f5ch29 - 776 CHAPTER 29 THE MAGNETIC FIELD CHAPTER 29...

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Unformatted text preview: 776 CHAPTER 29 THE MAGNETIC FIELD CHAPTER 29 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. (a) +z ; (b) x; (c) FB = 0 (a) 2, then tie of 1 and 3 (zero); (b) 4 (a) +z and z tie, then +y and y tie, then +x and x tie (zero); (b) +y (a) electron; (b) clockwise (a) all tie; (b) 1 and 4 tie , then 2 and 3 tie y Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. (a) all tie; (b) 1 and 2 (charge is negative) tie of (a), (b), and (c), and then (d) (zero) (a) no, v and FB must be perpendicular; (b) yes; (c) no, B and FB must be perpendicular (a) upper plate; (b) lower plate; (c) out of the page (a) FE ; (b) FB 2, 5, 6, 9, 10 (a) right; (b) right (a) a and c, a at higher potential; (b) b and d, b at higher potential; (c) no Hall voltage (a) negative; (b) equal; (c) equal; (d) half a circle into page: a, d, e; out of page: b, c, f (the particle is negatively charged) (a) B1 ; (b) B1 into page, B2 out of page; (c) less 1i, 2e, 3c, 4a, 5g, 6j , 7d, 8b, 9h, 10f , 11k all (b) CHAPTER 29 THE MAGNETIC FIELD 777 15. 16. 17. all tie downward (a) positive; (b) (1) and (2) tie, then (3), which is zero Solutions to Exercises & Problems One way to do this is through Eq. 29-2: FB = qv B, which gives [ ] ML= 2 M [B ] = [qFv] = (Q)(LTT) = QT : ][ = (a) Use eq. 29-3: FB = jqjvB sin = (+3:2 10 19 C)(550 m/s)(0:045 T)(sin 52 ) = 6:2 10 18 N. (b) a = FB =m = (6:2 10 18 N)=(6:6 10 27 kg) = 9:5 108 m/s2 . (c) Since it is perpendicular to v, FB does not do any work on the particle. Thus from the work-energy theorem both the kinetic energy and the speed of the particle remain unchanged. (a) FB; max = qjv Bjmax = qvB sin(90 ) = qvB = (1:60 10 19 C)(7:20 106 m/s) (83:0 10 3 T) = 9:56 10 14 N; FB; min = qjv Bjmin = qvB sin(0) = 0: (b) a = FB =me = qvB sin =me , so the angle between v and B is 1 1E 2E 3E m a (9:11 10 31 kg)(4:90 1014 m/s2 ) e 1 = sin qvB = sin (1:60 10 16 C)(7:20 106 m/s)(83:0 10 3 T) = 0:267 : 4E (a) The magnitude of the magnetic force on the proton is given by FB = evB sin , where v is the speed of the proton, B is the magnitude of the magnetic eld, and is the angle between the particle velocity and the eld when they are drawn with their tails at the same point. Thus :50 10 17 B v = eBFsin = (1:60 10 196C)(2:60 10N 3 T) sin 23:0 = 4:00 105 m/s : 778 CHAPTER 29 THE MAGNETIC FIELD (b) The kinetic energy of the proton is 1 1 K = 2 mv2 = 2 (1:67 10 27 kg)(4:00 105 m/s)2 = 1:34 10 This is (1:34 10 16 J)=(1:60 10 19 J/eV) = 835 eV. 5P 16 J : (a) The diagram shows the electron traveling to the north. The magnetic eld is into the page. The right-hand rule tells us that v B is to the west, but since the electron is negatively charged the magnetic force on it is to the east. W (b) Use F = me a, with F = evB sin . Here v is the speed of the electron, B is the magnitude of the magnetic eld, and is the angle between the electron velocity and the magnetic eld. The velocity and eld are perpendicular to each other so = 90 and sin = 1. Thus a = evB=me . The electron speed can be found from its kinetic energy. Since K = 1 mv2 , 2 N v E S v= Thus r 2K = m e s 2(12:0 103 eV)(1:60 10 9:11 10 31 kg 19 J/eV) = 6:49 107 m/s : = 6:27 1014 m/s2 : a = evB = (1:60 10 m e 19 C)(6:49 107 m/s)(55:0 10 6 T) 9:11 10 31 kg (c) The electron follows a circular path. Its acceleration is given by v2 =R, where R is the radius of the path. Thus 2 7 2 R = va = (6:49 1014m/s)2 = 6:72 m : 6:27 10 m/s The solid curve on the diagram is the path. Suppose it N subtends the angle at the center. l (= 0:200 m) is the length of the tube and d is the de ection. The right triangle yields d = R R cos and l = R sin . Thus R cos = R d and R sin = l. Square both these equations and add the R (R results to obtain R2 = p d)2 + l2 , or d2 2Rd + l2 = 0. l The solution is d = R R2 l2 . The plus sign corresponds d E to an angle of 180 ; the minus sign corresponds to the correct solution. p Since l is much less than R, use the binomial theorem to expand R2 l2 . The rst two terms are R 1 l2 =R, so d = 1 l2 =R and when numerical values are substituted, d = 2 2 0:00298 m is obtained. CHAPTER 29 THE MAGNETIC FIELD 779 (a) FB = qv B = q(vx i + vy j) (Bx i + By j) = q(vx By vy Bx )k = ( 1:6 10 19 C)[(2:0 106 m/s)( 0:15 T) (3:0 106 m/s)(0:030 T)] = (6:2 10 14 N)k: So the magnitude of FB is 6:2 1014 N, and FB points in the positive z direction. (b) All you need to do is to change in sign in q. So now FB still has the same magnitude but points in the negative z diretion. Let B = By j + Bz k, then FB = qv B = q(vx i + vy j) (By j + Bz k) = q(vy Bz i vx Bz j + vx By k) = Fx i + Fy j. Thus qvy Bz = Fx , qvx Bz = Fy and qvx By = 0. The last equation gives By = 0. So 7P 6P Fk B = Bz k = x 8E 15 4: = ( 1:60 ( 102 10 k) N 3 m/s) = (0:75k) T : qvy 19 C)(35 10 (a) The net force on the proton is given by F = FE + FB = q E + q v B = (1:6 10 19 C)[(4:0 V/m) k + (2000 m/s) j ( 2:5 mT) i] = (1:4 10 18 N) k : (b) In this case F = FE + FB = qE + qv B = (1:6 10 19 C)[( 4:0 V/m) k + (2000 m/s) j ( 2:5 mT) i] = (1:6 10 19 N) k : = qE + qv B = (1:6 10 19 C)[(4:0 V/m) i + (2000 m/s) j ( 2:5 mT) i] = (6:4 10 19 N) i + (8:0 10 19 N) k : 19 N)2 + (8:0 10 19 N)2 = 1:0 10 18 N : (a) In this case F = FE + FB The magnitude of FB is now q p 2 2 2 FB = FBx + FBy + FBz = (6:4 10 9E (a) The total force on the electron is F = e(E + v B), where E is the electric eld, B is the magnetic eld and v is the electron velocity. The magnitude of the magnetic 780 CHAPTER 29 THE MAGNETIC FIELD force is evB sin , where is the angle between the velocity and the eld. Since the total force must vanish, B = E=v sin . The force is the smallest it can be when the eld is perpendicular to the velocity and = 90 . Then B = E=v. Use K = 1 mv2 to nd the 2 speed: v= Thus r 2K = m e s 2(2:5 103 eV)(1:60 10 9:11 10 31 kg 19 J/eV) = 2:96 107 m/s : The magnetic eld must be perpendicular to both the electric eld and the velocity of the electron. (b) A proton will pass unde ected if its velocity is the same as that of the electron. Both the electric and magnetic forces reverse direction, but they still cancel. (a) Let F = q(E + v B) = 0, we get vB sin(v; B) = E . So vmin = E=B = (1:50 103 V/m)=(0:400 T) = 3:75 103 m/s: Apply F = q(E + v B) = me a to solve for E: E= 11P* 10E E = 10 103 V/m = 3:4 10 4 T : B = v 2:96 107 m/s me a + B v q 31 kg)(2:00 1012 2 = (9:11 10 1:60 10 19 C m/s ) i + (400 T) i [(12:0 km/s) j + (15:0 km/s)k] = ( 11:4 i 6:00 j + 4:80 k)(V/m) : (a) Let F = q(E + v B) = 0. Note that v ? B so jv Bj = vB . Thus 100 V=(20 10 3 m) B=E =p E =p v 2me K 2(9:11 10 31 kg)(1:0 103 V)(1:60 10 = 2:7 10 4 T : 13P 19 C) 12P Since the total force, given by F = e(E + v B), vanishes, the electric eld E must be perpendicular to both the particle velocity v and the magnetic eld B. The magnetic CHAPTER 29 THE MAGNETIC FIELD 781 eld is perpendicular to the velocity so v B has magnitude vB and the magnitude of the electric eld is given by E = vB . Since the particle has charge e and is accelerated through p a potential dierence V , 1 mv2 = eV and v = 2eV=m. Thus 2 eV 10 19 C)(10 103 V) E = B 2m = (1:2 T) 2(1::60u)(1:661 10 27 kg/u) = 6:8 105 V/m : (6 0 14P r s Use Eq. 29-12 to solve for V : (23 A)(0:65 T) iB V = nle = (8:47 1028 =m3 )(150 m)(1:6 10 15E 19 C) = 7:4 10 6 V : From F = q(E + v B) = 0 we get v = E=B . Also note that J = nev, so J = ne(E=B ); or n = JB=eE . (a) Use Eq. 29-10: vd = E=B = (10 10 6 V=1:0 10 2 m)=(1:5 T) = 6:7 10 4 m/s. (b) From the result of 15E 16P iB i n = JB = AeE = Aev eE 3: A = (1:0 10 2 m)(10 10 6 m)(10:60 10 = 2:8 1029 = m3 : 19 C)(6:7 10 4 m/s) a y (c) Refer to the diagram to the right. If the electrons move out of the page and the magnetic eld B is in the positive x direction, then Va > Vb . x b B=ne: Since J = Ec = Ec = and J = neE=B (see 15E), we get Ec = = neE=B , i.e., E=Ec = 17P 782 CHAPTER 29 THE MAGNETIC FIELD (b) E = B = 0:65 T 3 28 = m3 )(1:60 10 19 C)(1:69 10 8 m) = 2:84 10 : Ec ne (8:47 10 For a free charge q inside the metal strip with velocity v we have F = q(E + v B), so (3 10 9 V E v = B = jVx By j=dxy = (1:20 10 :90T)(0:850V) 10 2 m) = 0:382 m/s : 3 19E 18P (a) In the accelerating process the electron loses potential energy eV and gains the same amount of kinetic energy. Since it starts from rest, 1 me v2 = eV and 2 v= r 2eV = m e s 2(1:60 10 19 C)(350 V) = 1:11 107 m/s : 9:11 10 31 kg (b) The electron travels with constant speed around a circle. The magnetic force on it has magnitude FB = evB and its acceleration is v2 =R, where R is the radius of the circle. Newton's second law yields evB = me v2 =R, so 31 107 R = me v = (9:11 10 19kg)(1:11 10 3m/s) = 3:16 10 4 m : eB (1:60 10 C)(200 T) 20E Since the magnetic eld is perpendicular to the particle velocity the magnitude of the magnetic force is given by FB = evB and the acceleration of the electron has magnitude a = FB =me , where v is the speed of the electron and me is its mass. B is the magnitude of the magnetic eld. Since the electron is traveling with uniform speed in a circle, its acceleration is a = v2 =r, where r is the radius of the circle. Thus evB=me = v2 =r and 10 m/s) B = me v = (9:11 10 10kg)(1:3 35 m) = 2:1 10 5 T : 19 C)(0: er (1:60 31 6 (a) Use Eq. 29-16 to calculate r: 8 10 31 kg)(10%)(3: r = me v = (9:11 (1:60 10 19 C)(000 10 m/s) = 3:4 10 4 m : qB :50 T) 21P CHAPTER 29 THE MAGNETIC FIELD 783 (b) 10 10 1 K = 2 me v2 = (9:11 2(1:6 kg)(3:0J/eV) m/s) = 2:6 103 eV : 19 10 31 7 2 Use Eq. 29-16 to calculate B : 27 107 m/s) B = me v = (1:67 10 19kg)(1::0 106 m) = 1:6 10 8 T : qr (1:60 10 C)(6 37 22E (a) From K = 1 me v2 we get 2 3 eV)(1: 10 v = 2K = 2(1:20 10 :11 106031 kg me 9 23E r s 19 eV/J) = 2:05 107 m/s : (b) From r = me v=qB we get 31 107 m/s) B = me v = (9:11 10 19kg)(2:05 10 2 m) = 4:67 10 4 T : qr (1:60 10 C)(25:0 (c) (d) T = 1=f = (1:31 107 Hz) 1 = 7:63 10 8 s: 24E :07 107 v f = 2r = 22(25:0 10m/s = 1:31 107 Hz : 2 m) The period of revolution for the iodine ion is T = 2r=v = 2m=Bq, which gives 3 T)(1:60 10 19 C)(1 29 3 m = BqT = (45:0 10(7)(2)(1:66 10 27 kg=:u) 10 s) = 127 u : 2 (a) 25E 2 m)(1 19 eB v = rqB = 4200 u = 2(4:50 10 u)(1:66:60 10 kgC)(1:20 T) = 2:60 106 m/s : m : (4:00 10 27 =u) (b) T = 2r=v = 2(4:50 10 2 m)=(2:60 106 m/s) = 1:09 10 7 s. 784 CHAPTER 29 THE MAGNETIC FIELD (c) (d) V = K=q = 1:40 105 eV=2e = 7:00 104 V: (a) (b) 26E 27 kg u)(2:60 6 2 1 K = 2 m v2 = (4:00 u)(1:66 :10 10 =19 J/eV) 10 m/s) = 1:40 105 eV : 2(1 60 10 6 T)(1: 10 Bq f = 2m = (35:0 (9:11 106031 kg) 2 e 19 C) = 9:78 105 Hz : p p me v = 2me K = 2(9:11 10 31 kg)(100 eV)(1:60 10 19 J/eV) = 0:964 m : r = qB qB (1:60 10 19 C)(35:0 10 6 T) Orient the magnetic eld so it is perpendicular to the plane of page. Then the electron will travel with constant speed around a circle in the plane of the page. The magnetic force on an electron has magnitude FB = evB , where v is the speed of the electron and B is the magnitude of the magnetic eld. If r is the radius of the circle, the acceleration of the electron has magnitude a = v2 =r. Newton's second law yields evB = me v2 =r, so the radius of the circle is given by r = me v=eB . The kinetic energy of the electron is K = 1 me v2 , so 2 p v = 2K=me . Thus r r me 2K = 2me K : r = eB m e2 B 2 e This must be less than d, so r 2me K d or 27E If the electrons are to travel as shown in Fig. 29-40 the magnetic eld must be out of the page. Then the magnetic force is toward the center of the circular path. Let vk = v cos . The electron will proceed with a uniform speed vk in the direction of B while undergoing uniform cirular motion with frequency f in the direction perpendicular to B : f = eB=2me . The distance d is then v d = v T = k = (v cos )2me = 2(1:5 107 m/s)(9:11 10 31 kg)(cos 10 ) (1:60 10 19 C)(1:0 10 3 T) k 28P e2 B 2 r me B 2e2 dK : 2 f eB = 0:53 m : CHAPTER 29 THE MAGNETIC FIELD 785 From 26E, part (b) we see that r = 2mK=qB , or K = (rqB )2 =2m / q2 m 1 . Thus (a) K = (q =qp )2 (mp =m )Kp = (2)2 (1=4)Kp = Kp = 1:0 MeV; (b) Kd = (qd =qp )2 (mp =md )Kp = (1)2 (1=2)Kp = 1:0 MeV=2 = 0:50 MeV: (a) Since K = qV we p Kp = Kdp 1 K (as q = 2Kd = 2Kp ). have =2 (b) and (c) Since r = 2mK=qB / mK=q, we have 30P 29P p s md Kd qp rp = (2:00 u)Kp r = 10p2 cm = 14 cm ; rd = m K q (1:00 u)Kp p p p d s s p p r = m K qq rp = (1:(4:00 u)K 2) erep = 10 2 cm = 14 cm : mp Kp 00 u)(K = 2 s Use the result of the previous problem: r / mK=qB . Thus 31P p r m K qp r = 4:0 u erp = r ; r = m K q p p 1:0 u 2e p p s r md Kd qp r = 2:0 u erd = p2r : rd = m K q d p 1:0 u e p p d s The equation of motion for the proton is F = q v B = q (vx i + vy j + vz k) B i = qB (vz j 32 dv dv dv = mp a = mp dtx i + dty j + dtz k : vy k) 8 dvx > =0 > dt > > > < dvy > dt = !vz > > dv > z > = !v ; : y dt where ! = eB=mp . The solution is vx = v0x , vy = v0y cos !t and vz = v0y sin !t; i.e., v(t) = v0x i + v0y cos(!t) j v0y (sin !t)k. Thus 786 CHAPTER 29 THE MAGNETIC FIELD (a) From m = B 2 qx2 =8V we have m = (B 2 q=8V )(2xx). Here x = 8V m=B 2 q, which we sbustitute into the experssion for m to obatin 33P p r B 2 q s 8mV m = 8V 2 B 2 q x = B mq x : 2V (b) 2V x = m mq B 66 = (37 u 35 u)(1::50 T 10 0 = 8:2 10 3 m : 34P 27 kg=u) s s 2(7:3 103 V) (36 u)(1:66 10 27 kg=u)(1:60 10 19 C) (a) Solve B from m = B 2 qx2 =8V : V B = 8qxm : 2 Evaluate this expression using x = 2:00 m: s 3 V)(3 92 10 25 B = 8(100:20 10 10 19 :C)(2:00 m)2kg) = 0:495 T : (3 (b) Let N be the number of ions that are separated by the machine per unit time. The current is i = qN and the mass that is separated per unit time is M = mN , where m is the mass of a single ion. M has the value s 10 6 M = 100 3600 s kg = 2:78 10 8 kg/s : Since N = M=m we have 19 8 i = qM = (3:20 103:92C)(2:78 10 kg/s) = 2:27 10 2 A : m 10 25 kg (c) Each ion deposits an energy of qV in the cup, so the energy deposited in time t is given by E = NqV t = iqV t = iV t : q CHAPTER 29 THE MAGNETIC FIELD 787 For t = 1:0 h, E = (2:27 10 2 A)(100 103 V)(3600 s) = 8:17 106 J : To obtain the second expression, i=q was substituted for N . The condition for the ion beam to be undeviated is v = E=B . Then as the ion beam enters the magnetic eld B0 we have r = mv=(qB 0 ) = mE=(qBB 0 ), or q=m = E=(rBB 0 ). 36P 35P (a) If v is the speed of the positron then v sin is the component of its velocity in the plane that is perpendicular to the magnetic eld. Here is the angle between the velocity and the eld (89 ). Newton's second law yields eBv sin = me (v sin )2 =r, where r is the radius of the orbit. Thus r = (me v=eB ) sin . The period is given by 10 31 T = v 2r = 2me = (1:2(9:11 19 C)(0:kg)T) = 3:6 10 sin eB 60 10 10 10 s : The equation for r was substituted to obtain the second expression for T . (b) The pitch is the distance traveled along the line of the magnetic eld in a time interval of one period. Thus p = vT cos . Use the kinetic energy to nd the speed: K = 1 me v2 2 means v= Thus r 2K = m e s 2(2:0 103 eV)(1:60 10 9:11 10 31 kg 19 J/eV) = 2:651 107 m/s : p = (2:651 107 m/s)(3:58 10 31 10 s) cos 89 = 1:7 10 4 m : 7 (c) The orbit radius is R = me v sin = (9:11 10 :60 kg)(2:651 1010m/s) sin 89 = 1:5 10 3 m : eB (1 10 19 C)(0: T) 37P (a) q, from conservation of charges. (b) Each of the two particles will move in the same circular path, initially going in the opposite direction. After traveling half of the circular path they will collide. So the time is given by t = T=2 = m=Bq. 788 CHAPTER 29 THE MAGNETIC FIELD (a) The radius r of the circular orbit is given by r = p=eB , where B is the magnitude of the magnetic eld. The relp ativisitic expression p = mp v= 1 v2 =c2 must be used for the magnitude of the momentum. Here v is the speed of the proton, mp is its mass, and c is the speed of light. Hence r = pmp v 2 2 : eB 1 v =c Square both sides and solve for v. The result is 38P N B F v : v = q reBc m2 c2 + r2 e2 B 2 p S Substitute r = 6:37 106 m (the radius of the Earth), e = 1:6022 10 19 C (the charge on a proton), B = 41 10 6 T, mp = 1:6726 10 27 kg (the mass of a proton), and c = 2:9979 108 m/s to obtain v = 2:9977 108 m/s. The gure to the right depicts the path of the electron in the case of minumum B . When the electron is just about to strike the top plate its velocity v must be parallel to the plate or it will hit the plate. The magnetic force FB exerted on the plate is now directed away from the top plate, with magnitude FB = eBv (Note that B ? v). Here v satis es K = 1 me v 2 = eV , or v = p2eV=me . Now, FB 2 must be greater or equal to the electric force FE = eE = eV=d exerted on the electron, which tends to push the electron toward the top plate. Thus 39P FE v V FB d path of electron or eV FB = eBv = eB 2m FE = eE = eV ; d e r e B medV : 2 2 r (a) 40E :2 T) qB : 10 19 fosc = 2m = (1260 67 10C)(1kg) = 1:8 107 Hz : 27 (1: p CHAPTER 29 THE MAGNETIC FIELD 789 (b) From r = mp v=qB = 2mp K=qB we have 2 [(0: :60 10 K = (rqB ) = 2(1:6750 m)(127 kg)(1:60 C)(1:19T)] = 1:7 107 eV : 2mp 10 10 J/eV) 2 19 2 p (a) (b) 41E m/s) 10 r = mp v = (1:67 (1:60 kg)(3:00 10 T) =10 = 0:22 m : 19 C)(1:4 qB 10 27 8 v 7 m/s fosc = 2r = 3:00 (0:10 m) = 2:1 107 Hz : 2 22 (a) Since K = 1 mv2 = 1 m(2Rfosc )2 / m, 2 2 42P m 1 1 Kp = mp Kd = 2 Kd = 2 (17 MeV) = 8:5 MeV : d 1 1 Bp = 2 Bd = 2 (1:6 T) = 0:80 T : (b) (c) Since K / B 2 =m, 0 Kp = md Kd = 2Kd = 2(17 MeV) = 34 MeV : mp (d) Since fosc = Bq=(2m) / m 1 , m fosc; d = m d fosc; p = 2(12 106 s 1 ) = 2:4 107 Hz: P m K = m Kd = 2Kd = 2(17 MeV) = 34 MeV ; m d q 1 d B = m q Bd = 2 2 (1:6 T) = 1:6 T ; d 0 (Since B = Bd = 1:6 T) ; K = K = 34 MeV (e) Now 790 CHAPTER 29 THE MAGNETIC FIELD and q m 2 d fosc; = a m fosc, d = 2 4 (12 106 s 1 ) = 1:2 107 Hz : d The speed of the deuteron before it breaks up is 43P rqB = (50 10 2 m))(1:60 10 19 C)(1:5 T) = 3:6 105 m/s : vd = m 2(1:66 10 27 kg) d 2 2 2 Since Kd = 1 md vd = Kn + Kp = 1 mn vn + 1 mp vp and md vd = mn vn + mp vp (conservation 2 2 2 of linear momentum), we nd vp vn vd , where we noted that mn mp md =2. So the neutron will proceed with speed vn vd = 3:6 105 m/s, moving in a straight line tangent to the original path of the deuteron (since qn = 0); while the proton will move in a circle with radius m q 1 p d rp = m q rd = 2 (1)(50 cm) = 25 cm : d p 44P Its average energy during the accelerating process is 8:3 MeV. The radius of the orbit is p given by r = mv=qB , where v is the deuteron's speed. Since this is given by v = 2K=m, the radius is r m 2K = 1 p2Km : r = qB m qB For the average energy 6 19 J/eV)(3 r = 2(8:3 10 eV)(1:60 10 10 C)(1:57 T):34 10 (1:60 19 Approximate the total distance by the number of revolutions times the circumference of the orbit corresponding to the average energy. This should be a good approximation since the deuteron receives the same energy each revolution and its period does not depend on its energy. The deuteron accelerates twice in each cycle and each time it receives an energy of qV = 80 103 eV. Since its nal energy is 16:6 MeV the number of revolutions it makes is 16:6 106 eV n = 2(80 103 eV) = 104 : p 27 kg) = 0:375 m : The total distance traveled is about n2r = (104)(2)(0:375) = 2:4 102 m. CHAPTER 29 THE MAGNETIC FIELD 791 The magnitude of the magnetic force on the wire is given by FB = iLB sin , where i is the current in the wire, L is the length of the wire, B is the magnitude of the magnetic eld, and is the angle between the current and the eld. In this case = 70 . Thus 45E FB = (5000 A)(100 m)(60:0 10 6 T) sin 70 = 28:2 N : Apply the right-hand rule to the vector product FB = iL B to show that the force is to the west. The magnetic force on the wire must be upward and have a magnitude equal to the gravitational force mg on the wire. Apply the right-hand rule to show that the current must be from left to right. Since the eld and the current are perpendicular to each other the magnitude of the magnetic force is given by FB = iLB , where L is the length of the wire. The condition that the tension in the supports vanish is iLB = mg, which yields 46P mg 8 m/s2 i = LB = (0:0130 kg)(9::440 T) ) = 0:467 A : (0:620 m)(0 47E FB = iBL sin = (13:0 A)(1:50 T)(1:80 m)(sin 35:0 ) = 20:1 N : 48P FB = iL B = iL i (By j + Bz k) = iL( Bz j + By k) = (0:50 A)(0:50 m)[ (0:010 T) j + (0:0030 T)k] = ( 2:5 10 3 j + 0:75 10 3 k) N : The magnetic force on the wire is FB = idB , pointing to the left. Thus v = at = (FB =m)t = idBt=m, to the left (i.e., away from the generator). (a) Since B is uniform, FB 50P 49P = Z wire Z idL B = i wire dL B = iLab B ; 792 CHAPTER 29 THE MAGNETIC FIELD where we noted that wire dL = Lab , Lab being the displacement vector from a to b. (b) Now Lab = 0, so FB = iLab B = 0: Use dFB = idL B, where dL = dx i and B = Bx i + By j. Thus FB 51P R idL B = idx i (Bx i + By j) = i By dxk xi xi Z 3:0 2 dx) (m mT) k = ( 0:35 N) k : (8:0x = ( 5:0 A) = 1:0 Z Z xf Z xf (a) From FB = iLB we get 52P FB 10 3 N i = LB = (3:0 m)(1010 10 6 T) = 3:3 108 A : (b) P = i2 R = (3:3 108 A)2 (1:0 ) = 1:0 1017 W. (c) It is totally unrealistic because of the huge current and the consequent high power loss. The magnetic force must push horizontally on the rod to overcome the force of friction. But it can be oriented so it also pulls up on the rod and thereby reduces both the normal force and the force of friction. Suppose the magnetic eld makes the angle with the vertical. The diagram to the right shows the view from the end of the FB B sliding rod. The forces are also shown: FB is the force of the N magnetic eld if the current is out of the page, mg is the force of gravity, N is the normal force of the stationary rails on the rod, and f is the force of friction. Notice that the magnetic f force makes the angle with the horizontal. When the rod is on the verge of sliding, the net force acting on it is zero and the magnitude of the frictional force is given by f = s N , where s is the coecient of static friction. The magnetic eld is perpendicular to the wire so the magnitude of the magnetic force is given by FB = iLB , where i is the current in the rod and L is the length of the rod. mg The vertical component of Newton's second law yields 53P N + iLB sin mg = 0 CHAPTER 29 THE MAGNETIC FIELD 793 and the horizontal component yields iLB cos s N = 0 : Solve the second equation for N and substitute the resulting expression into the rst equation, then solve for B . You should get mg B = iL(cos s+ sin ) : s The minimum value of B occurs when cos + s sin is a maximum. Set the derivative of cos + s sin equal to zero and solve for . You should get = tan 1 s = tan 1 (0:60) = 31 . Now evaluate the expression for the minimum value of B : 0:60(1:0 kg)(9:8 m/s2 ) Bmin = (50 A)(1:0 m)(cos 31 + 0:60 sin 31 ) = 0:10 T : 54P (a) Denote the vectors representing the three sids of the triangle as a, b, and c. The direction of each vector is so chosen that it is in the same direction as the current ow. Choose a coordinate system as shown. Then B = B i. We have y B b 8 F = ic B = (ic) i B i = 0; > c > > F = ib B = ibB sin k > b > > > > = (4:00 A)(1:20 m)(0:0750 T)(50:0=130)k < > = (0:138 N)k ; > > > F = ia B = ( iaB sin )k = ( ibB sin )k > a > > > : = FB = (0:138 N)k : a i c O x (b) The sum of forces on the three sides is Fa + Fb + Fc = ( iaB sin )k +(ibB sin )k = 0, where we noted that a sin = b sin = ab=c. This is in fact a special case of 50P, part (b). The situation is shown in the left diagram in the next page. The y axis is along the hinge and the magnetic eld is in the positive x0 direction. A torque around the hinge is associated with the wire opposite the hinge and not with the other wires. The force on this wire is in the positive z 0 direction and has magnitude F = NibB , where N is the number of turns. 55P 794 CHAPTER 29 THE MAGNETIC FIELD y a x' a x b F i B x' x z z' F z' The right diagram shows the view from above. The magnitude of the torque is given by = Fa cos = NibBa cos = 20(0:10 A)(0:10 m)(0:50 10 3 T)(0:050 m) cos 30 = 4:33 10 3 N m : Use the right-hand rule to show that the torque is directed downward, in the negative y direction. 56P If N closed loops are formed from the wire of length L, the circumference of each loop is L=N , the radius of each loop is R = L=2N , and the area of each loop is A = R2 = (L=2N )2 = L2 =4N 2 . For maximum torque, orient the plane of the loops parallel to the magnetic eld, so the dipole moment is perpendicular to the eld. The magnitude of the torque is then L2 2 = NiAB = (Ni) 4N 2 B = iL B : 4N To maximize the torque take N to have the smallest possible value, 1. Then = iL2 B=4. Replace the current loop of arbitrary shape with an assembly of adjacent long, thin, rectangular loops, each with N turns and carrying a current i owing in the same sense as the original loop of arbitrary shape. See the gure to the right. As the widths of these rectangles shrink to in nitesimally small values the assembly gives a current distribution equivalent to that of the original loop. The magnitude of the torque exerted by B on the nth rectangular loop of area An is given by n = NiB sin An . 57P CHAPTER 29 THE MAGNETIC FIELD 795 Thus for the whole assembly = X n n = NiB X n An = NiAB sin : The total magnetic force on the loop L is FB 58P =i I where we noted that H L (dL B) = i( I L dL) B = 0 ; L dL = 0. I If B is not a constant, however, then the equality (dL B) = ( I L L dL) B is not necessarily valid so FB is not always zero. Use max = j Bjmax = B = ia2 B , and note that i = qf = qv=2a: So 59P qv 1 max = 2a a2 B = 2 qvaB : Consider an in nitesimal segment of the loop, of length B ds. The magnetic eld is perpendicular to the segment so the magnetic force on it is has magnitude dF = iB ds. The diagram shows the direction of the force for the segment on the far right of the loop. The horizontal compodF nent of the force has magnitude dFh = (iB cos ) ds and points inward toward the center of the loop. The vertical i component has magnitude dFv = (iB sin ) ds and points upward. Now sum the forces on all the segments of the loop. The horizontal component of the total force vanishes since each segment of wire can be paired with another, diametrically opposite, segment. The horizontal components of these forces are both toward the center of the loop and thus in opposite directions. The vertical component of the total force is 60P Fv = iB sin ds = (iB sin )2a : Z 796 CHAPTER 29 THE MAGNETIC FIELD Notice the i, B , and have the same value for every segment and so can be factored from the integral. (a) The current in the galvanometer should be 1:62 m A when the potential dierence across the resistor-galvanometer combination is 1:00 V. The potential dierence across the galvanometer alone is iRg = (1:62 10 3 A)(75:3 ) = 0:122 V so the resistor must be in series with the galvanometer and the potential dierence across it must be 1:00 V 0:122 V = 0:878 V. The resistance should be R = (0:878 V)=(1:62 10 3 A) = 542 . (b) The current in the galvanometer should be 1:62 m A when the total current in the resistor-galvanometer combination is 50:0 m A. The resistor should be in parallel with the galvanometer and the current through it should be 50 m A 1:62 m A = 48:38 m A. The potential dierence across the resistor is the same as that across the galvanometer, 0:122 V, so the resistance should be R = (0:122 V)=(48:38 10 3 A) = 2:52 . 62P 61P In the diagram to the right is the magnetic dipole moment of the wire loop and B is the magnetic eld. Since the plane of the loop is parallel to the incline the dipole moment is normal to the incline. The forces acting on the cylinder are the force of gravity mg, acting downward from the center of mass, the normal force of the incline N , acting perpendicularly to the incline through the center of mass, and the force of friction f , acting up the incline at the point of contact. Take the x axis to be positive down the incline. Then the x component of Newton's second law for the center of mass yields B mg sin f = ma : For purposes of calculating the torque take the axis of the cylinder to be the axis of rotation. The magnetic eld produces a torque with magnitude B sin and the force of friction produces a torque with magnitude fr, where r is the radius of the cylinder. The rst tends to produce an angular acceleration in the counterclockwise direction and the second tends to produce an angular acceleration in the clockwise direction. Newton's second law for rotation about the center of the cylinder, = I, gives fr B sin = I : Since you want the current that holds the cylinder in place, set a = 0 and = 0, then use one equation to eliminate f from the other. You should obtain mgr = B . The loop is CHAPTER 29 THE MAGNETIC FIELD 797 rectangular with two sides of length L and two of length 2r, so its area is A = 2rL and the dipole moment is = NiA = 2NirL. Thus mgr = 2NirLB and mg = (0:250 kg)(9:8 m/s2 ) = 2:45 A : i = 2NLB 2(10:0)(0:100 m)(0:500 T) 63E (a) The magnitude of the magnetic dipole moment is given by = NiA, where N is the number of turns, i is the current in each turn, and A is the area of a loop. In this case the loops are circular, so A = r2 , where r is the radius of a turn. Thus 2 30 A m2 i = Nr2 = (160)(:)(0:0190 m)2 = 12:7 A : (b) The maximum torque occurs when the dipole moment is perpendicular to the eld (or the plane of the loop is parallel to the eld). It is given by max = B = (2:30 Am2 )(35:0 10 3 T) = 8:05 10 2 N m. (a) From = NiA = ir2 we get 64E :00 1022 i = r2 = 8(3500 103J/T2 = 2:08 109 A: m) 65E (a) = NAi = r2 i = (0:150 m)2 (2:60 A) = 0:184 A m2 . (b) = j Bj = B sin = (0:184 A m2 )(12:0 T)(sin 41:0 ) = 1:45 N m: (a) The area A of the current loop is A = 1 (30 cm)(40 cm) = 6:0 102 cm2 , so 2 66E = iA = (5:0 A)(6:0 10 2 m2 ) = 0:30 A m2 : (b) = B sin = (0:30 A m2 )(80 103 T)(sin 90 ) = 2:4 10 2 N m : 798 CHAPTER 29 THE MAGNETIC FIELD (a) Use = B. From the diagram to the right it is clear that points at the 20-minute mark. So the time interval is 20 min. (b) The torque is given by 67E 5 min B r = j Bj = B sin( ; B) = NiAB sin 90 = Nir2 B = 6(2:0 A)(0:15 m)2 (70 10 3 T) = 5:9 10 2 N m : 68E 20 min i (a) = (b) Now X n 2 2 in An = r1 i1 + r2 i2 = (7:00 A)[(0:300 m)2 + (0:200 m)2 ] = 2:86 A m2 : 2 2 = r1 i1 r2 i2 = (7:00 A)[(0:300 m)2 (0:200 m)2 ] = 1:10 A m2 : The magnetic dipole moment is = (0:60 i 0:80 j), where = NiA = Nir2 = 1(0:20 A)(0:080 m)2 = 4:02 10 4 A m2 . Here i is the current in the loop, N is the number of turns, A is the area of the loop, and r is its radius. (a) The torque is 69P = B = (0:60 i 0:80 j) (0:25 i + 0:30 k) = [(0:60)(0:30)( i k) (0:80)(0:25)( j i) (0:80)(0:30)( j k)] = [ 0:18 j + 0:20 k 0:24 i] : Here i k = j, j i = k, and j k = i were used. We also used i i = 0. Substitute the value for to obtain = [ 0:97 10 4i 7:2 10 4 j + 8:0 10 4 k] N m : (b) The potential energy of the dipole is given by U = B = (0:60 i 0:80 j) (0:25 i + 0:30 k) = (0:60)(0:25) = 0:15 = 6:0 10 4 J : CHAPTER 29 THE MAGNETIC FIELD 799 Here i i = 1, i k = 0, j i = 0, and j k = 0 were used. Let a = 30:0 cm, b = 20:0 cm, and c = 10:0 cm. From the hint given we write 70P ~ Thus = (0:150)2 + (0:300)2 A m2 = 0:335 A m2 , and is in the yz plane at angle to the +y direction, where p = 1 + 2 = iab( k) + iab( j) = ia(c j bk) = (5:00 A)(0:300 m)[(0:100 m) j (0:200 m)k] = (0:150 j 0:300k) A m2 : 0:300 y 1 = tan = tan 0:150 = 63:4 : x 1 ...
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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.

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