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© 2000 by Harcourt, Inc. All rights reserved. Chapter 32 Solutions *32.1 ε = L I t = (3.00 × 10 – 3 H) 1.50 A 0.200 A 0.200 s = 1.95 × 10 –2 V = 19.5 mV 32.2 Treating the telephone cord as a solenoid, we have: L = µ 0 N 2 A l = (4π × 10 7 T m / A)(70.0) 2 ( π )(6.50 × 10 3 m) 2 0.600 m = 1.36 H µ 32.3 ε =+ L I t = (2.00 H) 0.500 A 0.0100 s = 100 V 32.4 L = µ 0 n 2 A l so n = L µ 0 A l = 7.80 × 10 3 turns/m 32.5 L = N Φ B I →Φ B = LI N = 240 nT · m 2 (through each turn) 32.6 ε = L dI dt where L = µ 0 N 2 A l Thus, ε = µ 0 N 2 A l dI dt = 4π × 10 7 T mA ( ) 300 ( ) 2 π × 10 4 m 2 ( ) 0.150 m 10.0 A s ( ) = 2.37 mV 32.7 ε back = –ε = L dI dt = L d dt (I max sin ω t) = Lω I max cos ω t = (10.0 × 10 -3 )(120π )(5.00) cos ω t ε back = (6.00π ) cos (120π t) = (18.8 V) cos (377t)
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2000 by Harcourt, Inc. All rights reserved.Chapter 32 Solutions*32.1 = L It = (3.00 10 3 H) 1. 50 A 0. 200 A0. 200 s|.`, = 1.95 102 V = 19.5 mV 32.2 Treating the telephone cord as a solenoid, we have:L = 0N2Al = (4 107 T m / A)(70.0)2()(6. 50 103 m)20.600 m = 1.36 H 32.3 = + L It|.`,= (2.00 H)0. 500 A0.0100 s|.

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= 100 V 32.4 L = 0n2A l so n =L0A l= 7.80 103 turns/ m 32.5 L =N BI B=LIN= 240 nT m2 (through each turn) 32.6 LdIdt where L 0N2AlThu s, 0N2Al|.

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(1. 22 A) 1 e350( ) 1.22 A (c) I Imaxet/ and 0.160 1. 22et/ so t = ln(0.131) = 58.1 ms Chapter 32 Solutions 251 2000 by Harcourt, Inc. All rights reserved.32.30 (a) For a series connection, both inductors carry equal currents at every instant, so dI/ dt is t hesame for both. The voltage across the pair is LeqdIdt= L1dIdt+ L2dIdtso Leq= L1+ L2 (b) LeqdIdt= L1dI1dt= L2dI2dt= VLwhere I = I1+ I2 and dIdt=dI1dt+dI2dtThus, VLLeq= VLL1+ VLL2 and 1Leq=1L1+1L2 (c) LeqdIdt+ ReqI = L1dIdt+ IR1+ L2dIdt+ IR2Now I and dI/dt are separate quantities under our control, so functional equality requires both Leq= L1+ L2 and Req= R 1+ R 2 (d) V = LeqdIdt+ ReqI = L1dI1dt+ R1I1= L2dI2dt+ R2I2 where I = I1+ I2 and dIdt=dI1dt+dI2dtWe may choose to keep the currents constant in time. Then, 1Req=1R 1+1R 2We may choose to make the current swing through 0. Then, 1Leq=1L1+1L2 This equivalent coil with resistance will be equivalentto the pair of real inductors for all other currents as well. 32.31 L = N BI = 200(3.70 10 4)1.75 = 42.3 mH so U = 12 LI 2 = 12 (0.423 H)(1.75 A) 2 = 0.0648 J 32.32 (a) The magnetic energy density is given byu = B 220 = (4.50 T)22(1.26 10 6 T m/ A) = 8.06 106 J/ m3 (b) The magnetic energy stored in the field equals u times the volume of the solenoid (t hevolume in which B is non-zero).U = uV = (8.06 106 J/ m3) (0. 260 m)(0.0310 m)2[ ] 6.32 kJ 252 Chapter 32 Solutions32.33 L 0N2Al 0(68.0)2(0.600 102)20.0800 8. 21 H U 12LI212(8. 21 106H)(0.770 A)2 2.44 J 32.34 (a) U = 12 LI 2 = 12 L 2R|.`,2L28R2 (0.800)(500)28(30.0)2 = 27.8 J (b) I = R|.`,1 e(R/ L)t[ ]so 2R R|.`,1 e(R/ L)t[ ]e(R/ L)t12 RLt ln 2 so t LRln 2 0.80030.0ln 2 18.5 ms 32.35 u = 0 E 22 = 44.2 nJ/ m3 u = B 220 = 995 J/ m3 *32.36 (a) U = 12 LI 2 = 12 (4.00 H)(0.500 A) 2 = 0.500 J (b)dUdt = LI = (4.00 H)(1.00 A) = 4.00 J/ s = 4.00 W (c) P = (V)I = (22.0 V)(0.500 A) = 11.0 W 32.37 From Equation 32.7, I R1 e Rt L( )(a) The maximum current, after a long time t , is I R 2.00 A.At that time, the inductor is fully energized and P I(V) (2.00 A)(10.0 V) = 20.0 W (b) Plost I2R (2.00 A)2(5.00 ) = 20.0 W (c) Pinductor I(Vdrop) 0 (d) U LI22 (10.0 H)(2.00 A)22 20.0 J Chapter 32 Solutions 253 2000 by Harcourt, Inc. All rights reserved.32.38 We have u e0E22and u B220Therefore e0E22 B220so B2 e00E2 B E e00 6.80 105 V/ m3.00 108 m / s 2.27 10 3 T 32.39 The total magnetic energy is the volume integral of the energy density, u B220Because B changes with position, u is not constant. For B B0R / r ( )2, u B0220|.

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Rr|.`,4Next, we set up an expression for the magnetic energy in a spherical shell of radius r andthickness dr. Such a shell has a volume 4 r 2 dr, so the energy stored in it is dU u 4 r2dr( ) 2B02R40|.

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drr2We integrate this expression for r = R to r = to obtain the total magnetic energy outside t hesphere. This gives U 2 B20 R 30 = 2 (5.00 105 T)2(6.00 106 m)3(1.26 10 6 T m/ A) = 2.70 1018 J 32.40 I1(t ) Imaxetsin t with Imax 5.00 A, 0.0250 s1, and 377 rad s . dI1dt Imaxet sin t + cos t ( )At t 0.800 s , dI1dt 5.00 A s ( )e0.0200 0.0250 ( )sin 0.800 377 ( ) ( ) + 377cos 0.800 377 ( ) ( ) [ ] dI1dt 1.85 103 A sThus, 2 MdI1dt: M 2dI1dt +3. 20 V1.85 103 A s 1.73 mH 254 Chapter 32 Solutions32.41 2 MdI1dt (1.00 104 H)(1.00 104 A/ s) cos(1000t ) 2( )max 1.00 V 32.42 M = 2dI1dt=96.0 mV1. 20 A/ s= 80.0 mH 32.43 (a) M NBBAIA700(90.0 106)3. 50 18.0 mH (b) LA AIA400(300 106)3. 50 34.3 mH (c) B MdIAdt (18.0 mH)(0. 500 A/ s) 9.00 mV 32.44 M N212I1N2B1A1( )I1N2 0n1I1( )A1 [ ]I1 N2 0n1A1 M (1.00) 4 107 T m A( )70.00.0500 m|.`, 5.00 103 m( )2

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2 1. 58 104rad / sTherefore, fd d2 2.51 kHz (b) Rc 4LC 69.9 32.57 (a) 0 1LC 1(0. 500)(0.100 106) 4.47 krad/ s (b) d 1LC R2L|.`,2 4.36 krad/ s (c) 0 = 2.53% lower 32.58 Choose to call positive current clockwise in Figure 32.19. It drains charge from the capacitoraccording to I = dQ/dt. A clockwise trip around the circuit then gives +QC IR LdIdt= 0 +QC+dQdtR + LddtdQdt= 0, identical with Equation 32.29.32.59 (a) Q QmaxeRt2L cos dt so Imax eRt2L 0. 500 eRt2Land Rt2L ln(0. 500) t 2LRln 0. 500 ( ) 0.6932LR|.`, (b) U0 Qmax2 and U 0. 500U0so Q = 0.500 Qmax = 0.707Qmax t 2LRln(0.707) = 0. 3472LR|.`, (half as long) Chapter 32 Solutions 259 2000 by Harcourt, Inc. All rights reserved.32.60 With Q Qmax at t 0, the charge on the capacitor at any time is Q Qmaxcos t where 1 LC . The energy stored in the capacitor at time t is then U Q22C Qmax22Ccos2t U0cos2t .When U 14U0, cos t 12and t 13 rad Therefore, tLC 3or t2LC 29The inductance is then: L 9t22C 32.61 (a) L LdIdt 1.00 mH ( )d 20.0t ( )dt 20.0 mV (b) Q I dt0t 20.0t ( )dt0t 10.0t2 VC QC 10.0t21.00 106 F 10.0 MV s2( )t2 (c) When Q22C 12LI2, or 10.0t2( )22 1.00 106( ) 121.00 103( )20.0t ( )2,then 100t4 400 109( )t2. The earliest time this is true is at t 4.00 109 s 63.2 s 32.62 (a) L= LdIdt= Lddt(Kt ) = LK (b) I =dQdt, so Q I dt0t Kt dt0t 12Kt2and VC QC Kt22C (c) When 12C VC( )2=12LI2, 12CK2t44C2|.

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=12L K2t2( )Thu s t = 2 LC 260 Chapter 32 Solutions32.63 12Q2C=12CQ2|.`,2+12LI2so I =3Q24CLThe flux through each turn of the coil is B=LIN= Q2N3LC where N is the number of turns.32.64 Equation 30.16: B 0NI2r(a) B BdA 0NI2rhdrab 0NIh2drrab 0NIh2lnba|.`, L NBI 0N2h2lnba|.`, (b) L 0(500)2(0.0100)2ln12.010.0|.`, 91.2 H (c) Lappx 0N22AR|.`, 0(500)222.00 104 m20.110|.

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90.9 H *32.65 (a) At the center, B N0IR22(R2+ 02)3/ 2 N0I2RSo the coil creates flux through itself B BA cos N0I2R R2cos 0 2N0IRWhen the current it carries changes, L NdBdt N 2N0RdIdt LdIdtso L 2N20R (b) 2 r 3(0.3 m), so r 0.14 m; L 2 12 .|,`4 107 T mA 0.14 m = 2.8 107 H ~ 100 nH (c)LR 2.8 107 V s/ A270 V/ A = 1.0 10 9 s ~ 1 ns Chapter 32 Solutions 261 2000 by Harcourt, Inc. All rights reserved.32.66 (a) If unrolled, the wire forms the diagonal of a 0.100 m(10.0 cm) rectangle as shown. The length of this rectangleis L 9.80 m ( )2 0.100 m ( )2 L 0.100 m9.80 mThe mean circumference of each turn is C r 2 , where r 24.0 + 0.6442 mm is the meanradius of each turn. The number of turns is then: N LC 9.80 m ( )2 0.100 m ( )2224.0 + 0.6442|.`, 103 m 127 (b) R lA 1.70 108 m( )10.0 m ( ) 0. 322 103 m( )2 0.522 (c) L N2A l 8000 l LC|.`,2 r ( )2 L 800 4 107( )0.100 m9.80 m ( )2 0.100 m ( )2 24.0 + 0.644 ( ) 103 m|.

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224.0 + 0.6442|.`, 103 m

]]]2 L 7.68 102 H 76.8 mH 32.67 From Amperes law, the magnetic field at distance r R is found as: B 2r ( ) 0J r2( ) 0IR2|.

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r2( ), or B 0Ir2R2The magnetic energy per unit length within the wire is then Ul B2202r dr ( )0R 0I24R4r3dr0R 0I24R4R44|.

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= 0I216 This is independent of the radius of the wire.262 Chapter 32 Solutions32.68 The primary circuit (containing the battery and solenoid) is a nRL circuit with R 14.0 , and L 0N2Al 4 107( )12 500 ( )21.00 104( )0.0700 0. 280 H(a) The time for the current to reach 63.2% of the maxi mu mvalue is the time constant of the circuit: LR 0. 280 H14.0 0.0200 s 20.0 ms (b) The solenoid's average back emf is L L It|.`, LIf 0t|.

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where If 0.632Imax 0.632 VR|.`, 0.63260.0 V14.0 |.`, 2.71 AThus, L 0. 280 H ( )2.71 A0.0200 s|.`, 37.9 V (c) The average rate of change of flux through each turn of the overwrapped concentric coil is t hesame as that through a turn on the solenoid: Bt 0n I ( )At 4 107 T m A( )12500 0.0700 m ( ) 2.71 A ( ) 1.00 104 m2( )0.0200 s = 3.04 mV (d) The magnitude of the average induced emf in the coil is L N B t ( ) and magnitude ofthe average induced current is I LR NRBt|.`, 82024.0 3.04 103 V( ) 0.104 A 104 mA 32.69 Left-hand loop: E (I + I2)R 1 I2R2 0Outside loop: E (I + I2)R 1 LdIdt 0Eliminating I 2 gives E I R LdIdt 0This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7: I t ( ) E R(1 e R t L)But R R1R2/ R1 + R2( ) and E R2E/ R1 + R2( ), so E R ER 2/ (R 1 + R2)R 1R2/ (R1 + R2) ER 1Thu s I(t ) = ER 1(1 e R t L)Chapter 32 Solutions 263 2000 by Harcourt, Inc. All rights reserved.32.70 When switch is closed, steady current I0 1. 20 A. Wh e nthe switch is opened after being closed a long time, t hecurrent in the right loop is I I0eR2t Lso eRt LI0Iand RtL lnI0I|.`,Therefore, L R2tln I0I ( ) 1.00 ( ) 0.150 s ( )ln 1. 20 A 0. 250 A ( ) 0.0956 H 95.6 mH 32.71 (a) While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of t hecircuit. Immediately after the switch is opened, a 9.00 mA current will flow around the out erloop of the circuit. Applying Kirchhoffs loop rule to this loop gives: +0 2.00 + 6.00 ( ) 103 [ ]9.00 103 A( ) 0 +0 72 0 . V with end at the higher potential b (b) (c) After the switch is opened, the current around the outer loop decays as I ImaxeRt L with Imax 9.00 mA, R 8.00 k, and L 0. 400 HThus, when the current has reached a value I 2.00 mA, the elapsed time is: t LR|.`,lnImaxI|.`, 0. 400 H8.00 103 |.`,ln9.002.00|.

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7. 52 105 s 75.2 s 264 Chapter 32 Solutions32.72 (a) The instant after the switch is closed, the situation is as shown i nthe circuit diagram of Figure (a). The requested quantities are: IL 0, IC 0R, IR 0R VL 0, VC 0, VR 0 (b) After the switch has been closed a long time, the steady-stateconditions shown in Figure (b) will exist. The currents andvoltages are: IL 0, IC 0, IR 0 VL 0, VC 0, VR 0 IR = 0+-0Q = 0VC = 0IR = 0/ R+-IL = 0 VL = 0VR = 0Figure (a) +-0Q = C0VC = 0+-IL = 0 VL = 0VR = 0Figure (b) IC = 0/ R32.73 When the switch is closed, asshown in Figure (a), the currentin the inductor is I :12.0 7.50I 10.0 = 0 I = 0.267 AWhen the switch is opened, t heinitial current in the inductorremains at 0.267 A.IR = V: (0.267 A)R 80.0 VR 300 (a) (b)Goal Solution To prevent damage from arcing in an electric motor, a discharge resistor is sometimes placed in parallelwith the armature. If the motor is suddenly unplugged while running, this resistor limits the voltagethat appears across the armature coils. Consider a 12.0-V dc motor with an armature that has a resistanceof 7.50 and an inductance of 450 mH. Assume that the back emf in the armature coils is 10.0 V whenthe motor is running at normal speed. (The equivalent circuit for the armature is shown in FigureP32.73.) Calculate the maximum resistance R that limits the voltage across the armature to 80.0 V whenthe motor is unplugged.Chapter 32 Solutions 265 2000 by Harcourt, Inc. All rights reserved.G: We should expect R to be significantly greater than the resistance of the armature coil, for otherwise alarge portion of the source current would be diverted through R and much of the total power wouldbe wasted on heating this discharge resistor. O: When the motor is unplugged, the 10-V back emf will still exist for a short while because the motorsinertia will tend to keep it spinning. Now the circuit is reduced to a simple series loop with an emf,inductor, and two resistors. The current that was flowing through the armature coil must now flowthrough the discharge resistor, which will create a voltage across R that we wish to limit to 80 V. Astime passes, the current will be reduced by the opposing back emf, and as the motor slows down, t heback emf will be reduced to zero, and the current will stop.A : The steady-state coil current when the switch is closed is found from applying Kirchhoff's loop rule t othe outer loop: + 12.0 V I 7. 50 ( ) 10.0 V 0so I 2.00 V7. 50 0. 267 AWe then require that VR 80.0 V 0. 267 A ( )Rso R VRI 80.0 V0. 267 A 300 L: As we expected, this discharge resistance is considerably greater than the coils resistance. Note thatwhile the motor is running, the discharge resistor turns P (12 V)2300 0. 48 W of power int oheat (or wastes 0.48 W). The source delivers power at the rate of about P IV 0. 267 A + 12 V/ 300 ( ) [ ]12 V ( ) 3.68 W, so the discharge resistor wastes about 13% of t hetotal power. For a sense of perspective, this 4-W motor could lift a 40-N weight at a rate of 0.1 m/ s. 32.74 (a) L1 0N12Al14 107 T m A( )1000 ( )21.00 104 m2( )0. 500 m 2. 51 104 H 251 H (b) M N22I1N21I1N2BAI1N2 0N1 l1( )I1 [ ]AI1 0N1N2Al1 M 4 107 T m A( )1000 ( ) 100 ( ) 1.00 104 m2( )0. 500 m 2. 51 105 H 25.1 H (c) 1 MdI2dt, or I1R1 MdI2dt and I1 dQ1dt MR1dI2dt Q1 MR1dI20tf MR1I2f I2i ( ) MR10 I2i( ) M I2iR1 Q1 2. 51 105 H( )1.00 A ( )1000 2. 51 108 C 25.1 nC 266 Chapter 32 Solutions32.75 (a) It has a magnetic field, and it stores energy, so L = 2UI 2 is non-zero.(b) Every field line goes through the rectangle between the conductors.(c) = LI so L I 1IBday awa L 1Ix dyawa 0I2 y + 0I2 w y ( )|.

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2I0Ix2 ydy 20x2ln yawaThu s L 0xlnw aa|.`,32.76 For an RL circuit, I(t ) ImaxeRLt: I(t )Imax 1 109 eRLt 1RLt RLt 109so Rmax (3.14 108)(109)(2. 50 yr)(3.16 107 s / yr) 3. 97 1025 (If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm2, itsresistance would be at least 10 6 ).32.77 (a) UB = 12 LI 2 = 12 (50.0 H)(50.0 10 3 A) 2 = 6.25 1010 J (b) Two adjacent turns are parallel wires carrying current in the same direction. Since the loopshave such large radius, a one-meter section can be regarded as straight. Then one wire creates a field of B = 0I2 r This causes a force on the next wire of F = IlB sin giving F = Il0I2 rsin 90 0lI22 rSolving for the force, F = (4 107 N/ A2) (1.00 m)(50.0 10 3 A) 2(2)(0.250 m) = 2000 N Chapter 32 Solutions 267 2000 by Harcourt, Inc. All rights reserved.32.78 P I V ( ) I PV 1.00 109 W200 103 V 5.00 103 AFrom Amperes law, B 2r ( ) 0Ienclosed or B 0Ienclosed2r(a) At r a 0.0200 m, Ienclosed 5.00 103 A and B 4 107 T m A( )5.00 103 A( )2 0.0200 m ( ) 0.0500 T 50.0 mT (b) At r b 0.0500 m, Ienclosed I 5.00 103 A and B 4 107 T m A( )5.00 103 A( )2 0.0500 m ( ) 0.0200 T 20.0 mT (c) U u dV B r ( ) [ ]22 rldr ( )20rarb 0I2l4drrab 0I2l4lnba|.`, U 4 107 T m A( )5.00 103 A( )21000 103 m( )4ln5.00 cm2.00 cm|.

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2. 29 106 J 2.29 MJ (d) The magnetic field created by the inner conductor exerts a force of repulsion on the current i nthe outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a smallrectangular section of the outer cylinder of length l and width w. It carries a current of 5.00 103 A( )w2 0.0500 m ( )|.

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and experiences an outward force F IlBsin 5.00 103 A( )w2 0.0500 m ( ) l 20.0 103 T( )sin 90.0The pressure on it is P FA Fwl5.00 103 A( )20.0 103 T( )2 0.0500 m ( ) 318 Pa 268 Chapter 32 Solutions*32.79 (a) B 0NIl 4 107 T m A( )1400 ( ) 2.00 A ( )1. 20 m 2. 93 103 T (upward) (b) u B2202. 93 103 T( )22 4 107 T m A( ) 3. 42 Jm31 N m1 J|.

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3. 42 Nm2 3.42 Pa (c) To produce a downward magnetic field, the surface of the super conductormust carry a clockwise current.(d) The vertical component of the field of the solenoid exerts an inward force on t hesuperconductor. The total horizontal force is zero. Over the top end of the solenoid, its fielddiverges and has a radially outward horizontal component. This component exerts upwardforce on the clockwise superconductor current. The total force on the core is upward . Youcan think of it as a force of repulsion between the solenoid with its north end pointing up,and the core, with its north end pointing down.(e) F PA 3. 42 Pa ( ) 1.10 102 m( )2

]]] 1. 30 103 N Note that we have not proven that energy density is pressure. In fact, it is not in some cases;see problem 12 in Chapter 21.