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32 INDUCTANCE CHAPTER AND MAGNETIC ENERGY ActivPhysics can help with these problems: Activity 14.1 Section 32-1: Problem 1. Two coils have a mutual inductance of 2.0 H. If current in the first coil is changing at the rate of 60 A/s, what is the emf in the second coil? Mutual Inductance Solution M ( dI1=dt ) (2 H)(60 A/s) From Equation 32-2, E2 emf opposes the process which creates it.) 120 V. (The minus sign, Lenz's law, signifies that an induced Problem 2. A 500-V emf appears in a coil when the current in an adjacent coil changes at the rate of 3.5 A/ms. What is the mutual inductance of the coils? Solution Equation 32-2 gives M coupled.) E2= dI1=dt ) ( 500 V= 3.5 A /ms) ( 143 mH. (The sign of M depends on how the coils are Problem 3. The current in one coil is given by I I p sin 2 f t, where I p 75 mA, f in a second coil if the mutual inductance between the coils is 440 mH. 60 Hz, and t time. Find the peak emf Solution Suppose I1 I p sin 2 f t in Equation 32-2. Then E2 M dI1= dt 2 f MI p cos 2 f t, and the peak emf (when 60 Hz)( 440 mH)(75 mA) 12.4 V. cos 2 ft 1) is 2 f MI p (2 Problem 4. Two coils have a mutual inductance of 580 mH. One coil is supplied with a current given by I is in amperes and t in seconds. What is the induced emf in the other coil at time t 2.5 s? 3t 2 2t 4, where I Solution Since dI1=dt 6t 2 (in A/s), Equation 32-2 gives, for t comment in solution to Problem 1). 2.5 s, E2 (580 mH)(6 2.5 2)(A/s) 7.54 V (see Problem 5. An alternating current given by Ip sin 2 f t is supplied to one of two coils whose mutual inductance is M. (a) Find an expression for the emf in the second coil. (b) When I p 1.0 A and f 60 Hz, the peak emf in the second coil is measured at 50 V. What is the mutual inductance? Solution (a) From Equation 32-2, E2 M dI1= dt M2 f I p cos(2 f t ). (b) The peak value of the cosine is 1, so M E2 p= f I p 50 V= 2 2 ( 60 Hz)(1 A) 133 mH. (From the information given, only the magnitude of M can be determined; its sign depends on how the coils are coupled.) 758 CHAPTER 32 Problem 6. Find the mutual inductance of the two-coil system described in Problem 19 of Chapter 31. Solution The mutual inductance of this configuration of two coils is calculated in Example 32-1 (simply replace 2 by N2): M (4 10 7 H/m)(5)(5000= m) 1 (30 cm)2 111 mH. 2 4 . 0 N 2 ( N1=1 ) A1 . The data from Problem 31-19 then yields M Problem 7. Two long solenoids of length both have n turns per unit length. They have circular cross sections with radii R and 2R, respectively. The smaller solenoid is mounted inside the larger one, with their axes coinciding. Find the mutual inductance of this arrangement, neglecting any nonuniformity in the magnetic field near the ends. Solution All of the flux from the smaller solenoid (number one) links the larger solenoid (number two), so 2 2 2 B,2 N2 B1 A1 N2 ( 0 n1 I1 )( R ) R I1 , since both solenoids have the same number of turns and length. Dividing by I1 gives 0n M (see Equation 32-1). Note that the sign of M depends on the relative direction of the windings in the two solenoids. N1 B2 A1 and (Alternatively, only a fraction A1=A2 of the flux from the larger solenoid links the smaller solenoid, so B,1 M I B,1= 2 is the same.) Problem 8. Coils A and B have mutual inductance 25 mH. At time t 0 the current in coil A is zero. Subsequently a time-varying current is supplied to A, and the induced emf in coil B is given by E 50 0.2t , with E in V and t in ms. Find an expression for the time-varying current in coil A. Solution Equation 32-2 specifies EB [(2 A /ms)t we find I A (t ) M dI A=dt, so dI A=dt [50 V (0.2 V/ms)t ]= 25 mH). Integrating, and using I A (0) ( 0, (4 A /(ms) 2 )t 2 ]. The direction of IA depends on how the coils are coupled. Problem 9. A rectangular loop of length and width w is located a distance a from a long, straight, wire, as shown in Fig. 32-20. What is the mutual inductance of this arrangement? FIGURE 32-20 Problem 9. Solution When current I1 flows to the left in the wire, the flux through the loop is B,2 ( 0 I1= ) a w dr= ( 0 I1= ) 2 r 2 a ln(1 w=a) (see Example 31-2). Then Equation 32-1 gives M =I1 ( 0 = ) ln(1 w=a). (In calculating the flux, 2 B,2 the normal to the loop area was taken into the page, so the positive sense of circulation around the loop is CW. This determines the direction of the induced emf E2 in Equation 32-2.) z CHAPTER 32 759 Problem 10. Two wire loops of radii a and b lie in the same plane and have a common center. Find the mutual inductance of this arrangement, assuming b a. Hint: With b a, the magnetic field will be essentially uniform over the smaller loop. See Example 30-1. Solution If b a is assumed, the magnetic field from the large loop is essentially uniform over the small loop, and equal to the value at the center of the large loop, Bb 2 0 I b = b. Therefore, the flux through the small loop (due to current in the large loop) is B, a Bb Aa difficult to calculate 2 ( 0 Ib= b)( a2 ), and the mutual inductance is M 2 I 2 B, a= b 0 a = b. (Note: It would be more I B, b= a , from the dipole field of the small loop, but the result for M would be the same.) Section 32-2: Problem Self-Inductance 11. What is the self-inductance of a solenoid 50 cm long and 4.0 cm in diameter that contains 1,000 turns of wire? Solution 2 Equation 32-4 gives L 0 N A= approximation is valid here.) (4 10 7 H /m)(10 3 ) 2 (2 cm) 2= 50 cm) ( 316 mH. (The long thin solenoid . Problem 12. The current in an inductor is changing at the rate of 100 A/s, and the inductor emf is 40 V. What is its self-inductance? Solution From Equation 32-5, L E= dI=dt ) ( 40 V= 100 A /s) ( 0.4 H. Problem 13. A 2.0-A current is flowing in a 20-H inductor. A switch is opened, interrupting the current in 1.0 ms. What emf is induced in the inductor? Solution Assume that the current changes uniformly from 2 A to zero in 1 ms (or consider average values). Then dI=dt and Equation 32-5 gives E (20 H)( 20 A /ms) 40 kV. (The emf opposes the decreasing current.) 2 A /ms, Problem 14. A 60-mA current is flowing in a 100-mH inductor. Over a period of 1.0 ms the current is reversed, going steadily to 60 mA in the opposite direction. What is the inductor emf during this time? Solution Equation 32-5 gives EL L dI dt (0.1 H) [( 60) 60]mA (1 ms) 12 V. ( EL opposes the reversal of current.) 760 CHAPTER 32 Problem 15. A cardboard tube measures 15 cm long by 2.2 cm in diameter. How many turns of wire must be wound on the full length of the tube to make a 5.8-mH inductor? Solution From Equation 32-4, N L= 0 A [(5.8 mH)(15 cm)= 4 ( 10 7 H/m) (11 cm)2 ]1=2 . 135 103 turns. . Problem 16. The current in a 2.0-H inductor is given by I 3t 2 expression for the magnitude of the inductor emf. 15t 8, where t is in seconds and I in amperes. Find an Solution From Equation 32-5, E L dI=dt 2(6t 15), where E is in volts and t is in seconds. Problem 17. The emf in a 50-mH inductor has magnitude E 0.020t, with t in seconds and E in volts. At t 0 the inductor current is 300 mA. (a) If the current is increasing, what will be its value at t 3.0 s? (b) Repeat for the case when the current is decreasing. Solution (a) E has the opposite sign to dI=dt in Equation 32-5. When I is increasing, dI=dt dI dt E L 0.3 A 0, E is negative, E E . Thus, (0.02 V/s)t (0.05 H) F.4 A It, 0 HsK 2 and 0, E I E 1 A 0.4 2 t 2 2 s 0, so dI=dt F I H K I0 . (0.4 A /s 2 )t, and I At t 3 s, I (0.2 A /s 2 )(3 s) 2 2.1 A. (b) For dI=dt (0.2 A /s 2 )t 2 I0 . At t 3 s in this case, I increasing in absolute value.) 1.5 A. (After t 3=2 s, the current reverses direction and begins Problem 18. The current in a 40-mH inductor is given by I I0 e bt, where I0 the inductor emf at (a) t 0, (b) t 25 ms, and (c) t 50 ms? 10 A and b 20 s 1 . What is the magnitude of Solution The magnitude of the inductor emf is E (10 A)(20 s ) 1 L dI=dt LI0 be bt, from Equation 32-5. (a) At t 0, E(0) ( 40 mH) 8 V, so (b) at t 25 ms, E(t ) (8 V)e 0.5 4.85 V, and (c) E(50 ms) 2.94 V. Problem 19. A 2,000-turn solenoid is 65 cm long and has cross-sectional area 30 cm2. What rate of change of current will produce a 600-V emf in this solenoid? Solution 2 (4 10 7 H /m)(2000) 2 (30 cm 2 )= 65 cm) ( The self-inductance (of this long, thin solenoid) is L 0 N A= E =L (600 V)= 23.2 mH) 25.9 A /ms. ( (see Equation 32-4), so Equation 32-5 gives dI=dt 23.2 mH CHAPTER 32 761 Problem 20. You have a plastic rod 20 cm long and 1.5 cm in diameter. What inductance will you get if you wind the entire rod with a single layer of (a) 22-gauge (0.64-mm-diameter) and (b) 34-gauge (0.16-mm-diameter) wire? Assume adjacent turns are touching, separated only by a negligible thickness of enamel insulation. Solution The number of turns of wire is N Equation 32-4 gives L 0.64 mm, L (64= ) 2 16 =d , where 20 cm is the length of the coil and d is the diameter of the wire. Then d2 1 0 (= ) ( 4 7 D )= 2 1 4 ( 2 10 H /m)(20 cm)(1.5 cm=0.64 mm) = 0 ( D d ) 2 2 , where D 1.5 cm is the diameter of the rod. (a) For d 0.16 mm, L (108 H) 108 H, while (b) for d 1.73 mH. Problem 21. The emf in a 50-mH inductor is given by E Ep sin t, where Ep 75 V and in the inductor? (Assume the current swings symmetrically about zero.) 140 s 1 . What is the peak current Solution From Equation 32-5, dI=dt (Ep=L) sin t, so integration yields I (t ) (Ep= L) cos t. (Since I ( t ) is symmetric about I 0, the constant of integration is zero.) The peak current is I p Ep= L 75 V= 140 s ( 1 50 mH) 10.7 A. Problem 22. A coaxial cable consists of an inner conductor of radius a and outer conductor of radius b, as shown in Fig. 32-21. Current flows along one conductor and back along the other. Show that the inductance per unit length of the cable is 0 2 ln(b=a). FIGURE 32-21 Problem 22 Solution. Solution 2 The magnetic field strength between the conductors (a r b) is like that from a long straight wire, B 0 I= r (see Section 30-4). If we divide a rectangular area of length between the conductors into strips of width dr, then the flux b ( 0 I=2 ) b dr=r ( 0 I=2 ) ln(b=a) . Therefore, Equation 32-3 gives linking this length of cable is B a B dr a L= ( 0=2 ) ln(b=a). z z 762 CHAPTER 32 Section 32-3: Problem Inductors in Circuits 23. Show that the inductive time constant has the units of seconds. Solution A henry is a volt second/ampere (see Equation 32-5), so the units of L L=R are H / V s/ A s. Problem 24. What inductance should you put in series with a 100resistor to give a time constant of 2.2 ms? Solution Since the inductive time constant in an RL circuit is L L=R, L LR (2.2 ms)(100 ) 220 mH. Problem 25. The current in a series RL circuit rises to 20% of its final value in 31 s. If L . 18 mH, what is the resistance R? . Solution The buildup of current in an RL circuit with a battery is given by Equation 32-8, I (t ) I (1 e Rt=L ), where I ( L=t ) ln(1 I=I ) (1.8 mH= . s) ln(1 20%) 130 . 31 the final current. Solving for R, one finds R E0=R is Problem 26. The current in a series RL circuit rises to half its final value in 7.6 s. What is the time constant? Solution From Equation 32-8, L L=R t=ln(1 I=I ) 7.6 s=ln(1 1 2) 7.6 s=ln 2 11.0 s (see previous solution). Problem 27. A 10-H inductor is wound of wire with resistance 2.0 . If the inductor is connected across an ideal 12-V battery, how long will it take the current to reach 95% of its final value? Solution Reference to the solution to Problem 25 shows that t percentage I=I is independent of E0 . ) ( L=R) ln(1 I=I ) (10 H=2 ) ln(1 0.95) 15.0 s. (The Problem 28. In a series RL circuit like Fig. 32-8a, E0 switch been closed? 45 V, R 3.3 , and L 2.1 H. If the current is 9.5 A, how long has the Solution As in the previous problem, t (2.1 H= .3 ) ln(1 9.5= .6) 3 13 ( L=R) ln(1 0.759 s. I=I ). Here, I E0=R 45 V= .3 3 13.6 A, so t Problem 29. In Fig. 32-8a, take R 2.5 k and E0 50 V. When the switch is closed, the current through the inductor rises to 10 mA in 30 s. (a) What is the inductance? (b) What will be the current in the circuit after many time constants? CHAPTER 32 763 Solution ), the exponential term in Equation 32-8 is negligible. Thus, I (b) After a long time (t E0=R 50 V=2.5 k 1 20 mA. (a) The current has risen to half its final value in 30 s. Thus (Equation 32-8 again), 2 1 e Rt=L , or L Rt=ln 2 (2.5 k )(30 s)=ln 2 108 mH. Problem 30. A series RL circuit like Fig. 32-8a has E0 60 V, R 22 , and L 1.5 H. Find the rate of change of the current (a) immediately after the switch is closed and (b) 0.10 s later. Solution From Equations 32-5, and 7, dI=dt dI=dt ( 40 A /s)e ( 22 )( 0.1 s)= 1.5 H ) ( (E0=L)e Rt=L . (a) For t 0, dI=dt E0=L 60 V= .5 H 1 40 A/s. (b) For t 01 s, . 9.23 A /s. Problem 31. In Fig. 32-8a, take R 100 , L 2.0 H, and E0 12 V. At 20 ms after the switch is closed, what are (a) the circuit current, (b) the inductor emf, (c) the resistor voltage, (d) the rate of change of the circuit current, and (e) the power dissipation in the resistor? Solution 100 20 ms, and the final current is E0=R 12 V= 100 120 mA. After one time (a) The time constant is L=R 2 H= 1 constant, Equation 32-8 gives I 120 mA(1 e ) 75.9 mA. (b) The voltage drop across the inductor is (from EL E0 e Rt=L . After one time constant, VL (12 V)e 1 4.41 V. (Note that Equation 32-7) VL E0 IR VL VR E0 is Kirchhoff's loop law.) (c) VR IR (75.9 mA)(100 ) 7.59 V. (Alternatively, VR E0 VL 12 V 4.41 V. ) (d) From Equation 32-5 and the loop law, VL EL L dI=dt. After one time constant, dI=dt VL =L 2 2 4.41 V=2 H 2.21 A/s. (e) PR I R (75.9 mA) (100 ) 575 mW. Problem 32. Show that a series RL circuit reaches 99% of its final current in approximately 5 time constants. Solution E0=R is the final current), t= From Equation 32-8, if I 0.99 I (where I 1 e 5 99.3%.) constants. (For t 5 L , I=I L ln(1=0.01) 4.61, or t 4.61 time Problem 33. Resistor R2 in Fig. 32-22 is to limit the emf that develops when the switch is opened. What should be its value in order that the inductor emf not exceed 100 V? FIGURE 32-22 Problem 33. 764 CHAPTER 32 Solution As explained in Example 32-6, when the switch is opened (after having been closed a long time), the voltage across R2 (which equals the inductor emf ) is V2 I2 R2 E0 R2=R1 . If we choose to limit this to no more than 100 V, then R2 (100 V)(180 )=45 V 400 . Problem 34. In Fig. 32-11a take E0 12 V, R 2.7 , and L 20 H. Initially the switch is in position B and there is no current anywhere. At t 0 the switch is thrown to position A, and at t 10 s it is thrown back to position B. Find the inductor cursent at (a) t 5.0 s and (b) t 15 s. Solution (a) When the current is building up from zero, Equation 32-8 gives I (5 s) ( 2. 7 (12 V=2.7 )(10 s)= 20 H ) ( )(1 e ( 2. 7 )( 5 s)= 20 H ) ( ) 2.18 A. ) 3.29 A. This current decays (b) At t 10 s, the current has built up to I (10 s) (12 V=2.7 )(1 e when the switch is thrown back to B. Equation 32-9 (where t is the time since 10 s) gives I (3.29 A)e (2.7 )(15 s 10 s)=(20 H) 168 A. . Problem 35. A 5.0-A current is flowing through a nonideal inductor with L 500 mH. If the inductor is suddenly short-circuited, the inductor current drops to 2.5 A in 6.9 ms. What is the resistance of the inductor? Solution A real inductor can be represented by a resistance, R, in series with an inductance, L. When short-circuited, the inductor constitutes an LR circuit without a battery, and the current decays exponentially with time constant L=R, I I0 e Rt=L (see Equation 32-9). When the given data is substituted, we can solve for the resistance, I L 500 mH R ln 0 ln 2 50.2 . t I 6.9 ms FI F I F HK GJ H HK I K Problem 35 Solution. Problem 36. In Fig. 32-23, take E0 12 V, R1 4.0 , R2 8.0 , R3 2.0 , and L 2.0 H. What is the current I2 (a) immediately after the switch is first closed and (b) a long time after the switch is closed? (c) After a long time the switch is again opened. Now what is I2? FIGURE 32-23 Problem 36. CHAPTER 32 765 Solution (a) As explained in Example 32-6, the inductor current is zero just after the switch is closed. At this instant, the currents can be determined from the circuit with the inductance open-circuited, so that its branch can be removed. Thus, I3 0, and I1 I2 E= R1 R2 ) 12 V= 4 8) ( ( 1 A. Problem 36 Solution (a) (b) After the currents have been flowing a long time, they reach steady values ( dI=dt 0) , and the voltage across the inductance is zero. The currents can be found by short-circuiting the inductance (see Example 32-6 again, and refer to Chapter 28 if necessary): I1 I2 I3 R1 R3 R2 R3 I1 R2=( R2 E R2 R3=( R2 I1 1 5 I1 R3 ) 12 A 4 8 2= 10 2.14 A, 0.429 A, and 4 5 I1 R3 ) 1.71 A. Problem 36 Solution (b) (c) When the switch is reopened, no current flows through the battery's branch, I1 0, which can be removed from the circuit to calculate I2 I3 at this instant. The induced emf acts to keep the current flowing at its value in part (b) (as explained in Example 32-6), so I2 I3 1.71 A. Problem 36 Solution (c) Problem 37. In Fig. 32-24, take E0 20 V, R1 10 , R2 5.0 , and assume the switch has been open for a long time. (a) What is the inductor current immediately after the switch is closed? (b) What is the inductor current a long time after the switch is closed? (c) If after a long time the switch is again opened, what will be the voltage across R1 immediately afterward? 766 CHAPTER 32 FIGURE 32-24 Problem 37. Solution (a) If the switch has been open a long time, a steady current flows through the inductance ( dI L =dt 0). When the switch is 10 2 A. (Of course, I1 (0) I L (0), and closed (at t 0 ), IL cannot change instantaneously, so I L (0) E=R1 20 V= ), the currents are steady again and EL 0 (the inductance behaves like a I2 (0) 0. ) (b) After another long time (t 1 1 short circuit). The resistors are in parallel; therefore I L ( ) E(1=R1 1=R2 ) 20 V 1 10 6 A. (c) When the 5 c h switch is again opened, the current through R2 is zero, but I L cannot change instantly, so I L voltage across R1 is V1 I1 R1 (6 A)(10 ) 60 V. I1 IL ( ) 6 A. Thus, the Section 32-4: Problem Magnetic Energy 38. How much energy is stored in a 5.0-H inductor carrying 35 A? Solution Equation 32-10 gives U 1 2 LI 2 1 2 (5 H)(35 A)2 3.06 kJ. Problem 39. What is the current in a 10-mH inductor when the stored energy is 50 J ? Solution From Equation 32-10, I 2U=L 2(50 J)= mH 10 0.1 A. Problem 40. A 220-mH inductor carries 350 mA. How much energy must be supplied to the inductor in raising the current to 800 mA? Solution From Equation 32-10, U 1 2 L( I 2 f Ii2 ) 1 2 (220 mH)[(800 mA)2 (350 mA)2 ] 56.7 mJ. Problem 41. A 12-V battery, 5.0- resistor, and 18-H inductor are connected in series and allowed to reach a steady state. (a) What is the energy stored in the inductor? (b) Once in the steady state, over what time interval is the energy dissipated in the resistor equal to that stored in the inductor? CHAPTER 32 767 Solution (a) The steady state (i.e., final) current in an RL circuit with emf E0 is I E0 =R, so the energy stored in the inductor (Equation 32-10) is UL 1 LI 2 1 (18 H)(12 V= )2 518 J. (b) Energy is dissipated in the resistor at the rate 5 . 2 2 PR I 2 R (the joule heat), so the time interval queried is t U L= R P 1 2 LI 2=I 2 R L= R 2 18 H= 2 ( 5 ) 18 s. . Problem 42. A battery, switch, resistor, and inductor are connected in series. When the switch is closed the current rises to half its steady-state in value 1.0 ms. How long does it take for the magnetic energy in the inductor to rise to half its steady-state value? Solution According to Equations 32-8 and 10, U= U Rtu=L I=I 1 e Rt=L . When the stored energy is half its steady-state value, ( 1= 2 1 e , or tu ( L=R) ln[ 2= 2 1)] 1.28 L=R. The current is half its steady-state value when t1 ( L=R) ln 2 1 ms (see solution to Problem 29). Dividing these results, we find tu (1.28=ln 2) ms 1.77 ms. Problem 43. The current in a 2.0-H inductor is decreased linearly from 5.0 A to zero over 10 ms. (a) What is the average rate at which energy is being extracted from the inductor during this time? (b) Is the instantaneous rate constant? Solution 2 2 1 1 0 in t 10 ms, so the rate of decrease is U= t (a) The energy falls from Ui 25 J to U f 2 LI 2 (2 H)(5 A) 25 J= ms 10 2.5 kW. (b) The discussion in the text leading to Equation 32-10 shows that the instantaneous power is PL LI ( dI=dt ), so even if dI=dt is constant, I and PL are not. Problem 44. When a nonideal 1.0-H inductor is short-circuited, its magnetic energy drops to one-fourth of its original value in 3.6 s. What is its resistance? Solution If U 1 2 LI 2 1 4 U0 L 1 4 dLI i, then I 1 2 2 0 1 2 I 0 . The time required for the current to decay to half its original value (the L ln 2= 1=2 t (1 H)(ln 2)= 3.6 s) ( 193 m . "half-life") is t1=2 ln 2 (see solution to Problem 35) so R Problem 45. The current in a 2.0-H inductor is increasing. At some instant, the current is 3.0 A and the inductor emf is 5.0 V. At what rate is the inductor's magnetic energy increasing at this instant? Solution The rate at which energy is stored in an inductor is PL LI ( dI=dt ) (see the discussion of "Magnetic Energy in an Inductor" EL , and PL I EL leading to Equation 32-10). When the current is increasing, as for this inductor, L(dI=dt ) (3 A )(5 V ) 15 W. Problem 46. A 500-turn solenoid is 23 cm long, 1.5 cm in diameter, and carries 65 mA. How much magnetic energy does it contain? 768 CHAPTER 32 Solution Combining Equations 32-4 and 10, we find U (23 cm) 0.510 J. 1 2 ( 0N 2 A= )I 2 (2 10 7 H/m)( 1 2 500 15 cm . 65 mA)2 Problem 47. A superconducting solenoid with inductance L 3.5 H carries 1.8 kA. Copper is embedded in the coils to carry the current in the event of a quench (see Example 32-7). (a) What is the magnetic energy in the solenoid? (b) What is the maximum resistance of the copper that will limit the power dissipation to 100 kW immediately after a loss of superconductivity? (c) With this resistance, how long will it take the power to drop to 50 kW? Solution (a) In its superconducting state, the solenoid's stored energy (Equation 32-10) is U 1 2 LI 2 1 2 (3.5 H)(18 kA)2 . 2 5.67 MJ. (b) The current in an inductor cannot change instantaneously, so if the power dissipated in the copper ( I R) just after a (. sudden loss of superconductivity must be less than 100 kW, then R 100 kW= 18 kA) 2 30.9 m . (c) The power drops to one half its maximum original value, when the current drops to 1= 2 times its initial value. From Equation 32-9, the decay time is t ( L=R) ln( I0=I ) (3.5 H= .9 m ) ln 2 39.3 s. 30 Problem 48. Show that the quantity B 2 =2 0 has the units of energy density ( J/m 3 ). Solution The units of B 2 =2 0 ( are T 2= N/A 2 ) ( N/A m) 2= N/A 2 ) ( ( N/m 2 )(m /m) J/m 3 . Problem 49. The Alcator fusion experiment at MIT has a 50-T magnetic field. What is the magnetic energy density in Alcator? Solution ( From Equation 32-11, u B (50 T) 2= 8 gasoline; see Appendix C.) 10 7 N/A 2 ) 995 MJ/m 3 . (This is about 2.8% of the energy density content of Problem 50. What is the magnetic field strength in a region where the magnetic energy density is 7.8 J/cm 3? Solution From Equation 32-11, B 2 0 uB (8 10 7 H/m)(7.8 J/cm 3 ) 4.43 T. Problem 51. The magnetic field of a neutron star is about 10 8 T. How does the energy density in this field compare with the energy density stored in (a) gasoline and (b) pure uranium-235 (mass density 19 10 3 kg /m 3 ) ? Consult Appendix C. Solution ( 10 7 H/m) 3.98 10 21 J/m 3 (see The energy density in a field of this strength is u B B2=2 0 (10 8 T) 2= 8 11 . Equation 32-11). This is about (a) 11 10 times the energy density content of gasoline ( 44 MJ/kg 800 kg /m 3 10 3 3.52 10 J/m ) , and (b) 2600 times that of pure U 235 (8 1013 J/kg 19 10 3 kg /m 3 1.52 1018 J/m 3 ). CHAPTER 32 769 Problem 52. A loop of magnetic field arches above the Sun's surface, forming a tube approximately 10 5 km long and 104 km in diameter. If the magnetic field strength in the tube is 50 G, what is its magnetic energy content? Solution The energy content is the average energy density times the volume of the magnetic field region, U uB Volume (B = 2 2 0) 1 4 d 2 (5 mT 10 km) (10 km)= 32 10 ( 4 2 5 7 H/m) 7.81 10 22 J. Problem 53. A single-turn loop of radius R carries current I. How does the magnetic energy density at the loop center compare with that of a long solenoid of the same radius, carrying the same current, and consisting of n turns per unit length? Solution ( The energy density at the center of the loop is u Bloop) B2=2 0 ( 0 I=2 R) 2=2 ( 32-11). In a long thin solenoid of the same radius, u Bsolenoid ) ( 0 nI ) 2=2 0 is 1=4n 2 R 2 . 0 2 2 0I 2 = R2 (see Equations 30-3 and 8 0n ( ( I =2, so the ratio of u Bloop) to u Bsolenoid ) Problem j 54. A magnetic field is given by B B0 ( x=a) 2 , where B0 and a are constants. Find an expression for the magnetic energy in a cube of side a with one corner at the origin and sides extending along the coordinate axes. Solution Divide the cube into volume elements consisting of slices parallel to the y-z plane with dV length). Then U a 2 dx (where a is the side z cube u B dV 2 B0 2 0 a2 z a 0 x 4 dx 2 B0 a 3 . 10 0 Problem 54 Solution. Problem 55. A toroidal coil has inner radius R and a square cross section of side (Fig. 32-25). It is wound with N turns of wire, and carries a current I. Show that the magnetic energy in the toroid is given by U 0N 2 2 I 4 ln R F I. HR K 770 CHAPTER 32 FIGURE 35-25 Problem 55. Solution 2 The magnetic field inside a cylindrically symmetric toroid was found by using Ampere's law, B 0 NI= r, where r is the distance from the central axis (see Equation 30-12). The energy in this magnetic field can be found by integrating the energy density, u B B 2=2 0 , over the volume of the toroid, with volume elements consisting of cylindrical shells of radius r, appropriate height, and thickness dr. For a toroid of square cross section of side , and inner radius R, dV 2 r dr (see figure), and U z toroid u B dV z R R 1 2 0 F NI I 2 GrJ 2 H K 2 0 r dr 0N 2 I 2 4 z R R dr r 0N 2 I 2 4 ln 1 F I. H RK B toroid An alternative approach is to use the flux linkage of the toroid from the solution to Problem 31-30, N (recall that a R, b R , and B was the flux through one turn), and U 1 I toroid . 2 LI Problem 56. A toroid of inner radius 1.5 m and square cross section is wound with 2,500 turns. What must be the length of its cross-sectional square if the toroid contains 80 J of magnetic energy at a current of 63 A? (See the preceding problem.) Solution The energy in the magnetic field of such a toroid was calculated in the previous problem, U ( 0=4 ) N 2 I 2 ln(1 =R), so ln(1 =R) (80 J)=[(10 7 H /m)(2500) 2 (63 A) 2 ] 3.22 cm. An approximate solution for R can be found by expanding the logarithm: ' 3.22 cm 150 cm 22.0 cm. A more exact solution to the transcendental equation can be found numerically or by using Newton's method (with 22.0 cm as a first guess) with the result that 22.79 cm 22.80 cm. Paired Problems Problem 57. Two coils have mutual inductance M. The current supplied to coil A is given by I magnitude of the induced emf in coil B. bt 2 . Find an expression for the Solution From Equation 32-2, EB M ( dI A=dt ) M (2bt ), where the negative value means the emf opposes the change in current. Problem 58. Two coils have mutual inductance M, and a time-varying current is supplied to coil A; at time t 0 that current is CHAPTER 32 771 zero. The magnitude of the induced emf in coil B is given by E b t . Find an expression for the current in coil A. Solution From Equation 32-2, dI A=dt EB=M bt 1=2=M. Integrating, and using the condition I A (0) 2bt 3=2= M. (The minus sign gives directions of IA and EB consistent with Lenz's law.) 3 0, we find I A (t ) Problem 59. In the circuit of Fig. 32-8a, take E0 250 mA. Find the inductance. 5.0 V and R 18 . At 2.5 s after the switch is closed, the circuit current is . Solution Equation 32-8 for the build-up of current in the RL circuit gives L 1.8=5) 47.7 H. Rt=ln(1 IR=E0 ) (1.8 )(2.5 s)=ln(1 0.25 Problem 60. In the circuit of Fig. 32-8a, take E0 2.1 V. Find the inductance. 5.0 V and R 18 . At 2.5 s after the switch is closed, the inductor emf is . Solution Equation 32-7 gives L Rt=ln( E0=EL ) (1.8 )(2.5 s)=ln(5.0=2.1) 519 H. (Note that EL . 2.1 V.) Problem 61. In Fig. 32-13a, take E0 25 V, R1 1.5 , and R2 4.2 . What is the voltage across R2 (a) immediately after the switch is first closed and (b) a long time after the switch is closed? (c) Long after the switch is closed it is again opened. Now what is the voltage across R2? Solution The circuit is analyzed in Example 32-6 at the instants of time mentioned in this problem, so all that is necessary here is ( 4.39 A, to find numerical values. (a) At the moment the switch is first closed, I E0=( R1 R2 ) 25 V= 1.5 4.2) and V2 (4.39 A)(4.2 ) 18.4 V (b) When the current is steady, the current through R2 is zero (since the inductor is an ideal one). (c) When the switch is reopened, V2 E0 R2=R1 (25 V)( 4.2=1.5) 70.0 V, momentarily. Problem 62. In Fig. 32-13, take R2 battery emf E0 ? 5R1 . If the maximum possible value for the inductor emf in this circuit is 300 V, what is the Solution The current in the inductor (I), and its derivative ( dI=dt ), are both exponential, and thus have maximum magnitudes just after the switch is closed or opened (refer to Example 32-6 and Fig. 32-14). Since the inductor is in parallel with R2, EL L(dI=dt ) I2 R2 , and the maximum values of inductor emf are E0 R2=( R1 R2 ) (when the switch is closed) and E0 R2=R1 (when the switch is reopened). For R2 5R1 , the latter is greater ( 5E0 =6 versus 5E0 ), so 300 V 5E0 , or E0 60 V. Problem 63. A wire of radius R carries a current I distributed uniformly over its cross section. Find an expression for the magnetic energy per unit length in the region from R to 100R. 772 CHAPTER 32 Solution The magnetic field from a long straight wire has cylindrical symmetry. A thin coaxial cylindrical shell of length , radius r, 2 and thickness dr, has volume dV 2 r dr and magnetic energy density u B B 2=2 0 . Outside the wire, B 0 I= r (see Equation 30-8), so the energy per unit length in a volume extending between r R and 100R is U z z u B dV R 100 R 8 2 0I 2 2 r 2 r dr 0I 2 4 ln 100. Problem 64. A wire of radius R carries a current I distributed uniformly over its cross section. Find an expression for the total magnetic energy per unit length within the wire. Solution The field inside such a wire is B solution) 2 0 Ir= R 2 (see Equation 30-9) so the energy within a unit length is (see previous U z R 0 8 0I 2 2 2 r 2 r dr R4 0I 2 4 r4 4 R4 R 0I 2 0 16 . Supplementary Problems Problem 65. (a) Use the result of Problem 55 to determine the inductance of a toroid. (b) Show that your result reduces to the inductance of a long solenoid when R . Solutioo (a) Since U 1 2 LI 2 , dividing the expression for U in Problem 55 by 2 2 2 1 2 I 2 , we find L ( 0N 2 =2 ) ln(1 =R). (b) For A is the cross-sectional area, 2 R Solenoid is the length, R, ln(1 =R) =R, so L 0 N =2 R . Since and n N= Solenoid is the number of turns per unit length of the toroidal solenoid, this result is approximately the same as Equation 32-4. Problem 66. Two long, flat parallel bars of width w and spacing d carry equal but opposite currents I, as shown in Fig. 32-26. (a) Use Ampre's law to find the magnetic field between the bars. Take d w so you can neglect fringing fields. (b) Use your result to find the magnetic energy per unit length stored between the bars. (c) Compare your result in (b) with the expression U 1 LI 2 to find an expression for the inductance per unit length of the bars. 2 Solution (a) The magnetic field between the bars is approximately uniform and directed horizontally to the right in Fig. 32-26. The w current in each bar (assumed uniform) contributes equally to the field (see Equation 30-10), so B 2( 1 0 I=w) 0 I= . 2 Outside the bars, the field is approximately zero. (See also Chapter 30, Problem 39.) (b) The product of the volume between a unit length of bars, V= wd , and the magnetic energy density, u B B2=2 0 (from Equation 32-11), gives the energy 2 2 LI 2=2, so the stored per unit length: U= ( 0 I=w) 2 wd=2 0 0 I d= w. (c) In terms of the inductance, U= w expression in part (b) implies that L= 0 d= . CHAPTER 32 773 FIGURE 32-26 Problem 66 Solution. Problem 77. (a) Use Equation 32-9 to write an expression for the power dissipation in the resistor as a function of time, and (b) integrate from t 0 to t to show that the total energy dissipated is equal to the energy initially stored in the 2 inductor, namely, 1 LI0 . 2 Solution Equation 32-9 gives the current decaying through a resistor connected to an inductor carrying an initial current I0 . (a) The 2 instantaneous power dissipated in the resistor is PR I 2 R I0 Re 2 Rt=L . (b) In a time interval dt, the energy dissipated is dU PR dt, so the total energy dissipated is U z 0 2 I0 Re 2 Rt=L dt 2 I0 R e 2 Rt=L ( 2 R=L) 0 2 I0 RL 2R 1 2 LI0 . 2 This is precisely the energy initially stored in the inductor. Problem 68. (a) Find the magnetic energy density as a function of radial distance for the coaxial cable of Problem 22. (b) Integrate over the volume between the cable conductors to show that the total energy per unit length of the cable is U 0 4 I 2 ln(b=a). Hint: Your volume element should be a cylindrical shell of radius r, thickness dr, and length . What is its volume dV? (c) Use the expression U 1 LI 2 to find the inductance per unit length, and show that your result agrees with that of 2 Problem 22. Solution (a) The magnetic field strength between the conductors is B Problem 40). Thus, Equation 32-11 gives u B ( 0 I=2 r ) 2=2 2 r dr, so the magnetic field energy it contains is dU for the energy is U 2 0 I= 0 r, a 2 8 0I = r 2 2 b (see Section 30-4 or Chapter 30, 0I 2 r . (b) The volume of the cylindrical shell is u B dV 0I 2 ( dr r 0I 2 = 8 0I 2 2 r )(2 r dr ) b FI. HK a dr=4 r. The integral z dU 4 z b a 2 4 ln (c) Comparison with the equation U 1 LI 2 shows that the inductance per unit length of cable is L= 2 which is the expression obtained in Problem 22 by calculation of the flux. ( 2 0= ) ln(b=a), 774 CHAPTER 32 Problem 69. An electric field and a magnetic field have the same energy density. Obtain an expressioo for the ratio E=B, and evaluate this ratio numerically. What are its units? Is your answer close to any of the fundamental constants listed inside the front cover? Solution The combination of Equations 26-3 and 32-11 implies that if uE Numerically, (9 9 0 4 2 2 10 7 7 N/A and (1=4 2 8 2 0) 9 10 N m /C , so 1= 9 1 2 2 0E 2 2 uB B2= 2 0 0 0, then E=B 1= 0 0. 10 N m /C )= 10 ( N/A ) 3 10 m/s, which is, in fact, the speed of light (see Section 34-5). Problem 70. Two long, straight, parallel wires are a distance d apart. The wires have radius a, where a d . Current flows down one wire and back along the other. Find the inductance per unit length of the parallel wires. Assume the wire radius is so small that you can neglect the magnetic flux within the wires themselves. Solution 2 x, directed out of the page for x a. The magnetic field from the first wire (whose axis is through x 0 ) is B1 0 I= 2 x ), also out of the page for (See Equation 30-8.) The field from the second wire (axis through x d) is B2 0 I= ( d x d a. If we divide an area of length , between the wires, into strips of width dx parallel to the y-axis, the total flux linking the circuit per unit length is B z z d a B1 dx ln ex d a 0 B2 dx ln( d x) 0 0I 2 F G H d a z z x dx 0 0I d a dx d x a d 0I 2 d a d a j 2 Fd ln Ha ln I K I J K 0I ln d . a (The upper limit of the first integral and the lower limit of the second presume that a d. ) Comparison with Equation 32-3 gives an inductance per unit length of L= ( 0= ) ln( d=a). Problem 70 Solution. CHAPTER 32 775 Problem 71. The switch in the circuit of Fig. 32-27 is closed at time t 0, at which instant the inductor current is zero. Write the loop and node laws for this circuit, and show that they are satisfied if the inductor current is given by I ( 0ER1 ) = R||t=L (1 e ) , where R|| is the resistance of R1 and R2 were they connected in parallel. Solutiooa The node law (applied to either node, with currents as shown superimposed on Fig. 32-27) gives I1 I2 I L 0. Two independent loop equations (for both loops containing E0 ) are E0 I1 R1 I2 R2 0 and E0 I1 R1 EL 0. Here, EL L( dI L=dt ) is the inductor's induced emf. Note that I1 and I2 can be determined from IL, since I1 (E0 EL )=R1 and I2 I1 I L . These currents automatically satisfy the node equation and the second loop equation, so we oeed only check that the expression given for I L satisfies the first loop equation. The equation can be written entirely in terms of IL: 0 E0 I1 R1 I2 R2 E0 (E0 EL ) ( I1 I L ) R2 EL (E0 EL )( R2=R1 ) I L R2 EL (1 R2=R1 ) I L R2 E0 R2=R1 . ) If we divide by R2, substitute for EL , and use R|| R1 R2=( R1 R2 ), the equation becomes ( L=R|| )( dI L= dt I L E0=R1 0. Substituting the given expression for I L , we, indeed, find that L R|| F E IF R e GR J L H H K 0 || 1 R||t=L I E KR 0 1 (1 e R||t=L ) E0 R1 0. FIGURE 32-27 Problem 71 Solution. Problem 72. Earth's magnetic field ends abruptly on the sunward side at approximately the point where the magnetic energy density has dropped to the same value as the kinetic energy density in the solar wind. Near Earth, the solar wind contains about 5 protons and 5 electrons per cubic centimeter, and flows at 400 km/s. Treating Earth's field as that of a dipole with dipole moment 8 10 22 J/T, estimate the distance to the point above the equator where the field ends. Solution Let the number density of particles of mass mp, moving with speed v p , be np. Then the kinetic energy density is uk 6.7 1 2 n p mp v2 p 10 1 2 2 p vp , where p n p m p is the particle mass density. (Since 1 2 6 e 1836, p= 3 the kinetic energy density 27 of electrons can be neglected compared to that of protons.) Thus, uk (5 10 10 m )(167 10 . kg)(4 105 m/s)2 J/m 3 . The dipole field strength in an equatorial plane (perpendicular to the magnetic moment) can be found, by 4 3 analogy with the corresponding electric dipole field strength in Equation 23-5a, to be B 0 = r . Thus, the energy 2 = 32 2 r 6 . density in the earth's magnetic field, at points above the equator, is u B B2=2 0 ( 0 =4 r 3 ) 2=2 0 0 The distance at which uk equals uB is r F G 32 H 0 2 I L10 N/A )(8 ( J M 8 (6.7 10 u K N 2 1=6 7 2 k 10 22 J/T) 2 10 J/m 3 ) O 5.8 P Q 1=6 10 4 km, or about 9.1RE from Earth's center. ... View Full Document