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Unformatted text preview: CHAPTER 32 MAGNETISM AND MATTER 865 CHAPTER 32 Answer to Checkpoint Questions 1. (d), (b), (c), (a) (zero) 2. (a) 2; (b) 1 3. (a) away; (b) away; (c) less 4. (a) toward; (b) toward; (c) less 5. a, c, b, d (zero) 6. tie of b, c, and d, then a Answer to Questions 1. (a) a, c, f ; (b) bar gh 2. (a) and (c) 3. supplied 4. b 5. (a) all down; (b) 1 up, 2 down, 3 zero 6. (a) increase; (b) increase 7. (a) 1 up, 2 up, 3 down; (b) 1 down, 2 up, 3 zero 8. (a) 1 down, 2 down, 3 up; (b) 1 up, 2 up, 3 zero 9. (a) rightward; (b) leftward 10. counterclockwise 11. (a) decreasing; (b) decreasing 12. (a) counterclockwise; (b) clockwise; (c) no direction (no induced B) 13. (a) tie of a and b, then c, d; (b) none (plate lacks circular symmetry, so B is not 14. 15.
tangent to a circular loop); (c) none (a) rightward; (b) leftward; (c) into 1=4 866 CHAPTER 32 MAGNETISM AND MATTER 16. 1, a; 2, b; 3, c and d Solutions to Exercises & Problems (a) 1E (b) The sign of B dA for every dA on the side of the paper cylinder is negative. (c) No, because Gauss's law for magnetism applies to an enclosed surface only.H In fact if we include the top and bottom of the cylinder to form an enclosed surface S then s B dA = 0 will be valid, as the ux through the open end of the cylinder near the magnet is positive. Use 6 =1 Bn = 0 to obtain n B6 =
5 X 2E P n=1 Bn = ( 1 Wb + 2 Wb 3 Wb + 4 Wb 5 Wb) = +3 Wb : 3P Use Gauss' law for magnetism: B dA = 0. Write B dA = 1 + 2 + C , where 1 is the magnetic ux through the rst end mentioned, 2 is the magnetic ux through the second end mentioned, and C is the magnetic ux through the curved surface. Over the rst end the magnetic eld is inward, so the ux is 1 = 25:0 Wb. Over the second end the magnetic eld is uniform, normal to the surface, and outward, so the ux is 2 = AB = r2 B , where A is the area of the end and r is the radius of the cylinder. It value is 2 = (0:120 m)2 (1:60 10 3 T) = +7:24 10 5 Wb = +72:4 Wb : H H CHAPTER 32 MAGNETISM AND MATTER 867 Since the three uxes must sum to zero, C = 1 2 = 25:0 Wb 72:4 Wb = 47:4 Wb : The minus sign indicates that the ux is inward through the curved surface. From Gauss' law for magnetism the ux through S1 is equal to that through S2 , the portion of the xz plane that lies within the cylinder. Here the normal direction of S2 is in the positive yaxis. So B (S1 ) = B (S2 ) =
Z 4P* S1 r S2 r Z r 0 i 1 =2 r 2 2r =2 Z r B (x)L dx
r x Bleft (x)L dx x L dx = 0iL ln 3 : The horizontal component of the Earth's magnetic eld is given by Bh = B cos i , where B is the magnitude of the eld and i is the inclination angle. Thus 16 B B = cosh = cos T = 55 T : i 73 The ux through Arizona is = Br A = (43 10 6 T)(295; 000 km2 )(103 m/km)2 = 1:3 107 Wb ; inward. By Gauss' law this is equal to the negative value of the ux 0 through the rest of the surface of the Earth. So 0 = 1:3 107 Wb, outward.
2 (a) From = iA = iRe we get 5E 6E 7E 1022 J/T i = R2 = 8:037 106 m)2 = 6:3 108 A : (6: e 868 CHAPTER 32 MAGNETISM AND MATTER (b) Yes, because far away from the Earth the elds of both the Earth itself and the current loop are dipole elds. If these two dipoles cancel out then the net eld will be zero. (c) No, because the eld of the current loop is not that of a magnetic dipole in the region close to the loop. (a) 8P
0 cos 2 + 0 sin 2 B= h v m m 4r3 2r3 0 qcos2 + 4 sin2 = 0 q1 + 3 sin2 ; = 4r3 m m 4r3 m
q B2 + B2 = s where cos2 m + sin2 m = 1 was used. (b) B ( =2r3 ) sin tan i = Bv = ( 0=4r3 ) cos m = 2 tan m : h
0 m (a) At the magnetic equator m = 0, so
7 m/A)(8: 22 2 0 B = 4r3 = (4 10 4T(6:37 1000m)310 A m ) = 3:10 10 5 T 6 9P and i = tan 1 (2 tan m ) = tan 1 (0) = 0: (b) At m = 60 : and i = tan 1 (2 tan 60 ) = 74 . (c) At the north magnetic pole m = 90:0 , so and i = tan 1 (2 tan 90 ) = 90 . 0 q1 + 3 sin2 = (3:10 10 5 )p1 + 3 sin2 60 = 5:6 10 5 T B = 4r3 m 0 q1 + 3 sin2 = (3:1 10 5 )p1 + 3(1:00)2 = 6:20 10 5 T B = 4r3 m 10P (a) At a distance r from the center of the Earth, the magnitude of the magnetic eld is given by 0 q1 + 3 sin2 ; B = 4r3 m CHAPTER 32 MAGNETISM AND MATTER 869 where is the Earth's dipole moment and m is the magnetic latitude. The ratio of the eld magnitudes for two dierent distances at the same latitude is
3 B2 = r1 : 3 B1 r2 Take B1 to be at the surface and B2 to be half of B1 . Set r1 equal to the radius Re of the Earth and r2 equal to Re + h, where h is altitude at which B is half its value at the surface. Thus 3 1 = Re : 2 (R + h)3 Take the cube root of both sides and solve for h. You should get
e h= 21=3 1 Re = 21=3 1 (6370 km) = 1660 km : (b) Use the expression for B obtained in 8P, part (a). For maximum B set sin m = 1. Also, r = 6370 km 2900 km = 3470 km. Thus 0 q1 + 3 sin2 Bmax = 4r3 m 7 m/A)(8: 1022 2 p = (4 10 4T(3:47 1000m)3 A m ) 1 + 3(1)2 = 3:83 10 4 T : 6 (c) The angle between the magnetic axis and the rotational axis of the Earth is 11:5 , so m = 90:0 11:5 = 78:5 at the Earth's north geographic pole. Also r = Re = 6370 km. Thus 0 q1 + 3 sin2 B = 4R3 m E p (4 10 7 T m/A)(8:0 1022 J/T) 1 + 3 sin2 78:5 = 6:11 10 5 T ; = 4(6:37 106 m)3 and i = tan 1 (2 tan 78:5 ) = 84:2 . A plausible explaination to the discrepancy between the calculated and measured values of the Earth's magnetic eld is that the formulas we obtained in 8P are based on dipole approximation, which does not accurately represent the Earth's actual magnetic eld distribution on or near its surface. (Incidentally, the dipole approximation does get better when we calculate the Earth's magnetic eld far from its center.) 11E Use Eq. 327 to obtain U = (s;z B ) = B s;z , where s;z = eh=4me = B (see Eqs. 324 and 325). Thus U = B [B ( B )] = 2B B = 2(9:27 10
24 J/T)(0:25 T) = 4:6 10 24 J : 870 CHAPTER 32 MAGNETISM AND MATTER Use Eq. 3211: orb;z = ml B . (a) For ml = 1, orb;z = (1)(9:27 10 24 J/T) = 9:27 10 24 J/T. (b) For ml = 2, orb;z = ( 2)(9:27 10 24 J/T) = 1:85 10 23 J/T. (a) 12E 13E
19 C)(8:99 109 2 2 e E = 4 r2 = (1:60 10 (5:2 10 11 m)2 N m = C ) = 5:3 1011 N/C : 0 7 0 p 4 B = 2 3 = (4 10 2T :m/A)(1:11 10 r (5 2 10 m)3 26 J/T) (b) Use Eq. 3029: = 2:0 10 2 T: (c) From Eq. 3210, orb = eh=4me = B = 9:27 10 24 J/T = 6:6 102 : p p p 1:4 10 26 J/T (a) Since ml = 0, Lorb;z = ml h=2 = 0. (b) Since ml = 0, orb;z = ml B = 0. (c) Since ml = 0, from Eq. 3212 U = orb;z Bext = ml B Bext = 0. (d) Regardless of the value of ml , for the spin part 14E U = s;z B = B B = (9:27 10
(d) Now ml = 3 so
l Lorb;z = mh = ( 3)(6:6262 10 2 24 J/T)(35 mT) = 3:2 10 25 J : 34 J s) = 3:16 10 34 J s and orb;z = ml B = ( 3)(9:27 10 U = orb;z Bext = (2:78 10 24 J/T) = 2:78 10 23 J/T : The potential energy associated with the electron's orbital magnetic moment is now
23 J/T)(35 10 3 T) = 9:73 10 25 J ; while the potential energy associated with the electron spin, being independent of ml , remains the same: 3:2 10 25 J. CHAPTER 32 MAGNETISM AND MATTER 871 For a given value of l, ml varies from l to +l. Thus in our case l = 3, and the number of dierent ml 's is 2l + 1 = 2(3) + 1 = 7. (a) Since Lorb;z / ml , there are a total of seven dierent values of Lorb;z . (b) Similarly, since orb;z / ml , there are also a total of seven dierent values of orb;z . (c) Since Lorb;z = ml h=2, the greatest allowed value of Lorb;z is given by jml jmax h=2 = 3h=2; while the least allowed value is given by jml jmin h=2 = 0. (d) Similar to part (c), since orb;z = ml B , the greatest allowed value of orb;z is given by jml jmax B = 3eh=4me ; while the least allowed value is given by jml jmin B = 0. (e) From Eqs. 323 and 329 the z component of the net angular momentum of the electron is given by l Lnet;z = Lorb;z + Ls;z = mh + ms h : 2 2 For the maximum value of Lnet;z let ml = [ml ]max = 3 and ms = 1 . Thus 2 : 1 [Lnet;z ]max = 3 + 2 2h = 325h : (f) Since the maximum value of Lnet;z is given by [mJ ]max h=2 with [mJ ]max = 3:5 (see the last part above), the number of allowed values for the z component of Lnet;z is given by 2[mJ ]max + 1 = 2(3:5) + 1 = 8. (a) and (b) The potential energy of the atom in association with the presence an external magnetic eld Bext is given by Eqs. 3211 and 3212: U = orb Bext = orb;z Bext = ml B Bext : For level E1 there is no change in energy as a result of the introduction of Bext so U / ml = 0, meaning that ml = 0 for this level. For level E2 the single level splits into a triplet (i.e., three separate ones) in the presence of Bext , meaning that there are three dierent values of ml . The middle one in the triplet is unshifted from the original value of E2 so its ml must be equal to 0. The other two in the triplet then correspond to ml = 1 and m1 = +1, respectively. (c) For any pair of adjacent levels in the triplet jml j = 1. Thus the spacing is given by U = j( ml B B )j = jml jB B = B B = (5:79 10 5 eV/T)(0:50 T) = 2:9 10 5 eV ; which is equivalent to 4:6 10 24 J. 15E 16P 17P (a) The period of rotation is T = 2=! and in this time all the charge passes any xed point near the ring. The average current is i = q=T = q!=2 and the magnitude of the magnetic dipole moment is 1 = iA = q! r2 = 2 q!r2 : 2 872 CHAPTER 32 MAGNETISM AND MATTER (b) Curl the ngers of your right hand in the direction of rotation. Since the charge is positive your thumb points in the direction of the dipole moment. It is the same as the direction of the angular momentum of the ring. 18E
(a) S S i B (b) Since the material is diamagnetic its magnetic moment is opposed to the direction of B, so points away from the bar magnet. (c) In order to produce a magnetic moment which points away from the bar magnet the current in the ring must be running clockwise when viewed along the symmetry axis from the location of the bar magnet. (d) Consider the magnetic force dFB exerted by the bar magnet on an innitesimal segment dL of the ring. Use dFB = idL B to convince yourself that the net magnetic force on the ring points away from the magnet. An electric eld with circular eld lines is induced as the magnetic eld is turned on. Suppose the magnetic eld increases linearly from zero to B in time t. According to Eq. 3224 the magnitude of the electric eld at the orbit is given by 19P r r E = 2 dB = 2 B ; dt t
where r is the radius of the orbit. The induced electric eld is tangent to the orbit and changes the speed of the electron, the change in speed being given by erB t = 2m : 2 t e The average current associated with the circulating electron is i = ev=2r and the dipole moment is 2 ev = 1 evr : = Ai = r 2r 2 eE e v = at = m t = m e e r B CHAPTER 32 MAGNETISM AND MATTER 873 The change in the dipole moment is
2 2 1 erB = 1 er v = 2 er 2m = e4r B : 2 me e Let E = 3 kT = j B ( B)j = 2B to obtain 2 4B = 4(1:0 10 23 J/T)(0:50 T) = 0:48 K : T = 3k 3(1:38 10 23 J/K) 20E 21E The magnetization is the dipole moment per unit volume, so the dipole moment is given by = MV , where M is the magnetization and V is the volume of the cylinder. Use V = r2 L, where r is the radius of the cylinder and L is its length. Thus = Mr2 L = (5:30 103 A/m)(0:500 10 2 m)2 (5:00 10 2 m) = 2:08 10 2 J/T : 22E This problem is analogous to 18E. (a) S S i B (b) Since the material is paramagnetic its magnetic moment is in the same direction with B, so points toward the bar magnet. (c) In order to produce a magnetic moment which points toward the bar magnet the current in the ring must be running counterclockwise when viewed along the axis from the location of the bar magnet. 874 CHAPTER 32 MAGNETISM AND MATTER (d) Consider the magnetic force dFB exerted by the bar magnet on an innitesimal segment dL of the ring. Use dFB = idL B to convince yourself that the net magnetic force on the ring points toward the magnet. For the measurements carried out the largest ratio of the magnetic eld to the temperature is (0:50 T)=(10 K) = 0:050 T/K. Look at Fig. 3210 to see if this is in the region where the magnetization is a linear function of the ratio. It is quite close to the origin, so we conclude that the magnetization obeys Curie's law. (a) From Fig. 3210 we see that B=T = 0:50 when M=Mmax = 50%. So B = 0:50 T = (0:50)(300 K) = 150 K. (b) Similarly, now B=T = 2:0 so B = (2:0)(300) = 600 T. (c) Not yet. (a) Again from Fig. 3210, for M=Mmax = 50% we have B=T = 0:50. So T = B=0:50 = 2=0:50 = 4 K. Now B=T = 2:0 so T = 2=2:0 = 1 K. (a) A charge e traveling with uniform speed v around a circular path of radius r takes time T = 2r=v to complete one orbit, so the average current is The magnitude of the dipole moment is this times the area of the orbit: Since the magnetic force, with magnitude evB , is centripetal, Newton's law yields evB = me v2 =r, so r = me v : eB Thus 1 (ev) me v = 1 1 m v2 = Ke : = 2 eB B 2 e B 23P 24P 25P 26P e i = T = 2ev : r = 2ev r2 = evr : r 2 The magnetic force ev B must point toward the center of the circular path. If the magnetic eld is into the page, for example, the electron will travel clockwise around the circle. Since the electron is negative, the current is in the opposite direction, counterclockwise and, by the righthand rule for dipole moments, the dipole moment is out of the page. That is, the dipole moment is directed opposite to the magnetic eld vector. CHAPTER 32 MAGNETISM AND MATTER 875 counterclockwise around a circular orbit and the current is in the same direction. Thus the dipole moment is again out of the page, opposite to the magnetic eld. (c) The magnetization is given by M = e ne + i ni , where e is the dipole moment of an electron, ne is the electron concentration, i is the dipole moment of an ion, and ni is the ion concentration. Since ne = ni , we may write n for both concentrations. Substitute e = Ke =B and i = Ki =B to obtain 21 3 n M = B (Ke + Ki ) = 5:3 110 T m 6:2 10 20 J + 7:6 10 21 J = 310 A/m : :2 (a) (b) Notice that the charge canceled in the derivation of = Ke =B . Thus the relation = Ki =B holds for a positive ion. If the magnetic eld is into the page, the ion travels 27P NP () NP ( ) = N eB=KT e B=KT = N tanh B : M = P () + P ( ) kT eB=KT + e B=KT (b) For B kT (i.e., B=kT 1) we have eB=kT 1 B=kT , so B N[(1 + B=kT ) (1 B=kT )] = N2 B : M = N tanh kT (1 + B=kT ) + (1 B=kT ) kT (c) For B kT we have tanh(B=kT ) 1, so B N : M = N tanh kT (d) The following is a plot of tanh(x) as a function of x for = 1. Notice the resemblence between this plot and Fig. 3210. By adjusting properly you can t the curve in Fig. 310 with a tanh function.
1 tanh(x) ) 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 x 876 CHAPTER 32 MAGNETISM AND MATTER The Curie temperature for iron is 770 C. If x is the depth at which the temperature has this value, then 10 C + (30 C/km)x = 770 C or 10 x = 770 CC/km C = 25 km : 30 28E (a) The eld of a dipole along its axis is given by Eq. 30{29: where is the dipole moment and z is the distance from the dipole. Thus
7 m 23 B = (4 10 2T(10=A)(1:5 10 J/T) = 3:0 10 6 T : 10 9 m) (b) The energy of a magnetic dipole in a magnetic eld B is given by U = B = B cos , where is the angle between the dipole moment and the eld. The energy required to turn it endforend (from = 0 to = 180 ) is 29E 0 B = 2 z3 ; U = 2B = 2(1:5 10 23 J/T)(3:0 10 6 T) = 9:0 10 29 J = 5:6 10 10 eV : The mean kinetic energy of translation at room temperature is about 0:04 eV. Thus if dipoledipole interactions were responsible for aligning dipoles, collisions would easily randomize the directions of the moments and they would not remain aligned. The saturation magnetization corresponds to complete alignment of all atomic dipoles and is given by Msat = n, where n is the number of atoms per unit volume and is the magnetic dipole moment of an atom. The number of nickel atoms per unit volume is n = =m, where is the density of nickel and m is the mass of a single nickel atom, calculated using m = M=NA , where M is the atomic mass of nickel and NA is the Avogadro constant. Thus
3 23 n = NA = (8:90 g/cm )(6::02 10 atoms/mol) M 58 71 g/mol = 9:126 1022 atoms/cm3 = 9:126 1028 atoms/m3 : 30E The dipole moment of a single atom of nickel is A/m 4 70 sat = Mn = 9::126 10 28 m3 = 5:15 10 10
5 24 A m2 : CHAPTER 32 MAGNETISM AND MATTER 877 (a) The number of iron atoms in the iron bar is
3 cm)(1 0 2 ) N = (55(7:9 g/cm )(5:0:022 :10cm mol) = 4:3 1023 : 23 = :847 g/mol)=(6 31E Thus the dipole moment of the iron bar is = (2:1 10 23 J/T)(4:3 1023 ) = 8:9 A m2 : (b) = B sin 90 = (8:9 A m2 )(1:57 T) = 13 N m : 32P (a) If the magnetization of the sphere is saturated the total dipole moment is total = N, where N is the number of iron atoms in the sphere and is the dipole moment of an iron atom. We wish to nd the radius of an iron sphere with N iron atoms. The mass of such a sphere is Nm, where m is the mass of an iron atom. It is also given by 4R3 =3, where is the density of iron and R is the radius of the sphere. Thus Nm = 4R3 =3 and Substitute this into total = N to obtain Solve for R and obtain The mass of an iron atom is N = 4R : 3m
3 3 total = 4R : 3m 3mtotal 1=3 : R = 4 m = 56 u = (56 u)(1:66 10
So
" 27 kg/u) = 9:30 10 26 kg :
#1=3 26 22 J/T) = 1:8 105 m : R = 3(9:30 10 kg)(8:0 10 23 4(14 103 kg/m3 )(2:1 10 J/T) (b) The volume of the sphere is Vs = 43 R3 = 43 (1:82 105 m)3 = 2:53 1016 m3 and the volume of the Earth is Ve = 43 (6:37 106 m)3 = 1:08 1021 m3 ; 878 CHAPTER 32 MAGNETISM AND MATTER so the fraction of the Earth's volume that is occupied by the sphere is 2:53 1016 m3 = 2:3 10 5 : 1:08 1021 m3 From the way the wire is wound it is clear that P2 is the magnetic north pole while P1 is the south pole. (a) The de
ection will be toward P1 (away from the magnetic north pole). (b) As the electromagnet is turned on the magnetic us through the aluminum changes abruptly, causing a strong induced current which produces a magnetic eld opposite to that of the electromagnet. As a result the aluminum sample will be pushed toward P1 , away from the magnetic north pole of the bar magnet. As reaches a constant value, however, the induced current disappears and the aluminum sample, being paramagnetic, will move slightly toward P2 , the magnetic north pole of the electromagnet. (c) A magnetic north pole will now be induced on the side of the sample closer to P1 , and a magnetic south pole will appear on the other side. If the eld of the electromagnet is stonger near P1 then the sample will move toward P1 . The interacting potenial energy between the magnetic dipole of the compass and the Earth's magnetic eld is U = Be = Be cos , where is the angle between and Be . For small angle 2 = 1 k2 B ; U () = Be cos Be 1 2 e 2 where k = Be . Conservation of energy for the compass then gives 33P 34P 1 I d 2 + 1 k2 = const. : 2 dt 2 This is to be compared with the following expression for the mechanical energy of a springmass system: 1 m dx 2 + 1 kx2 = const. ; 2 dt 2 which yields ! = k=m. So by analogy, in our case
r r p B ! = k = Be = ml2 =e12 ; I I
which gives
2 !2 2 10 2 2 = ml B = (0:050 kg)(4:0 10 m) (45 rad/s) = 8:4 102 J/T : 6 T) 12 12(16 s e CHAPTER 32 MAGNETISM AND MATTER 879 (a) The magnitude of the toroidal eld is given by B0 = 0 nip , where n is the number of turns per unit length of toroid and ip is the current required to produce the eld (in the absence of the ferromagnetic material). Use the average radius (r = 5:5 cm) to calculate n: turns n = 2N = 2(5400 10 2 m) = 1:16 103 turns/m : r :5 Thus
3T B ip = 0 = (4 10 7 0:20 10 :16 103 m 1 ) = 0:14 A : T m/A)(1 0n 35P (b) If is the magnetic ux through the secondary coil then the magnitude of the emf induced in that coil is E = N (d=dt) and the current in the secondary is is = E =R, where R is the resistance of the coil. Thus i = N d :
s R dt The charge that passes through the secondary when the primary current is turned on is N Z d dt = N Z d = N : q = is dt = R dt R 0 R The magnetic eld through the secondary coil has magnitude B = B0 + BM = 801B0 , where BM is the eld of the magnetic dipoles in the magnetic material. The total eld is perpendicular to the plane of the secondary coil, so the magnetic ux is = AB , where A = 801r2 B0 : The radius r is (6:0 cm 5:0 cm)=2 = 0:50 cm and = 801(0:50 10 2 m)2 (0:20 10 3 T) = 1:26 10 5 Wb : Thus Z is the area of the Rowland ring (the eld is inside the ring, not in the region between the ring and coil). If r is the radius of the ring's cross section then A = r2 . Thus 5 q = 50(1:26 8:010 Wb) = 7:9 10 5 C : 36E Let R be the radius of a capacitor plate and r be the distance from axis of the capacitor. For points with r R the magnitude of the magnetic eld is given by B = 020 r dE dt 880 CHAPTER 32 MAGNETISM AND MATTER and for r R it is There are two values of r for which B = Bmax =2: one less than R and one greater. To nd the one that is less than R, solve The maximum magnetic eld occurs at points for which r = R and its value is given by either of the formulas above: Bmax = 0 0 R dE : 2 dt 2 B = 0 0rR dE : 2 dt 0 0 r dE = 0 0 R dE 2 dt 4 dt for r. The result is r = R=2 = (55:0 mm)=2 = 27:5 mm. To nd the one that is greater than R, solve 0 0 R2 dE = 0 0 R dE 2r dt 4 dt for r. The result is r = 2R = 2(55:0 mm) = 110 mm. 37E Use the result of part (b) in Sample Problem 324:
2 B = 0 0rR dE 2 dt (for r R) to solve for dE=dt: dE = 2Br dt 0 0 R2 2(2:0 7 T 3 = (4 10 7 T m/A)(8:10 10)(6:0C2=10 m2m) :0 10 3 m)2 12 85 N )(3 13 V/m s : = 2:4 10 38P (a) Use the result of part (a) in Sample Problem 324: where r = 0:80R and B = 020 r dE dt (for r R) ; dE = d V = 1 d V e t= = V0 e t= : dt dt d d dt 0 d CHAPTER 32 MAGNETISM AND MATTER 881 Here V0 = 100 V. Thus V0 e t= = 0 0 V0 r e t= B (t) = 020 r d 2d 7 T m/A)(8:85 10 12 C2=N m2 )(100 V)(0:80)(16 mm) e t=12 ms = (4 10 2(12 10 3 s)(5:0 mm) = (1:2 10 13 T) e t=12 ms :
The minus sign here is insignicant. (b) At time t = 3 , B (t) = (1:2 10
13 T)e 3= = 5:9 10 15 T. Apply Eq. 3228 to a circular loop of radius R centered at the center of the plates: I B ds = 2RB = dE = d (R2E ) = R2 dE
0 0 39P = Thus 0 0 R2 d d [V cos(!t + )] = dt m dt 0 0 dt 0 0 0 0 R2 Vm ! d dt sin(!t + ) : 2 RV Bmax = 0 02R Vm ! [sin(!t + )]max = 0 0d m f = RV2m f Rd cd 3 m)(150 V)(60 Hz) (30 10 = (3:0 108 m/s)2 (5:0 10 3 m) = 1:9 10 12 T : 1 (b) For r R B (r) / r, and for r R B (r) / r is shown below.
Bmax(r) /Bmax (R)
1 (see Eqs. 3238 and 3239). The plot 1/2 1/3 r (mm)
0 30 60 90 120 882 CHAPTER 32 MAGNETISM AND MATTER 40E The displacement current is given by id = 0 A dE ; dt
where A is the area of a plate and E is the magnitude of the electric eld between the plates. The eld between the plates is uniform, so E = V=d, where V is the potential dierence across the plates and d is the plate separation. Thus id = 0dA dV : dt
Now 0 A=d is the capacitance C of a parallelplate capacitor without a dielectric, so id = C dV : dt
Let the area plate be A and the plate separation be d. Use Eq. 3232: d d id = 0 ddtE = 0 dt (AE ) = 0 A dt V = 0dA dV ; d dt or 41E So we need to change the voltage dierence across the capacitor at the rate of 7:5 105 V/s. dV = id d = id = 1:5 A = 7:5 105 V/s : dt 0 A C 2:0 10 6 F 42E Consider an area A, normal to a uniform electric eld E. The displacement current density is uniform and normal to the area. Its magnitude is given by Jd = id =A. For this situation so id = 0 A dE ; dt
1 Jd = A 0 A dE = 0 dE : dt dt Use Eq. 3236: 43E id = 0 A dE : dt CHAPTER 32 MAGNETISM AND MATTER 883 Note that here A is the area over which a changing electric eld is present. In this case r > R so A = R2 . Thus dE = id = id = 2:0 A 12 2 (8:85 10 12 C2=N m2 )(0:10 m)2 = 7:2 10 V/m s : dt 0 A 0 R 44E Let the area of each circular plate be A and that of the central circular section be a, then A = R2 = 4 : a (R=2)2
Thus from Eqs. 3236 and 3237 the total discharge current is given by i = id = 4(2:0 A) = 8:0 A. 45P From Eq. 3232 d id = 0 ddtE = 0 A dE = 0 A dt (4:0 105 ) (6:0 104 t) dt 4 V/m s) = 0 A(6:0 10 = (8:85 10 12 C2=N m2 )(4:0 10 2 m2 )(6:0 104 V/m s) = 2:1 10 8 A : (b) If one draws a counterclockwise circular loop s around the plates then according to Eq. 3223 I B ds = 0id < 0 ; which means that B ds < 0. Thus B must be clockwise. (a) From Sample Problem 324 we know that B / r for r R and B / r 1 for r R. So the maximum value of B occurs at r = R, and there are two possible values of r at which the magnetic eld is 75% of Bmax . Denote these two values as r1 and r2 , where r1 < R and r2 > R. Then 0:75Bmax =Bmax = r1 =R, or r1 = 0:75R; and 0:75Bmax =Bmax = (r2 =R) 1 , or r2 = R=0:75 = 1:3R. (b) From Eqs. 3237 and 3243
7 m/A)(6 0 id Bmax = 2R = 20 i = (4 10(0T040 m) :0 A) = 3:0 10 5 T : R 2 : s 46P 884 CHAPTER 32 MAGNETISM AND MATTER (a) Use 47P H B ds = 0Ienclosed to nd
Jd r2 1 enclosed B = 0 I2r = 0 (2r ) = 2 0 Jd r = 1 (1:26 10 6 H/m)(20 A/m2 )(50 10 3 m) = 6:3 10 7 T : 2 (b) From we get id = Jd r2 = 0 ddtE = 0 r2 dE dt 20 A/m2 dE = Jd = 12 12 F/m = 2:3 10 V/m s : dt 0 8:85 10 (a) 48P
jid j = dE 0 dt = dE 0 A dt = (8:85 10 (b) id / dE=dt = 0: (c) 5 12 F/m)(1:6 m2 ) 4:5 10 N/C 6:0 105 N/C = 0:71 A: 4:0 10 6 s jid j = dE 0 A dt = (8:85 10 12 F/m)(1:6 m2 ) 4:0 105 N/C 15 10 6 s 10 10 6 s = 1:1 A : (a) At any instant the displacement current id in the gap between the plates equals the conduction current i in the wires. Thus id = i = 2:0 A. (b) 49P dE = 1 dE = id = 2:0 A 11 0 dt 12 F/m)(1:0 m)2 = 2:3 10 V/m s : dt 0 A 0 A (8:85 10
(c) 2 0 = id area enclosed by the path = (2:0 A) 0:50 m = 0:50 A : id area of each plate 1:0 m CHAPTER 32 MAGNETISM AND MATTER 885 (d) I B ds = 0i0d = (1:26 10 6 H/m)(0:50 A) = 6:3 10 7 T m : (a) (b) 50P 10 8 E = J = J = i = (1:62 5:00 10 m)(100 A) = 0:324 V/m : 6 m2 A d di id = 0 ddtE = 0 A dE = 0 A dt i = 0 dt dt A 12 F)(1:62 10 8 )(2000 A/s) = 2:87 10 = (8:85 10 16 A : 18 : (c) B ( due to id ) = 0 id =2r = id = 2:87 10 16 A = 2:87 10 B ( due to i) 0 i=2r i 100 A (a) At any instant the displacement current id in the gap between the plates equals the conduction current i in the wires. Thus imax = id max = 7:60 A. (b) Since id = 0 (dE =dt), 6A 7: dE = id max = 8:8560 1012 F/m = 8:59 105 V m/s : dt max 10 0 (c) According to 40E id = C dV = 0dA dV : dt dt Now the potential dierence across the capacitor is the same in magnitude as the emf of the generator, so V = Em sin !t and dV=dt = !Em cos !t. Thus and This means 51P id = 0 A!Em cos !t d id max = 0 A!Em : d d = 0iA!Em = (8:85 10 d max = 3:39 10 3 m ; where A = R2 was used. 12 F/m) (0:180 m)2 (130 rad/s)(220 V) 7:60 10 6 A 886
H CHAPTER 32 MAGNETISM AND MATTER (d) Use the AmpereMaxwell law in the form B ds = 0 Id , where the path of integration is a circle of radius r between the plates and parallel to them. Id is the displacement current through the area bounded by the path of integration. Since the displacement current density is uniform between the plates Id = (r2 =R2 )id , where id is the total displacement current between the plates and R is the plate radius. The eld lines are circles centered on the axis of the plates, so B is parallel to ds. The eld has constant magnitude around the H circular path, so B ds = 2rB . Thus 2rB = 0 R2 id and
0 id B = Rr : 2 2 12 T : 2 r The maximum magnetic eld is given by 7 6 i Bmax = 02d max r = (4 10 T m/A)(7:6 10 A)(0:110 m) = 5:16 10 R2 2(0:180 m)2 ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
 Fall '07
 GROUPTEST
 Magnetism

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