HW-19Solutions-03-24-08

# HW-19Solutions-03-24-08 - Lehigh University HW-19 Solutions...

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1 Figure 32-35 Lehigh University Physics 21, Spring 2008 March 24, 2008 HW-19 Solutions 19-1. (HRW 32-24) Figure 32-35 a shows the current i that is produced in a wire of resistivity . The magnitude of the current versus time t is shown in Fig. 32-35 b. The vertical axis scale is set by , and the horizontal axis scale is set by . Point P is at radial distance 9.00 mm from the wire's center. Determine the magnitude of the magnetic field at point P due to the actual current i in the wire at (a) , (b) , and (c) . Next, assume that the electric field driving the current is confined to the wire. Then determine the magnitude of the magnetic field at point P due to the displacement current i d in the wire at (d) , (e) , and (f) . At point P at , what is the direction (into or out of the page) of (g and (h) ? Solution: 24. (a) Fig. 32-35 indicates that i = 4.0 A when t = 20 ms. Thus, B i = μ o i /2 π r = 0.089 mT. (b) Fig. 32-35 indicates that i = 8.0 A when t = 40 ms. Thus, B i 0.18 mT. (c) Fig. 32-35 indicates that i = 10 A when t > 50 ms. Thus, B i 0.220 mT. (d) Eq. 32-4 gives the displacement current in terms of the time-derivative of the electric field: i d = ε o A ( dE/dt ), but using Eq. 26-5 and Eq. 26-10 we have E = ρ i/A (in terms of the real current); therefore, i d = ε o ρ ( di/dt ). For 0 < t < 50 ms, Fig. 32-35 indicates that di/dt = 200 A/s. Thus, B id = μ o i d /2 π r = 6.4 × 10 22 T.

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## This note was uploaded on 03/30/2008 for the course PHYSIC 2 taught by Professor Kim during the Spring '08 term at Lehigh University .

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HW-19Solutions-03-24-08 - Lehigh University HW-19 Solutions...

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