Homework 10 Solutions

# 5 103 s c h 75 the flux over the loop cross section

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Unformatted text preview: enerating thermal energy is Pthermal = i 2 R = ε2 R 2 c 1 − e−t τ L h 2 R= ε2 c1 − e h . R −t τ L 2 We equate this to dUB/dt, and solve for the time: ε2 (1 − e ) R −t τ L 2 = ε2 R (1 − e ) e −t τ L −t τ L t = τ L ln 2 = ( 37.0 ms ) ln 2 = 25.6 ms. ⇒ bg bg U bt g = U bt → ∞g = Li . This gives ibt g = i 1 64. Let U B t = 2 Li 2 t . We require the energy at time t to be half of its final value: 1 2 1 4 B 2 f 1 − e−t τ L = 1 2 2 . But i (t ) = i f (1 − e − t /τ L ) , so f 1⎞ ⎛ = − ln ⎜1 − ⎟ = 1.23. τL 2⎠ ⎝ t ⇒ 65. (a) The energy delivered by the battery is the integral of Eq. 28-14 (where we use Eq. 30-41 for the current): ∫ t 0 Pbattery dt = ∫ t ε2 0 (1 − e R − Rt L L ) dt = εR ⎡t + R ( e ⎢ ⎣ 2 ( − Rt L ⎤ − 1) ⎥ ⎦ )⎤ ⎥ ⎡ ( 5.50 H ) e−( 6.70 Ω)( 2.00 s) 5.50 H − 1 ⎢ 2.00 s + 6.70 Ω ⎢ 6.70 Ω ⎣ = 18.7 J. (10.0 V ) = 2 ⎥ ⎦ (b) The energy stored in the magnetic field is given by Eq. 30-49: 2 2 2 2 ⎛ 10.0 V ⎞ 1 1 ⎛ε ⎞ 1 ⎡1 − e −( 6.70 Ω )( 2.00 s ) 5.50 H ⎤ U B = Li 2 ( t ) = L ⎜ ⎟ (1 − e − Rt L ) = ( 5.50 H ) ⎜ ⎟⎣ ⎦ 2 2 ⎝R⎠ 2 ⎝ 6.70 Ω ⎠ = 5.10 J . (c) The difference of the previous two results gives the amount “lost” in the resistor: 18.7 J – 5.10 J = 13.6 J. 66. (a) The magnitude of the magnetic field at the center of the loop, using Eq. 29-9, is ( 4π ×10 H m ) (100 A ) = 1.3×10 B= = 2R 2 ( 50 ×10 m ) μ 0i −7 −3 −3 T. 1193 Φ 21 = Mi1 ( 3.0mH ) ( 6.0mA ) = = 90nWb . N2 200 (d) The mutually induced emf is di1 = ( 3.0mH ) ( 4.0 A s ) = 12mV. dt ε 21 = M 73. (a) Equation 30-65 yields ε1 M= di2 dt = 25.0 mV = 167 mH . . 15.0 A s (b) Equation 30-60 leads to b gb g . N 2 Φ 21 = Mi1 = 167 mH 3.60 A = 6.00 mWb . 74. We use ε2 = –M di1/dt ≈ M|Δi/Δt| to find M: M= ε Δi1 Δt = 30 × 103 V = 13 H . 6.0 A 2.5 × 10−3 s c h 75. The flux over the loop cross section due to the current i in the wire is given by Φ=∫ a +b a Bwireldr = ∫ a +b a Thus, M= μ0il μ il ⎛ b ⎞ dr = 0 ln ⎜ 1 + ⎟ . 2π r 2π ⎝ a ⎠ FG H IJ K NΦ Nμ 0l b ln 1 + . = 2π i a From the formula for M obtained above, we have M= (100 ) ( 4π ×10−7 H m ) ( 0.30 m ) 2π ⎛ 8.0 ⎞ −5 ln ⎜1 + ⎟ = 1.3 ×10 H . ⎝ 1.0 ⎠ 76. (a) The coil-solenoid mutual inductance is M = M cs = 2 N Φ cs N ( μ0is nπ R ) = = μ0π R 2 nN . is is (b) As long as the magnetic field of the solenoid is entirely contained within the cross section of the coil we have Φsc = BsAs = BsπR2, regardless of the shape, size, or possible lack of close-packing of the coil. 1194 CHAPTER 30 77. (a) We assume the current is changing at (nonzero) rate di/dt and calculate the total emf across both coils. First consider the coil 1. The magnetic field due to the current in that coil points to the right. The magnetic field due to the current in coil 2 also points to the right. When the current increases, both fields increase and both changes in flux contribute emf’s in the same direction. Thus, the induced emf...
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