Homework 10 Solutions

The magnitude of the induced emf divided by the loop

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Unformatted text preview: nt (in time), Faraday’s law leads to 1175 Φ B = ∫ d Φ B = ∫ ( 4t 2 y ) dy = 2t 2 3 0 . Thus, Faraday’s law yields ε= dΦ B = 4t dt 3 . At t = 2.5 s, the magnitude of the induced emf is 8.0 × 10–5 V. (b) Its “direction” (or “sense’’) is clockwise, by Lenz’s law. 28. (a) We assume the flux is entirely due to the field generated by the long straight wire (which is given by Eq. 29-17). We integrate according to Eq. 30-1, not worrying about the possibility of an overall minus sign since we are asked to find the absolute value of the flux. r +b / 2 ⎛ μ i ⎞ μ ia ⎛ r + b / 2 ⎞ 0 | ΦB | = ∫ (a dr ) = 0 ln ⎜ ⎟. r − b / 2 ⎜ 2π r ⎟ 2π ⎝ r −b/2 ⎠ ⎝ ⎠ When r = 1.5b , we have | ΦB | = (4p × 10 -7 T ⋅ m A)(4.7A)(0.022m) ln(2.0) = 1.4 × 10−8 Wb. 2π (b) Implementing Faraday’s law involves taking a derivative of the flux in part (a), and recognizing that dr / dt = v . The magnitude of the induced emf divided by the loop resistance then gives the induced current: iloop = ε R =− μ0ia d ⎛ r + b / 2 ⎞ μ0iabv ln ⎜ ⎟= 2πR dt ⎝ r − b / 2 ⎠ 2πR[r 2 − (b / 2) 2 ] (4π ×10−7 T ⋅ m A)(4.7A)(0.022m)(0.0080m)(3.2 ×10−3 m/s) = 2π (4.0 ×10−4 Ω)[2(0.0080m) 2 ] = 1.0 ×10−5 A. 29. (a) Equation 30-8 leads to ε = BLv = (0.350 T)(0.250 m)(0.55 m / s) = 0.0481 V . (b) By Ohm’s law, the induced current is i = 0.0481 V/18.0 Ω = 0.00267 A. By Lenz’s law, the current is clockwise in Fig. 30-50. 1176 CHAPTER 30 (c) Equation 26-27 leads to P = i2R = 0.000129 W. 2 30. Equation 26-28 gives ε /R as the rate of energy transfer into thermal forms (dEth /dt, which, from Fig. 30-51(c), is roughly 40 nJ/s). Interpreting ε as the induced emf (in absolute value) in the single-turn loop (N = 1) from Faraday’s law, we have ε= d Φ B d ( BA) dB = =A . dt dt dt Equation 29-23 gives B = μoni for the solenoid (and note that the field is zero outside of the solenoid, which implies that A = Acoil), so our expression for the magnitude of the induced emf becomes di dB d ε=A = Acoil ( μ0 nicoil ) = μ0 nAcoil coil . dt dt dt where Fig. 30-51(b) suggests that dicoil/dt = 0.5 A/s. With n = 8000 (in SI units) and Acoil = π(0.02)2 (note that the loop radius does not come into the computations of this problem, just the coil’s), we find V = 6.3 μV. Returning to our earlier observations, we can now solve for the resistance: 2 R = ε /(dEth /dt) = 1.0 mΩ. 31. Thermal energy is generated at the rate P = ε2/R (see Eq. 26-28). Using Eq. 27-16, the resistance is given by R = ρL/A, where the resistivity is 1.69 × 10–8 Ω·m (by Table 27-1) and A = πd2/4 is the cross-sectional area of the wire (d = 0.00100 m is the wire thickness). The area enclosed by the loop is 2 L 2 Aloop = πrloop = π 2π FG IJ HK since the length of the wire (L = 0.500 m) is the circumference of the loop. This enclosed area is used in Faraday’s law (where we ignore minus signs in the interest of finding the magnitudes of t...
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