Capítulo 32 (5th Edition)

of 27 /27
© 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)

Embed Size (px)

Transcript of Capítulo 32 (5th Edition)

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|.

`,

= 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|.

`,

dIdt 4 107 T m A( )300 ( )2 104 m2( )0.150 m10.0 A s ( ) = 2.37 mV 32.7 back = = L dIdt = L ddt (Imax sin t) = L Imax cos t = (10.0 10-3)(120 )(5.00) cos t back = (6.00 ) cos (120 t) = (18.8 V) cos (377t) Chapter 32 Solutions 243 2000 by Harcourt, Inc. All rights reserved.*32.8 From = L It|.`,, we have L = t|.`, = 24.0 10 3 V10.0 A/ s = 2.40 10 3 HFrom L = NBI , we have B = LIN = (2.40 10 3 H)(4.00 A)500 = 19.2 T m2 32.9 L = 0N 2A l = 0(420)2(3.00 10 4)0.160 = 4.16 10 4 H = L dIdt dIdt = L = 175 10 6 V4.16 10 4 H = 0.421 A/ s 32.10 The induced emf is LdIdt, where the self-inductance of a solenoid is given by L 0N2Al. Thus, dIdt L l0N2A 32.11 = L dIdt = (90.0 10-3) ddt (t 2 6t) V(a) At t = 1.00 s, = 360 mV (b) At t = 4.00 s, = 180 mV (c) = (90.0 10-3)(2t 6) = 0 when t = 3.00 s 32.12 (a) B = 0nI = 0 .|,`4500.120 (0.0400 mA) = 188 T (b) B = BA = 3.33 10-8 T m2 (c) L = NBI = 0.375 mH 244 Chapter 32 Solutions(d) B and B are proportional to current; L is independent of current Chapter 32 Solutions 245 2000 by Harcourt, Inc. All rights reserved.32.13 (a) L = 0N 2A l = 0(120)2 ( 5.00 103)20.0900 = 15.8 H (b) B = m0 B L = mN 2A l = 800(1.58 10 5 H) = 12.6 mH 32.14 L = NBI = NBAI N AI 0NI2 R = 0N2A2 R 32.15 = 0ekt= LdIdt dI = 0LektdtIf we require I 0 as t , the solution is I = 0kLekt=dqdt Q = I dt= 0kLekt0dt = 0k2L Q = 0k2L 32.16 I R(1 eRt/ L) 0. 900 R R1 eR(3.00 s)/ 2.50 H[ ] exp R(3.00 s)2. 50 H|.`, 0.100 R 2. 50 H3.00 sln 10.0 1.92 246 Chapter 32 Solutions32.17 = LR = 0.200 s: IImax = 1 et/(a) 0.500 = 1 et/ 0.200 t = ln 2.00 = 0.139 s (b) 0.900 = 1 et/ 0.200 t = ln 10.0 = 0.461 s Figure for GoalSolutionGoal Solution A 12.0-V battery is about to be connected to a series circuit containing a 10.0- resistor and a 2.00-Hinductor. How long will it take the current to reach (a) 50.0% and (b) 90.0% of its final value? G: The time constant for this circuit is L R 0. 2 s, which means that in 0.2 s, the current will reach1/ e = 63% of its final value, as shown in the graph to the right. We can see from this graph that t hetime to reach 50% of Imax should be slightly less than the time constant, perhaps about 0.15 s, and t hetime to reach 0. 9Imax should be about 2.5 = 0.5 s. O: The exact times can be found from the equation that describes the rising current in the above graphand gives the current as a function of time for a known emf, resistance, and time constant. We settime t 0 to be the moment the circuit is first connected.A : At time t , I t ( ) (1 et/ )Rwhere, after a long time, Imax (1 e)R RAt I t ( ) 0. 500Imax, 0. 500 ( )R (1 et/ 0.200 s)Rso 0. 500 1 et/ 0.200 sIsolating the constants on the right, ln et/ 2.00 s( ) ln 0. 500 ( )and solving for t , t0. 200 s 0.693 or t 0.139 s(b) Similarly, to reach 90% of Imax, 0. 900 1 et/ and t ln 1 0. 900 ( )Thus, t 0. 200 s ( )ln 0.100 ( ) 0. 461 sL: The calculated times agree reasonably well with our predictions. We must be careful to avoidconfusing the equation for the rising current with the similar equation for the falling current.Checking our answers against predictions is a safe way to prevent such mistakes.Chapter 32 Solutions 247 2000 by Harcourt, Inc. All rights reserved.32.18 Taking L R, I I0et/ : dIdt I0et/ 1|.`, IR + LdIdt 0 will be true if I0Ret/ + L I0et/ ( ) 1|.`, 0Because L R, we have agreement with 0 = 0*32.19 (a) L R = 2.00 10 3 s = 2.00 ms (b) I Imax1 et/ ( ) 6.00 V4.00 |.`,1 e0.250/ 2.00( ) 0.176 A (c) Imax = R = 6.00 V4.00 = 1.50 A (d) 0.800 = 1 et/ 2.00 ms t = (2.00 ms) ln(0.200) = 3.22 ms *32.20 I = R (1 et/ ) = 1209.00 (1 e1.80/ 7.00) = 3.02 AVR = IR = (3.02)(9.00) = 27.2 VVL = VR = 120 27.2 = 92.8 V 32.21 (a) VR IR (8.00 )(2.00 A) 16.0 V and VL VR 36.0 V 16.0 V 20.0 VTherefore, VRVL16.0 V20.0 V 0.800 (b) VR IR (4. 50 A)(8.00 ) 36.0 V VL VR 0 Figure for GoalSolution248 Chapter 32 SolutionsGoal Solution For the RL circuit shown in Figure P32.19, let L = 3.00 H, R = 8.00 , and = 36.0 V. (a) Calculate the ratioof the potential difference across the resistor to that across the inductor when I = 2.00 A. (b) Calculate t hevoltage across the inductor when I = 4.50 A.G: The voltage across the resistor is proportional to the current, IR VR , while the voltage across t heinductor is proportional to the rate of change in the current, L LdI dt . When the switch is firstclosed, the voltage across the inductor will be large as it opposes the sudden change in current. As t hecurrent approaches its steady state value, the voltage across the resistor increases and the inductorsemf decreases. The maximum current will be / R = 4.50 A, so when I = 2.00 A, the resistor andinductor will share similar voltages at this mid-range current, but when I = 4.50 A, the entire circuitvoltage will be across the resistor, and the voltage across the inductor will be zero. O: We can use the definition of resistance to calculate the voltage across the resistor for each current.We will find the voltage across the inductor by using Kirchhoff's loop rule.A : (a) When I 2.00 A, the voltage across the resistor is VR IR 2.00 A ( ) 8.00 ( ) 16.0 VKirchhoff's loop rule tells us that the sum of the changes in potential around the loop must be zero: VR L 36.0 V 16.0 V L 0 so L 20.0 V and VRL16.0 V20.0 V 0.800(b) Similarly, for I 4. 50 A, VR IR 4. 50 A ( ) 8.00 ( ) 36.0 V VR L 36.0 V 36.0 V L 0 so L 0L: We see that when I 2.00 A, VR < L, but they are similar in magnitude as expected. Also aspredicted, the voltage across the inductor goes to zero when the current reaches its maximum value.A worthwhile exercise would be to consider the ratio of these voltages for several different times afterthe switch is reopened.*32.22 After a long time, 12.0 V = (0.200 A)R Thus, R = 60.0 . Now, = LR givesL = R = (5.00 10 4 s)(60.0 V/ A) = 30.0 mH 32.23 I Imax1 et/ ( ): dIdt Imaxet/ ( ) 1|.`, = LR = 15.0 H30.0 = 0.500 s :dIdt = RL Imax et/ and Imax = R(a) t = 0: dIdt = RL Imax e0 = L = 100 V15.0 H = 6.67 A/ s (b) t = 1.50 s: dIdt = L et/ = (6.67 A/ s)e 1.50/ (0.500) = (6.67 A/ s)e3.00 = 0.332 A/ s Chapter 32 Solutions 249 2000 by Harcourt, Inc. All rights reserved.32.24 I Imax1 et/ ( ) 0. 980 1 e3.0010 3/ 0.0200 e3.0010 3/ 3.00 103ln(0.0200) 7.67 104s L R, so L R (7.67 104)(10.0) 7.67 mH 32.25 Name the currents as shown. By Kirchhoffs laws: I1 I2 + I3(1) +10.0 V 4.00I1 4.00I2 0 (2) +10.0 V 4.00I1 8.00I3 1.00 ( )dI3dt 0 (3)From (1) and (2), +10.0 4.00I1 4.00I1 + 4.00I3 0 and I1 0. 500I3 + 1. 25 AThen (3) becomes 10.0 V 4.00 0. 500I3 + 1. 25 A ( ) 8.00I3 1.00 ( )dI3dt 0 1.00 H ( ) dI3dt ( ) + 10.0 ( ) I3 5.00 VWe solve the differential equation using Equations 32.6 and 32.7: I3t ( ) 5.00 V10.0 1 e 10.0 ( )t 1.00 H[ ] = 0. 500 A ( ) 1 e10t/ s[ ] I1 1. 25 + 0. 500I3 1. 50 A 0. 250 A ( )e10t/ s 32.26 (a) Using RC LR, we get R LC 3.00 H3.00 106 F 1.00 103 1.00 k (b) RC 1.00 103 ( )3.00 106 F( ) 3.00 103 s 3.00 ms 250 Chapter 32 Solutions32.27 For t 0, the current in the inductor is zero. At t 0, it starts t ogrow from zero toward 10.0 A with time constant L R 10.0 mH ( ) 100 ( ) 1.00 104 s .For 0 t 200 s, I Imax1 et/ |.`,= 10.00 A ( ) 1 e10000t/ s( ) At t 200 s, I 10.00 A ( ) 1 e2.00( ) 8.65 AThereafter, it decays exponentially as I I0e t , so for t 200 s , I 8.65 A ( )e10000 t 200 s ( ) s 8.65 A ( )e10000t s +2.00 8.65e2.00 A( )e10000t s = 63. 9 A ( )e10000t s 32.28 (a) I = R = 12.0 V12.0 = 1.00 A (b) Initial current is 1.00 A, : V12 = (1.00 A)(12.00 ) = 12.0 V V1200 = (1.00 A)(1200 ) = 1.20 kV VL = 1.21 kV (c) I = Imax eRt/ L:dIdt = Imax RL eRt / Land L dIdt = VL = Imax ReRt/ LSolving 12.0 V = (1212 V)e1212t/ 2.00so 9.90 10 3 = e 606tThu s, t = 7.62 ms 32.29 LR 0.1404. 90 28.6 ms; Imax R 6.00 V4. 90 1. 22 A(a) I Imax1 et/ ( )so 0. 220 1. 22 1 et/ ( ) et/ 0.820 t ln(0.820) 5.66 ms (b) I Imax1 e10.00.0286|.

`,

(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|.

`,

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|.

`,

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

]]] 138 nH 32.45 B at center of (larger) loop: B1 0I12R(a) M 2I1B1A2I1(0I1/ 2R)(r2)I1 022rR (b) M 0(0.0200)22(0. 200) 3.95 nH Chapter 32 Solutions 255 2000 by Harcourt, Inc. All rights reserved.*32.46 Assume the long wire carries current I. Then the magnitude of the magnetic field itgenerates at distance x from the wire is B 0I 2x, and this field passes perpendicularlythrough the plane of the loop. The flux through the loop is B B dA BdA B ldx ( ) 0I l2dxx0.400 mm1.70 mm 0Il2ln1.700. 400|.`,The mutual inductance between the wire and the loop is then M N212I1N20I l2Iln1.700. 400|.`, N20l21. 45 ( ) 1(4 107T m A)(2.70 103m)21. 45 ( ) M 7.81 1010 H 781 pH 32.47 With I I1 + I2, the voltage across the pair is: V L1dI1dt MdI2dt L2dI2dt MdI1dt LeqdIdtSo, dI1dt= VL1+ML1dI2dtand L2dI2dt+M V ( )L1+M2L1dI2dt= V(a) (b) (L1L2 + M2)dI2dt V(L1 M) [1]By substitution, dI2dt= VL2+ML2dI1dtleads to ( L1L2+ M2)dI1dt= V (L2 M) [2]Adding [1] to [2], ( L1L2+ M2)dIdt= V (L1+ L2 2M)So, Leq= VdI / dt= L1L2 M2L1+ L2 2M 32.48 At different times, UC( )max UL( )max so 12C V ( )2[ ]max12LI2( )max Imax CL V ( )max 1.00 106F10.0 103H40.0 V ( ) 0.400 A 256 Chapter 32 Solutions32.49 12C V ( )2[ ]max12LI2( )maxso VC( )max LCImax 20.0 103H0. 500 106F0.100 A ( ) 20.0 V 32.50 When the switch has been closed for a long time, battery, resistor,and coil carry constant current Imax= / R. When the switch isopened, current in battery and resistor drops to zero, but the coilcarries this same current for a moment as oscillations begin in t heLC loop.We interpret the problem to mean that the voltage amplitude ofthese oscillations is V, in 12C V ( )212LImax2.Then, L C V ( )2Imax2 C V ( )2R22 0. 500 106 F( )150 V ( )2250 ( )250.0 V ( )2 0.281 H 32.51 C =1(2 f )2L=1(2 6. 30 106)2(1.05 106)= 608 pF Goal Solution A fixed inductance L = 1.05 H is used in series with a variable capacitor in the tuning section of a radio.What capacitance tunes the circuit to the signal from a station broadcasting at 6.30 MHz?G: It is difficult to predict a value for the capacitance without doing the calculations, but we might expecta typical value in the F or pF range.O: We want the resonance frequency of the circuit to match the broadcasting frequency, and for a simpleRLC circuit, the resonance frequency only depends on the magnitudes of the inductance andcapacitance.A : The resonance frequency is f0 12 LCThu s, C 1(2 f0)2L 1(2)(6. 30 106 Hz)[ ]2(1.05 106 H) 608 pFL: This is indeed a typical capacitance, so our calculation appears reasonable. However, you probablywould not hear any familiar music on this broadcast frequency. The frequency range for FM radiobroadcasting is 88.0 108.0 MHz, and AM radio is 535 1605 kHz. The 6.30 MHz frequency falls in t heMaritime Mobile SSB Radiotelephone range, so you might hear a ship captain instead of Top 40tunes! This and other information about the radio frequency spectrum can be found on the Nat ionalTelecommunications and Information Administration (NTIA) website, which at the time of t hisprinting was at http:/ / www.ntia.doc.gov/ osmhome/ allochrt.html Chapter 32 Solutions 257 2000 by Harcourt, Inc. All rights reserved.32.52 f =12 LC: L =1(2 f )2C=1(2 120)2(8.00 106)= 0.220 H 32.53 (a) f 12 LC 12 (0.0820 H)(17.0 106F) 135 Hz (b) Q Qmaxcos t (180 C) cos(847 0.00100) 119 C (c) I dQdt Qmaxsin t (847)(180) sin (0.847) 114 mA 32.54 (a) f =12 LC=12 (0.100 H)(1.00 106F)= 503 Hz (b) Q = C = (1.00 106F)(12.0 V) = 12.0 C (c) 12C2=12LImax2 Imax= CL = 12 V1.00 106F0.100 H= 37.9 mA (d) At all times U = 12C2=12(1.00 106F)(12.0 V)2= 72.0 J 32.55 1LC 13. 30 H ( ) 840 1012 F( ) 1.899 104 rad s Q Qmaxcos t , I dQdt Qmaxsin t(a) UC Q22C 105 106[ ]cos 1.899 104 rad s( )2.00 103 s( )[ ] ( )22 840 1012( ) 6.03 J (b) UL 12LI212L2Qmax2sin2t ( ) Qmax2sin2t ( )2C UL 105 106 C( )2sin21.899 104 rad s( )2.00 103 s( )[ ]2 840 1012 F( ) 0.529 J (c) Utotal UC +UL 6.56 J 258 Chapter 32 Solutions32.56 (a) d 1LC R2L|.`,212. 20 103( )1.80 106( ) 7.602 2. 20 103( )|.

`,

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|.

`,

=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|.

`,

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|.

`,

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|.

`,

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

`,

= 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|.

`,

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|.

`,

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 ( )|.

`,

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|.

`,

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 ( )|.

`,

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|.

`,

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.