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 work at the rate P = Fv = (2:87 10 8 N)(5:00 m/s) = 1:44 10 7 W. This is the same as the rate at which thermal energy is generated in the rod. All the energy supplied by the agent is converted to thermal energy. (a) The eld point is inside the solenoid, so Eq. 31-27 applies. The magnitude of the induced electric eld is 1 E = 2 dB r = 1 (6:5 10 3 T/s)(0:0220 m) = 7:15 10 5 V/m : dt 2 (b) Now the eld point is outside the solenoid and Eq. 31{29 applies. The magnitude ofthe induced eld is 2 1 2 E = 1 dB R = 2 (6:5 10 3 T/s) (0::0600 m) = 1:43 10 4 V/m : 2 dt r (0 0820 m) 40E 41E I d 2 E ds = dB1 = dt (B1A1) = A1 dB1 = r1 dB1 dt dt dt 1 2 E ds = dB2 = r2 dB2 dt dt 2 = (0:200 m)2 ( 8:50 10 3 T/s) = 1:07 10 3 V ; I = (0:300 m)2 ( 8:50 10 3 T/s) = 2:40 10 3 V ; CHAPTER 31 INDUCTION AND INDUCTANCE 845 and I 3 E ds = E ds 1 I I 2 E ds = 1:07 10 3V ( 2:4 10 3 V) = 1:33 10 3 V : 42P The magnetic eld B can be expressed as B (t) = B0 + B1 sin(!t + 0 ) ; where B0 = (30:0 T + 29:6 T)=2 = 29:8 T and B1 = (30:0 T 29:6 T)=2 = 0:200 T. Then from Eq. 31-27 1 rd E = 1 dB r = 2 dt [B0 + B1 sin(!t + 0 )] = 2 B1 !r cos(!t + 0 ) : 2 dt Thus 1 Emax = 2 B1 (2f )r = 1 (0:200 T)(2)(15 Hz)(1:6 10 2 m = 0:15 V/m : 2 The induced electric eld E as a function of r is given by E (r) = (r=2)(dB=dt). So 43P ac = aa = eE = 2er dB m m dt 19 2 3 :0 = (1:60 10 C)(511 10 27m)(10 10 T/s) = 4:4 107 m/s2 : 2(9: 10 kg) aa points to the right and ac points to the left. At point b we have ab / rb = 0: Use Faraday's law in the form E ds = (dB =dt). Integrate around the dotted path shown in Fig. 31{61. At all points on the upper and lower sides the electric eld is either perpendicular to the side or else it vanishes. Assume it vanishes at all points on the right side (outside the capacitor). On the left side it is parallel to the side and has constant H magnitude. Thus direct integration yields E ds = EL, where L is the length of the left side of the rectangle. The magnetic eld is zero and remains zero, so dB =dt = 0. Faraday's law leads to a contradiction: EL = 0, but neither E nor L is zero. There must be an electric eld along the right side of the rectangle. 44P H 846 CHAPTER 31 INDUCTION AND INDUCTANCE 45E Since N B = Li, where N is the number of turns, L is the inductance, and i is the current, 3 3 B = Li = (8:0 10 H)(5:0 10 A) = 1:0 10 7 Wb : N 400 (a) (b) 46E B = NBA = NB (r2 ) = (30:0)(2:60 10 3 T)()(0:100 m)2 = 2:45 10 3 Wb : 10 3 L = iB = 2:45 :80 A Wb = 6:45 10 4 H/m : 3 (a) N = 2:0 m=2:5 mm = 800. (b) L=l = 0 n2 A = (4 10 7 T m/A)(800=2:0 m)2 ()(0:040 m)2 =4 = 2:5 10 4 H: If the solenoid is long and thin, then when it is bent into a toroid (b a)=a 1. Thus 47E 48P 0 N 2 h ln b = 0 N 2 h ln 1 + b a 0 N 2 h(b a) : Ltoroid = 2 a 2 b 2b Since A = h(b a) is the cross-sectional area and l = 2b is the length of the toroid, we may rewrite the expression for L toroid above as Ltoroid 0 N 2 A = n2 A ; 0 l l2 which indeed reduces to that of a long solenoid. Note that the approximation ln(1+ x) x was used for jxj 1. (a) Suppose we divide the one-turn solenoid into N small circular loops placed along the width W of the copper strip. Each loop carries a current i = i=N . Then the magnetic eld inside the solenoid is B = 0 ni = 0 (N=W )(i=N ) = 0 i=W: (b) 2 2 2 L = iB = Ri B = R (i0 i=W ) = 0 R : W 49P CHAPTER 31 INDUCTION AND INDUCTANCE 847 The area of integartion for the calculation of the magnetic ux is bounded by the two dotted lines and the boundaries of the wires. If the origin is taken to be on the axis of the right-hand wire and r measures distance from that axis, it extends from r = a to r = d a. Consider the right-hand wire rst. In the region of integation the eld it produces is into the page and has magnitude B = 0 i=2r. Divide the region into strips of length l and width dr, as shown. The ux through the strips a distnace r from the axis of the wire is d = Bl dr adn the ux through the entire region is 50P a dr l r d 0 il Z d = 2 a a dr = 0 il ln d a : r 2 a The other wire produces the same result, so the total ux through the dotted rectangle is 0 total = il ln d a a : The inductance is total diveded by i: 0 L = total = l ln d a a : i (a) Since E enhances i, i must be decreasing. (b) From E = L di=dt we get 51E E 17 V L = di=dt = 2:5 kA/s = 6:8 10 4 H : 52E Since E = L(di=dt), we may obtain the desired induced emf by setting di = E = 60 V = 5:0 A/s : dt L 12 H You might, for example, uniformly reduce the current from 2:0 A to zero in 40 ms. 848 CHAPTER 31 INDUCTION AND INDUCTANCE (a) 53E L = n2 A = (4 10 7 T m/A)(100 turns=cm)2 ()(1:6 cm)2 = 0:10 H/m : 0 l (b) The induced emf per meter is di E = L dt = (0:10 H/m)(13 A=s) = 1:3 V/m : (a) 54E 3 E L = di=dt = 3:05 10 V = 6:0 10 4 H : :0 A/s 4 A)(6:0 N = iL = (8:040 10610 H) = 120 : Wb B (b) From L = N B =i we get Use E = L di=dt. (a) For 0 < t < 2 ms 55P :0 i E = L t = (4:6 :H)(710A3 s 0) = 1:6 104 V : 20 (b) For 2 ms < t < 5 ms 6 0A : i E = L t = (4:(5H)(5:2:0)10 730sA) = 3:1 103 V : :0 (c) For 5 ms < t < 6 ms i 5 E = L t = (4::60H)(0:0)10:0 3A) = 2:3 104 V : (6 5 s (a) 56P Leq = total = 1 + 2 = 1 + 2 = L1 + L2 : i i i i CHAPTER 31 INDUCTION AND INDUCTANCE 849 (b) If the separation is not large enough then the magnetic eld in one inductor may produce a ux in the other, which would render the expression = Li invalid. (c) For N inductors in series N N N total = 1 X = X n = X L : Leq = i i n=1 n n=1 i n=1 n (a) In this case itotal = i1 + i2 . But Vtotal = =Leq and i1; 2 = =L1; 2 so =Leq = =L1 + =L2 , or 1 = 1 + 1 : Leq L1 L2 (b) The reason is the same as in part (b) of the previous problem. (c) For N inductors in parallel you can easily generize the expression above to N 1 =X 1 : L L 57P eq n=1 n Suppose that i(t0 ) = i0 =3 at t = t0 . Write i(t0 ) = i0 (1 e 58E t0 =L ), which gives 5 L = ln(1 t0 i=i ) = ln(1:001s=3) = 12:3 s : 0 59E Starting with zero current at time t = 0, when the switch is closed, the current in an RL series circuit at a later time t is given by where L is the inductive time constant, E is the emf, and R is the resistance. You want to calculate the time t for which i = 0:9990E =R. This means so E i= R 1 e t=L ; E E 0:9990 R = R 1 e 0:9990 = 1 e t=L ; t=L 850 CHAPTER 31 INDUCTION AND INDUCTANCE = 0:0010 : Take the natural logarithm of both sides to obtain (t= ) = ln(0:0010) = 6:91. Thus t = 6:91L . That is, 6:91 inductive time constants must elapse. or e t=L 60E The current in the circuit is given by i = i0 e t=L ; where i0 is the initial current (at time t = 0) and L (= L=R) is the inductive time constant. Solve for L . Divide by i0 and take the natural logarithm of both sides of the resulting equation to obtain ln ii = t : t 1 L = ln [i=i ] = ln [(10 10 :0 s =(1:0 A)] = 0:217 s : 3 A) 0 Thus R = L=L = (10 H)=(0:217 s) = 46 . Write i = i0 e This yields 0 L 61E t=L and note that i = 10% i0 . Solve for t: L i0 t = L ln ii0 = R ln ii0 = 2::00 H ln 10:0% i = 1:54 s : 3 00 0 (a) Immediately after the switch is closed E EL = iR. But i = 0 at this instant so EL = E . (b) EL (t) = E e t=L = E e 2:0L =L = E e 2:0 = 0:135E : (c) Solve EL (t) = E e t=L for t=L : t=L = ln(E =EL ) = ln 2: So t = L ln 2 = 0:693L . 62E 63E where L = L=R. You want to nd the time for which i = 0:800E =R. This means 0:800 = 1 e t=L (a) If the battery is switched into the circuit at time t = 0, then the current at a later time t is given by E i = R 1 e t=L ; CHAPTER 31 INDUCTION AND INDUCTANCE 851 = 0:200 : Take the natural logarithm of both sides to obtain (t=L ) = ln(0:200) = 1:609. Thus or e t=L 6 t = 1:609L = 1:609L = 1:609(6:30 10 H) = 8:45 10 9 s : R 1:20 103 (b) At t = 1:0L the current in the circuit is E i= R 1 e 1:0 = 14:0 V 1:20 103 1 e 1:0 = 7:37 10 3 A : (a) L = =i = 26 10 3 Wb=5:5 A = 4:7 10 3 H: (b) Use Eq. 31-45 to solve for t: 64E L t = L ln 1 iR = R ln 1 iR E E 3H 10 = 4:7 0:75 ln 1 (2:5 A)(0:75 ) = 2:4 10 3 s : 6:0 V Apply the loop theorem E (t) L(di=dt) = iR and solve for E (t): 65P d di E (t) = L dt + iR = L dt (3:0 + 5:0t) + (3:0 + 5:0t)R = (6:0 H)(5:0 A/s) + (3:0 A + 5:0t)(4:0 ) = (42 + 20t) V ; where t is in seconds. (a) and (b) Write VL (t) = E e t=l . Consider the rst two data points, (VL1 ; t1 ) and (VL2 ; t2 ), satisfying VLi = E e ti =L (i = 1; 2). We have VL1 =VL2 = E e (t1 t2 )=L ; which gives :0 ms ms t t L = ln(1V =V2 ) = 1ln(13:8=2:0:2) = 3:6 ms : 18 2 1 (18:2 V)e1:0 ms=3:6 ms 66P = 24 V. You can easily check that the values of So E = VL1 et1 =L = L and E are consistent with the rest of the date points. 852 CHAPTER 31 INDUCTION AND INDUCTANCE Take the time derivative of both sides of Eq. 31-45: 67P di = d E 1 e Rt= = E e RT =L dt R dt L 3 = 50:045:0 V 3 H e (180 )(1:2010 s)=(50:010 10 3 H) = 12:0 A/s : (a) Since the inner circumference of the toroid is l = 2a = 2(10 cm) = 62:8 cm, the number fo turns of the toroid is roughly N = 62:8 cm=1:0 mm = 628. Thus 0 N 2 H ln b (4 10 7 T m/A)(628)2 (0:12 m 0:10 m) ln 12 cm L = 2 a 2 10 cm 4 H: = 2:9 10 (b) Since the total length l of the wire is l = (628)4(2:0 cm) = 50 m, the resistance of the wire is R = (50 m)(0:02 = m) = 1:0 : Thus 10 4 L L = R = 2:9 :0 H = 2:9 10 4 s : 1 (a) The inductor prevents a fast build-up of the current through it, so immediately after the switch is closed the current in the inductor is zero. This means E 100 i1 = i2 = R + R = 10:0 + V :0 = 3:33 A : 20 1 2 (b) A long time later the current reaches steady state and no longer changes. The emf across the inductor is zero and the circuit behaves as if it were replaced by a wire. The current in R3 is i1 i2 . Kirchho's loop rule gives E i1 R1 i2 R2 = 0 and E i1 R1 (i1 i2 )R3 = 0 : Solve these simultaneously for i1 and i2 . The results are ( 3 i1 = R R E+R2 + R+)R R R1 R3 2 3 1 2 (100 V)(20:0 = (10:0 )(20:0 ) + (10:0 )(30+ 30:0 ) :0 )(30:0 ) :0 ) + (20 = 4:55 A 68P 69P CHAPTER 31 INDUCTION AND INDUCTANCE 853 and E i2 = R R + R R3 + R R 1 2 1 R3 2 3 = 2:73 A : (100 = (10:0 )(20:0 ) + (10:0V)(30::0 ) + (20:0 )(30:0 ) )(30 0 ) (c) The left-hand branch is now broken. If its inductance is zero the current immediately drops to zero when the switch is opened. That is, i1 = 0. The current in R3 changes only slowly because there is an inductor in its branch. Immediately after the switch is opened it has the same value as it had just before the switch was opened. That value is 4:55 A 2:73 A = 1:82 A. The current in R2 is the same as that in R3 , 1:82 A. (d) There are no longer any sources of emf in the circuit, so all currents eventually drop to zero. (a) When swith S is just closed (case I), V1 = E and i1 = E =R1 = 10 V=5:0 = 2:0 A. After a long time (case II) we still have V1 = E , so i1 = 2:0 A: (b) Case I: since now EL = E , i2 = 0; case II: since now EL = 0, i2 = E =R2 = 10 V=10 = 1:0 A. (c) Case I: i = i1 + i2 = 2:0 A + 0 = 2:0 A; case II: i = i1 + i2 = 2:0 A + 1:0 A = 3:0 A. (d) Case I: since EL = E , V2 = E EL = 0; case II: since EL = 0, V2 = E EL = E = 10 V: (e) Case I: EL = E = 10 V; case II: EL = 0: (f) Case I: di2 =dt = EL =L = E =L = 10 V=5:0 H = 2:0 A/s; case II: di2 =dt = EL =L = 0. For t < 0, no current goes through L2 , so i2 = 0 and i1 = E =R. As the switch is opened there will be a very brief sparking across the gap. i1 drops while i2 increases, both very quickly. The loop rule can be written as 70P 71P consider the situation shortly after t = 0. Since the sparking is very brief, we can reasonably assume that both i1 and i2 get equalized quickly, before they can change appreciably from their respective initial values. Thus the loop rule requires that L1 (di1 =dt), which is large and negative, must roughly cancel L2 (di2 =dt), which is large and positive: E i1 R L1 di1 L2 R L2 di2 = 0 ; dt dt where the initial value of i1 at t = 0 is given by E =R and that of i2 at t = 0 is 0. Now L1 di1 L2 di2 : dt dt 854 CHAPTER 31 INDUCTION AND INDUCTANCE Let the common value reached by i1 and i2 be i, then di1 i1 = i E =R dt t t and The equations above yield or di2 i2 = i 0 : dt t t E = L (i 0) ; L1 i R 2 L E i = L R E+1L R = L L1 L R : 2 1 1 2 1+ 2 (a) Before the fuse blows, the current through the resistor remains zero. Apply the loop theorem to the battery-fuse-inductor loop: E L di=dt = 0. So i = E t=L. As the fuse blows at t = t0 , i = i0 = 3:0 A. Thus 72P t0 = i0 L = (3:0 A)(5:0 H) = 1:5 s : E 10 V (b) i 3.0 A 0.67 A 1.5 s (fuse blows) t 73P (a) Assume i is from left to right through the closed switch. Let i1 be the current in the resistor and take it to be downward. Let i2 be the current in the inductor and also take it to be downward. The junction rule gives i = i1 +i2 and the loop rule gives i1 R L(di2 =dt) = 0. CHAPTER 31 INDUCTION AND INDUCTANCE 855 This equation is similar to Eq. 31-48, and its solution is the function given as Eq. 31-49: According to the junction rule, (di1 =dt) = (di2 =dt). Substitute into the loop equation to obtain L di1 + i1 R = 0 : dt i1 = i0 e Rt=L ; where i0 is the current through the resistor at t = 0, just after the switch is closed. Now just after the switch is closed the inductor prevents the rapid build-up of current in its branch, so at that time i2 = 0 and i1 = i. Thus i0 = i, so i1 = ie and (b) When i2 = i1 , so Rt=L i2 = i i1 = i 1 e e Rt=L Rt=L : =1 e Rt=L Rt=L ; e Take the natural logarithm of both sides and use ln(1=2) = ln 2 to obtain (Rt=L) = ln 2 or L t = R ln 2 : (a) Use UB = 1 Li2 : 2 1 = 2: 74P 2(25 J) L = 2UB = (60:0:0 103 A)2 = 13:9 H: 2 i 10 (b) Since UB / i2 , for UB to increase by a factor of 4 i must increase by a factor of 2, i.e., i should be increased to 2(60:0 mA) = 120 mA. 3 Let UB (t) = 1 Li2 (t) = 1 UB (t ! 1) = 1 Li2 . This gives i(t) = i0 = 2. But i(t) = 2 2 4 0 p t=L ), so 1 t=L = 1= 2. Solve for t: e i0 (1 e 1 t = L ln 1 p = 1:23L : 2 75E p 856 CHAPTER 31 INDUCTION AND INDUCTANCE (a) i0 = E =R = 100 V=10 = 10 A. (b) UB = 1 Li2 = 1 (2:0 H)(10 A)2 = 100 J: 2 0 2 (a) 76E 77E dUB = d 1 Li2 = Li di = Li 1 e t=L d hi 1 e t=L i 0 dt dt 2 dt dt 0 2 LE 2 = Li0 e t=L 1 e t=L = (L=R)R2 e Rt=L 1 e t=L L (100 V)2 e (10 )(0:10 s)=2:0 H h1 e (10 )(0:10 s)=2:0 H i = 10 = 2:4 102 W : (b) Pthermal = i2 R = i2 1 e 0 2h t=L 2 V) = (100 1 e 10 = 1:5 102 W : i (10 )(0:10 s)=2:0 H 2 R= E 1 e R 2 Rt=L 2 (c) Pbattery = iE = E i0 1 e V) = (100 1 e 10 = 3:9 102 W : 2h t=L (10 )(0:10 s)=2:0 H E 2 1 e =R i Rt=L Note that the law of conservation of energy requirement Pbattery = dUB =dt + Pthermal is satis ed. Use the results of the last problem. Let Pthermal = dUB =dt, or 78P E 2 1 e R and solve for t: Rt=L 2 E2 e = R Rt=L 1 e Rt=L L t = R ln 2 = L ln 2 = (37:0 ms) ln 2 = 25:6 ms : CHAPTER 31 INDUCTION AND INDUCTANCE 857 79P (a) If the battery is applied at time t = 0 the current is given by where E is the emf of the battery, R is the resistance, and L is the inductive time constant. In terms of R and the inductance L, L = L=R. Solve the current equation for the time constant. First obtain e t=L = 1 iR ; E then take the natural logarithm of both sides to obtain E i= R 1 e t=L ; t = ln 1 iR : L E Since 3 A)(10 3 ln 1 iR = ln 1 (2:00 10 50:0 V :0 10 ) = :5108 ; E the inductive time constant is L = t=0:5108 = (5:00 10 3 s)=(0:5108) = 9:79 10 3 s and the inductance is L = L R = (9:79 10 3 s)(10:0 103 ) = 97:9 H : (b) The energy stored in the coil is 1 1 UB = 2 Li2 = 2 (97:9 H)(2:00 10 3 A)2 = 1:96 10 4 J : (a) 80P Ebattery = Z t (10:0 V)2 2:00 s + (5:50 H) e = 6:70 = 18:7 J : 0 Pbattery dt = Z t 0 E 2 1 e R Rt=L E 2 t + L e dt = R R 1 Rt=L 1 (6:70 )(2:00 s)=5:50 H 6:70 (b) 1 1 E 2 Einductor = UB (t) = 2 Li2 (t) = 2 L R (1 e Rt=L )2 1 (5:50 H) 10:0 V 2 1 e (6:70 )(2:00 s)=5:50 H 2 =2 6:70 = 5:10 J : 858 CHAPTER 31 INDUCTION AND INDUCTANCE (c) Edissipated = Ebattery Einductor = 18:7 J 5:10 J = 13:6 J: (a) The inductance of the solenoid is 81P L = 0 n2 Al = (4 10 7 T m/A)(3000 turns=0:800 m)2 ()(5:00 10 2 m)2 (0:800 m) = 0:111 H : Thus 1 E 2 UB (t) = 1 Li2 (t) = 2 L R 1 e 2 (0:111 H) 12:0 V 2 h1 e = 2 10:0 = 1:05 10 2 J : (b) Rt=L 2 i (10:0 )(5:0010 3 s)=0:111 H 2 Ebattery E 2 t + L e = R R Rt=L 1 h 2 :111 H = (12::00V) 5:00 10 3 s + 010:0 10 = 1:41 10 2 J : e (10:0 )(5:0010 3 s)=0:111 H 1 i Suppose that the switch has been in position a for a long time, so the current has reached the steady-state value i0 . The energy stored in the inductor is UB = 1 Li2 . Now the switch 2 0 is thrown to position b at time t = 0. Thereafter the current is given by 82P i = i0 e t=L ; where L is the inductive time constant, given by = L=R. The rate at which thermal energy is generated in the resistor is given by P = i2 R = i2 Re 0 Over a long time period the energy dissipated is 2t=L : 1 2t=L E= Z 1 0 P dt = i2 R 0 Z 1 0 1 e 2t=L dt = 2 i2 RL e 0 0 1 = 2 i2 RL : 0 CHAPTER 31 INDUCTION AND INDUCTANCE 859 Substitute L = L=R to obtain the same as the total energy originally stored in the inductor. (a) At any point the magnetic energy density is given by uB = B 2 =20 , where B is the magnitude of the magnetic eld at that point. Inside a solenoid B = 0 ni, where n is the number of turns per unit length and i is the current. For the solenoid of this problem n = (950)=(0:850 m) = 1:118 103 m 1 . The magnetic energy density is 1 1 uB = 2 0 n2 i2 = 2 (4 10 7 T m/A)(1:118 103 m 1 )2 (6:60 A)2 = 34:2 J/m3 : (b) Since the magnetic eld is uniform inside an ideal solenoid, the total energy stored in the eld is UB = uB V , where V is the volume of the solenoid. V is calculated as the product of the cross-sectional area and the length. Thus 1 E = 2 Li2 ; 0 83E UB = (34:2 J/m3 )(17:0 10 4 m2 )(0:850 m) = 4:94 10 2 J : The magnetic energy stored in the toroid is given by UB = 1 Li2 , where L is its inductance 2 and i is the current. The energy is also given by UB = uB V , where uB is the average energy density and V is the volume. Thus 84E i= r 2uB V = L s 2(70:0 J/m3 )(0:0200 m3 ) = 5:58 A : 90:0 10 3 H Let uE = 1 0 E 2 = uB = 1 B 2 =0 and solve for E : 2 2 85E E = pB = p (8:85 10 0 0 Use 1 ly = 9:46 1015 m. 0:50 T 8 12 F/m)(4 10 7 T m/A) = 1:5 10 V/m: 86E B2 :46 1015 3 10 T)2 UB = V uB = V = (92(4 10 )7(10m/A) = 3 1036 J : 2 0 T 860 CHAPTER 31 INDUCTION AND INDUCTANCE Use Eq. 31-53 to solve for L: 87E b b 0 i2 L = 2iU = i2 4 l ln a = 0 l ln a : 2 2 2 (a) The energy needed is 88E UE = uE V = 1 0 E 2 V = 1 (8:85 10 2 2 (b) The energy needed is 12 F/m)(100 kV/m)2 (10 cm)3 = 4:4 10 5 J : 2 1 UB = uB V = 2 B 2 V = 2(4 (1:0 T) m/A) (10 cm)3 = 4:0 102 J : 10 7 T 0 (b) Obviously, since UB > UE greater amounts of energy can be stored in the magneticc eld. (a) (b) 89E 10 T m/A)(100 0 B = Ri = (4 2(50 10 3 m) A) = 1:3 10 3 T : 2 7 3 2 B2 3 uB = 2 = 2(4(1: 107 T T) = 0:63 J/m3 : 10 m/A) 0 90P (b) Evaluate the integral UB = uB dV over the volume of the toroid. A circular strip with radius r, height h, and thickness dr has volume dV = 2rh dr, so (a) The energy density at any point is given by uB = B 2 =20 , where B is the magnitude of the magnetic eld. The magnitude of the eld inside a toroid, a distance r from the center, is given by Eq. 30{26: B = 0 iN=2r, where N is the number of turns and i is the current. Thus 1 0 iN 2 = 0 i2 N 2 : uB = 2 2r 82 r2 R 0 0 i2 N 2 2h Z b dr = 0 i2 N 2 h ln b : UB = 82 4 a a r CHAPTER 31 INDUCTION AND INDUCTANCE 861 Numerically, 7 2 2 3 95 mm UB = (4 10 T m/A)(0:500A) (1250) (13 10 m) ln 52 mm 4 = 3:06 10 4 J : (c) The inductance is given by Eq. 31{37: b N2 L = 02 h ln a so the energy is given by 22 1 b UB = 2 Li2 = 0 Ni h ln a : 4 This the exactly the same as the expression found in part (b) and yields the same numerical result. (a) 91P B2 1 0 uB = 2 = 2 Ri 2 0 0 2 2 2 10 7 = 0 i = (4 8(2:5 T m/A)(10 A) = 1:0 J/m3 : R2 10 3 m=2)2 (b) 1 E 2 = 0 J 2 = 0 iR 2 uE = 2 0 2 2 l 1 = 2 (8:85 10 12 F/m)[(10 A)(3:3 =103 m)]2 = 4:8 10 15 J/m3 : Here we used J= = i=A and R=l = =A = 1=A to obtain J= = iR=l: (a) 92P 6 2 B2 uB = 2e = 2(4(50 107 TT) = 1:0 10 3 J/m3 : 10 m/A) 0 2 (b) The volume of the shell of thickness h is V 4Re h, where Re is the radius of the Earth. So UB V uB 4(6:4 106 m)2 (16 103 m)(1:0 10 3 J/m3 ) = 8:4 1015 J : 862 CHAPTER 31 INDUCTION AND INDUCTANCE (a) (b) 93E 25 0 mV 1 M = jdiE=dtj = 15::0 A/s = 1:67 mH : 1 2 = Mi1 = (1:67 mH)(3:60 A) = 6:00 mWb : (a) 94E 12 = L1 i1 = (25 mH)(6:0 mA) = 1:5 Wb : N 100 (b) E1 = L1 di1 = (25 mH)(4:0 A/s) = 100 mV : dt 2 1 21 = Mi1 = (3:0 mH)(6:0 mA) = 90 nWb : N 200 Use E2 = M di1 =dt to nd M : 95E E21 = M di1 = (3:0 mH)(4:0 A/s) = 12 mV : dt 3 V M = di E=dt = 6:0 A30 510 10 3 s) = 13 H : =(2: 1 (a) Assume the current is changing at the rate di=dt and calculate the total emf across both coils. First consider the left-hand coil. The magnetic eld due to the current in that coil points to the left. So does the magnetic eld due to the current in coil 2. When the current increases both elds increase and both changes in ux contribute emf's in the same direction. Thus the emf in coil 1 is di E1 = (L1 + M ) dt : The magnetic eld in coil 2 due to the current in that coil points to the left, as does the eld in coil 2 due to the current in coil 1. The two sources of emf are again in the same direction and the emf in coil 2 is di E2 = (L2 + M ) dt : The total emf across both coils is di E = E1 + E2 = (L1 + L2 + 2M ) dt : This is exactly the emf that would be produced if the coils were replaced by a single coil with inductance Leq = L1 + L2 + 2M . 96P CHAPTER 31 INDUCTION AND INDUCTANCE 863 (b) Reverse the leads of coil 2 so the current enters at the back of coil rather than the front as pictured in the diagram. Then the eld produced by coil 2 at the site of coil 1 is opposite the eld produced by coil 1 itself. The uxes have opposite signs. An increasing current in coil 1 tends to increase the ux in that coil but an increasing current in coil 2 tends to decrease it. The emf across coil 1 is Similarly the emf across coil 2 is The total emf across both coils is di E1 = (L1 M ) dt : di E2 = (L2 M ) dt : This the same as the emf that would be produced by a single coil with inductance Leq = L1 + L2 2M . di E = (L1 + L2 2M ) dt : 97P As long as the magnetic eld of the solenoid is entirely contained within the cross-section of the coil we always have sc = Bs As = Bs R2 , regardless of the shape, size, or possible lack of close-packing of the coil. Use the expression for the ux over the toroid cross-section derived in Sample Problem 31-6 to obtain N b b M = Mct = Ncict = Nc 0 i2t t h ln a = 0 N1N2 h ln a ; it 2 t where Nt = N1 and Nc = N2 . Assume the current in solenoid 1 is i and calculate the ux linkage in solenoid 2. The mutual inductance is this ux linkage divided by i. The magnetic eld inside solenoid 1 is parallel to the axis and has uniform magnitude B = 0 in1 , where n1 is the number of 2 turns per unit length of the solenoid. The cross-sectional area of the solenoid is R1 and since the eld is normal to a cross section the ux through a cross section is 2 = AB = R1 0 n1 i : 2 M = Mcs = Nics = N (0 iis nR ) = 0 R2 nN: s s 98P 99P 864 CHAPTER 31 INDUCTION AND INDUCTANCE Since the magnetic eld is zero outside the solenoid, this is also the ux through a cross section of solenoid 2. The number of turns in a length l of solenoid 2 is N2 = n2 l and the ux linkage is 2 N2 = n2 lR1 0 n1 i : The mutual inductance is 2 M = N2 = R1 l0 n1 n2 : i M does not depend on R2 because there is no magnetic eld in the region between the solenoids. Changing R2 does not change the ux through solenoid 2, but changing R1 does. (a) The ux over the loop cross section due to the current i in the wire is given by = Z a+b a 100P Bwire (r)l dr = Z a+b a 0 il dr = 0 il ln 1 + b : 2r 2 a Thus (b) From the formula for M obtained above b M = Ni = N0 l ln 1 + a : 2 7T 80 M = (100)(4 10 2 m/A)(0:30 m) ln 1 + 1::0 = 1:3 10 5 H : ...
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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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