Chapter 30

30.1: a) ε 2 = M (di1 / dt ) = (3.25 × 10 −4 H) (830 A /s) = 0.270 V, and is constant. b) If the second coil has the same changing current, then the induced voltage is the same and ε 1 = 0.270 V. 30.2: For a toroidal solenoid, M = N 2 Φ B2 / i1 , and Φ B2 = µ 0 N 1i1 A/ 2πr. So, M = µ 0 AN1 N 2 / 2πr. 30.3: a) M = N 2 Φ B2 / i1 = (400) (0.0320 Wb) / (6.52 A) = 1.96 H. b) When i2 = 2.54 A, Φ B1 = i2 M / N 1 = (2.54 A) (1.96 H) / (700) = 7.11 × 10 −3 Wb. 30.4: a) M = ε 2 / ( di / dt ) = 1.65 × 10−3 V / (−0.242 A / s) = 6.82 × 10−3 H. b) N 2 = 25, i1 = 1.20 A, ⇒ Φ B2 = i1 M / N 2 = (1.20 A) (6.82 × 10 −3 H) / 25 = 3.27 × 10 −4 Wb. c) di 2 / dt = 0.360 A / s and ε 1 = Mdi2 / dt = (6.82 × 10 −3 H) (0.360 A / s) = 2.45 mV. 30.5: 1 H = 1 Wb / A = 1 Tm 2 / A = 1 Nm / A 2 = 1 J / A 2 = 1 (J / AC)s = 1 (V / A)s = 1 Ωs. 30.6: For a toroidal solenoid, L = NΦ B / i = ε / (di / dt ). So solving for N we have: N = εi / Φ B (di / dt ) =

(12.6 × 10 −3 V) (1.40 A) = 238 turns. (0.00285 Wb) (0.0260 A / s)

30.7: a) ε = L(di1 / dt ) = (0.260 H) (0.0180 A / s) = 4.68 × 10 −3 V. b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the + terminal at a. (500 μ0 ) (1800) 2 (4.80 × 10 −5 m 2 ) 30.8: a) LK m = K m μ0 N A/ 2πr = = 0.130 H. 2π (0.120 m) 1 1 b) Without the material, L = LK m = (0.130 H) = 2.60 × 10 −4 H. Km 500 2

30.9: For a long, straight solenoid: L = NΦ B / i and Φ B = μ0 NiA/ l ⇒ L = μ0 N 2 A/ l. 30.10: a) Note that points a and b are reversed from that of figure 30.6. Thus, according V −V 1.04 V to Equation 30.8, dtdi = b L a = −0.260 H = − 4.00 A / s. Thus, the current is decreasing. b) From above we have that di = ( − 4.00 A / s)dt. After integrating both sides of this expression with respect to t , we obtain ∆i = ( − 4.00 A / s) ∆t ⇒ i = (12.0 A) − (4.00 A/s) (2.00 s) = 4.00 A. 30.11: a) L = ε / (di / dt ) = (0.0160 V) / (0.0640 A/s) = 0.250 H. b) Φ B = iL / N = (0.720 A) (0.250 H) / (400) = 4.50 × 10−4 Wb. 1 2 LI = (12.0 H) (0.300 A) 2 / 2 = 0.540 J. 2 2 b) P = I R = (0.300 A) 2 (180 Ω) = 16.2 W. c) No. Magnetic energy and thermal energy are independent. As long as the current is constant, U = constant. 30.12: a) U =

1 2 μ0 N 2 Al 2 30.13: U = LI = 2 4πr 4πrU 4π (0.150 m) (0.390 J) ⇒N= = = 2850 turns. 2 μ0 AI μ0 (5.00 × 10 −4 m 2 ) (12.0 A) 2

30.14: a) U = Pt = (200 W) (24 h/day × 3600 s / h) = 1.73 × 10 7 J. 1 2 2U 2(1.73 × 10 7 J) b) U = LI ⇒ L = 2 = = 5406 H. 2 I (80.0 A) 2

30.15: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability µ is used in place of µ 0 .

(0.560 T) 2 B2 30.16: a) free space: U = uV = V= (0.0290 m 3 ) = 3619 J. 2µ 0 2µ 0 b) material with K m = 450 ⇒ U = uV =

(0.560 T) 2 B2 V= (0.0290 m 3 ) = 8.04 J. 2K m µ 0 2(450) µ 0

2 µ 0U 2 µ 0 (3.60 × 10 6 J) U B2 30.17: a) u = = ⇒ Volume = = = 25.1 m 3 . 2 2 Vol 2µ 0 B (0.600 T) 2 µ 0U 2 µ 0 (3.60 × 10 6 J) b) B = = = 141.4 T 2 ⇒ B = 11.9 T. 3 Vol (0.400 m) 2

30.18: a) B =

µ 0 NI µ 0 (600) (2.50 A) = = 4.35 mT. 2πr 2π (0.0690 m)

b) From Eq. (30.10), u =

(4.35 × 10 −3 T ) 2 B2 = = 7.53 J / m 3 . 2µ 0 2µ 0

c) Volume V = 2πrA = 2π (0.0690 m) (3.50 × 10 −6 m 2 ) = 1.52 × 10 −6 m 3 . d) U = uV = (7.53 J / m 3 ) (1.52 × 10 −6 m 3 ) = 1.14 × 10 −5 J. µ N 2 A µ 0 (600) 2 (3.50 × 10 −6 m 2 ) e) L = 0 = = 3.65 × 10 −6 H. 2πr 2π (0.0690 m) 1 1 U = LI 2 = (3.65 × 10 −6 H) (2.50 A) 2 = 1.14 × 10 −5 J same as (d). 2 2 di ε − iR di 6.00 V = . When i = 0 ⇒ = = 2.40 A / s. dt L dt 2.50 H di 6.00 V − (0.500 A) (8.00 Ω) b) When i = 1.00 A ⇒ = = 0.800 A/s. dt 2.50 H ε 6.00 V c) At t = 0.200 s ⇒ i = (1 − e −( R / L ) t ) = (1 − e −(8.00 Ω / 2.50 H) (0.250 s) ) = 0.413 A. R 8.00 Ω ε 6.00 V d) As t → ∞ ⇒ i → = = 0.750 A. R 8.00 Ω

30.19: a)

30 V 30.20: (a) imax = 1000 Ω = 0.030 A = 30 mA, long after closing the switch.

b) − i = imax (1 − e −t / ( L / R ) ) = 0.030 A1 − e 10 μs    = 0.0259 A 20 μs

VR = Ri = (1000 Ω) (0.0259 A) = 26 V VL = ε Battery − VR = 30 V − 26 V = 4.0 V (or, could use VL = L dtdi at t = 20 µs) c)

30.21: a) i = ε / R (1 − e − t /τ ), τ = L / R imax = ε / R so i = imax / 2 when (1 − e −t / τ ) = 12 , and e −t / τ = − t / τ = ln ( 12 ) and t =

1 2

L ln2 (ln 2) (1.25 × 10 −3 H) = = 17.3 μs R 50.0 Ω

b) U = 12 Li 2 ; U max = 12 Li 2 max U = 12 U max when i = i max / 2 1 − e − t / τ = 1 2 so e −t /τ = 1 − 1/ 2 = 0.2929 t = − L ln(0.2929) / R = 30.7 μs

2(0.260 J) 1 2 2U LI ⇒ I = = = 2.13 A 2 L 0.115 H ⇒ ε = IR = (2.13 A) (120 Ω) = 256 V. 1 1 1 11  b) i = Ie −( R / L )t and U = Li 2 = Li 2 e −2 ( R / L ) t = U 0 =  LI 2  2 2 2 22  1 ⇒ e − 2( R / L )t = 2 0.115 H L 1 1 ⇒t =− ln   = − ln   = 3.32 × 10 −4 s. 2R  2  2(120 Ω)  2 

30.22: a) U =

30.23: a) I 0 =

60 V ε = = 0.250 A. R 240 Ω −4

b) i = I 0 e − ( R / L ) t = (0.250 A) e − (240 Ω / 0.160 H) (4.00 × 10 s ) = 0.137 A. c) Vcb = Vab = iR = (0.137 A) (240 Ω) = 32.9 V, and c is at the higher potential. d)

(0.160 H)  1  i 1 L 1 = = e −( R / L ) t1 / 2 ⇒ t1 / 2 = − ln   = − ln   = 4.62 × 10 −4 s. I0 2 R 2 (240 Ω) 2

30.24: a) At t = 0 ⇒ v ab = 0 and vbc = 60 V. b) As t → ∞ ⇒ v ab → 60 V and vbc → 0. c) When i = 0.150 A ⇒ vab = iR = 36.0 V and vbc = 60.0 V − 36.0 V = 24.0 V.

30.25: a) P = εi = εI 0 (1 − e −( R / L ) t ) =

ε2 (6.00 V) 2 (1 − e −( R / L ) t ) = (1 − e −(8.00 Ω / 2.50 H ) t ) R 8.00 Ω

⇒ P = (4.50 W ) (1 − e − ( 3.20 s b) PR = i 2 R =

−1

).

ε2 (6.00 V) 2 (1 − e −( R / L ) t ) 2 = (1 − e −(8.00 Ω / 2.50 H ) t ) 2 R 8.00 Ω

⇒ PR = (4.50 W ) (1 − e − (3.20 s c) PL = iL

)t

−1

)t

)2.

2 di ε ε  ε = (1 − e −( R / L ) t ) L  e −( R / L ) t  = (e −( R / L ) t − e −2 ( R / L ) t ) dt R L  R

⇒ PL = (4.50 W ) (e − ( 3.20 s

−1

)t

− e − ( 6.40 s

−1

)t

).

d) Note that if we expand the exponential in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor.

30.26: When switch 1 is closed and switch 2 is open: i di ′ di di R R t L + iR = 0 ⇒ =−i ⇒∫ = − ∫ dt ′ I 0 i′ dt dt L L 0 R ⇒ ln(i / I 0 ) = − t ⇒ i = I 0 e −t ( R / L ) . L 30.27: Units of L / R = H / Ω = (Ω s) / Ω = s = units of time.

1 = 2πf LC 1 1 ⇒L= = = 2.37 × 10 −3 H. 2 2 2 6 2 −12 4π f C 4π (1.6 × 10 ) ( 4.18 × 10 F) 1 1 b) C max = = = 3.67 × 10 −11 F. 2 2 2 5 2 −3 4π f min L 4π (5.40 × 10 ) (2.37 × 10 H)

30.28: a) ω =

2π = 2π LC = 2π (1.50 H) (6.00 × 10 −5 F) ω = 0.0596 s, ω = 105 rad s.

30.29: a) T =

b) Q = CV = (6.00 × 10 −5 F)(12.0 V) = 7.20 × 10 −4 C. 1 1 c) U 0 = CV 2 = (6.00 × 10 −5 F)(12.0 V) 2 = 4.32 × 10 −3 J. 2 2 d) At t = 0, q = Q = Q cos(ωt + φ ) ⇒ φ = 0.   0.0230 s  t = 0.0230 s, q = Q cos(ωt ) = (7.20 × 10 −4 C) cos  (1.50 H)(6.00 × 10 −5 F)    −4 = −5.43 × 10 C. Signs on plates are opposite to those at t = 0. dq e) t = 0.0230 s, i = = −ωQ sin( ωt ) dt   7.20 × 10 −4 C 0.0230 s  = −0.0499A. ⇒i=− sin  −5 −5   (1.50H)(6.00 × 10 H)  (1.50 H) (6.00 × 10 H)  Positive charge flowing away from plate which had positive charge at t = 0. q 2 (5.43 × 10 −4 C) 2 f) Capacitor: U C = = = 2.46 × 10 −3 J. −5 2C 2(6.00 × 10 F) 1 1 Inductor: U L = Li 2 = (1.50 H) (0.0499 A) 2 = 1.87 × 10 −3 J. 2 2

30.30: (a) Energy conservation says U L (max) = U C (max) 1 2 1 Li max = CV 2 2 2 18 × 10 −6 F imax = V C L = (22.5V) = 0.871A 12 × 10 −3 H The charge on the capacitor is zero because all the energy is in the inductor. (b)

2π = 2π LC ω 1 1 π at 1 4 period: T = (2π LC ) = (12 × 10 −3 H) (18 ×10 −6 F) 4 4 2 −4 = 7.30 × 10 s 3 at 3 4 period: T = 3(7.30 × 10 −4 s) = 2.19 × 10 −3 s 4 T=

(c) q 0 = CV = (18µF) (22.5 V) = 405µC

30.31: C =

Q 150 × 10 −9 C = = 30.0 µF V 4.29 × 10 −3 V

For an L-C circuit, ω = 1 LC and T = 2π ω = 2π LC L=

(T 2π ) 2 = 0.601 mH C

30.32: ω =

1 (0.0850 H) (3.20 × 10 −6 F)

a) imax = ωQmax ⇒ Qmax =

= 1917 rad/s

imax 8.50 × 10 −4 A = = 4.43 × 10 −7 C ω 1917 rad s

 5.00 × 10 − 4 A   b) From Eq. 31.26 q = Q − LCi = (4.43 × 10 C) −  −1 1917 s   −7 = 3.58 × 10 C. 2

30.33: a)

2

−7

2

2

d 2q 1 di + q = 0 ⇒ q = LC = (0.640 H)(3.60 × 10 −6 F) (2.80 A s) 2 LC dt dt = 6.45 × 10 −6 C.

q 8.50 × 10 −6 C b) ε = = = 2.36 V. C 3.60 × 10 −6 F

30.34: a) imax = ωQmax ⇒ Qmax =

imax = imax LC . ω

⇒ Qmax = (1.50 A) (0.400H)(2.50 × 10 −10 F) = 1.50 × 10 −5 C. (1.50 × 10 −5 C) 2 Q 2 max ⇒ U max = = = 0.450 J 2C 2(2.50 × 10 −10 F) 2ω 1 1 b) 2 f = = = = 3.18 × 10 4 s −1 2π π LC π (0.400 H) (2.50 × 10 −10 F) (must double the frequency since it takes the required value twice per period).

30.35: [ LC ] = H ⋅ F = H ⋅

C C Ω C 1 = Ω ⋅ s ⋅ = ⋅ ⋅ s2 = ⋅ A ⋅ s2 = s2 ⇒ V V V s A

[

]

LC = s.

d 2q 1 + q = 0. We will solve the equation using: 2 LC dt dq d 2q q = Q cos(ωt + φ ) ⇒ = −ωQ sin( ωt + φ ) ⇒ 2 = −ω 2Q cos(ωt + φ ). dt dt 2 d q 1 Q 1 ⇒ 2 + q = −ω 2 Q cos(ωt + φ ) + cos(ωt + φ ) = 0 ⇒ ω 2 = ⇒ω= dt LC LC LC

30.36: Equation (30.20) is

1 . LC

1 q 2 1 Q 2 cos 2 (ωt + φ ) = . 2 C 2 C 1 1 1 Q 2 sin 2 (ωt + φ ) U L = Li 2 = Lω 2 Q 2 sin 2 (ωt + φ ) = , since ω 2 = 2 2 2 C

30.37: a) U C =

1 2 1 = 2 1 = 2

b) U Total = U C + U L =

1 . LC

Q2 1 cos 2 (ωt + φ ) + Lω 2Q 2 sin 2 (ωt + φ ) C 2 2 Q 1  1  2 2 cos 2 (ωt + φ ) + L Q sin (ωt + φ ) C 2  LC  Q2 (cos2 (ωt + φ ) + sin 2 (ωt + φ )) C 1 Q2 = ⇒ U Total is a constant. 2 C

30.38: a) q = Ae − ( R / 2 L ) t cos(ω′t + φ ) dq R −( R / 2 L ) t ⇒ = −A e cos(ω′t + φ ) − ω′ Ae −( R / 2 L ) t sin( ω′t + φ ). dt 2L 2

d 2q R −( R / 2 L ) t  R ⇒ 2 = A  e −( R / 2 L ) t cos(ω′t + φ ) + 2ω′ A e sin( ω′t + φ ) dt 2L  2L  − ω′ 2 Ae − ( R / 2 L ) t cos(ω′t + φ ).   R 2 d 2 q R dq q R2 1  2  ⇒ 2 + + = q   − ω′ − 2 + =0   2L  dt L dt LC 2L LC   1 R2 2 ′ ⇒ω = − LC 4 L2 dq b) At t = 0, q = Q, i = = 0: dt dq R ⇒ q = Acos φ = Q and =− Acos φ − ω′Asin φ = 0 dt 2L Q QR R ⇒ A= and − − ω′Q tan φ = − cos φ 2L 2 Lω′ =−

R 2 L 1 / LC − R 2 / 4 L2

.

1 , we find: C d 2 x b dx kx d 2 q R dq q a) Eq. (13.41): 2 + + = 0 → Eq.(30.27) : 2 + + = 0. dt m dt m dt L dt LC

30.39: Subbing x → q, m → L, b → R, k →

k b2 1 R2 ′ − → Eq .( 30 . 28 ) : ω = − . m 4m 2 LC 4 L2 c) Eq. (13.42): x = Ae − ( b / 2 m ) t cos(ω′t + φ ) → Eq.(30.28) : q = Ae − ( R / 2 L ) t cos(ω′t + φ ). b) Eq. (13.43): ω′ =

 L  L H Ω⋅s Ω⋅V 30.40:   = = = = Ω2 ⇒   = Ω. A C  F C V  C

30.41: ω ′ 2 =

1 R2 1 1  1 1  1 − 2 = ⇒ R 2 = 4 L2  − −  ⇒ R = 2L LC 4 L 6 LC LC 6 LC  LC 6 LC 

⇒ R = 2(0.285 H)

1 1 − = 45.4 Ω . −4 (0.285 H) (4.60 × 10 F) 6(0.285 H) (4.60 × 10 −4 F)

30.42: a) When R = 0, ω0 =

1 = LC

1 (0.450 H) (2.50 × 10 −5 F)

= 298 rad s.

ω (1 LC − R 2 4 L2 ) R 2C b) We want = 0.95 ⇒ = 1− = (0.95) 2 ω0 1 LC 4L ⇒R=

4L (1 − (0.95) 2 ) = C

4(0.450 H) (0.0975) = 83.8 Ω. (2.50 × 10 −5 F)

30.43: a)

b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph.

di d = − L ((0.124 A) cos[(240 π s)t ] dt dt ⇒ ε = + (0.250 H) (0.124 A) (240 π ) sin(( 240 π s)t ) = +(23.4 V) sin ((240 π s)t ).

30.44: a) ε = − L

b) ε max = 23.4 V; i = 0, since the emf and current are 90° out of phase. c) imax = 0.124 A; ε = 0, since the emf and current are 90° out of phase.

b b b  μ Ni  μ Nih dr μ0 Nih Φ B = ∫ B (hdr ) = ∫  0 (hdr ) = 0 = ln (b a ). ∫ 2 π r 2 π r 2 π  a a a 2 NΦ B μ 0 N h b) L = = ln (b a ). i 2π μ0 N 2 h  b − a  b − a (b − a) 2 c) ln (b / a) = ln ( 1 − (b − a ) / a ) ≈ + + ⋅ ⋅ ⋅ ⇒ L ≈  . a 2a 2 2π  a 

30.45: a)

2

30.46: a) M = b) ε 2 = N 2

N2 N A N A μ N IA μ NN A μ N N πr Φ B2 = 2 2 Φ B1 = 2 2 0 1 1 = 0 1 2 2 = 0 1 2 2 . I I A1 IA1 l1 l1 l1

dΦ B2 dt

2

= N2

μ0 N1 A2 di1 μ0 N1 N 2 π r2 di1 = . l1 dt l1 dt 2

μ N N π r di2 di di c) ε1 = M 12 2 = M 2 = 0 1 2 2 . dt dt l1 dt

30.47: a) ε = − L

di ⇒ L = ε /(di / dt ) = (30.0 V) /(4.00 A / s) = 7.5 H. dt

dΦ ⇒ Φ f − Φ i = ε∆t ⇒ Φ f = (30.0 V)(12.0 s) = 360 Wb. dt di c) PL = Li = (7.50 H)(48.0 A)(4.00 A /s) = 1440 W. dt P PR = i 2 R = (48.0 A) 2 (60.0 Ω) = 138240 W ⇒ L = 0.0104. PR b) ε =

di d = (3.50 × 10 −3 H) ((0.680 A) cos(π t / 0.0250 s)) dt dt π ⇒ ε max = (3.50 × 10 −3 H)(0.680 A) = 0.299 V. 0.0250 s Li max (3.50 × 10 −3 H)(0.680 A) = = 5.95 × 10 −6 Wb. b) Φ B max = N 400 di c) ε (t ) = − L = − (3.50 × 10 −3 H)(0.680 A)(π / 0.0250 s)sin (π t / 0.0250 s). dt ⇒ ε (t ) = − (0.299 V) sin((125.6 s −1 )t ) ⇒ ε (0.0180 s) = − (0.299 V)sin ((125.6 s −1 )(0.0180 s)) ⇒ ε (t ) = 0.230 V

30.48: a) ε = L

di1 di di + L2 2 = Leq , dt dt dt di di di but i1 = i2 = i for series components so 1 = 2 = , thus L1 + L2 = Leq dt dt dt di1 di 2 di b) Parallel: Now L1 = L2 = Leq , where i = i1 + i2 . dt dt dt L Leq di di di di di1 di2 eq di So = + . But 1 = and 2 = dt dt dt dt L2 dt dt L2 dt

30.49: a) Series: L1

−1

1 di Leq di Leq di 1   . ⇒ = + ⇒ Leq =  + dt L1 dt L2 dt  L1 L2 

  μi B ⋅ di = μ0 I encl ⇒ B 2πr = μ0i ⇒ B = 0 . 2πr μ0 i b) dΦ B = BdA = ldr. 2πr b b μ il dr μ0il c) Φ B = ∫ dΦ B = 0 ∫ = ln ( b a ). 2π a r 2π a

30.50: a)

∫

μ NΦ B = l 0 ln( b a). i 2π μ0 li 2 1 2 1 μ0 2 e) U = Li = l ln( b a)i = ln( b a ). 2 2 2π 4π d) L =

30.51: a)

∫

  μi B ⋅ dl = μ0 I encl ⇒ B 2π r = μ0i ⇒ B = 0 . 2πr 2

µ i 2l B2 1  µ 0i    (l 2π rdr ) = 0 dr. b) u = ⇒ dU = udV = u (l 2π rdr ) = 2µ 0 2 µ 0  2π r  4π r b

b

μ0i 2l dr μ0i 2l c) U = ∫ dU = = ln (b / a ). 4π ∫a r 4π a μ 1 2U d) U = Li 2 ⇒ L = 2 = l 0 ln (b / a), which is the same as in Problem 30.50. 2 i 2π

30.52: a) L1 = L2 =

N1Φ B1 i1 N 2 Φ B2 i2

2 N1 A  μ0 N1i1  μ0 N1 A  = = , i1  2π r  2π r

=

2 N 2 A  μ 0 N 2 i2  μ 0 N 2 A   = i2  2π r  2π r

2

2 2  μ N N A μ N A μ0 N 2 A b) M =  0 1 2  = 0 1 = L1 L2 . 2π r 2π r  2π r  2

30.53: u B = u E ⇒

ε0 E 2 B2 = ⇒ B = ε 0 μ0 E 2 = 2 2 μ0

= ε0 μ0 (650 V/m) = 2.17 × 10 −6 T.

μ0 ε 0 E

12.0 V V = = 1860 Ω. i f 6.45 × 10 −3 A Rt − Rt b) i = i f (1 − e −( R / L ) t ) ⇒ = − ln (1 − i / i f ) ⇒ L = L ln (1 − i / i f )

30.54: a) R =

⇒L=

− (1860 Ω)(7.25 × 10 −4 s) = 0.963 H. ln (1 − (4.86 / 6.45))

30.55: a) After one time constant has passed: 6.00 V ε i = (1 − e −1 ) = (1 − e −1 ) = 0.474 A R 8.00 Ω 1 2 1 ⇒ U = Li = (2.50 H)(0.474 A) 2 = 0.281 J. 2 2 Or, using Problem (30.25(c)): 3/7

U = ∫ PL dt = (4.50 W ) ∫ (e −( 3.20 ) t − e −( 6.40 ) t )dt. 0

 (1 − e ) (1 − e −2 ) = (4.50 W )  − 6.40  3.20 −1

  = 0.281 J 

L/R

L L  b) U tot = (4.50 W ) ∫ (1 − e −( R / L ) t )dt =(4.50 W ) + (e −1 − 1)  R R  0 2.50 H −1 ⇒ U tot = (4.50 W ) e = 0.517 J 8.00 Ω L/R

c) U R = (4.50 W ) ∫ (1 − 2e −( R / L ) t + e − 2 ( R / L ) t )dt 0

 L 2 L −1  L = (4.50 W ) + (e − 1) − (e −2 − 1)  R 2R R  2.50 H ⇒ U R = (4.50 W ) (0.168) = 0.236 J. 8.00 Ω The energy dissipated over the inductor (part (a)), plus the energy lost over the resistor (part (c)), sums to the total energy output (part (b)).

2

2

 60 V  1 2 1 ε  1  = 5.00 × 10 −3 J. 30.56: a) U = Li0 = L  = (0.160 H) 2 2 R 2  240 Ω  b) i = ⇒

dU L ε −( R / L ) t di R di ε 2 −2( R / L )t e ⇒ =− i⇒ = iL = − Ri 2 = e R dt L dt dt R dU L (60 V ) 2 − 2 ( 240 / 0.160 )( 4.00×10 − 4 ) =− e = − 4.52 W. dt 240 Ω

c) In the resistor: dU R ε 2 −2( R / L )t (60 V) 2 −2 ( 240 / 0.160 )( 4.00×10−4 ). 2 =i R= e = e = 4.52 W. dt R 240 Ω ε 2 −2 ( R / L ) t d) PR (t ) = i R = e R 2

⇒ UR =

∞

ε 2 −2 ( R / L ) t ε 2 L (60 V) 2 (0.160 H) e = = = 5.00 × 10 −3 J, R ∫0 R 2R 2(240 Ω) 2

which is the same as part (a).

30.57: Multiplying Eq. (30.27) by i, yields: i 2 R + Li

2 di q di q dq d 1  d 1 q   − i = i 2 R + Li + = i 2 R +  Li 2  +  dt C dt C dt dt  2  dt  2 C  = PR + PL + PC = 0.

That is, the rate of energy dissipation throughout the circuit must balance over all of the circuit elements.

3T  2π 3T   3π ⇒ q = Q cos(ω t ) = Q cos  = Q cos 8 T 8   4

30.58: a) If t = 1

⇒i=

( Q2 − q2 ) =

1

( Q2 − Q2 2) =

LC LC 2 1 1 Q 1 Q2 q2 ⇒ U E = Li 2 = L = = = UB. 2 2 2 LC 2 2C 2C b) The two energies are next equal when q =

 Q = 2 

Q2 2 LC

Q 5π 5T ⇒ ωt = ⇒t = . 8 8 2

30.59: VC = 12.0 V; U C = 12 CVC2 so C = 2U C / VC2 = 2(0.0160 J) /(12.0 V) 2 = 222 μF f =

1 1 so L = (2πf ) 2 C 2π LC

f = 3500 Hz gives L = 9.31μ H

30.60: a) Vmax =

Q 6.00 × 10 −6 C = = 0.0240 V. C 2.50 × 10 −4 F

1 2 Q2 Limax = ⇒ imax = b) 2 2C c) U max =

Q = LC

6.00 × 10 −6 −4

(0.0600 H)(2.50 × 10 F)

= 1.55 × 10 −3 A

1 2 1 Li max = (0.0600 H)(1.55 × 10 −3 A) 2 = 7.21 × 10 −8 J. 2 2 2

d) If i =

3 Q = 5.20 × 10 −6 C. 4 1 1 q2 = Li 2 + for all times. 2 2 C

⇒q= U max

1 1 3 imax ⇒ U L = U max = 1.80 × 10 −8 J ⇒ U C = U max 2 4 4

 3   Q  4  2  = q = 2C 2C

30.61: The energy density in the sunspot is u B = B 2 / 2 μ0 = 6.366 × 10 4 J /m 3 . The total energy stored in the sunspot is U B = u BV . The mass of the material in the sunspot is m = ρV . K = U B so

1 2 mv = U B ; 2

1 ρVv 2 = u BV 2

The volume divides out, and v = 2u B / ρ = 2 × 10 4 m /s 30.62: (a) The voltage behaves the same as the current. Since V R ∝ i, the scope must be across the 150 Ω resistor. (b) From the graph, as t → ∞, V R → 25V, so there is no voltage drop across the inductor, so its internal resistance must be zero. V R = Vmax (1 − e −t / r ) when t = τ , VR = Vmax (1 − 1e ) ≈ 0.63Vmax . From the graph, when V = 0.63 Vmax = 16V, t ≈ 0.5 ms = τ

L / R = 0.5 ms → L = (0.5 ms ) (150Ω) = 0.075 H (c) Scope across the inductor:

30.63: a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid. b) At t → ∞ the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the solenoid has some resistance R L . The final voltage across the solenoid is IRL , where I is the final current in the circuit. c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit. d) As t → ∞, ε − IR − IRL = 0

ε = 50 V and from the graph I R L = 15 V (the final voltage across the inductor), so I R = 35 V and I = (35 V) /R = 3.5 A e) I RL = 15 V, so RL = (15 V) / (3.5 A) = 4.3Ω ε − VL − iR = 0, where VL includes the voltage across the resistance of the solenoid. VL = ε − iR, i =

ε R (1 − e −t /τ ), so VL = ε[1 − (1 − e −t / τ )] Rtot Rtot

ε = 50 V, R = 10 Ω, Rtot = 14.3 Ω, so when t = τ , Vl = 27.9 V From the graph, V L has this value when t = 3.0 ms (read approximately from the graph), so τ = L / Rtot = 3.0 ms. Then L = (3.0 ms )(14.3 Ω) = 43 mH.

30.64: circuit is

(a) Initially the inductor blocks current through it, so the simplified equivalent

i=

ε 50 V = = 0.333 A R 150 Ω

V1 = (100 Ω)(0.333 A) = 33.3 V V4 = (50 Ω)(0.333 A) = 16.7 V V3 = 0 since no current flows through it. V2 = V4 = 16.7 V (inductor in parallel with 50 Ω resistor) A1 = A3 = 0.333 A, A2 = 0 (b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it. Simplified circuit becomes

50 V = 0.385 A 130 Ω V1 = (100 Ω)(0.385 A) = 38.5 V ; V2 = 0 i = ε /R =

V3 = V4 = 50 V − 38.5 V = 11.5 V i1 = 0.385 A, i2 = i3 =

11.5 V = 0.153 A 75 Ω

11.5 V = 0.230 A 50 Ω

30.65: a) Just after the switch is closed the voltage V5 across the capacitor is zero and there is also no current through the inductor, so V3 = 0. V2 + V3 = V4 = V5 , and since V5 = 0 and V3 = 0, V4 and V2 are also zero. V4 = 0 means V3 reads zero. V1 then must equal 40.0 V, and this means the current read by A1 is ( 40.0 V) / (50.0 Ω) = 0.800 A. A2 + A3 + A4 = A1 , but A2 = A3 = 0 so A4 = A1 = 0.800 A. A1 = A4 = 0.800 A; all other ammeters read zero. V1 = 40.0 V and all other voltmeters read zero. b) After a long time the capacitor is fully charged so A4 = 0. The current through the inductor isn’t changing, so V2 = 0. The currents can be calculated from the equivalent circuit that replaces the inductor by a short-circuit.:

I = (40.0 V) / (83.33 Ω) = 0.480 A; A1 reads 0.480 A V1 = I (50.0 Ω) = 24.0 V The voltage across each parallel branch is 40.0 V − 24.0 V = 16.0 V V2 = 0, V3 = V4 = V5 = 16.0 V V3 = 16.0 V means A2 reads 0.160 A. V4 = 16.0 V means A3 reads 0.320 A. A4 reads zero. Note that A2 + A3 = A1 . c) V5 = 16.0 V so Q = CV = (12.0 µF)(16.0 V) = 192 µC d) At t = 0 and t → ∞, V2 = 0. As the current in this branch increases from zero to 0.160 A the voltage V2 reflects the rate of change of current.

30.66: (a) Initially the capacitor behaves like a short circuit and the inductor like an open circuit. The simplified circuit becomes

i=

75 V ε = = 0.500 A R 150 Ω

V1 = Ri = (50 Ω)(0.50 A) = 25.0 V V3 = 0, V4 = (100 Ω)(0.50 A) = 50.0 V V2 = V4 (in parallel) = 50.0 V A1 = A3 = 0.500 A, A2 = 0 (b) Long after S is closed, capacitor stops all current. Circuit becomes

V3 = 75.0 V and all other meters read zero. (c) q = CV = (75 nF) (75 V) = 5630 nC, long after S is closed.

30.67: a) Just after the switch is closed there is no current through either inductor and they act like breaks in the circuit. The current is the same through the 40.0 Ω and 15.0 Ω resistors and is equal to (25.0 V) (40.0 Ω + 15.0 Ω) = 0.455 A. A1 = A4 = 0.455 A; A2 = A3 = 0. b) After a long time the currents are constant, there is no voltage across either inductor, and each inductor can be treated as a short-circuit . The circuit is equivalent to:

I = (25.0 V) ( 42.73 Ω) = 0.585 A A1 reads 0.585 A. The voltage across each parallel branch is 25.0 V − (0.585 A)(40.0 Ω) = 1.60 V. A2 reads (1.60 V) (5.0 Ω) = 0.320 A. A3 reads (1.60 V) 10.0 Ω) = 0.160 A. A4 reads (1.60 V) (15.0 Ω) = 0.107 A. 30.68: (a) τ = L R = 1025mH Ω = 0.40 ms since 0.50 s >> τ , steady state has been reached, for all practical purposes. i = ε R = 50 V 25 Ω = 2.00 A The upper limit of the energy that the capacitor can get is the energy stored in the inductor initially. Q2 1 U C = U L → max = Li02 → Qmax = i0 LC 2C 2 Qmax = (2.00 A) (10 × 10 −3 H) (20 × 10 −6 F) = 0.90 × 10 −3 C (b) Eventually all the energy in the inductor is dissipated as heat in the resistor. 1 1 U R = U L = Li02 = (10 × 10 −3 H) (2.00 A) 2 2 2 = 2.0 × 10 −2 J

30.69: a) At t = 0, all the current passes through the resistor R1 , so the voltage v ab is the total voltage of 60.0 V. b) Point a is at a higher potential than point b. c) vcd = 60.0 V since there is no current through R2 . d) Point c is at a higher potential than point b. e) After a long time, the switch is opened, and the inductor initially maintains the 60.0 V ε current of i R2 = = = 2.40 A. Therefore the potential between a and b is R2 25.0 Ω vab = − iR1 = − ( 2.40 A) (40.0 Ω) = − 96.0 V. f) Point b is at a higher potential than point a. g) vcd = − i ( R1 + R2 ) = − (2.40 A) (40 Ω + 25 Ω ) = − 156 V h) Point d is at a higher potential than point c. 30.70: a) Switch is closed, then at some later time: di di = 50.0 A/s ⇒ v cd = L = (0.300 H) (50.0 A/s) = 15.0 V. dt dt 60.0 V The top circuit loop: 60.0 V = i1 R1 ⇒ i1 = = 1.50 A. 40.0 Ω 45.0 V The bottom loop: 60 V − i2 R2 − 15.0 V = 0 ⇒ i2 = = 1.80 A. 25.0 Ω 60.0 V b) After a long time: i 2 = = 2.40 A, and immediately when the switch is 25.0 Ω opened, the inductor maintains this current, so i1 = i2 = 2.40 A.

30.71: a) Immediately after S1 is closed, i0 = 0, v ac = 0, and vcb = 36.0 V, since the inductor stops the current flow. 36.0 V ε b) After a long time, i0 = = = 0.180 A , R0 + R 50 Ω + 150 Ω v ac = i0 R0 = (0.18 A) (50 Ω) = 9.00 V, and v cb = 36.0 V − 9.00 V = 27.0 V. −1 (R L )t ε c) i (t ) = (1 − e − total ) ⇒ i (t ) = (0.180 A) (1 − e −( 50 s ) t ), Rtotal

(

) and = 36.0 V − (9.00 V) (1 − e

vac (t ) = i (t ) R0 = (9.00 V) 1 − e − ( 50 s

−1

)t

−1

)

(

− ( 50 s ) t v cb (t ) = ε − i (t ) R0 = (9.00 V) 3 + e − ( 50 s Below are the graphs of current and voltage found above.

−1

)t

).

30.72: a) Immediately after S 2 is closed, the inductor maintains the current i = 0.180 A through R. The Kirchoff’s Rules around the outside of the circuit yield: ε + ε L − iR − i0 R0 = 36.0 V + (0.18) (150) − (0.18) (150) − i0 (50) = 0 36 V = 0.720 A, vac = (0.72 A) (50 V) = 36.0 V and vcb = 0. 50 Ω b) After a long time, vac = 36.0 V, and vcb = 0. Thus ε 36.0 V i0 = = = 0.720 A, R0 50 Ω iR = 0, and is 2 = 0.720 A −1 ε −( R L ) t c) i0 = 0.720 A, iR (t ) = e ⇒ iR (t ) = (0.180 A)e −(12.5 s ) t , and Rtotal ⇒ i0 =

−1

(

is 2 (t ) = (0.720 A) − (0.180 A)e − (12.5 s ) t = (0.180 A) 4 − e − (12.5 s Below are the graphs of currents found above.

−1

)t

)

30.73: a) Just after the switch is closed there is no current in the inductors. There is no current in the resistors so there is no voltage drop across either resistor. A reads zero and V reads 20.0 V. b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the inductors can be replaced by short-circuits. The circuit becomes equivalent to

I a = (20.0 V) (75.0 Ω) = 0.267 A The voltage between points a and b is zero, so the voltmeter reads zero. c) Use the results of problem 30.49 to combine the inductor network into its equivalent:

R = 75.0 Ω is the equivalent resistance. Eq.(30.14) says i = (ε R )(1 − e −t τ ), with τ = L / R = (10.8 mH) (75.0 Ω) = 0.144 ms ε = 20.0 V , R = 75.0 Ω, t = 0.115 ms, so i = 0.147 A VR = iR = (0.147 A)(75.0 Ω) = 11.0 V 20.0 V − VR − VL = 0 so VL = 20.0 V − VR = 9.0 V

30.74: (a) Steady state: i =

ε 75.0 V = = 0.600 A R 125 Ω

(b) Equivalent circuit: 1 1 1 = + C s 25 µF 35µF C s = 14.6 µF

Energy conservation:

q2 1 2 = Li0 2C 2

q = i0 LC = (0.600 A) (20 × 10 −3 H) (14.6 × 10 −6 F ) = 3.24 × 10 −4 C

1 1 π T = (2π LC ) = LC 4 4 2 π t= (20 × 10 −3 H) (14.6 × 10 −6 F) = 8.49 × 10 −4 s 2 t=

30.75: a) Using Kirchhoff’s Rules: ε − i1 R1 = 0 ⇒ i1 =

ε , and R1

di2 ε − i2 R2 = 0 ⇒ i2 = (1 − e −( R2 L ) t ). dt R2 ε ε b) After a long time, i1 = still, and i2 = . R1 R2 ε −(( R1 +R2 ) L ) t c) After the switch is opened, i1 = i2 = e , and the current drops off. R2

ε−L

V 2 (120 V) 2 = = 360 Ω. If the switch is opened, P 40 W and the current is to fall from 0.600A to 0.150 A in 0.0800 s, then: i2 = (0.600 A)e − ( ( R1 + R2 ) L ) t ⇒ 0.150 A = (0.600 A)e − ( ( 360 Ω + R2 ) 22.0 H )(0.0800 s ) 22.0 H ⇒ ln( 4.00) = 360 Ω + R2 ⇒ R2 = 21.0 Ω 0.0800 s ⇒ ε = i2 R2 = (0.600 A)(21.2 Ω) = 12.7 V. ε 12.7 V e) Before the switch is opened, i0 = = = 0.0354 A R1 360 Ω d) A 40-W light bulb implies R =

di1 di di di di +L2 2 + M 21 1 + M 12 2 ≡ Leq . dt dt dt dt dt di di1 di 2 But i = i1 + i 2 ⇒ = + and M 12 = M 21 ≡ M . dt dt dt di di So ( L1 +L2 + 2 M ) = Leq , dt dt or Leq = L1 +L2 + 2M . 30.76: Series: L1

di1 di di + M 12 2 = Leq dt dt dt di2 di1 di and L2 + M 21 = Leq , dt dt dt di di di with 1 + 2 = and M 12 = M 21 ≡ M . dt dt dt di di di To simplify the algebra let A = 1 , B = 2 , and C = . dt dt dt So L1 A +MB = Leq C , L2 B +MA =Leq C , A + B = C. Now solve for A and B in terms of C. ⇒ ( L1 − M )A + ( M −L2 ) B = 0 using A = C − B. Parallel : We have L1

⇒ ( L1 − M ) (C − B) + ( M − L2 ) B = 0 ⇒ ( L1 − M )C − ( L1 −M ) B + ( M −L2 ) B = 0 ( M −L1 ) C. (2M − L1 − L2 ) ( M − L1 )C (2 M − L1 − L2 ) − M + L1 But A = C − B = C − = C, (2M − L1 − L2 ) (2 M − L1 − L2 ) M − L2 or A = C. Substitute A in B back into original equation. 2M − L1 − L2 L ( M − L2 )C M ( M − L1 ) So 1 + C = Leq C 2 M − L1 − L2 (2 M − L1 − L2 ) ⇒ (2M −L1 − L2 ) B = ( M − L1 )C ⇒ B =

⇒

M 2 −L1 L2 C = Leq C. 2 M − L1 −L2

Finally, Leq =

L1 L2 − M 2 L1 +L2 − 2 M

30.77: a) Using Kirchhoff’s Rules on the top and bottom branches of the circuit: di ε ε − i1 R1 − L 1 = 0 ⇒ i1 = (1 − e −( R1 L )t ). dt R1 ε − i2 R2 − ⇒ q2 = ∫

t

0

q2 di i ε −(1 R2C ) t = 0 ⇒ − 2 R2 − 2 = 0 ⇒ i2 = e ) C dt C R2 t

ε i2 dt ′ = − R2 Ce −(1 / R2C ) t ′ = εC (1 − e −(1 / R2C ) t ). R2 0

ε ε 0 48.0 V (1 − e 0 ) = 0, i2 = e = = 9.60 × 10 −3 A. R1 R2 5000 Ω ε ε 48.0 V ε −∞ c) As t → ∞ : i1 (∞) = (1 − e −∞ ) = = = 1.92 A, i2 = e = 0. R1 R1 25.0 Ω R2 A good definition of a “long time” is many time constants later. R ε ε −(11 R2C )t d) i1 = i2 ⇒ (1 − e −( R1 L ) t ) = e ⇒ (1 − e −( R1 L ) t ) = 1 e −(1 R2C )t . R1 R2 R2 b) i1 (0) =

x 2 x3 + +  , we find : 2 3! 2  R1 1  R1  2 R  t t2 t −   t +  = 1 1 − + −  2 2 L 2 L  R2  RC 2 R C 

Expanding the exponentials like e x = 1 + x +

R R  R ⇒ t  1 + 21  + O(t 2 ) + ⋅ ⋅ ⋅ = 1 , if we have assumed that t << 1. Therefore: R2  L R2 C    LR2 C  1  1 =  ⇒t ≈ 2 R2  (1 L) + (1 R2 C )   L + R2 2 C   (8.0 H)(5000 Ω)(2.0 × 10 −5 F)   = 1.6 × 10 −3 s. ⇒ t =  2 −5 8 . 0 H + ( 5000 Ω ) ( 2 . 0 × 10 F )  

ε 48 V (1 − e −( R1 L ) t ) = (1 − e −( 25 8) t ) = 9.4 × 10 −3 A. R1 25 Ω f) We want to know when the current is half its final value. We note that the current i 2 is very small to begin with, and just gets smaller, so we ignore it and find: ε i1 2 = 0.960 A = i1 = (1 − e −( R1 L ) t ) = (1.92 A)(1 − e −( R1 L ) t ). R1 L 8.0 H ⇒ e −( R1 L ) t = 0.500 ⇒ t = ln( 0.5) = ln( 0.5) = 0.22 s R1 25 Ω e) At t = 1.57 × 10 −3 s : i1 =

30.78: a) Using Kirchoff’s Rules on the left and right branches: di di Left: ε − (i1 + i2 ) R − L 1 = 0 ⇒ R (i1 + i2 ) + L 1 = ε. dt dt q q Right: ε − (i1 + i2 ) R − 2 = 0 ⇒ R (i1 + i2 ) + 2 = ε. C C ε b) Initially, with the switch just closed, i1 = 0, i2 = and q2 = 0. R c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat tedious exercise in bookkeeping that is left to the reader. We will show that the initial conditions are satisfied: ε − βt ε At t = 0, q2 = e sin( ωt ) = sin( 0) = 0 ωR ωR ε ε i1 (t ) = (1 − e − βt [(2ωRC ) −1 sin( ωt ) + cos(ωt )] ⇒ i1 (0) = (1 − [cos(0)]) = 0. R R 1 1 d) When does i2 first equal zero? ω = − = 625 rad/s LC (2 RC ) 2 ε i2 (t ) = 0 = e −bt [−(2ωRC ) −1 sin( ωt ) + cos(ωt )] ⇒ − (2ωRC ) −1 tan(ωt ) + 1 = 0 R ⇒ tan(ωt ) = + 2ωRC = + 2(625 rad/s)(400 Ω)(2.00 × 10 −6 F) = + 1.00. ⇒ ωt = arctan(+1.00) = + 0.785 ⇒ t =

0.785 = 1.256 × 10 −3 s. 625 rad/s

30.79: a) Φ B = BA = BL AL + BAir AAir = μ0 Ni[( D − d ) + Kd ] ⇒L=

μ0 Ni Kμ Ni (( D −d )W ) + 0 (dW ) = W W

NΦ B d d = μ0 N 2[( D − d ) +Kd ] = L0 −L0 + L f = L0 i D D

 L f − L0  d +  D 

 L − L0   D, where L0 = μ0 N 2 D, and L f = Kμ0 N 2 D. ⇒d = L − L  0   f d  b) Using K = χ m + 1 we can find the inductance for any height L = L0 1 + χ m  . D  __________________________________________________________________ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury d=D 4 0.63024 H 0.63000 H d=D 2 0.63048 H 0.62999 H d = 3D 4 0.63072 H 0.62999 H 0.63096 H 0.62998 H d=D __________________________________________________________________ Where are used the values χ m (O 2 ) = 1.52 × 10 −3 and χ m (Hg) = − 2.9 × 10 −5. d) The volume gauge is much better for the liquid oxygen than the mercury because there is an easily detectable spread of values for the liquid oxygen, but not for the mercury.