ch32-p029 - 29(a At any instant the displacement current id...

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29. (a) At any instant the displacement current i d in the gap between the plates equals the conduction current i in the wires. Thus i max = i d max = 7.60 µ A. (b) Since i d = ε 0 ( d Φ E / dt ), d dt i E d Φ F H G I K J == × × max . .. 0 6 12 5 760 10 859 10 A 8.85 10 F m Vms (c) According to Problem 32-13, the displacement current is iC dV dt A d dV dt d 0 . Now the potential difference across the capacitor is the same in magnitude as the emf of the generator, so V = m sin ω t and dV / dt = ωε m cos t . Thus, 0m (/ ) c o s d iA dt = and max 0 m /. d d = This means () ( ) ( )( ) 2 12 3 6 max 8.85 10 F m 0.180 m 130rad s 220 V 3.39 10 m, 7.60 10 A d A d i εωε ×π = × × where A = π R 2 was used. (d) We use the Ampere-Maxwell law in the form G G Bd s I d ⋅= z 0 , where the path of integration is a circle of radius r between the plates and parallel to them. I d is the displacement current through the area bounded by the path of integration. Since the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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