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Unformatted text preview: ECE 329 Homework 6 Solution Due: July 6, 2011, 5PM 1. Given the time-varying magnetic field B = B ( t cos( t ) x + sin( t ) z ) Wb / m 2 , we can apply Faradays law to compute the emf E around the following closed paths. Since the closed paths are not varying in time and the magnetic field B is independent of position, we can rewrite Faradays law as follows E = C E d l =- d B dt S d S =- d B dt n Area , where n is the unit vector normal to the surface, and d B dt = B ((cos( t )- t sin( t )) x + cos( t ) z ) . a) For the rectangular path shown in the figure below, x y 1m 1m 1m 1m n = z z the area of the enclosed surface S is 1 m 2 and the unit vector normal to S is n = z. Therefore, the electromotive force is E =- d B dt z =- B cos( t ) V . b) If we consider the same rectangle of part (a), but the direction of the path is reversed, we have that n =- z. x y 1m 1m 1m 1m n =- z z 1 As a result, the electromotive force is E = d B dt z = B cos( t ) V ....
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This note was uploaded on 11/06/2011 for the course ECE 329 taught by Professor Kim during the Summer '08 term at University of Illinois, Urbana Champaign.
- Summer '08