Homework 10 Solutions

2 dt 2 dt 2 we note that 2f and that the factor in

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Unformatted text preview: he quantities): dΦ B dB L2 dB ε= = Aloop = dt dt 4 π dt where the rate of change of the field is dB/dt = 0.0100 T/s. Consequently, we obtain ε2 2 ( L2 / 4π ) 2 (dB / dt ) 2 d 2 L3 ⎛ dB ⎞ (1.00 × 10−3 m) 2 (0.500 m)3 2 P= = = ( 0.0100 T/s ) ⎜ ⎟= −8 2 R 64πρ ⎝ dt ⎠ 64π (1.69 ×10 Ω ⋅ m) ρ L /(π d / 4) = 3.68 ×10−6 W . 32. Noting that |ΔB| = B, we find the thermal energy is 1180 CHAPTER 30 38. From the “kink” in the graph of Fig. 30-55, we conclude that the radius of the circular region is 2.0 cm. For values of r less than that, we have (from the absolute value of Eq. 30-20) d Φ B d ( BA) dB E (2π r ) = = =A = π r 2a dt dt dt which means that E/r = a/2. This corresponds to the slope of that graph (the linear portion for small values of r) which we estimate to be 0.015 (in SI units). Thus, a = 0.030 T/s. 39. The magnetic field B can be expressed as bg b g B t = B0 + B1 sin ωt + φ 0 , where B0 = (30.0 T + 29.6 T)/2 = 29.8 T and B1 = (30.0 T – 29.6 T)/2 = 0.200 T. Then from Eq. 30-25 E= FG IJ HK b g b g 1 dB rd 1 r= B0 + B1 sin ωt + φ 0 = B1ωr cos ωt + φ 0 . 2 dt 2 dt 2 We note that ω = 2πf and that the factor in front of the cosine is the maximum value of the field. Consequently, 1 1 Emax = B1 ( 2π f ) r = ( 0.200 T )( 2π )(15 Hz ) (1.6 ×10−2 m ) = 0.15 V/m. 2 2 40. Since NΦB = Li, we obtain c hc h 8.0 × 10−3 H 5.0 × 10−3 A Li = = 10 × 10−7 Wb. ΦB = . 400 N 41. (a) We interpret the question as asking for N multiplied by the flux through one turn: c h b gc hb gb g 2 Φ turns = NΦ B = NBA = NB πr 2 = 30.0 2.60 × 10−3 T π 0100 m = 2.45 × 10−3 Wb. . (b) Equation 30-33 leads to L= NΦ B 2.45 × 10−3 Wb = = 6.45 × 10−4 H. 380 A . i 42. (a) We imagine dividing the one-turn solenoid into N small circular loops placed along the width W of the copper strip. Each loop carries a current Δi = i/N. Then the magnetic field inside the solenoid is 1182 CHAPTER 30 or | di / dt | = 5.0A/s. We might, for example, uniformly reduce the current from 2.0 A to zero in 40 ms. 45. (a) Speaking anthropomorphically, the coil wants to fight the changes—so if it wants to push current rightward (when the current is already going rightward) then i must be in the process of decreasing. (b) From Eq. 30-35 (in absolute value) we get L= ε di / dt 17 V = 6.8 × 10−4 H. 2.5kA / s = 46. During periods of time when the current is varying linearly with time, Eq. 30-35 (in absolute values) becomes | ε |= L | Δi / Δt | . For simplicity, we omit the absolute value signs in the following. (a) For 0 < t < 2 ms, ε=L (b) For 2 ms < t < 5 ms, ε=L b gb g 4.6 H 7.0 A − 0 Δi . = = 16 × 104 V. −3 Δt 2.0 × 10 s b gb g 4.6 H 5.0 A − 7.0 A Δi = 31 × 103 V. . = −3 Δt 5.0 − 2.0 10 s (c) For 5 ms < t < 6 ms, ε=L b b g gb g 4.6 H 0 − 5.0 A Δi = = 2.3 × 104 V. −3 Δt 6.0 − 5.0 10 s b g 47. (a) Voltage is proportional to inductance (by Eq. 30-35) just as, for resistors, it is proportional to resistance. Since the (independent) voltages for series elements add (V1 + V2), then inductances in series must add, Leq = L1 + L2 , just as was th...
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