2)(1/2 + 1)= 54.7= 125.3 .E47-49 All of the statements are true.E47-50 There are n possible values for l (start at 0!). For each value of l there are 2l + 1 possiblevalues for ml . If n = 1, the sum is 1. If n = 2, the sum is 1 + 3 = 4. If n = 3, the sum is 1 + 3 + 5 = 9.The pattern is clear, the sum is n2 . But there are two spin states, so the number of states is 2n2 .P47-1We can simplify the energy expression asE = E0 n 2 + n2 + n 2xyzwhere E0 =h2.8mL2To nd the lowest energy levels we need to focus on the values of nx , ny , and nz .It doesnt take much imagination to realize that the set (1, 1, 1) will result in the smallest valuefor n2 + n2 + n2 . The next choice is to set one of the values equal to 2, and try the set (2, 1, 1).xyzThen it starts to get harder, as the next lowest might be either (2, 2, 1) or (3, 1, 1). The only wayto nd out is to try. Ill tabulate the results for you:nx12232ny11212nz11112n2 + n2 + n2xyz3691112Mult.13331nx33434ny22132nz12111n2 + n2 + n2xyz1417181921Mult.63336We are now in a position to state the ve lowest energy levels. The fundamental quantity isE0 =(hc)2(1240 eV nm)2== 6.02106 eV.2 L28mc8(0.511106 eV)(250 nm)2The ve lowest levels are found by multiplying this fundamental quantity by the numbers in thetable above.P47-2 (a) Write the states between 0 and L. Then all states, odd or even, can be written withprobability distribution function2nxP (x) = sin2,LL270we nd the probability of nding the particle in the region 0 x L/3 isL/3P=013=nx2cos2dx,LLsin(2n/3)1.2n/3(b) If n = 1 use the formula and P = 0.196.(c) If n = 2 use the formula and P = 0.402.(d) If n = 3 use the formula and P = 0.333.(e) Classically the probability distribution function is uniform, so there is a 1/3 chance of ndingit in the region 0 to L/3.P47-3 The region of interest is small compared to the variation in P (x); as such we can approximate the probability with the expression P = P (x)x.(b) Evaluating,P===24xsin2x,LL24(L/8)sin2(0.0003L),LL0.0006.(b) Evaluating,P===P47-424xsin2x,LL24(3L/16)sin2(0.0003L),LL0.0003.(a) P = , or2P = A2 e2mx0/h.(b) Integrating,1= A20e2mx2/hdx,= A20= A2042mhh2mh2m= A0 .(c) x = 0.P47-5We will want an expression ford20 .dx2271pi,2eu du,Doing the math one derivative at a time,d20dx2dd0 ,dx dx2d=A0 (2mx/h)emx /h ,dx22= A0 (2mx/h)2 emx /h + A0 (2m/h)emx /h ,===2(2mx/h)2 (2m/h) A0 emx/h,2(2mx/h) (2m/h) 0 .In the last line we factored out 0 . This will make our lives easier later on.Now we want to go to Schrdingers equation, and make some substitutions.oh2 d 2 0 + U 08 2 m dx2h2(2mx/h)2 (2m/h) 0 + U 08 2 mh2 2(2mx/h)2 (2m/h) + U8 m= E0 ,= E0 ,= E,where in the last line we divided through by 0 . Now for some algebra,Uh2(2mx/h)2 (2m/h) ,8 2 mm 2 x2h= E+.24= E+But we are given that E = h/4, so this simplies tom 2 x22U=which looks like a harmonic oscillator type potential.P47-6 Assume the electron is originally in the state n. The classical frequency of the electron isf0 , wheref0 = v/2r.According to electrostatics and uniform circular motion,mv 2 /r = e2 /4 0 r2 ,orv=Thenf0 =e4e2=.4 2 h2 n 22 0 hn0e2==4 0 mre2 1 me2me42E1= 2 3 3 =2 0 hn 2 0 h2 n24 0h nhn3Here E1 = 13.6 eV.Photon frequency is related to energy according to f = Enm /h, where Enm is the energy oftransition from state n down to state m. Thenf=E1h11 22nm272,where E1 = 13.6 eV. Combining the fractions and letting m = n , where is an integer,f====E1 m2 n2,h m2 n2E1 (n m)(m + n),hm2 n2E1 (2n + ),h (n + )2 n2E1 (2n),h (n)2 n22E1 = f0 .hn3P47-7 We need to use the reduced mass of the muon since the muon and proton masses are soclose together. Then(207)(1836)me = 186me .m=(207) + (1836)(a) Apply Eq. 47-20 1/2:a = a0 /(186) = (52.9 pm)/(186) = 0.284 pm.(b) Apply Eq. 47-21:E = E1 (186) = (13.6 eV)(186) = 2.53 keV.(c) = (1240 keV pm)/(2.53 pm) = 490 pm.P47-8(a) The reduced mass of the electron ism=(1)(1)me = 0.5me .(1) + (1)The spectrum is similar, except for this additional factor of 1/2; hencepos = 2H .(b) apos = a0 /(186) = (52.9 pm)/(1/2) = 105.8 pm. This is the distance between the particles,but they are both revolving about the center of mass. The radius is then half this quantity, or52.9 pm.P47-9 This problem isnt really that much of a problem. Start with the magnitude of a vectorin terms of the components,L2 + L2 + L2 = L2 ,xyzand then rearrange,L2 + L2 = L2 L2 .xyzAccording to Eq. 47-28 L2 = l(l + 1)h2 /4 2 , while according to Eq. 47-30 Lz = ml h/2. Substitutethat into the equation, andL2 + L2 = l(l + 1)h2 /4 2 m2 h2 /4 2 = l(l + 1) m2xyllTake the square root of both sides of this expression, and we are done.273h2.4 2The maximum value for ml is l, while the minimum value is 0. Consequently,L2 + L2 =xyl(l + 1) m2 h/2 ll(l + 1) h/2,andL2 + L2 =xyP47-10Thenl(l + 1) m2 h/2 ll h/2.Assume that (0) is a reasonable estimate for (r) everywhere inside the small sphere.2 =e01=.a3a300The probability of nding it in a sphere of radius 1.11015 m is1.11015 m0P47-11Then4r2 dr4 (1.11015 m)3== 1.21014 .3a03 (5.291011 m)3Assume that (0) is a reasonable estimate for (r) everywhere inside the small sphere.2 =(2)2 e01=.32a38a300The probability of nding it in a sphere of radius 1.11015 m is1.11015 m0P47-124r2 dr1 (1.11015 m)3== 1.51015 .8a36 (5.291011 m)30(a) The wave function squared is2 =e2r/a0a30The probability of nding it in a sphere of radius r = xa0 isxa0P=04r2 e2r/a0 dr,a30x4x2 e2x dx,=0= 1 e2x (1 + 2x + 2x2 ).(b) Let x = 1, thenP = 1 e2 (5) = 0.323.P47-13We want to evaluate the dierence between the values of P at x = 2 and x = 2. ThenP (2) P (1)==1 e4 (1 + 2(2) + 2(2)2 ) 1 e2 (1 + 2(1) + 2(1)2 ) ,5e2 13e4 = 0.439.274P47-14Using the results of Problem 47-12,0.5 = 1 e2x (1 + 2x + 2x2 ),ore2x = 1 + 2x + 2x2 .The result is x = 1.34, or r = 1.34a0 .P47-15The probability of nding it in a sphere of radius r = xa0 isxa0P=0==r2 (2 r/a0 )2 er/a0 dr8a301 x 2x (2 x)2 ex dx8 01 ex (y 4 /8 + y 2 /2 + y + 1).The minimum occurs at x = 2, soP = 1 e2 (2 + 2 + 2 + 1) = 0.0527.275E48-1 The highest energy x-ray photon will have an energy equal to the bombarding electrons,as is shown in Eq. 48-1,hcmin =eVInsert the appropriate values into the above expression,min =(4.14 1015 eV s)(3.00 108 m/s)1240 109 eV m=.eVeVThe expression is then1240 109 V m1240 kV pm=.VVSo long as we are certain that the V will be measured in units of kilovolts, we can write this asmin =min = 1240 pm/V.E48-2 f = c/ = (3.00108 m/s)/(31.11012 m) = 9.6461018 /s. Plancks constant is thenh=E(40.0 keV)== 4.141015 eV s.f(9.6461018 /s)E48-3 Applying the results of Exercise 48-1,V =(1240kV pm)= 9.84 kV.(126 pm)E48-4 (a) Applying the results of Exercise 48-1,min =(1240kV pm)= 35.4 pm.(35.0 kV)(b) Applying the results of Exercise 45-1,K =(1240keV pm)= 49.6 pm.(25.51 keV) (0.53 keV)(c) Applying the results of Exercise 45-1,K =(1240keV pm)= 56.5 pm.(25.51 keV) (3.56 keV)E48-5 (a) Changing the accelerating potential of the x-ray tube will decrease min . The newvalue will be (using the results of Exercise 48-1)min = 1240 pm/(50.0) = 24.8 pm.(b) K doesnt change. It is a property of the atom, not a property of the accelerating potentialof the x-ray tube. The only way in which the accelerating potential might make a dierence is ifK < min for which case there would not be a K line.(c) K doesnt change. See part (b).276E48-6 (a) Applying the results of Exercise 45-1,(1240keV pm)= 64.2 keV.(19.3 pm)E =(b) This is the transition n = 2 to n = 1, soE = (13.6 eV)(1/12 1/22 ) = 10.2 eV.E48-7 Applying the results of Exercise 45-1,E =(1240keV pm)= 19.8 keV.(62.5 pm)E =(1240keV pm)= 17.6 keV.(70.5 pm)andThe dierence isE = (19.8 keV) (17.6 keV) = 2.2 eV.E48-8 Since E = hf = hc/, and = h/mc = hc/mc2 , thenE = hc/ = mc2 .or V = E /e = mc2 /e = 511 kV.E48-9 The 50.0 keV electron makes a collision and loses half of its energy to a photon, then thephoton has an energy of 25.0 keV. The electron is now a 25.0 keV electron, and on the next collisionagain loses loses half of its energy to a photon, then this photon has an energy of 12.5 keV. On thethird collision the electron loses the remaining energy, so this photon has an energy of 12.5 keV. Thewavelengths of these photons will be given by=(1240 keV pm),Ewhich is a variation of Exercise 45-1.E48-10 (a) The x-ray will need to knock free a K shell electron, so it must have an energy of atleast 69.5 keV.(b) Applying the results of Exercise 48-1,min =(1240kV pm)= 17.8 pm.(69.5 kV)(c) Applying the results of Exercise 45-1,K =(1240keV pm)= 18.5 pm.(69.5 keV) (2.3 keV)Applying the results of Exercise 45-1,K =(1240keV pm)= 21.3 pm.(69.5 keV) (11.3 keV)277E48-11 (a) Applying the results of Exercise 45-1,EK =(1240keV pm)= 19.7 keV.(63 pm)Again applying the results of Exercise 45-1,EK =(1240keV pm)= 17.5 keV.(71 pm)(b) Zr or Nb; the others will not signicantly absorb either line.E48-12 Applying the results of Exercise 45-1,K =(1240keV pm)= 154.5 pm.(8.979 keV) (0.951 keV)Applying the Bragg reection relationship,d=(154.5 pm)== 282 pm.2 sin 2 sin(15.9 )E48-13 Plot the data. The plot should look just like Fig 48-4. Note that the vertical axis iswhich is related to the wavelength according to f = c/.f,E48-14 Remember that the m in Eq. 48-4 refers to the electron, not the nucleus. This meansthat the constant C in Eq. 48-5 is the same for all elements. Since f = c/, we have1=2Z2 1Z1 12.For Ga and Nb the wavelength ratio is thenNb=Ga(31) 1(41) 12= 0.5625.E48-15 (a) The ground state question is fairly easy. The n = 1 shell is completely occupied bythe rst two electrons. So the third electron will be in the n = 2 state. The lowest energy angularmomentum state in any shell is the s sub-shell, corresponding to l = 0. There is only one choice forml in this case: ml = 0. There is no way at this level of coverage to distinguish between the energyof either the spin up or spin down conguration, so well arbitrarily pick spin up.(b) Determining the conguration for the rst excited state will require some thought. We couldassume that one of the K shell electrons (n = 1) is promoted to the L shell (n = 2). Or we couldassume that the L shell electron is promoted to the M shell. Or we could assume that the L shellelectron remains in the L shell, but that the angular momentum value is changed to l = 1. Thequestion that we would need to answer is which of these possibilities has the lowest energy.The answer is the last choice: increasing the l value results in a small increase in the energy ofmulti-electron atoms.E48-16 Refer to Sample Problem 47-6:r1 =a0 (1)2(5.291011 m)== 5.751013 m.Z(92)278E48-17 We will assume that the ordering of the energy of the shells and sub-shells is the same.That ordering is1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p< 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s.If there is no spin the s sub-shell would hold 1 electron, the p sub-shell would hold 3, the d sub-shell5, and the f sub-shell 7. inert gases occur when a p sub-shell has lled, so the rst three inert gaseswould be element 1 (Hydrogen), element 1 + 1 + 3 = 5 (Boron), and element 1 + 1 + 3 + 1 + 3 = 9(Fluorine).Is there a pattern? Yes. The new inert gases have half of the atomic number of the original inertgases. The factor of one-half comes about because there are no longer two spin states for each setof n, l, ml quantum numbers.We can save time and simply divide the atomic numbers of the remaining inert gases in half:element 18 (Argon), element 27 (Cobalt), element 43 (Technetium), element 59 (Praseodymium).E48-18 The pattern is2 + 8 + 8 + 18 + 18 + 32 + 32+?or2(12 + 22 + 22 + 33 + 33 + 42 + 42 + x2 )The unknown is probably x = 5, the next noble element is probably118 + 2 52 = 168.E48-19 (a) Apply Eq. 47-23, which can be written asEn =(13.6 eV)Z 2.n2For the valence electron of sodium n = 3,Z=(5.14 eV)(3)2= 1.84,(13.6 eV)while for the valence electron of potassium n = 4,Z=(4.34 eV)(4)2= 2.26,(13.6 eV)(b) The ratios with the actual values of Z are 0.167 and 0.119, respectively.E48-20 (a) There are three ml states allowed, and two ms states. The rst electron can be inany one of these six combinations of M1 and m2 . The second electron, given no exclusion principle,could also be in any one of these six states. The total is 36. Unfortunately, this is wrong, becausewe cant distinguish electrons. Of this total of 36, six involve the electrons being in the same state,while 30 involve the electron being in dierent states. But if the electrons are in dierent states,then they could be swapped, and we wont know, so we must divide this number by two. The totalnumber of distinguishable states is then(30/2) + 6 = 21.(b) Six. See the above discussion.279E48-21 (a) The Bohr orbits are circular orbits of radius rn = a0 n2 (Eq. 47-20). The electron isorbiting where the force ise2Fn =,24 0 rnand this force is equal to the centripetal force, somv 2e2=.2rn4 0 rnwhere v is the velocity of the electron. Rearranging,e2.4 0 mrnv=The time it takes for the electron to make one orbit can be used to calculate the current,i=qee==t2rn /v2rne2.4 0 mrnThe magnetic moment of a current loop is the current times the area of the loop, so = iA =e2r2 ,4 0 mrn ne2rnwhich can be simplied to=e2e2rn .4 0 mrnBut rn = a0 n2 , so=nThis might not look right, but a0 ==n0he22e2a0 e2.4 0 m/me2 , so the expression can simplify toh2=n4 2 m2eh4m= nB .(b) In reality the magnetic moments depend on the angular momentum quantum number, notthe principle quantum number. Although the Bohr theory correctly predicts the magnitudes, it doesnot correctly predict when these values would occur.E48-22 (a) Apply Eq. 48-14:Fz = zdBz= (9.271024 J/T)(16103 T/m) = 1.51025 N.dz(b) a = F/m, z = at2 /2, and t = y/vy . Thenz =F y2(1.51025 N)(0.82 m)2== 3.2105 m.22mvy2(1.671027 kg)(970 m/s)2E48-23 a = (9.271024 J/T)(1.4103 T/m)/(1.71025 kg) = 7.6104 m/s2 .280E48-24 (a) U = 2B, orU = 2(5.79105 eV/T)(0.520 T) = 6.02105 eV.(b) f = E/h = (6.02105 eV)(4.141015 eV s) = 1.451010 Hz.(c) = c/f = (3108 m/s)/(1.451010 Hz) = 2.07102 m.E48-25 The energy change can be derived from Eq. 48-13; we multiply by a factor of 2 becausethe spin is completely ipped. ThenE = 2z Bz = 2(9.271024 J/T)(0.190 T) = 3.521024 J.The corresponding wavelength is=hc(6.631034 J s)(3.00108 m/s)== 5.65102 m.E(3.521024 J)This is somewhere near the microwave range.E48-26 The photon has an energy E = hc/. This energy is related to the magnetic eld in thevicinity of the electron according toE = 2B,soB=hc(1240 eV nm)== 0.051 T.22(5.79105 J/T)(21107 nm)E48-27 Applying the results of Exercise 45-1,E=(1240 eV nm)= 1.55 eV.(800nm)The production rate is thenR=(5.0103 W)= 2.01016 /s.(1.55 eV)(1.61019 J/eV)E48-28 (a) x = (3108 m/s)(121012 s) = 3.6103 m.(b) Applying the results of Exercise 45-1,E=(1240 eV nm)= 1.786 eV.(694.4nm)The number of photons in the pulse is thenN = (0.150J)/(1.786 eV)(1.61019 J/eV) = 5.251017 .E48-29 We need to nd out how many 10 MHz wide signals can t between the two wavelengths.The lower frequency isf1 =c(3.00 108 m/s)== 4.29 1014 Hz.1700 109 m)281The higher frequency isf1 =c(3.00 108 m/s)== 7.50 1014 Hz.1400 109 m)The number of signals that can be sent in this range isf2 f1(7.50 1014 Hz) (4.29 1014 Hz)== 3.21 107 .(10 MHz)(10 106 Hz)Thats quite a number of television channels.E48-30 Applying the results of Exercise 45-1,E=(1240 eV nm)= 1.960 eV.(632.8nm)The number of photons emitted in one minute is thenN=(2.3103 W)(60 s)= 4.41017 .(1.960 eV)(1.61019 J/eV)E48-31 Apply Eq. 48-19. E1 3 E1 1 = 2(1.2 eV).. The ratio is then5n1 3= e(2.4 eV)/(8.6210 eV/K)(2000 K) = 9107 .n1 1E48-32 (a) Population inversion means that the higher energy state is more populated; this canonly happen if the ratio in Eq. 48-19 is greater than one, which can only happen if the argument ofthe exponent is positive. That would require a negative temperature.(b) If n2 = 1.1n1 then the ratio is 1.1, soT =E48-33(2.26 eV= 2.75105 K.(8.62105 eV/K) ln(1.1)(a) At thermal equilibrium the population ratio is given byN2eE2 /kT= E /kT = eE/kT .N1e 1But E can be written in terms of the transition photon wavelength, so this expression becomesN2 = N1 ehc/kT .Putting in the numbers,5N2 = (4.01020 )e(1240 eVnm)/(582 nm)(8.6210 eV/K)(300 K)) = 6.621016 .Thats eectively none.(b) If the population of the upper state were 7.01020 , then in a single laser pulseE=Nhc(6.631034 J s)(3.00108 m/s)= (7.01020 )= 240 J.(582109 m)282E48-34 The allowed wavelength in a standing wave chamber are n = 2L/n. For large n we canwrite2L2L 2Ln+1 = 2.n+1nnThe wavelength dierence is then22L = 2 = n ,n2Lwhich in this case is(533109 m)2 == 1.71012 m.2(8.3102 m)E48-35 (a) The central disk will have an angle as measured from the center given byd sin = (1.22),and since the parallel rays of the laser are focused on the screen in a distance f , we also haveR/f = sin . Combining, and rearranging,1.22f .d(b) R = 1.22(3.5 cm)(515 nm)/(3 mm) = 7.2106 m.(c) I = P/A = (5.21 W)/(1.5 mm)2 = 7.37105 W/m2 .(d) I = P/A = (0.84)(5.21 W)/(7.2 m)2 = 2.71010 W/m2 .R=E48-36P48-1 Let 1 be the wavelength of the rst photon. Then 2 = 1 + 130 pm. The total energytransfered to the two photons is thenhc hcE1 + E2 =+= 20.0 keV.12We can solve this for 1 ,20.0 keV11=+,hc11 + 130 pm21 + 130 pm=,1 (1 + 130 pm)which can also be written as211 (1 + 130 pm) = (62 pm)(21 + 130 pm),+ (6 pm)1 (8060 pm2 ) = 0.This equation has solutions1 = 86.8 pm and 92.8 pm.Only the positive answer has physical meaning. The energy of this rst photon is thenE1 =(1240 keV pm)= 14.3 keV.(86.8 pm)(a) After this rst photon is emitted the electron still has a kinetic energy of20.0 keV 14.3 keV = 5.7 keV.(b) We found the energy and wavelength of the rst photon above. The energy of the secondphoton must be 5.7 keV, with wavelength2 = (86.8 pm) + 130 pm = 217 pm.283P48-2Originally,1=1(2.73108 m/s)2 /(3108 m/s)2= 2.412.The energy of the electron isE0 = mc2 = (2.412)(511 keV) = 1232 keV.Upon emitting the photon the new energy isE = (1232 keV) (43.8 keV) = 1189 keV,so the new gamma factor is = (1189 keV)/(511 keV) = 2.326,and the new speed isv = c 1 1/(2.326)2 = (0.903)c.P48-3 Switch to a reference frame where the electron is originally at rest.Momentum conservation requires0 = p + pe = 0,while energy conservation requiresmc2 = E + Ee .Rearrange toEe = mc2 E .Square both sides of this energy expression and2E 2E mc2 + m2 c42E 2E mc2p2 c2 2E mc22= Ee = p2 c2 + m2 c4 ,e= p2 c2 ,e= p2 c2 .eBut the momentum expression can be used here, and the result is2E mc2 = 0.Not likely.P48-4 (a) In the Bohr theory we can assume that the K shell electrons see a nucleus with chargeZ. The L shell electrons, however, are shielded by the one electron in the K shell and so they seea nucleus with charge Z 1. Finally, the M shell electrons are shielded by the one electron in theK shell and the eight electrons in the K shell, so they see a nucleus with charge Z 9.The transition wavelengths are then1==EE0 (Z 1)2=hchc2E0 (Z 1) 3.hc411 2221and1=EE0=hchc=E0 (Z 9)2 8.hc928411 2321,,The ratio of these two wavelengths is27 (Z 1)2=.32 (Z 9)2Note that the formula in the text has the square in the wrong place!P48-5(a) E = hc/; the energy dierence is thenE1112 2 1= hc.2 1hc=.2 1= hc,Since 1 and 2 are so close together we can treat the product 1 2 as being either 2 or 2 . Then12E =(1240 eV nm)(0.597 nm) = 2.1103 eV.(589 nm)2(b) The same energy dierence exists in the 4s 3p doublet, so =(1139 nm)2(2.1103 eV) = 2.2 nm.(1240 eV nm)P48-6 (a) We can assume that the K shell electron sees a nucleus of charge Z 1, since theother electron in the shell screens it. Then, according to the derivation leading to Eq. 47-22,rK = a0 /(Z 1).(b) The outermost electron sees a nucleus screened by all of the other electrons; as such Z = 1,and the radius isr = a0P48-7 We assume in this crude model that one electron moves in a circular orbit attracted tothe helium nucleus but repelled from the other electron. Look back to Sample Problem 47-6; weneed to use some of the results from that Sample Problem to solve this problem.The factor of e2 in Eq. 47-20 (the expression for the Bohr radius) and the factor of (e2 )2 in Eq.47-21 (the expression for the Bohr energy levels) was from the Coulomb force between the singleelectron and the single proton in the nucleus. This force isF =e2.4 0 r2In our approximation the force of attraction between the one electron and the helium nucleus isF1 =2e2.4 0 r2The factor of two is because there are two protons in the helium nucleus.There is also a repulsive force between the one electron and the other electron,F2 =e2,4 0 (2r)2285where the factor of 2r is because the two electrons are on opposite side of the nucleus.The net force on the rst electron in our approximation is thenF1 F2 =2e2e2,4 0 r24 0 (2r)2which can be rearranged to yieldF net =e24 0 r2214=e24 0 r274.It is apparent that we need to substitute 7e2 /4 for every occurrence of e2 .(a) The ground state radius of the helium atom will then be given by Eq. 47-20 with theappropriate substitution,240hr== a0 .2 /4)m(7e7(b) The energy of one electron in this ground state is given by Eq. 47-21 with the substitutionof 7e2 /4 for every occurrence of e2 , thenE=m(7e2 /4)249 me4=.8 4 h216 8 4 h200We already evaluated all of the constants to be 13.6 eV.One last thing. There are two electrons, so we need to double the above expression. The groundstate energy of a helium atom in this approximation isE0 = 249(13.6 eV) = 83.3eV.16(c) Removing one electron will allow the remaining electron to move closer to the nucleus. Theenergy of the remaining electron is given by the Bohr theory for He+ , and isE He+ = (4)(13.60 eV) = 54.4 eV,so the ionization energy is 83.3 eV - 54.4 eV = 28.9 eV. This compares well with the accepted value.P48-8Applying Eq. 48-19:T =(3.2 eV)= 1.0104 K.(8.62105 eV/K) ln(6.11013 /2.51015 )P48-9 sin r/R, where r is the radius of the beam on the moon and R is the distance to themoon. Then1.22(600109 m)(3.82108 m)r== 2360 m.(0.118 m)The beam diameter is twice this, or 4740 m.P48-10(a) N = 2L/n , orN=2(6102 m)(1.75)= 3.03105 .(694109 )286(b) N = 2nLf /c, sof =c(3108 m/s)== 1.43109 /s.2nL2(1.75)(6102 m)Note that the travel time to and fro is t = 2nL/c!(c) f /f is then(694109 )f=== 3.3106 .f2nL2(1.75)(6102 m)287E49-1(a) Equation 49-2 isn(E) =8 2m3/2 1/28 2(mc2 )3/2 1/2E=E .h3(hc)3We can evaluate this by substituting in all known quantities,8 2(0.511 106 eV)3/2 1/2n(E) =E= (6.81 1027 m3 eV3/2 )E 1/2 .(1240 109 eV m)3Once again, we simplied the expression by writing hc wherever we could, and then using hc =1240 109 eV m.(b) Then, if E = 5.00 eV,n(E) = (6.81 1027 m3 eV3/2 )(5.00 eV)1/2 = 1.52 1028 m3 eV1 .E49-2 Apply the results of Ex. 49-1:n(E) = (6.81 1027 m3 eV3/2 )(8.00 eV)1/2 = 1.93 1028 m3 eV1 .E49-3 Monovalent means only one electron is available as a conducting electron. Hence we needonly calculate the density of atoms:NA(19.3103 kg/m3 )(6.021023 mol1 )N=== 5.901028 /m3 .VAr(0.197 kg/mol)E49-4 Use the ideal gas law: pV = N kT . Thenp=E49-5NkT = (8.491028 m3 )(1.381023 J/ K)(297 K) = 3.48108 Pa.V(a) The approximate volume of a single sodium atom isV1 =(0.023 kg/mol)= 3.931029 m3 .(6.021023 part/mol)(971 kg/m3 )The volume of the sodium ion sphere is4(981012 m)3 = 3.941030 m3 .3The fractional volume available for conduction electrons isV2 =V 1 V2(3.931029 m3 ) (3.941030 m3 )== 90%.V1(3.931029 m3 )(b) The approximate volume of a single copper atom isV1 =(0.0635 kg/mol)= 1.181029 m3 .(6.021023 part/mol)(8960 kg/m3 )The volume of the copper ion sphere is4(961012 m)3 = 3.711030 m3 .3The fractional volume available for conduction electrons isV2 =V 1 V2(1.181029 m3 ) (3.711030 m3 )== 69%.V1(1.181029 m3 )(c) Sodium, since more of the volume is available for the conduction electron.288E49-6 (a) Apply Eq. 49-6:5p = 1/ e(0.0730 eV)/(8.6210 eV/K)(0 K) + 1 = 0.(b) Apply Eq. 49-6:5p = 1/ e(0.0730 eV)/(8.6210 eV/K)(320 K) + 1 = 6.62102 .E49-7 Apply Eq. 49-6, remembering to use the energy dierence:5p = 1/ e(1.1) eV)/(8.6210 eV/K)(273 K) + 1 = 1.00,5p = 1/ e(0.1) eV)/(8.6210 eV/K)(273 K) + 1 = 0.986,5p = 1/ e(0.0) eV)/(8.6210 eV/K)(273 K) + 1 = 0.5,5p = 1/ e(0.1) eV)/(8.6210 eV/K)(273 K) + 1 = 0.014,5p = 1/ e(1.1) eV)/(8.6210 eV/K)(273 K) + 1 = 0.0.(b) Inverting the equation,E,k ln(1/p 1)T =soT =(0.1 eV)= 700 K(8.62105 eV/K) ln(1/(0.16) 1)E49-8 The energy dierences are equal, except for the sign. Then1e+E/kt+1+1eE/kt+1eE/2kte+E/2kt+ E/2kte+E/2kt + eE/2kte+ e+E/2ktE/2kte+ e+E/2kteE/2kt + e+E/2ktE49-9= ,= ,= 1.The Fermi energy is given by Eq. 49-5,EF =h28m3n2/3,where n is the density of conduction electrons. For gold we have3n=(19.3 g/cm )(6.021023 part/mol)33= 5.901022 elect./cm = 59 elect./nm(197 g/mol)The Fermi energy is then(1240 eV nm)2EF =8(0.511106 eV)33(59 electrons/nm )2892/3= 5.53 eV.E49-10 Combine the results of Ex. 49-1 and Eq. 49-6:C Eno = E/kt.e+1Then for each of the energies we haveno =(6.811027 m3 eV3/2 ) (4 eV)= 1.361028 /m3 eV,e(3.06 eV)/(8.62105 eV/K)(1000 K) + 1no =(6.811027 m3 eV3/2 ) (6.75 eV)= 1.721028 /m3 eV,e(0.31 eV)/(8.62105 eV/K)(1000 K) + 1no =(6.811027 m3 eV3/2 ) (7 eV)= 9.021027 /m3 eV,e(0.06 eV)/(8.62105 eV/K)(1000 K) + 1no =(6.811027 m3 eV3/2 ) (7.25 eV)= 1.821027 /m3 eV,e(0.19 eV)/(8.62105 eV/K)(1000 K) + 1no =(6.811027 m3 eV3/2 ) (9 eV)= 3.431018 /m3 eV.e(1.94 eV)/(8.62105 eV/K)(1000 K) + 1E49-11 Solven2 (hc)28(mc2 )L2for n = 50, since there are two electrons in each level. ThenEn =(50)2 (1240 eV nm)2= 6.53104 eV.8(5.11105 eV)(0.12 nm)2Ef =E49-12 We need to be much higher than T = (7.06 eV)/(8.62105 eV/K) = 8.2104 K.E49-13Equation 49-5 isEF =h28mh28m33n2/3,and if we collect the constants,EF =2/3n3/2 = An3/2 ,where, if we multiply the top and bottom by c2A=(hc)28mc232/3=(1240 109 eV m)28(0.511 106 eV)32/3= 3.65 1019 m2 eV.E49-14 (a) Inverting Eq. 49-6,E = kT ln(1/p 1),soE = (8.62105 eV/K)(1050 K) ln(1/(0.91) 1) = 0.209 eV.Then E = (0.209 eV) + (7.06 eV) = 6.85 eV.(b) Apply the results of Ex. 49-1:n(E) = (6.81 1027 m3 eV3/2 )(6.85 eV)1/2 = 1.78 1028 m3 eV1 .(c) no = np = (1.78 1028 m3 eV1 )(0.910) = 1.62 1028 m3 eV1 .290E49-15 Equation 49-5 isEF =h28m3n2/3,and if we rearrange,E F 3/2 =3h3n,16 2m3/2Equation 49-2 is then8 2m3/2 1/23n(E) =E= nE F 3/2 E 1/2 .3h2E49-16 ph = 1 p, soph===11,eE/kT + 1eE/kT,eE/kT + 11.E/kT1+eE49-17 The steps to solve this exercise are equivalent to the steps for Exercise 49-9, except nowthe iron atoms each contribute 26 electrons and we have to nd the density.First, the density is=m(1.991030 kg)== 1.84109 kg/m34r3 /34(6.37106 m)3 /3Then3=(26)(1.84106 g/cm )(6.021023 part/mol)3= 5.11029 elect./cm ,(56 g/mol)=n5.1108 elect./nm3The Fermi energy is then(1240 eV nm)2EF =8(0.511106 eV)33(5.1108 elect./nm )2/3= 230 keV.E49-18 First, the density is=m2(1.991030 kg)== 9.51017 kg/m33 /34r4(10103 m)3 /3Thenn = (9.51017 kg/m3 )/(1.671027 kg) = 5.691044 /m3 .The Fermi energy is then(1240 MeV fm)2EF =8(940 MeV)33(5.69101 /fm )2912/3= 137 MeV.E49-19E49-20 (a) E F = 7.06 eV, sof=3(8.62105 eV K)(0 K)= 0,2(7.06 eV)(b) f = 3(8.62105 eV K)(300 K)/2(7.06 eV) = 0.0055.(c) f = 3(8.62105 eV K)(1000 K)/2(7.06 eV) = 0.0183.E49-21Using the results of Exercise 19,T =2f E F2(0.0130)(4.71 eV)== 474 K.3k3(8.62105 eV K)E49-22 f = 3(8.62105 eV K)(1235 K)/2(5.5 eV) = 0.029.E49-23 (a) Monovalent means only one electron is available as a conducting electron. Hence weneed only calculate the density of atoms:NNA(10.5103 kg/m3 )(6.021023 mol1 )=== 5.901028 /m3 .VAr(0.107 kg/mol)(b) Using the results of Ex. 49-13,E F = (3.65 1019 m2 eV)(5.901028 /m3 )2/3 = 5.5 eV.(c) v =2K/m, orv=2(5.5 eV)(5.11105 eV/c2 ) = 1.4108 m/s.(d) = h/p, or=(6.631034 J s)= 5.21012 m.(9.111031 kg)(1.4108 m/s)E49-24 (a) Bivalent means two electrons are available as a conducting electron. Hence we needto double the calculation of the density of atoms:NNA2(7.13103 kg/m3 )(6.021023 mol1 )=== 1.321029 /m3 .VAr(0.065 kg/mol)(b) Using the results of Ex. 49-13,E F = (3.65 1019 m2 eV)(1.321029 /m3 )2/3 = 9.4 eV.(c) v =2K/m, orv=2(9.4 eV)(5.11105 eV/c2 ) = 1.8108 m/s.(d) = h/p, or=(6.631034 J s)= 4.01012 m.(9.111031 kg)(1.8108 m/s)292E49-25 (a) Refer to Sample Problem 49-5 where we learn that the mean free path can bewritten in terms of Fermi speed v F and mean time between collisions as = v F .The Fermi speed isv F = c 2EF /mc2 = c 2(5.51 eV)/(5.11105 eV) = 4.64103 c.The time between collisions is=m(9.111031 kg)== 3.741014 s.228 m3 )(1.601019 C)2 (1.62108 m)ne(5.8610We found n by looking up the answers from Exercise 49-23 in the back of the book. The mean freepath is then = (4.64103 )(3.00108 m/s)(3.741014 s) = 52 nm.(b) The spacing between the ion cores is approximated by the cube root of volume per atom.This atomic volume for silver isV =(108 g/mol)(6.021023 part/mol)(10.53g/cm )= 1.711023 cm3 .The distance between the ions is thenl=3V = 0.257 nm.The ratio is/l = 190.E49-26 (a) For T = 1000 K we can use the approximation, so for diamond5p = e(5.5 eV)/2(8.6210 eV/K)(1000 K) = 1.41014 ,while for silicon,5p = e(1.1 eV)/2(8.6210 eV/K)(1000 K) = 1.7103 ,(b) For T = 4 K we can use the same approximation, but now Efunction goes to zero.kT and the exponentialE49-27 (a) E E F 0.67 eV/2 = 0.34 eV.. The probability the state is occupied is then5p = 1/ e(0.34) eV)/(8.6210 eV/K)(290 K) + 1 = 1.2106 .(b) E E F 0.67 eV/2 = 0.34 eV.. The probability the state is unoccupied is then 1 p, or5p = 1 1/ e(0.34) eV)/(8.6210 eV/K)(290 K) + 1 = 1.2106 .E49-28 (a) E E F 0.67 eV/2 = 0.34 eV.. The probability the state is occupied is then5p = 1/ e(0.34) eV)/(8.6210 eV/K)(289 K) + 1 = 1.2106 .293E49-29(a) The number of silicon atoms per unit volume is3n=(6.021023 part/mol)(2.33 g/cm )3= 4.991022 part./cm .(28.1 g/mol)If one out of 1.0eex7 are replaced then there will be an additional charge carrier density of334.991022 part./cm /1.0107 = 4.991015 part./cm = 4.991021 m3 .(b) The ratio is(4.991021 m3 )/(2 1.51016 m3 ) = 1.7105 .The extra factor of two is because all of the charge carriers in silicon (holes and electrons) are chargecarriers.E49-30 Since one out of every 5106 silicon atoms needs to be replaced, then the mass of phosphorus would be1 30m== 2.1107 g.5106 28E49-31 l =31/1022 /m3 = 4.6108 m.E49-32 The atom density of germanium isNNA(5.32103 kg/m3 )(6.021023 mol1 )=== 1.631028 /m3 .VAr(0.197 kg/mol)The atom density of the impurity is(1.631028 /m3 )/(1.3109 ) = 1.251019 .The average spacing isl=31/1.251019 /m3 = 4.3107 m.E49-33 The rst one is an insulator because the lower band is lled and band gap is so large;there is no impurity.The second one is an extrinsic n-type semiconductor: it is a semiconductor because the lowerband is lled and the band gap is small; it is extrinsic because there is an impurity; since the impuritylevel is close to the top of the band gap the impurity is a donor.The third sample is an intrinsic semiconductor: it is a semiconductor because the lower band islled and the band gap is small.The fourth sample is a conductor; although the band gap is large, the lower band is not completelylled.The fth sample is a conductor: the Fermi level is above the bottom of the upper band.The sixth one is an extrinsic p-type semiconductor: it is a semiconductor because the lower bandis lled and the band gap is small; it is extrinsic because there is an impurity; since the impuritylevel is close to the bottom of the band gap the impurity is an acceptor.E49-34 6.62105 eV/1.1 eV = 6.0105 electron-hole pairs.E49-35 (a) R = (1 V)/(501012 A) = 21010 .(b) R = (0.75 V)/(8 mA) = 90.294E49-36 (a) A region with some potential dierence exists that has a gap between the chargedareas.(b) C = Q/V . Using the results in Sample Problem 49-9 for q and V ,C=E49-37n0 eAd/2= 2 0 A/d.n0 ed2 /4 0(a) Apply that ever so useful formula=hc(1240 eV nm)== 225 nm.E(5.5 eV)Why is this a maximum? Because longer wavelengths would have lower energy, and so not enoughto cause an electron to jump across the band gap.(b) Ultraviolet.E49-38 Apply that ever so useful formulaE=hc(1240 eV nm)== 4.20 eV.(295 nm)E49-39 The photon energy isE=hc(1240 eV nm)== 8.86 eV.(140 nm)which is enough to excite the electrons through the band gap. As such, the photon will be absorbed,which means the crystal is opaque to this wavelength.E49-40P49-1We can calculate the electron density from Eq. 49-5,3/2=38mc2 E F(hc)2=38(0.511106 eV)(11.66 eV)(1240 eV nm)2=n181 electrons/nm .,3/2,3From this we calculate the number of electrons per particle,3(181 electrons/nm )(27.0 g/mol)3(2.70 g/cm )(6.021023 particles/mol)which we can reasonably approximate as 3.295= 3.01,P49-249-15,At absolute zero all states below E F are lled, an none above. Using the results of Ex.E av1n=En(E) dE,03 3/2 E F 3/2EFE dE,203 3/2 2 5/2EFEF ,253EF.5===P49-3EF(a) The total number of conduction electron isn=(0.0031 kg)(6.021023 mol1 )= 2.941022 .(0.0635 kg/mol)The total energy isE=3(7.06 eV)(2.941022 ) = 1.241023 eV = 2104 J.5(b) This will light a 100 W bulb fort = (2104 J)/(100 W) = 200 s.P49-4(a) First do the easy part: nc = N c p(E c ), soNc.e(E c E F )/kT + 1Then use the results of Ex. 49-16, and writenv = N v [1 p(E v )] =Nv.e(E v E F )/kT + 1Since each electron in the conduction band must have left a hole in the valence band, then these twoexpressions must be equal.(b) If the exponentials dominate then we can drop the +1 in each denominator, andNc(E c E F )/kTe=Nv(E v E F )/kTe,NcNv= e(E c 2E F +E v )/kT ,EF=1(E c + E v + kT ln(N c /N v )) .2P49-5 (a) We want to use Eq. 49-6; although we dont know the Fermi energy, we do knowthe dierences between the energies in question. In the un-doped silicon E E F = 0.55 eV for thebottom of the conduction band. The quantitykT = (8.62105 eV/K)(290 K) = 0.025 eV,which is a good number to remember at room temperature kT is 1/40 of an electron-volt.296Then1= 2.81010 .e(0.55 eV)/(0.025 eV) + 1In the doped silicon E E F = 0.084 eV for the bottom of the conduction band. Thenp=p=e(0.0841= 3.4102 .eV)/(0.025 eV) + 1(b) For the donor state E E F = 0.066 eV, sop=P49-61= 0.93.e(0.066 eV)/(0.025 eV) + 1(a) Inverting Eq. 49-6,E E F = kT ln(1/p 1),soE F = (1.1 eV 0.11 eV) (8.62105 eV/K)(290 K) ln(1/(4.8105 ) 1) = 0.74 eVabove the valence band.(b) E E F = (1.1 eV) (0.74 eV) = 0.36 eV, sop=e(0.361= 5.6107 .eV)/(0.025 eV) + 1P49-7 (a) Plot the graph with a spreadsheet. It should look like Fig. 49-12.(b) kT = 0.025 eV when T = 290 K. The ratio is thenife(0.5 eV)/(0.025 eV) + 1= (0.5 eV)/(0.025 eV)= 4.9108 .ire+1P49-8297E50-1 We want to follow the example set in Sample Problem 50-1. The distance of closestapproach is given byd===qQ,4 0 K(2)(29)(1.601019 C)24(8.851012 C2 /Nm2 )(5.30MeV)(1.60 1013 J/MeV),1.571014 m.Thats pretty close.E50-2 (a) The gold atom can be treated as a point particle:=q1 q2,4 0 r2=(2)(79)(1.601019 C)2,4(8.851012 C2 /Nm2 )(0.16109 m)2=F1.4106 N.(b) W = F d, sod=(5.3106 eV)(1.61019 J/eV)= 6.06107 m.(1.4106 N)Thats 1900 gold atom diameters.E50-3 Take an approach similar to Sample Problem 50-1:K===qQ,4 0 d(2)(79)(1.601019 C)24(8.851012 C2 /Nm2 )(8.781015 m)(1.602.6107 eV.E50-4 All are stable exceptE50-5 1019 J/eV)88Rb and239Pb.We can make an estimate of the mass number A from Eq. 50-1,R = R0 A1/3 ,where R0 = 1.2 fm. If the measurements indicate a radius of 3.6 fm we would have3A = (R/R0 )3 = ((3.6 fm)/(1.2 fm)) = 27.E50-6E50-7 The mass number of the sun isA = (1.991030 kg)/(1.671027 kg) = 1.21057 .The radius would beR = (1.21015 m)31.21057 = 1.3104 m.298,E50-8 239 Pu is composed of 94 protons and 239 94 = 145 neutrons. The combined mass of thefree particles isM = Zmp + N mn = (94)(1.007825 u) + (145)(1.008665 u) = 240.991975 u.The binding energy is the dierenceE B = (240.991975 u 239.052156 u)(931.5 MeV/u) = 1806.9 MeV,and the binding energy per nucleon is then(1806.9 MeV)/(239) = 7.56 MeV.E50-9 62 Ni is composed of 28 protons and 62 28 = 34 neutrons. The combined mass of thefree particles isM = Zmp + N mn = (28)(1.007825 u) + (34)(1.008665 u) = 62.513710 u.The binding energy is the dierenceE B = (62.513710 u 61.928349 u)(931.5 MeV/u) = 545.3 MeV,and the binding energy per nucleon is then(545.3 MeV)/(62) = 8.795 MeV.E50-10 (a) Multiply each by 1/1.007825, som1H = 1.00000,m12C = 11.906829,andm238U = 236.202500.E50-11 (a) Since the binding energy per nucleon is fairly constant, the energy must be proportionalto A.(b) Coulomb repulsion acts between pairs of protons; there are Z protons that can be chosen asrst in the pair, and Z 1 protons remaining that can make up the partner in the pair. That makesfor Z(Z 1) pairs. The electrostatic energy must be proportional to this.(c) Z 2 grows faster than A, which is roughly proportional to Z.E50-12 Solve(0.7899)(23.985042) + x(24.985837) + (0.2101 x)(25.982593) = 24.305for x. The result is x = 0.1001, and then the amount29926Mg is 0.1100.E50-13 The neutron conned in a nucleus of radius R will have a position uncertainty on theorder of x R. The momentum uncertainty will then be no less thanp hh.2x2RAssuming that p p, we haveh,2Rand then the neutron will have a (minimum) kinetic energy ofpEp2h2.2 mR22m8E(hc)2.28 2 mc2 R0 A2/3But R = R0 A1/3 , soFor an atom with A = 100 we getE(1240 MeV fm)2= 0.668 MeV.MeV)(1.2 fm)2 (100)2/38 2 (940This is about a factor of 5 or 10 less than the binding energy per nucleon.E50-14 (a) To remove a proton,E = [(1.007825) + (3.016049) (4.002603)] (931.5 MeV) = 19.81 MeV.To remove a neutron,E = [(1.008665) + (2.014102) (3.016049)] (931.5 MeV) = 6.258 MeV.To remove a proton,E = [(1.007825) + (1.008665) (2.014102)] (931.5 MeV) = 2.224 MeV.(b) E = (19.81 + 6.258 + 2.224)MeV = 28.30 MeV.(c) (28.30 MeV)/4 = 7.07 MeV.E50-15 (a) = [(1.007825) (1)](931.5 MeV) = 7.289 MeV.(b) = [(1.008665) (1)](931.5 MeV) = 8.071 MeV.(c) = [(119.902197) (120)](931.5 MeV) = 91.10 MeV.E50-16 (a) E B = (ZmH + N mN m)c2 . Substitute the denition for mass excess, mc2 = Ac2 + ,andEB(b) For197= Z(c2 + H ) + N (c2 + N ) Ac2 ,= ZH + N N .Au,E B = (79)(7.289 MeV) + (197 79)(8.071 MeV) (31.157 MeV) = 1559 MeV,and the binding energy per nucleon is then(1559 MeV)/(197) = 7.92 MeV.300E50-17 The binding energy of63Cu is given byM = Zmp + N mn = (29)(1.007825 u) + (34)(1.008665 u) = 63.521535 u.The binding energy is the dierenceE B = (63.521535 u 62.929601 u)(931.5 MeV/u) = 551.4 MeV.The number of atoms in the sample isn=(0.003 kg)(6.021023 mol1 )= 2.871022 .(0.0629 kg/mol)The total energy is then(2.871022 )(551.4 MeV)(1.61019 J/eV) = 2.531012 J.E50-18 (a) For ultra-relativistic particles E = pc, so=(1240 MeV fm)= 2.59 fm.(480 MeV)(b) Yes, since the wavelength is smaller than nuclear radii.E50-19 We will do this one the easy way because we can. This method wont work except whenthere is an integer number of half-lives. The activity of the sample will fall to one-half of the initialdecay rate after one half-life; it will fall to one-half of one-half (one-fourth) after two half-lives. Sotwo half-lives have elapsed, for a total of (2)(140 d) = 280 d.E50-20 N = N0 (1/2)t/t1/2 , soN = (481019 )(0.5)(26)/(6.5) = 3.01019 .E50-21 (a) t1/2 = ln 2/(0.0108/h) = 64.2 h.(b) N = N0 (1/2)t/t1/2 , soN/N0 = (0.5)(3) = 0.125.(c) N = N0 (1/2)t/t1/2 , soN/N0 = (0.5)(240)/(64.2) = 0.0749.E50-22 (a) = (dN/dt)/N , or = (12/s)/(2.51018 ) = 4.81018 /s.(b) t1/2 = ln 2/, sot1/2 = ln 2/(4.81018 /s) = 1.441017 s,which is 4.5 billion years.301E50-23(a) The decay constant for67Ga can be derived from Eq. 50-8,ln 2ln 2== 2.461106 s1 .t1/2(2.817105 s)=The activity is given by R = N , so we want to know how many atoms are present. That can befound from1 atom1u= 3.0771022 atoms.3.42 g1.66051024 g66.93 uSo the activity isR = (2.461106 /s1 )(3.0771022 atoms) = 7.5721016 decays/s.(b) After 1.728105 s the activity would have decreased toR = R0 et = (7.5721016 decays/s)e(2.461106/s1 )(1.728105 s)= 4.9491016 decays/s.E50-24 N = N0 et , but = ln 2/t1/2 , soN = N0 e ln 2t/t1/2 = N0 (2)t/t1/2 = N0E50-25 The remaining22312t/t1/2isN = (4.71021 )(0.5)(28)/(11.43) = 8.61020 .The number of decays, each of which produced an alpha particle, is(4.71021 ) (8.61020 ) = 3.841021 .E50-26 The amount remaining after 14 hours ism = (5.50 g)(0.5)(14)/(12.7) = 2.562 g.The amount remaining after 16 hours ism = (5.50 g)(0.5)(16)/(12.7) = 2.297 g.The dierence is the amount which decayed during the two hour interval:(2.562 g) (2.297 g) = 0.265 g.E50-27(a) Apply Eq. 50-7,R = R0 et .We rst need to know the decay constant from Eq. 50-8,=ln 2ln 2== 5.618107 s1 .t1/2(1.234106 s)And the the time is found fromt1R= ln, R01(170 counts/s)= ln,(5.618107 s1 ) (3050 counts/s)=5.139106 s 59.5 days.302.Note that counts/s is not the same as decays/s. Not all decay events will be picked up by a detectorand recorded as a count; we are assuming that whatever scaling factor which connects the initialcount rate to the initial decay rate is valid at later times as well. Such an assumption is a reasonableassumption.(b) The purpose of such an experiment would be to measure the amount of phosphorus that istaken up in a leaf. But the activity of the tracer decays with time, and so without a correction factorwe would record the wrong amount of phosphorus in the leaf. That correction factor is R0 /R; weneed to multiply the measured counts by this factor to correct for the decay.In this case7 15R= et = e(5.61810 s )(3.00710 s) = 1.184.R0E50-28 The number of particles ofn = (0.15)147Sm is(0.001 kg)(6.021023 mol1 )= 6.1431020 .(0.147 kg/mol)The decay constant is = (120/s)/(6.1431020 ) = 1.951019 /s.The half-life ist1/2 = ln 2/(1.951019 /s) = 3.551018 s,or 110 Gy.E50-29 The number of particles ofn0 =239Pu is(0.012 kg)(6.021023 mol1 )= 3.0231022 .(0.239 kg/mol)The number which decay isn0 n = (3.0251022 ) 1 (0.5)(20000)/(24100) = 1.321022 .The mass of helium produced is thenm=(0.004 kg/mol)(1.321022 )= 8.78105 kg.(6.021023 mol1 )E50-30 Let R33 /(R33 + R32) = x, where x0 = 0.1 originally, and we want to nd out at whattime x = 0.9. Rearranging,(R33 + R32)/R33 = 1/x,soR32/R33 = 1/x 1.t/t1/2Since R = R0 (0.5)we can write a ratio11=x1 1 (0.5)t/t32 t/t33 .x0Put in some of the numbers, andln[(1/9)/(9)] = ln[0.5]twhich has solution t = 209 d.3031114.3 25.3,E50-31E50-32 (a) N/N0 = (0.5)(4500)/(82) = 3.01017 .(b) N/N0 = (0.5)(4500)/(0.034) = 0.E50-33The Q values areQ3Q4Q5= (235.043923 232.038050 3.016029)(931.5 MeV) = 9.46 MeV,= (235.043923 231.036297 4.002603)(931.5 MeV) = 4.68 MeV,= (235.043923 230.033127 5.012228)(931.5 MeV) = 1.33 MeV.Only reactions with positive Q values are energetically possible.E50-34 (a) For the14C decay,Q = (223.018497 208.981075 14.003242)(931.5 MeV) = 31.84 MeV.For the 4 He decay,Q = (223.018497 219.009475 4.002603)(931.5 MeV) = 5.979 MeV.E50-35 Q = (136.907084 136.905821)(931.5 MeV) = 1.17 MeV.E50-36 Q = (1.008665 1.007825)(931.5 MeV) = 0.782 MeV.E50-37 (a) The kinetic energy of this electron is signicant compared to the rest mass energy,so we must use relativity to nd the momentum. The total energy of the electron is E = K + mc2 ,the momentum will be given bypc ==E 2 m2 c4 = K 2 + 2Kmc2 ,(1.00 MeV)2 + 2(1.00 MeV)(0.511 MeV) = 1.42 MeV.The de Broglie wavelength is then=hc(1240 MeV fm)== 873 fm.pc(1.42 MeV)(b) The radius of the emitting nucleus isR = R0 A1/3 = (1.2 fm)(150)1/3 = 6.4 fm.(c) The longest wavelength standing wave on a string xed at each end is twice the length ofthe string. Although the rules for standing waves in a box are slightly more complicated, it is a fairassumption that the electron could not exist as a standing a wave in the nucleus.(d) See part (c).304E50-38 The electron is relativistic, sopc =E 2 m2 c4 ,=(1.71 MeV + 0.51 MeV)2 (0.51 MeV)2 ,= 2.16 MeV.This is also the magnitude of the momentum of the recoiling 32 S. Non-relativistic relations areK = p2 /2m, so(2.16 MeV)2K== 78.4 eV.2(31.97)(931.5 MeV)E50-39 N = mNA /Mr will give the number of atoms of = ln 2/t1/2 will give the decay constant. Combining,m=198Au; R = N will give the activity;Rt1/2 MrN Mr=.NAln 2NAThen for the sample in questionm=(250)(3.71010 /s)(2.693)(86400 s)(198 g/mol)= 1.02103 g.ln 2(6.021023 /mol)E50-40 R = (8722/60 s)/(3.71010 /s) = 3.93109 Ci.E50-41 The radiation absorbed dose (rad) is related to the roentgen equivalent man (rem) bythe quality factor, so for the chest x-ray(25 mrem)= 29 mrad.(0.85)This is well beneath the annual exposure average.Each rad corresponds to the delivery of 105 J/g, so the energy absorbed by the patient is(0.029)(105 J/g)12(88 kg) = 1.28102 J.E50-42 (a) (75 kg)(102 J/kg)(0.024 rad) = 1.8102 J.(b) (0.024 rad)(12) = 0.29 rem.E50-43 R = R0 (0.5)t/t1/2 , soR0 = (3.94 Ci)(2)(6.048105s)/(1.82105 s)= 39.4 Ci.E50-44 (a) N = mNA /MR , soN=(2103 g)(6.021023 /mol)= 5.081018 .(239 g/mol)(b) R = N = ln 2N/t1/2 , soR = ln 2(5.081018 )/(2.411104 y)(3.15107 s/y) = 4.64106 /s.(c) R = (4.64106 /s)/(3.71010 decays/s Ci) = 1.25104 Ci.305E50-45 The hospital uses a 6000 Ci source, and that is all the information we need to nd thenumber of disintegrations per second:(6000 Ci)(3.71010 decays/s Ci) = 2.221014 decays/s.We are told the half life, but to nd the number of radioactive nuclei present we want to know thedecay constant. Thenln 2ln 2=== 4.17109 s1 .t1/2(1.66108 s)The number of60Co nuclei is thenN=R(2.221014 decays/s)== 5.321022 .(4.17109 s1 )E50-46 The annual equivalent does is(12104 rem/h)(20 h/week)(52 week/y) = 1.25 rem.E50-47 (a) N = mNA /MR and MR = (226) + 2(35) = 296, soN=(1101 g)(6.021023 /mol)= 2.031020 .(296 g/mol)(b) R = N = ln 2N/t1/2 , soR = ln 2(2.031020 )/(1600 y)(3.15107 s/y) = 2.8109 Bq.(c) (2.8109 )/(3.71010 ) = 76 mCi.E50-48 R = N = ln 2N/t1/2 , soN=(4.6106 )(3.71010 /s)(1.28109 y)(3.15107 s/y)= 9.91021 ,ln 2N = mNA /MR , som=E50-49(40 g/mol)(9.91021 )= 0.658 g.(6.021023 /mol)We can apply Eq. 50-18 to nd the age of the rock,t===t1/2NFln 1 +,ln 2NI(4.47109 y)(2.00103 g)/(206 g/mol)ln 1 +ln 2(4.20103 g)/(238 g/mol)2.83109 y.306,E50-50 The number of atoms ofN=238U originally present is(3.71103 g)(6.021023 /mol)= 9.381018 .(238 g/mol)The number remaining after 260 million years isN = (9.381018 )(0.5)(260 My)/(4470 My) = 9.011018 .The dierence decays into lead (eventually), so the mass of lead present should bem=(206 g/mol)(0.371018 )= 1.27104 g.(6.021023 /mol)E50-51 We can apply Eq. 50-18 to nd the age of the rock,t===t1/2NFln 1 +,ln 2NI(150106 g)/(206 g/mol)(4.47109 y)ln 1 +ln 2(860106 g)/(238 g/mol),1.18109 y.Inverting Eq. 50-18 to nd the mass of40K originally present,NF= 2t/t1/2 1,NIso (since they have the same atomic mass) the mass ofm=40K is(1.6103 g)= 1.78103 g.2(1.18)/(1.28) 1E50-52 (a) There is an excess proton on the left and an excess neutron, so the unknown must bea deuteron, or d.(b) Weve added two protons but only one (net) neutron, so the element is Ti and the massnumber is 43, or 43 Ti.(c) The mass number doesnt change but we swapped one proton for a neutron, so 7 Li.E50-53 Do the math:Q = (58.933200 + 1.007825 58.934352 1.008665)(931.5 MeV) = 1.86 MeV.E50-54 The reactions are201Hg(, )197 Pt,Au(n, p)197 Pt,196Pt(n, )197 Pt,198Pt(, n)197 Pt,196Pt(d, p)197 Pt,198Pt(p, d)197 Pt.197307E50-55 We will write these reactions in the same way as Eq. 50-20 represents the reaction ofEq. 50-19. It is helpful to work backwards before proceeding by asking the following question: whatnuclei will we have if we subtract one of the allowed projectiles?The goal is 60 Co, which has 27 protons and 60 27 = 33 neutrons.1. Removing a proton will leave 26 protons and 33 neutrons, which isunstable.592. Removing a neutron will leave 27 protons and 32 neutrons, which isstable.593. Removing a deuteron will leave 26 protons and 32 neutrons, which isstable.Fe; but that nuclide isCo; and that nuclide is58Fe; and that nuclide isIt looks as if only 59 Co(n)60 Co and 58 Fe(d)60 Co are possible. If, however, we allow for the possibilityof other daughter particles we should also consider some of the following reactions.1. Swapping a neutron for a proton:60Ni(n,p)60 Co.2. Using a neutron to knock out a deuteron:61Ni(n,d)60 Co.3. Using a neutron to knock out an alpha particle:634. Using a deuteron to knock out an alpha particle:Cu(n,)60 Co.62Ni(d,)60 Co.E50-56 (a) The possible results are 64 Zn, 66 Zn, 64 Cu, 66 Cu,(b) The stable results are 64 Zn, 66 Zn, 61 Ni, and 67 Zn.61Ni,63Ni,65Zn, and67Zn.E50-57E50-58 The resulting reactions are194Pt(d,)192 Ir,196Pt(d,)194 Ir, and198Pt(d,)196 Ir.E50-59E50-60 Shells occur at numbers 2, 8, 20, 28, 50, 82. The shells occur separately for protons andneutrons. To answer the question you need to know both Z and N = A Z of the isotope.(a) Filled shells are 18 O, 60 Ni, 92 Mo, 144 Sm, and 207 Pb.(b) One nucleon outside a shell are 40 K, 91 Zr, 121 Sb, and 143 Nd.(c) One vacancy in a shell are 13 C, 40 K, 49 Ti, 205 Tl, and 207 Pb.E50-61 (a) The binding energy of this neutron can be found by considering the Q value of thereaction 90 Zr(n)91 Zr which is(89.904704 + 1.008665 90.905645)(931.5 MeV) = 7.19 MeV.89(b) The binding energy of this neutron can be found by considering the Q value of the reactionZr(n)90 Zr which is(88.908889 + 1.008665 89.904704)(931.5 MeV) = 12.0 MeV.This neutron is bound more tightly that the one in part (a).(c) The binding energy per nucleon is found by dividing the binding energy by the number ofnucleons:(401.007825 + 511.008665 90.905645)(931.5 MeV)= 8.69 MeV.91The neutron in the outside shell of 91 Zr is less tightly bound than the average nucleon in 91 Zr.308P50-1 Before doing anything we need to know whether or not the motion is relativistic. Therest mass energy of an particle ismc2 = (4.00)(931.5 MeV) = 3.73 GeV,and since this is much greater than the kinetic energy we can assume the motion is non-relativistic,and we can apply non-relativistic momentum and energy conservation principles. The initial velocityof the particle is thenv=2K/m = c 2K/mc2 = c 2(5.00 MeV)/(3.73 GeV) = 5.18102 c.For an elastic collision where the second particle is at originally at rest we have the nal velocityof the rst particle asv 1,f = v 1,im2 m1(4.00u) (197u)= (5.18102 c)= 4.97102 c,m2 + m1(4.00u) + (197u)while the nal velocity of the second particle isv 2,f = v 1,i2m12(4.00u)= (5.18102 c)= 2.06103 c.m2 + m1(4.00u) + (197u)(a) The kinetic energy of the recoiling nucleus isK=11mv 2 = m(2.06103 c)2 = (2.12106 )mc222= (2.12106 )(197)(931.5 MeV) = 0.389 MeV.(b) Energy conservation is the fastest way to answer this question, since it is an elastic collision.Then(5.00 MeV) (0.389 MeV) = 4.61 MeV.P50-2The gamma ray carries away a mass equivalent energy ofm = (2.2233 MeV)/(931.5 MeV/u) = 0.002387 u.The neutron mass would then bemN = (2.014102 1.007825 + 0.002387)u = 1.008664 u.P50-3 (a) There are four substates: mj can be +3/2, +1/2, -1/2, and -3/2.(b) E = (2/3)(3.26)(3.15108 eV/T)(2.16 T) = 1.48107 eV.(c) = (1240 eV nm)/(1.48107 eV) = 8.38 m.(d) This is in the radio region.P50-4 (a) The charge density is = 3Q/4R3 . The charge on the shell of radius r is dq = 4r2 dr.The potential at the surface of a solid sphere of radius r isV =qr2=.4 0 r3 0The energy required to add a layer of charge dq isdU = V dq =30942 r4dr,3 0which can be integrated to yieldU=(b) For23942 R53Q2=.3 020 0 RPu,U=3(94)2 (1.61019 C)= 1024106 eV.20(8.851012 F/m)(7.451015 m)(c) The electrostatic energy is 10.9 MeV per proton.P50-5 The decay rate is given by R = N , where N is the number of radioactive nuclei present.If R exceeds P then nuclei will decay faster than they are produced; but this will cause N to decrease,which means R will decrease until it is equal to P . If R is less than P then nuclei will be producedfaster than they are decaying; but this will cause N to increase, which means R will increase until itis equal to P . In either case equilibrium occurs when R = P , and it is a stable equilibrium becauseit is approached no matter which side is larger. ThenP = R = Nat equilibrium, so N = P/.P50-6 (a) A = N ; at equilibrium A = P , so P = 8.881010 /s.(b) (8.881010 /s)(1e0.269t ), where t is in hours. The factor 0.269 comes from ln(2)/(2.58) = .(c) N = P/ = (8.881010 /s)(3600 s/h)/(0.269/h) = 1.191015 .(d) m = N Mr /NA , orm=P50-7(1.191015 )(55.94 g/mol)= 1.10107 g.(6.021023 /mol)(a) A = N , soA=ln 2mNAln 2(1103 g)(6.021023 /mol)== 3.66107 /s.t1/2 Mr(1600)(3.15107 s)(226 g/mol)(b) The rate must be the same if the system is in secular equilibrium.(c) N = P/ = t1/2 P/ ln 2, som=P50-8(3.82)(86400 s)(3.66107 /s)(222 g/mol)= 6.43109 g.(6.021023 /mol) ln 2The number of water molecules in the body isN = (6.021023 /mol)(70103 g)/(18 g/mol) = 2.341027 .There are ten protons in each water molecule. The activity is thenA = (2.341027 ) ln 2/(11032 y) = 1.62105 /y.The time between decays is then1/A = 6200 y.310P50-9 Assuming the 238 U nucleus is originally at rest the total initial momentum is zero, whichmeans the magnitudes of the nal momenta of the particle and the 234 Th nucleus are equal.The particle has a nal velocity ofv=2K/m = c 2K/mc2 = c 2(4.196 MeV)/(4.0026931.5 MeV) = 4.744102 c.Since the magnitudes of the nal momenta are the same, theof(4.744102 c)The kinetic energy of theK=234(4.0026 u)(234.04 u)234Th nucleus has a nal velocity= 8.113104 c.Th nucleus is11mv 2 = m(8.113104 c)2 = (3.291107 )mc222= (3.291107 )(234.04)(931.5 MeV) = 71.75 keV.The Q value for the reaction is then(4.196 MeV) + (71.75 keV) = 4.268 MeV,which agrees well with the Sample Problem.P50-10(a) The Q value isQ = (238.050783 4.002603 234.043596)(931.5 MeV) = 4.27 MeV.(b) The Q values for each step areQ = (238.050783 237.048724 1.008665)(931.5 MeV) = 6.153 MeV,Q = (237.048724 236.048674 1.007825)(931.5 MeV) = 7.242 MeV,Q = (236.048674 235.045432 1.008665)(931.5 MeV) = 5.052 MeV,Q = (235.045432 234.043596 1.007825)(931.5 MeV) = 5.579 MeV.(c) The total Q for part (b) is 24.026 MeV. The dierence between (a) and (b) is 28.296 MeV.The binding energy for the alpha particle isE = [2(1.007825) + 2(1.008665) 4.002603](931.5 MeV) = 28.296 MeV.P50-11 (a) The emitted positron leaves the atom, so the mass must be subtracted. But thedaughter particle now has an extra electron, so that must also be subtracted. Hence the factor2me .(b) The Q value isQ = [11.011434 11.009305 2(0.0005486)](931.5 MeV) = 0.961 MeV.P50-12 (a) Capturing an electron is equivalent to negative beta decay in that the total numberof electrons is accounted for on both the left and right sides of the equation. The loss of the K shellelectron, however, must be taken into account as this energy may be signicant.(b) The Q value isQ = (48.948517 48.947871)(931.5 MeV) (0.00547 MeV) = 0.596 MeV.311P50-13The decay constant for90Sr isln 2ln 2== 7.581010 s1 .t1/2(9.15108 s)=The number of nuclei present in 400 g ofN = (400 g)so the overall activity of the 400 g of9090Sr is(6.021023 /mol)= 2.681024 ,(89.9 g/mol)Sr isR = N = (7.581010 s1 )(2.681024 )/(3.71010 /Ci s) = 5.49104 Ci.This is spread out over a 2000 km2 area, so the activity surface density is(5.49104 Ci)= 2.74105 Ci/m2 .(20006 m2 )If the allowable limit is 0.002 mCi, then the area of land that would contain this activity is(0.002103 Ci)= 7.30102 m2 .(2.74105 Ci/m2 )P50-14(a) N = mNA /Mr , soN = (2.5103 g)(6.021023 /mol)/(239 g/mol) = 6.31018 .(b) A = ln 2N/t1/2 , so the number that decay in 12 hours isln 2(6.31018 )(12)(3600 s)= 2.51011 .(24100)(3.15107 s)(c) The energy absorbed by the body isE = (2.51011 )(5.2 MeV)(1.61019 J/eV) = 0.20 J.(d) The dose in rad is (0.20 J)/(87 kg) = 0.23 rad.(e) The biological dose in rem is (0.23)(13) = 3 rem.P50-15(a) The amount of238N=U per kilogram of granite is(4106 kg)(6.021023 /mol)= 1.011019 .(0.238 kg/mol)The activity is thenA=ln 2(1.011019 )= 49.7/s.(4.47109 y)(3.15107 s/y)The energy released in one second isE = (49.7/s)(51.7 MeV) = 4.11010 J.The amount of232Th per kilogram of granite isN=(13106 kg)(6.021023 /mol)= 3.371019 .(0.232 kg/mol)312The activity is thenA=ln 2(3.371019 )= 52.6/s.(1.411010 y)(3.15107 s/y)The energy released in one second isE = (52.6/s)(42.7 MeV) = 3.61010 J.The amount of40K per kilogram of granite isN=(4106 kg)(6.021023 /mol)= 6.021019 .(0.040 kg/mol)The activity is thenA=ln 2(6.021019 )= 1030/s.(1.28109 y)(3.15107 s/y)The energy released in one second isE = (1030/s)(1.32 MeV) = 2.21010 J.The total of the three is 9.91010 W per kilogram of granite.(b) The total for the Earth is 2.71013 W.P50-16(a) Since only a is moving originally then the velocity of the center of mass isV =ma va + mX (0)ma= va.mX + mama + mXNo, since momentum is conserved.(b) Moving to the center of mass frame gives the velocity of X as V , and the velocity of a asva V . The kinetic energy is nowK cm1mX V 2 + ma (va V )2 ,22vam2m2aX=mX+ ma22(ma + mX )(ma + mX )22ma va ma mX + m2X=,2 (ma + mX )2mX= K lab.ma + mX=Yes; kinetic energy is not conserved.(c) va = 2K/m, sova =2(15.9 MeV)/(1876 MeV)c = 0.130c.The center of mass velocity isV = (0.130c)(2)= 2.83103 c.(2) + (90)Finally,K cm = (15.9 MeV)(90)= 15.6 MeV.(2) + (90)313,P50-17Let Q = K cm in the result of Problem 50-16, and invert, solving for K lab .P50-18(a) Removing a proton from209Bi:E = (207.976636 + 1.007825 208.980383)(931.5 MeV) = 3.80 MeV.Removing a proton from208Pb:E = (206.977408 + 1.007825 207.976636)(931.5 MeV) = 8.01 MeV.(b) Removing a neutron from209Pb:E = (207.976636 + 1.008665 208.981075)(931.5 MeV) = 3.94 MeV.Removing a neutron from208Pb:E = (206.975881 + 1.008665 207.976636)(931.5 MeV) = 7.37 MeV.314E51-1 (a) For the coal,m = (1109 J)/(2.9107 J/kg) = 34 kg.(b) For the uranium,m = (1109 J)/(8.21013 J/kg) = 1.2105 kg.E51-2 (a) The energy from the coal isE = (100 kg)(2.9107 J/kg) = 2.9109 J.(b) The energy from the uranium in the ash isE = (3106 )(100 kg)(8.21013 J) = 2.51010 J.E51-3(a) There are(1.00 kg)(6.021023 mol1 )= 2.561024(235g/mol)atoms in 1.00 kg of 235 U.(b) If each atom releases 200 MeV, then(200106 eV)(1.61019 J/ eV)(2.561024 ) = 8.191013 Jof energy could be released from 1.00 kg of 235 U.(c) This amount of energy would keep a 100-W lamp lit fort=(8.191013 J)= 8.191011 s 26, 000 y!(100 W)E51-4 2 W = 1.251019 eV/s. This requires(1.251019 eV/s)/(200106 eV) = 6.251010 /sas the ssion rate.E51-5 There areatoms in 1.00 kg of(1.00 kg)(6.021023 mol1 )= 2.561024(235g/mol)235U. If each atom releases 200 MeV, then(200106 eV)(1.61019 J/ eV)(2.561024 ) = 8.191013 Jof energy could be released from 1.00 kg of 235 U. This amount of energy would keep a 100-W lamplit for(8.191013 J)t== 8.191011 s 30, 000 y!(100 W)E51-6 There areatoms in 1.00 kg of(1.00 kg)(6.021023 mol1 )= 2.521024(239g/mol)239Pu. If each atom releases 180 MeV, then(180106 eV)(1.61019 J/ eV)(2.521024 ) = 7.251013 Jof energy could be released from 1.00 kg of239Pu.315E51-7 When the 233 U nucleus absorbs a neutron we are given a total of 92 protons and 142neutrons. Gallium has 31 protons and around 39 neutrons; chromium has 24 protons and around28 neutrons. There are then 37 protons and around 75 neutrons left over. This would be rubidium,but the number of neutrons is very wrong. Although the elemental identication is correct, becausewe must conserve proton number, the isotopes are wrong in our above choices for neutron numbers.E51-8 Beta decay is the emission of an electron from the nucleus; one of the neutrons changes intoa proton. The atom now needs one more electron in the electron shells; by using atomic masses (asopposed to nuclear masses) then the beta electron is accounted for. This is only true for negativebeta decay, not for positive beta decay.E51-9 (a) There are(1.0 g)(6.021023 mol1 )= 2.561021(235g/mol)atoms in 1.00 g of235U. The ssion rate isA = ln 2N/t1/2 = ln 2(2.561021 )/(3.51017 y)(365d/y) = 13.9/d.(b) The ratio is the inverse ratio of the half-lives:(3.51017 y)/(7.04108 y) = 4.97108 .E51-10 (a) The atomic number of Y must be 92 54 = 38, so the element is Sr. The mass numberis 235 + 1 140 1 = 95, so Y is 95 Sr.(b) The atomic number of Y must be 92 53 = 39, so the element is Y. The mass number is235 + 1 139 2 = 95, so Y is 95 Y.(c) The atomic number of X must be 92 40 = 52, so the element is Te. The mass number is235 + 1 100 2 = 134, so X is 134 Te.(d) The mass number dierence is 235 + 1 141 92 = 3, so b = 3.E51-11 The Q value isQ = [51.94012 2(25.982593)](931.5 MeV) = 23 MeV.The negative value implies that this ssion reaction is not possible.E51-12 The Q value isQ = [97.905408 2(48.950024)](931.5 MeV) = 4.99 MeV.The two fragments would have a very large Coulomb barrier to overcome.E51-13The energy released is(235.043923 140.920044 91.919726 21.008665)(931.5 MeV) = 174 MeV.E51-14 Since En > Eb ssion is possible by thermal neutrons.E51-15 (a) The uranium starts with 92 protons. The two end products have a total of 58 + 44 =102. This means that there must have been ten beta decays.(b) The Q value for this process isQ = (238.050783 + 1.008665 139.905434 98.905939)(931.5 MeV) = 231 MeV.316E51-16 (a) The other fragment has 92 32 = 60 protons and 235 + 1 83 = 153 neutrons. Thatelement is 153 Nd.(b) Since K = p2 /2m and momentum is conserved, then 2m1 K1 = 2m2 K2 . This means thatK2 = (m1 /m2 )K1 . But K1 + K2 = Q, soK1m2 + m1= Q,m2orK1 =with a similar expression for K2 . Then for83m2Q,m1 + m2GeK=K=while for153(c) For(153)(170 MeV) = 110 MeV,(83 + 153)(83)(170 MeV) = 60 MeV,(83 + 153)Nd83Ge,v=1532(110 MeV)c = 0.053c,(83)(931 MeV)v=while for2K=m2K=m2(60 MeV)c = 0.029c.(153)(931 MeV)NdE51-17 Since 239 Pu is one nucleon heavier than 238 U only one neutron capture is required. Theatomic number of Pu is two more than U, so two beta decays will be required. The reaction seriesis then238U+n 239U 239Np 239U,Np + + ,239Pu + + .239E51-18 Each ssion releases 200 MeV. The total energy released over the three years is(190106 W)(3)(3.15107 s) = 1.81016 J.Thats(1.81016 J)/(1.61019 J/eV)(200106 eV) = 5.61026ssion events. That requiresm = (5.61026 )(0.235 kg/mol)/(6.021023 /mol) = 218 kg.But this is only half the original amount, or 437 kg.E51-19 According to Sample Problem 51-3 the rate at which non-ssion thermal neutron captureoccurs is one quarter that of ssion. Hence the mass which undergoes non-ssion thermal neutroncapture is one quarter the answer of Ex. 51-18. The total is then(437 kg)(1 + 0.25) = 546 kg.317E51-20 (a) Qe = E/N , where E is the total energy released and N is the number of decays.This can also be written asP t1/2P t1/2 MrP==,Qe =Aln 2Nln 2NA mwhere A is the activity and P the power output from the sample. Solving,Qe =(2.3 W)(29 y)(3.15107 s)(90 g/mol)= 4.531013 J = 2.8 MeV.ln 2(6.021023 /mol)(1 g)(b) P = (0.05)m(2300 W/kg), som=(150 W)= 1.3 kg.(0.05)(2300 W/kg)E51-21 Let the energy released by one ssion be E1 . If the average time to the next ssion eventis tgen , then the average power output from the one ssion is P1 = E1 /tgen . If every ssion eventresults in the release of k neutrons, each of which cause a later ssion event, then after every timeperiod tgen the number of ssion events, and hence the average power output from all of the ssionevents, will increase by a factor of k.For long enough times we can writeP (t) = P0 k t/tgen .E51-22 Invert the expression derived in Exercise 51-21:k=PP0tgen /t=(350)(1200)(1.3103 s)/(2.6 s)= 0.99938.E51-23 Each ssion releases 200 MeV. Then the ssion rate is(500106 W)/(200106 eV)(1.61019 J/eV) = 1.61019 /sThe number of neutrons in transit is then(1.61019 /s)(1.0103 s) = 1.61016 .E51-24 Using the results of Exercise 51-21:P = (400 MW)(1.0003)(300 s)/(0.03 s) = 8030 MW.E51-25The time constant for this decay is=ln 2= 2.501010 s1 .(2.77109 s)The number of nuclei present in 1.00 kg isN=(1.00 kg)(6.021023 mol1 )= 2.531024 .(238 g/mol)The decay rate is thenR = N = (2.501010 s1 )(2.531024 ) = 6.331014 s1 .The power generated is the decay rate times the energy released per decay,P = (6.331014 s1 )(5.59106 eV)(1.61019 J/eV) = 566 W.318E51-26 The detector detects only a fraction of the emitted neutrons. This fraction isA(2.5 m2 )== 1.62104 .24R4(35 m)2The total ux out of the warhead is then(4.0/s)/(1.62104 ) = 2.47104 /s.The number of239Pu atoms isN=(2.47104 /s)(1.341011 y)(3.15107 s/y)A== 6.021022 .ln 2(2.5)Thats one tenth of a mole, so the mass is (239)/10 = 24 g.E51-27 Using the results of Sample Problem 51-4,t=sot=ln[R(0)/R(t)],5 8ln[(0.03)/(0.0072)]= 1.72109 y.(0.984 0.155)(1109 /y)E51-28 (a) (15109 W y)(2105 y) = 7.5104 W.(b) The number of ssions required isN=The mass of235(15109 W y)(3.15107 s/y)= 1.51028 .(200 MeV)(1.61019 J/eV)U consumed ism = (1.51028 )(0.235kg/mol)/(6.021023 /mol) = 5.8103 kg.E51-29and thenIf 238 U absorbs a neutron it becomes 239 U, which will decay by beta decay to rst 239 NpPu; we looked at this in Exercise 51-17. This can decay by alpha emission according to239239Pu 235 U + .E51-30 The number of atoms present in the sample isN = (6.021023 /mol)(1000 kg)/(2.014g/mol) = 2.991026 .It takes two to make a fusion, so the energy released is(3.27 MeV)(2.991026 )/2 = 4.891026 MeV.Thats 7.81013 J, which is enough to burn the lamp fort = (7.81013 J)/(100 W) = 7.81011 s = 24800 y.E51-31 The potential energy at closest approach isU=(1.61019 C)2= 9105 eV.4(8.851012 F/m)(1.61015 m)319E51-32 The ratio can be written asn(K1 )=n(K2 )K1 (K2 K1 )/kTe,K2so the ratio is(5000 eV) (3100 eV)/(8.62105 eV/K)(1.5107 K)e= 0.15.(1900 eV)E51-33 (a) See Sample Problem 51-5.E51-34 Add up all of the Q values in the cycle of Fig. 51-10.E51-35The energy released is(34.002603 12.0000000)(931.5 MeV) = 7.27 MeV.E51-36 (a) The number of particle of hydrogen in 1 m3 isN = (0.35)(1.5105 kg)(6.021023 /mol)/(0.001 kg/mol) = 3.161031(b) The density of particles is N/V = p/kT ; the ratio is(3.161031 )(1.381023 J/K)(298 K)= 1.2106 .(1.01105 Pa)E51-37 (a) There are(1.00 kg)(6.021023 mol1 )= 6.021026(1g/mol)atoms in 1.00 kg of 1 H. If four atoms fuse to releases 26.7 MeV, then(26.7 MeV)(6.021026 )/4 = 4.01027 MeVof energy could be released from 1.00 kg of 1 H.(b) There are(1.00 kg)(6.021023 mol1 )= 2.561024(235g/mol)atoms in 1.00 kg of235U. If each atom releases 200 MeV, then(200 MeV)(2.561024 ) = 5.11026 MeVof energy could be released from 1.00 kg of235U.E51-38 (a) E = mc2 , som =(3.91026 J/s)= 4.3109 kg/s.(3.0108 m/s)2(b) The fraction of the Suns mass lost is(4.3109 kg/s)(3.15107 s/y)(4.5109 y)= 0.03 %.(2.01030 kg)320E51-39 The rate of consumption is 6.21011 kg/s, the core has 1/8 the mass but only 35% ishydrogen, so the time remaining ist = (0.35)(1/8)(2.01030 kg)/(6.21011 kg/s) = 1.41017 s,or about 4.5109 years.E51-40 For the rst two reactions into one:Q = [2(1.007825) (2.014102)](931.5 MeV) = 1.44 MeV.For the second,Q = [(1.007825) + (2.014102) (3.016029)](931.5 MeV) = 5.49 MeV.For the last,Q = [2(3.016029) (4.002603) 2(1.007825)](931.5 MeV) = 12.86 MeV.E51-41 (a) Use mNA /Mr = N , so(3.3107 J/kg)1(0.012 kg/mol)= 4.1 eV.(6.021023 /mol) (1.61019 J/eV)(b) For every 12 grams of carbon we require 32 grams of oxygen, the total is 44 grams. The totalmass required is then 40/12 that of carbon alone. The energy production is then(3.3107 J/kg)(12/44) = 9106 J/kg.(c) The sun would burn for(21030 kg)(9106 J/kg)= 4.61010 s.(3.91026 W)Thats only 1500 years!E51-42 The rate of fusion events is(5.31030 W)= 4.561042 /s.(7.27106 eV)(1.61019 J/eV)The carbon is then produced at a rate(4.561042 /s)(0.012 kg/mol)/(6.021023 /mol) = 9.081016 kg/s.The process will be complete in(4.61032 kg)= 1.6108 y.(9.081016 kg/s)(3.15107 s/y)E51-43 (a) For the reaction d-d,Q = [2(2.014102) (3.016029) (1.008665)](931.5 MeV) = 3.27 MeV.(b) For the reaction d-d,Q = [2(2.014102) (3.016029) (1.007825)](931.5 MeV) = 4.03 MeV.(c) For the reaction d-t,Q = [(2.014102) + (3.016049) (4.002603) (1.008665)](931.5 MeV) = 17.59 MeV.321E51-44 One liter of water has a mass of one kilogram. The number of atoms of 2 H is(0.00015 kg)(6.021023 /mol)= 4.51022 .(0.002 kg/mol)The energy available is(3.27106 eV)(1.61019 J/eV)(4.51022 )/2 = 1.181010 J.The power output is then(1.181010 J)= 1.4105 W(86400 s)E51-45Assume momentum conservation, thenp = pn or v n /v = m /mn .The ratio of the kinetic energies is thenKnmn v 2mn== 4.2Km vmnThen K n = 4Q/5 = 14.07 MeV while K = Q/5 = 3.52 MeV.E51-46 The Q value isQ = (6.015122 + 1.008665 3.016049 4.002603)(931.5 MeV) = 4.78 MeV.Combine the two reactions to get a net Q = 22.37 MeV. The amount of 6 Li required isN = (2.61028 MeV)/(22.37 MeV) = 1.161027 .The mass of LiD required ism=P51-1(1.161027 )(0.008 kg/mol)= 15.4 kg.(6.021023 /mol)(a) Equation 50-1 isR = R0 A1/3 ,where R0 = 1.2 fm. The distance between the two nuclei will be the sum of the radii, orR0 (140)1/3 + (94)1/3 .The potential energy will beU====1 q1 q2,4 0 re2(54)(38),4 0 R0 (140)1/3 + (94)1/3(1.601019 C)2211,4(8.851012 C2 /Nm2 )(1.2 fm)253 MeV.(b) The energy will eventually appear as thermal energy.322P51-2 (a) Since R = R0 3 A, the surface area a is proportional to A2/3 . The fractional change insurface area is(a1 + a2 ) a0(140)2/3 + (96)2/3 (236)2/3== 25 %.a0(236)2/3(b) Nuclei have a constant density, so there is no change in volume.(c) Since U Q2 /R, U Q2 / 3 A. The fractional change in the electrostatic potential energy isU1 + U2 U0(54)2 (140)1/3 + (38)2 (96)1/3 (92)2 (236)1/3== 36 %.U0(92)2 (236)1/3P51-3(a) There are(2.5 kg)(6.021023 mol1 )= 6.291024(239g/mol)atoms in 2.5 kg of239Pu. If each atom releases 180 MeV, then(180 MeV)(6.291024 )/(2.61028 MeV/megaton) = 44 kilotonis the bomb yield.P51-4(a) In an elastic collision the nucleus moves forward with a speed ofv = v02mn,mn + mso the kinetic energy when it moves forward isK =4m2mn mm 2nv0=K,2 (m + mn )2(mn + m)2where we can write K because in an elastic collision whatever energy kinetic energy the nucleuscarries o had to come from the neutron.(b) For hydrogen,K4(1)(1)== 1.00.K(1 + 1)2For deuterium,K4(1)(2)== 0.89.K(1 + 2)2For carbon,K4(1)(12)== 0.28.K(1 + 12)2For lead,K4(1)(206)== 0.019.K(1 + 206)2(c) If each collision reduces the energy by a factor of 10.89 = 0.11, then the number of collisionsrequired is the solution to(0.025 eV) = (1106 eV)(0.11)N ,which is N = 8.323P51-5The radii of the nuclei are3R = (1.2 fm) 7 = 2.3 fm.The using the derivation of Sample Problem 51-5,K=P51-6(3)2 (1.61019 C)2= 1.4106 eV.16(8.851012 F/m)(2.31015 m)(a) Add up the six equations to get12C+1 H +13 N +13 C +1 H +14 N +1 H +15 O +15 N +1 H 13N + +13 C + e+ + +14 N + +15 O + +15 N + e+ + +12 C +4 He.Cancel out things that occur on both sides and get1H +1 H +1 H +1 H + e+ + + + + e+ + +4 He.(b) Add up the Q values, and then add on 4(0.511 MeV for the annihilation of the two positrons.P51-7 (a) Demonstrating the consistency of this expression is considerably easier than derivingit from rst principles. From Problem 50-4 we have that a uniform sphere of charge Q and radiusR has potential energy3Q2U=.20 0 RThis expression was derived from the fundamental expressiondU =1 q dq.4 0 rFor gravity the fundamental expression isdU =Gm dm,rso we replace 1/4 0 with G and Q with M . But like charges repel while all masses attract, so wepick up a negative sign.(b) The initial energy would be zero if R = , so the energy released isU=3GM 23(6.71011 Nm2 /kg2 )(2.01030 kg)2== 2.31041 J.5R5(7.0108 m)At the current rate (see Sample Problem 51-6), the sun would bet=(2.31041 J)= 5.91014 s,(3.91026 W)or 187 million years old.324P51-8(a) The rate of fusion events is(3.91026 W)= 9.31037 /s.(26.2106 eV)(1.61019 J/eV)Each event produces two neutrinos, so the rate is1.861038 /s.(b) The rate these neutrinos impinge on the Earth is proportional to the solid angle subtendedby the Earth as seen from the Sun:r2(6.37106 m)2== 4.51010 ,24R4(1.501011 m)2so the rate of neutrinos impinging on the Earth is(1.861038 /s)(4.51010 ) = 8.41028 /s.P51-9(a) Reaction A releases, for each d(1/2)(4.03 MeV) = 2.02 MeV,Reaction B releases, for each d(1/3)(17.59 MeV) + (1/3)(4.03 MeV) = 7.21 MeV.Reaction B is better, and releases(7.21 MeV) (2/02 MeV) = 5.19 MeVmore for each N .P51-10(a) The mass of the pellet ism=4(2.0105 m)3 (200 kg/m3 ) = 6.71012 kg.3The number of d-t pairs isN=(6.71012 kg)(6.021023 /mol)= 8.061014 ,(0.005 kg/mol)and if 10% fuse then the energy release is(17.59 MeV)(0.1)(8.061014 )(1.61019 J/eV) = 230 J.(b) Thats(230 J)/(4.6106 J/kg) = 0.05 kgof TNT.(c) The power released would be (230 J)(100/s) = 2.3104 W.325E52-1 (a) The gravitational force is given by Gm2 /r2 , while the electrostatic force is given byq 2 /4 0 r2 . The ratio is=4(8.851012 C2 /Nm2 )(6.671011 Nm2 /kg2 )(9.111031 kg)2,(1.601019 C)2=4 0 Gm2q22.41043 .Gravitational eects would be swamped by electrostatic eects at any separation.(b) The ratio is=4(8.851012 C2 /Nm2 )(6.671011 Nm2 /kg2 )(1.671027 kg)2,(1.601019 C)2=4 0 Gm2q28.11037 .E52-2 (a) Q = 938.27 MeV 0.511 MeV) = 937.76 MeV.(b) Q = 938.27 MeV 135 MeV) = 803 MeV.E52-3 The gravitational force from the lead sphere is4Gme RGme M=.R23Setting this equal to the electrostatic force from the proton and solving for R,R=16 23e22,0 Gme a0or3(1.61019 C)216 2 (8.851012 F/m)(6.671011 Nm2 /kg2 )(11350 kg/m3 )(9.111031 kg)(5.291011 m)2which means R = 2.851028 m.E52-4 Each takes half the energy of the pion, so=E52-5(1240 MeV fm)= 18.4 fm.(135 MeV)/2The energy of one of the pions will beE=(pc)2 + (mc2 )2 =(358.3 MeV)2 + (140 MeV)2 = 385 MeV.There are two of these pions, so the rest mass energy of the 0 is 770 MeV.E52-6 E = mc2 , so = (1.5106 eV)/(20 eV) = 7.5104 .The speed is given byv = c 1 1/ 2 c c/2 2 ,where the approximation is true for large . Thenv = c/2(7.5104 )2 = 2.7102 m/s.326E52-7 d = ct = hc/2E. Thend=(1240 MeV fm)= 2.16103 fm.2(91200 MeV)E52-8 (a) Electromagnetic.(b) Weak, since neutrinos are present.(c) Strong.(d) Weak, since strangeness changes.E52-9 (a) Baryon number is conserved by having two p on one side and a p and a 0 on theother. Charge will only be conserved if the particle x is positive. Strangeness will only be conservedif x is strange. Since it cant be a baryon it must be a meson. Then x is K + .(b) Baryon number on the left is 0, so x must be an anti-baryon. Charge on the left is zero, so xmust be neutral because n is neutral. Strangeness is everywhere zero, so the particle must be n.(c) There is one baryon on the left and one on the right, so x has baryon number 0. The chargeon the left adds to zero, so x is neutral. The strangeness of x must also be 0, so it must be a 0 .E52-10neutral.neutronpossibleThere are two positive on the left, and two on the right. The anti-neutron must then beThe baryon number on the right is one, that on the left would be two, unless the antihas a baryon number of minus one. There is no strangeness on the right or left, exceptthe anti-neutron, so it must also have strangeness zero.E52-11 (a) Annihilation reactions are electromagnetic, and this involves s.s(b) This is neither weak nor electromagnetic, so it must be strong.(c) This is strangeness changing, so it is weak.(d) Strangeness is conserved, so this is neither weak nor electromagnetic, so it must be strong.E52-12 (a) K0 e+ + e ,(b) K0 + + 0 ,(c) K0 + + + + ,(d) K0 + + 0 + 0 ,E52-13 (a) 0 p + + .(b) n p + e+ + e .(c) + + + + .(d) K + .E52-14E52-15 From top to bottom, they are ++ , + , 0 , + , 0 , 0 , , , , and .E52-16 (a) This is not possible.(b) uuu works.E52-17 A strangeness of +1 corresponds to the existence of an anti-quark, which has a chargesof +1/3. The only quarks that can combine with this anti-quark to form a meson will have charges of-1/3 or +2/3. It is only possible to have a net charge of 0 or +1. The reverse is true for strangeness-1.327E52-18 Put bars over everything. For the anti-proton, uudZ, for the anti-neutron, udd.quarksucdcsccccucdcsE52-19 Well construct a table:Q0-1-10011S00-10001C-1-1-10111particleD0DDscD0D+D+sE52-20 (a) Write the quark content out then cancel out the parts which are the same on bothsides:dds udd + d,uso the fundamental process iss u + d + u.(b) Write the quark content out then cancel out the parts which are the same on both sides:d ud + d,suso the fundamental process is u + d + u.s(c) Write the quark content out then cancel out the parts which are the same on both sides:ud + uud uus + u,sso the fundamental process isd + d s + .s(d) Write the quark content out then cancel out the parts which are the same on both sides: + udd uud + d,uso the fundamental process is u + u.E52-21 The slope is(7000 km/s)= 70 km/s Mpc.(100 Mpc)E52-22 c = Hd, sod = (3105 km/s)/(72 km/s Mpc) = 4300 Mpc.E52-23 The question should read What is the...The speed of the galaxy isv = Hd = (72 km/s Mpc)(240 Mpc) = 1.72107 m/s.The red shift of this would then be = (656.3nm)1 (1.72107 m/s)2 /(3108 m/s)2= 695 nm.1 (1.72107 m/s)/(3108 m/s)328E52-24 We can approximate the red shift as = 0 /(1 u/c),so0u=c 1=c 1(590 nm)(602 nm)= 0.02c.The distance isd = v/H = (0.02)(3108 m/s)/(72 km/s Mpc) = 83 Mpc.E52-25 The minimum energy required to produce the pairs is through the collision of two 140MeV photons. This corresponds to a temperature ofT = (140 MeV)/(8.62105 eV/K) = 1.621012 K.This temperature existed at a timet=(1.51010 s1/2 K)2= 86 s.(1.621012 K)2E52-26 (a) 0.002 m.(b) f = (3108 m/s)/(0.002 m) = 1.51011 Hz.(c) E = (1240 eV nm)/(2106 nm) = 6.2104 eV.E52-27 (a) Use Eq. 52-3:t=(1.51010 sK)2= 91012 s.(5000 K)2Thats about 280,000 years.(b) kT = (8.62105 eV/K)(5000 K) = 0.43 eV.(c) The ratio is(109 )(0.43 eV)= 0.457.(940106 eV)P52-1The total energy of the pion is 135 + 80 = 215 MeV. The gamma factor of relativity is = E/mc2 = (215 MeV)/(135 MeV) = 1.59,so the velocity parameter is=1 1/ 2 = 0.777.The lifetime of the pion as measured in the laboratory ist = (8.41017 s)(1.59) = 1.341016 s,so the distance traveled isd = vt = (0.777)(3.00108 m/s)(1.341016 s) = 31 nm.329P52-2(a) E = K + mc2 and pc =pc =E 2 (mc2 )2 , so(2200 MeV + 1777 MeV)2 (1777 MeV)2 = 3558 MeV.Thats the same asp=(3558106 eV)(1.61019 J/eV) = 1.901018 kg m/s(3108 m/s).(b) qvB = mv 2 /r, so p/qB = r. Thenr=P52-3(1.901018 kg m/s)= 9.9 m.(1.61019 C)(1.2 T)(a) Apply the results of Exercise 45-1:E=(1240 MeV fm)= (4.281010 MeV/K)T.(2898 m K)T(b) T = 2(0.511 MeV)/(4.281010 MeV/K) = 2.39109 K.P52-4(a) Since = 01 2,1we have1 2 1,1=0orz=1 2 + 1.1Now invert,z(1 ) + 1 =1 2,(z + 1)2 (1 )2 = 1 2 ,(z 2 + 2z + 1)(1 2 + 2 ) = 1 2 ,22(z + 2z + 2) 2(z 2 + 2z + 1) + (z 2 + 2z) = 0.Solve this quadratic for , and=z 2 + 2z.+ 2z + 2z2(b) Using the result,=(4.43)2 + 2(4.43)= 0.934.(4.43)2 + 2(4.43) + 2(c) The distance isd = v/H = (0.934)(3108 m/s)/(72 km/s Mpc) = 3893 Mpc.330P52-5(a) Using Eq. 48-19,E = kT lnn1.n2Here n1 = 0.23 while n2 = 1 0.23, thenE = (8.62105 eV/K)(2.7 K) ln(0.23/0.77) = 2.8104 eV.(b) Apply the results of Exercise 45-1:=P52-6(1240 eV nm)= 4.4 mm.(2.8104 eV)(a) Unlimited expansion means that v Hr, so we are interested in v = Hr. ThenHrH 2 r33H 2 /8G=2GM/r,= 2G(4r3 /3),= .(b) Evaluating,3[72103 m/s (3.0841022 m)]2 (6.021023 /mol)= 5.9/m3 .(0.001 kg/mol)8(6.671011 N m2 /kg2 )P52-7 (a) The force on a particle in a spherical distribution of matter depends only on the mattercontained in a sphere of radius smaller than the distance to the center of the spherical distribution.And then we can treat all that relevant matter as being concentrated at the center. If M is the totalmass, thenr3M = M 3,Ris the fraction of matter contained in the sphere of radius r < R. The force on a star of mass m adistance r from the center isF = GmM /r2 = GmM r/R3 .This force is the source of the centripetal force, so the velocity isv = ar = F r/m = r GM/R3 .The time required to make a revolution is thenT =2r= 2vR3 /GM .Note that this means that the system rotates as if it were a solid body!(b) If, instead, all of the mass were concentrated at the center, then the centripetal force wouldbeF = GmM/r2 ,sov=ar =F r/m =GM/r,and the period would beT =2r= 2v331r3 /GM .P52-8 We will need to integrate Eq. 45-6 from 0 to min , divide this by I(T ), and set it equal toz = 0.2109 . Unfortunately, we need to know T to perform the integration. Writing what we doknow and then letting x = hc/kT ,z===m2c2 hd15c2 h3,5 k4 T 45hc/kT 12 e015c2 h3 2k 4 T 4 xm x3 dx,x2 5 k 4 T 4 h3 c2 e 115 x3 dx. 4 xm ex 1The result is a small number, so we expect that xm is fairly large. We can then ignore the 1 inthe denominator and then writex3 ex dxz 4 /15 =xmwhich easily integrates toz 4 /15 xm 3 exm .The solution isx 30,soT =(2.2106 eV)= 8.5108 K.(8.62105 eV/K)(30)332...

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