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f5ch31 - CHAPTER 31 INDUCTION AND INDUCTANCE 829 CHAPTER 31...

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Unformatted text preview: CHAPTER 31 INDUCTION AND INDUCTANCE 829 CHAPTER 31 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. 7. (a) and (b) tie, then (c) (zero) c and d tie, then a and b tie b, out; c, out; d, into; e, into (d) and (e) (a) 2, 3, 1 (zero); (b) 2, 3, 1 a and b tie, then c b, then d and e tie, and then a and c tie (zero) Answer to Questions 1. (a) all tie (zero); (b) all tie (nonzero); (c) 3, then tie of 1 and 2 (zero) 2. (a) all tie (zero; (b) 2, then tie of 1 and 3 (zero) 3. out 4. (a) 2, 6; (b) 4; (c) 1, 3, 5, 7 5. (a) into; (b) counterclockwise; (c) larger 6. (a) clockwise; (b) increasing and then decreasing 7. (a) leftward; (b) rightward 8. clockwise 9. c, a, b 10. d and c tie, then b, a 11. (a) 1, 3, 2; (b) 1 and 3 tie, then 2 12. c, b, a 13. (a), then tie of (b) and (c) 14. (c), (a), (b) 15. (a) more; (b) same; (c) same; (d) same (zero) 830 CHAPTER 31 INDUCTION AND INDUCTANCE 16. (a) all tie (zero); (b) 1 and 2 tie, then 3; (c) all tie (zero) 17. a 2, b 4, c 1, d 3 18. (a) all tie (independent of the current); (b) 1, then 2 and 3 tie Solutions to Exercises & Problems 1E B = Z B dA = BA cos 57 = (4:2 10 6 T)(2:5 m2 )(cos 57 ) = 5:7 10 5 Wb : 2E d d E = dB = d(BA) = A dB = A dt (0 in) = A0 n dt (i0 sin !t) dt dt dt = A0 ni0 ! cos !t : The magnetic eld is normal to the plane of the loop and is uniform over the loop. Thus at any instant the magnetic ux through the loop is given by B = AB = r2 B , where A (= r2 ) is the area of the loop. According to Faraday's law the magnitude of the emf in the loop is E = dB = r2 dB = (0:055 m)2 (0:16 T/s) = 1:5 10 3 V : 3E dt dt 4E 5E (a) d E = dB = A dB = r2 dt (B0 e dt dt t= 2 0 ) = r B e t= : jEj = dB dt d = dt (6:0t2 + 7:0t) = 12t + 7:0 = 12(2:0) + 7:0 = 31 mV : CHAPTER 31 INDUCTION AND INDUCTANCE 831 (b) Use Lenz's law to determine the direction of the current ow. It is from right to left. Use E = dB =dt = r2 dB=dt. (a) For 0 < t < 2:0 s: 6E dt = (b) 2:0 s < t < 4:0 s : E / dB=dt = 0: (c) 4:0 s < t < 6:0 s : 0:5 T 2 dB = (0:12 m)2 2 E = r dt 6:0 s 4:0 s = 1:1 10 V : E= r2 dB (0:12 m)2 0:5 T = 1:1 10 2 V : 2:0 s 7E (a) The magnitude of the average induced emf is 2 dB B Eav = dt = t = BAi = (2:0 T)(0:20 m) = 0:40 V : t 0:20 s (b) av iav = ER = 20 0:40 V3 = 20 A : 10 (a) 8E L R = A = (1:68 10 8 m) (2:(0:10 m))2 =4 = 1:1 10 3 : 5 10 3 (b) Use i = jEj=R = jdB =dtj=R = (r2 =R)jdB=dtj: Thus 3 dB = iR = (10 A)(1:1 10 ) = 1:4 T/s : dt r2 (0:10 m)2 =4 9P (a) The emf as a function of time is given by dB = N d(BA) = NA d ( ni) = N d2 n di E = N dt dt dt 0 4 0 dt The plot is shown in the next page. d2 n d (3:0t + 1:0t2 ) = 1 d2 N n(3:0 + 2:0t) : = N 4 0 dt 0 4 832 CHAPTER 31 INDUCTION AND INDUCTANCE (mV) 4.4 1.2 0 1.0 2.0 3.0 4.0 t (s) (b) 2 ijt=2:0 s = Ejt=2:0 s = d N0 n4(3:0 + 2:0t) R R 2 m)2 (130)(1:26 10 6 T m/A)(2:2 turns/m)[3:0 + (2:0)(2:0)]( A/s) = (2:1 10 4(0:15 ) 2 A: = 5:8 10 The magnitude of the magnetic eld inside the solenoid is B = 0 nis , where n is the number of turns per unit length and is is the current. The eld is parallel to the solenoid 2 axis, so the ux through a cross section of the solenoid is B = As B = 0 rs nis , where As 2 (= rs ) is the cross-sectional area of the solenoid. Since the magnetic eld is zero outside the solenoid this is also the ux through the coil. The emf in the coil has magnitude 2 E = Nd = 0 rs Nn dis dt dt 10P and the current in the coil is E ic = R = 0 rs Nn dis ; R dt 2 where N is the number of turns in the coil and R is the resistance of the coil. According to Sample Problem 31{1 the current changes linearly by 3:0 A in 50 ms, so dis =dt = (3:0 A)=(50 10 3 s) = 60 A/s. Thus 7 2 1 m)2 ic = (4 10 T m/A)(0:016 (120)(220 10 m ) = 3:0 10 2 A : 5:3 CHAPTER 31 INDUCTION AND INDUCTANCE 833 11P Note that since B only appears inside the solenoid the area A should be the cross-sectional area of the solenoid, not the (larger) loops. d di E = dB = d(BA) = A dt (0 ni) = 0 nr2 dt dt dt 6 T m/A)(1:00 turns/m)( )(25 10 3 m)2 0:50 A 1:0 A = (1:26 10 10 10 3 s = 1:2 10 3 V : 12P Consider the cross-section of the toroid. The magnetic ux through the shaded area as shown is dB = B (r)dA = B (r)hdr. Thus from Eq. 31-35 B = b a O r r+dr h r B (r)hdr = a iNh b = 02 ln a : Z b Z b a 0 i0 Nh dr 2r cross-sectional view 13P 0 B = BA = 2i0 N A r 7 m/A)(0:800 2 2 00 = (4 10 T2(0:150 m + A)(500)(5:2) 10 m) 0:0500 m= 6 Wb : = 1:15 10 14P 2 E = dB = d(BA) = B dA = B d(r ) = 2rB dr dt dt dt dt dt = 2(0:12 m)(0:800 T)( 0:750 m/s) = 0:452 V : 15P The magnetic ux B through the loop is given by B = 2B (r2 =2)(cos 45 ) = r2 B= 2. Thus dB = d r2 B = r2 B p t E = dt dt p2 2 2 m)2 0 76 10 3 T 10 2 = (3:7 p 4:5 10 3 s = 5:1 10 V : 2 p 834 CHAPTER 31 INDUCTION AND INDUCTANCE The direction of the induced current is clockwise when viewed along the direction of B. 16P 17P Since B does not change, E = dB =dt = 0. (a) Use B ' 0 i=2R, where R is the radius of the large loop. Here i(t) = i0 + kt, where i0 = 200 A and k = ( 200 A 200 A)=1:00 s = 400 A/s. Thus 0 i0 = (4 10 7 T m/A)(200 A) = 1:26 10 4 T ; B jt=0 = 2R 2(1:00 m) 7 0+ B jt=0:500 s = 0 (i2R kt) = (4 10 T m/A)[200 Am) (400 A/s)(0:500 s)] = 0 ; 2(1:00 and 0 (i0 + kt) = (4 10 7 T m/A)[200 A (400 A/s)(1:00 s)] B jt=1:00 s = 2R 2(1:00 m) 4 T: = 1:26 10 (b) Let the area of the small loop be a. Then B = Ba, and Ba E (t) = dB = d(dt ) = dt 4 m2 ) = (2:00 10 a dB = a B dt t 4 T 1:26 10 4 T 1:26 10 = 5:04 10 8 V : 1:00 s 18P (a) In the region of the smaller loop the magnetic eld produced by the larger loop may be taken to be uniform and equal to its value at the center of the smaller loop, on the axis. Eq. 30{29, with z = x and much greater than R, gives 0 iR B = 2x3 2 for the magnitude. The eld is upward in the diagram. The magnetic ux through the smaller loop is the product of this eld and the area (r2 ) of the smaller loop: B = 0 ir3 R : 2x 2 2 CHAPTER 31 INDUCTION AND INDUCTANCE 835 (b) The emf is given by Faraday's law: E = dB = dt 0 ir2 R2 2 d 1 = dt x3 0 ir2 R2 2 3 dx = 30 ir2 R2 v : x4 dt 2x4 (c) The eld of the larger loop is upward and decreases with distance away from the loop. As the smaller loop moves away the ux through it decreases. The induced current will be directed so as to produce a magnetic eld that is upward through the smaller loop, in the same direction as the eld of the larger loop. It will be counterclockwise as viewed from above, in the same direction as the current in the larger loop. (a) The emf induced around the loop is given by Faraday's law: E = dB =dt and the current in the loop is given by i = E =R = (1=R)(dB =dt). The charge that passes through the resistor from time zero to time t is given by the integral 19P q= Z t 0 1 Z t dB dt = 1 Z B (t) d = 1 [ (0) (t)] : i dt = R B R B (0) B R B 0 dt All that matters is the change in the ux, not how it was changed. (b) If B (t) = B (0) then q = 0. This does not mean that the current was zero for any extended time during the interval. If B increases and then decreases back to its original value there is current in the resistor while B is changing. It is in one direction at rst, then in the opposite direction. When equal charge has passed through the resistor in opposite directions the net charge is zero. 20P From the result of the last problem A 1 q = R [B (0) B (t)] = R [B (0) B (t)] 3 2 = 1:20 13:10 m [1:60 T ( 1:60 T)] = 2:95 10 2 C : 0 21P Note that the axis of the coil is at 30 , not 70 , from the magnetic eld of the Earth. q(t) = N B = N [BA cos 70 ( BA cos 70 )] = 2WBA cos 30 R R R 4 T) (0:100 m)2 (cos 30 ) = 1:55 10 5 C : = 2(1000)(0:590 10 + 140 85:0 836 CHAPTER 31 INDUCTION AND INDUCTANCE 22P (a) Let L be the length of a side of the square circuit. Then the magnetic ux through the circuit is B = L2 B=2 and the induced emf has magnitude 2 Ei = dB = L dB : dt 2 dt Now B = 0:042 0:870t and dB=dt = 0:870 T/s. Thus Ei = (2:00 m) (0:870 T/s) = 1:74 V : 2 The magnetic eld is out of the page and decreasing so the induced emf is counterclockwise around the circuit, in the same direction as the emf of the battery. The total emf is E + Ei = 20:0 V + 1:74 V = 21:7 V. (b) The current is in the sense of the total emf, counterclockwise. 2 23P (a) The magnetic ux B through the loop is = Z 1 B dA = 4 r2B : Thus r2 dB = 4 dt jEj = = = 1 (0:10 m)2 (3:0 10 3 T/s) = 2:4 10 5 V : 4 (b) From c to b (following Lenz's law). dB dt d 1 2 r B dt 4 24P (a) and (b) Let A = r2 . The emf as a function of time is given by d d d E = dB = dt (B A) = BA dt cos(B; A) = BA dt cos(2ft + 0 ) dt = 92f )BA sin(2ft + 0 ) : 2 r2 Bf : Thus the frequency of E is f and the amplitude is Em = 2fBA = (2f )B (r2 =2) = CHAPTER 31 INDUCTION AND INDUCTANCE 837 25P (a) The area of the coil is A = ab. Suppose that at some instant of time the normal to the loop makes the angle with the magnetic eld. The magnetic ux through the loop is then B = NabB cos and the emf induced around the coil is d E = dB = dt (NabB cos ) = (NabB sin ) d : dt dt In terms of the frequency of rotation f and the time t, is given by = 2ft and d=dt = 2f . The emf is therefore E = 2fNabB sin(2ft) : This can be written E = E0 sin(2ft), where E0 = 2fNabB . (b) You want 2fNabB = 150 V. This means E0 150 V Nab = 2fB = 2(60:0 rev/s)(0:500 T) = 0:796 m2 : Any loop for which Nab = 0:796 m2 will do the job. An example is N = 100 turns, a = b = 8:92 cm. 26P Use the result obtained in 24P, part(b): Em = (2fBA)N = 2(1000=60 s)(3:50 T)(0:500 m)(0:300 m)(100) = 5:50 103 V : 27P (a) It is clear that the magnetic ux through areas 1 and 2 as shown cancel out. Thus B = B3 = = b a Z a b-a O Z a ib = 20 ln b a a ; and b r 2r bdr 0i b a B (r)(bdr) 2 b-a 3 2a-b 838 CHAPTER 31 INDUCTION AND INDUCTANCE E= = = = dB = d 0 ib ln a di = 0 b ln b a a dt dt dt 2 b a 2 0 b ln a d (4:50t2 10:0t) = 4:500 bt ln a 2 b a dt b a 12:0 cm (4:50)(1:26 10 6 T m/A)(0:160 m)(3:00 s) ln 16:0 cm 12:0 cm 7 V: 5:98 10 (b) From Lenz's law, the induced current in the loop is counterclockwise. 28P Use Faraday's law to nd an expression for the emf induced by the changing magnetic eld. First nd an expression for the magnetic ux through the loop. Since the eld depends on y but not on x, divide the area into strips of length L and width dy, parallel to the x axis. Here L is the length of one side of the square. At time t the ux through a strip with coordinate y is dB = BL dy = 4:0Lt2 y dy and the total ux through the square is B = Z L 0 4:0Lt2 y dy = 2:0L3 t2 : According to Faraday's law the magnitude of the emf around the square is d E = dB = dt 2:0L3 t2 = 4:0L3 t : dt At t = 2:5 s this is 4:0(0:020 m)3 (2:5 s) = 8:0 10 5 V. The externally-produced magnetic eld is out of the page and is increasing with time. The induced current produces a eld that is into the page, so it must be clockwise. The induced emf is also clockwise. 29P (a) Analogous to 27P, part (a), we have B = = Z loop Z r+ b 2 r 2 B (r)dA = Z r+ b 2 r 2 b B (r)adr b 0 ia dr = 0 ia ln 2r + b : 2r 2 2r b CHAPTER 31 INDUCTION AND INDUCTANCE 839 (b) Now ia @ E = dB = @@rB dr = v @r 20 ln 2r + b dt dt 2r b iaw 2 = 0 2r 1 b 2r 1 b = (40 iabv2 ) : + r2 b Thus E i = R = R2(40 iabvb2 ) : r2 (a) Suppose each wire has radius R and the distance between their axes is a. Consider a single wire and calculate the ux through a rectangular area with the axis of the wire along one side. Take this side to have length L and the other dimension of the rectangle to be a. The magnetic eld is everywhere perpendicular to the rectangle. First consider the part of the rectangle that is inside the wire. The eld a distance r from the axis is given by B = 0 ir=2R2 and the ux through the strip of length L and width dr at that distance is (0 ir=2R2 )L dr. Thus the ux through the area inside the wire is in = Z R 30P R dr L a Now consider the region outside the wire. There the eld is given by B = 0 i=2r and the ux through an in nitesimally thin strip is (0 i=2r)L dr. The ux through the whole region is Z a a 0 iL 0 iL out = 2 dr = 2 ln R : r R 0 0 iL r dr = 0 iL : 2R2 4 Now include the contribution of the other wire. Since the currents are in the same direction the two contributions have the same sign. They also have the same magnitude, so The total ux through the area bounded by the dotted lines is the sum of the two contributions: h a i 0 iL = 4 1 + 2 ln R : 0 iL h1 + 2 ln a i : total = 2 R 840 CHAPTER 31 INDUCTION AND INDUCTANCE The total ux per unit length is total = 0 i h1 + 2 ln a i = (4 10 7 T m/A)(10 A) 1 + 2 ln 20 mm L 2 R 2 1:25 mm = 1:3 10 5 Wb/m : (b) Again consider the ux of a single wire. The ux inside the wire itself is again in = 0 iL=4. The ux inside the region of the other wire is out = Z a a 0 iL dr = 0 iL ln a : r 2 a R R 2 Double this to include the ux of the other wire (inside the rst wire) and divide by L to obtain the ux per unit length. The total ux per unit length that is inside the wires is wires = 0 i 1 + 2 ln a L 2 a R 7 T m/A)(10 A) 20 mm (4 10 1 + 2 ln 20 mm 1:25 mm = 2 = 2:26 10 6 Wb/m : The fraction of the total ux that is inside the wires is 2:26 10 6 Wb/m = 0:17 : 1:31 10 5 Wb/m (c) The contributions of the two wires to the total ux have the same magnitudes but now the currents are in opposite directions, so the contributions have opposite signs. This means total = 0. 31E E 2 t = 1 dB 2 t = 1 A B 2 t = A2 B 2 : Uthermal = Pthermal t = R R dt R t Rt 32E Thermal energy is generated at the rate E 2 =R, where E is the emf in the wire and R is the resistance of the wire. The resistance is given by R = L=A, where is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. The resistivity can be found in Table 27{1. Thus m)(0:500 R = (1:69 :10 10 3 m)2 m) = 1:076 10 2 : (0 500 8 CHAPTER 31 INDUCTION AND INDUCTANCE 841 Faraday's law is used to nd the emf. If B is the magnitude of the magnetic eld through the loop, then E = A dB=dt, where A is the area of the loop. The radius r of the loop is r = L=2 and its area is r2 = L2 =42 = L2 =4. Thus E = L dB = (0:500 m) (10:0 10 3 T/s) = 1:989 10 4 V : 4 dt 4 The rate of thermal energy generation is 2 4 2 P = E = (11::989 10 2V) = 3:68 10 6 W : R 076 10 2 2 (a) The ux changes because the area bounded by the rod and rails increases as the rod moves. Suppose that at some instant the rod is a distance x from the right-hand end of the rails and has speed v. Then the ux through the area is B = BA = BLx, where L is the distance between the rails. According to Faraday's law the magnitude of the emf induced is E = dB =dt = BL (dx=dt) = BLv = (0:350 T)(0:250 m)(0:550 m/s) = 4:81 10 2 V. (b) Use Ohm's law. If R is the resistance of the rod then the current in the rod is i = E =R = (4:81 10 2 V)=(18:0 ) = 2:67 10 3 A. (c) The rate at which thermal energy is generated is 33E P = i2 R = (2:67 10 3 A)2 (18:0 ) = 1:28 10 4 W : (a) Let x be the distance from the right end of the rails to the rod. The area enclosed by the rod and rails is Lx and the magnetic ux through the area is B = BLx. The emf induced is E = dB =dt = BL dx=dt = BLv, where v is the speed of the rod. Thus E = (1:2 T)(0:10 m)(5:0 m/s) = 0:60 V. (b) If R is the resistance of the rod, the current in the loop is i = E =R = (0:60 V)=(0:40 ) = 1:5 A. Since the rod moves to the left in the diagram, the ux increases. The induced current must produce a magnetic eld that is into the page in the region bounded by the rod and rails. To do this the current must be clockwise. (c) The rate of generation of thermal energy by the resistance of the rod is P = E 2 =R = (0:60 V)2 =(0:40 ) = 0:90 W. (d) Since the rod moves with constant velocity the net force on it must be zero. This means the force of the external agent has the same magnitude as the magnetic force but is in the opposite direction. The magnitude of the magnetic force is FB = iLB = (1:5 A)(0:10 m)(1:2 T) = 0:18 N. Since the eld is out of the page and the current is upward through the rod the magnetic force is to the right. The force of the external agent must be 0:18 N, to the left. 34E 842 CHAPTER 31 INDUCTION AND INDUCTANCE (e) As the rod moves an in nitesimal distance dx the external agent does work dW = F dx, where F is the force of the agent. The force is in the direction of motion, so the work done by the agent is positive. The rate at which the agent does work is dW=dt = F dx=dt = Fv = (0:18 N)(5:0 m/s) = 0:90 W, the same as the rate at which thermal energy is generated. The energy supplied by the external agent is converted completely to thermal energy. Let Fnet = BiL mg = 0 and solve for i: 35P 1 v mg i = BL = jEj = R dB = B dA = B (Rt L) ; dt dt R R so mgR vt = B 2 L2 : 36P (b) (a) At time t the area of the closed triangular loop is A(t) = 1 (vt)(2vt) = v2 t2 . Thus 2 B jt=3:00 s = BA(t) = (0:350 T)[(5:20 m/s)(3:00 s)]2 = 85:2 T m2 : 22 E = dB = d(BA) = B dA = B d(vdtt ) = 2Bv2 t dt dt dt = 2(0:350 T)(5:20 m/s)2 (3:00 s) = 56:8 V : (c) From part (b) above we see that E = 2Bv2 t / t1 . Thus n = 1. 37P From [sin2 (2ft)]av = we obtain 1 Z T sin2 (2ft)dt = 1 T 2 0 2 2 2 E02 (150 V)2 av Pav = ER = E0 [sin (2ft)]av = 2R = 2(42:0 ) = 268 W : R 38P (a) Apply Newton's second law to the rod: m dv = iBL : dt CHAPTER 31 INDUCTION AND INDUCTANCE 843 Integrate to obtain v points away from the generator G. jEinduced j = v = iBLt : m (b) When the current i in the rod becomes zero, the rod will no longer be accelerated by a force F = iBL and will therefore reach a constant terminal velocity. This happens when jEinduced j = E , i.e., dB dt = d(BA) dt = dA B dt = BvL = E ; or v = E =BL, to the left. (c) In case (a) above electric energy is supplied by the generator and is transferred into the kinetic energy of the rod. In the current case the battery initially supplies electric energy to the rod, causing its kinetic energy to increase to a maximum value of 1 mv2 = 1 (E =BL)2 . 2 2 Afterwards, there is no further energy transfer from the battery to the rod, and the kinetic energy of the rod remains constant. (a) Let x be the distance from the right end of the rails to the rod and nd an expression for the magnetic ux through the area enclosed by the rod and rails. The magnetic eld is not uniform but varies with distance from the long straight wire. The eld is normal to the area and has magnitude B = 0 i=2r, where r is the distance from the wire and i is the current in the wire. Consider an in nitesimal strip of length x and width dr, parallel to the wire and a distance r from it. The area of this strip is A = x dr and the ux through it is dB = (0 ix=2r)dr. The total ux through the area enclosed by the rod and rails is 39P 0 ix Z a+L dr = 0 ix ln a + L : B = 2 r 2 a a According to Faraday's law the emf induced in the loop is d = 0 i dx ln a + L = 0 iv ln a + L E = dt 2 dt a 2 a 7 T m/A)(100 A)(5:00 m/s) 1:00 cm + 10:0 cm ln = (4 10 2 1:00 cm 4 V: = 2:40 10 (b) If R is the resistance of the rod then the current in the conducting loop is i` = E =R = (2:40 10 4 V)=(0:400 ) = 6:00 10 4 A. Since the ux is increasing the magnetic eld produced by the induced current must be into the page in the region enclosed by the rod and rails. This means the current is clockwise. (c) Thermal energy is being generated at the rate P = i2 R = (6:00 10 4 A)2 (0:400 ) = ` 1:44 10 7 W. 844 CHAPTER 31 INDUCTION AND INDUCTANCE (d) Since the rod moves with constant velocity the net force on it is zero. The force of the external agent must have the same magnitude as the magnetic force and must be in the opposite direction. The magnitude of the magnetic force on an in nitesimal segment of the rod, with length dr and a distance r from the long straight wire, is dFB = i` B dr = (0 i` i=2r) dr. The total magnetic force on the rod has magnitude 0 i` i Z a+L dr = 0 i` i ln a + L FB = 2 r 2 a a (4 10 7 T m/A)(6:00 10 4 A)(100 A) ln 1:00 cm + 10:0 cm = 2 1:00 cm 8 N: = 2:87 10 Since the eld is out of the page and the currentin the rod is upward in the diagram, this force is toward the right. The external agent must apply a force of 2:87 10 8 N, to the left. (e) The external agent does wor...
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