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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( ωt ) ˆ x + sin( ωt ) ˆ z ) Wb / m 2 , we can apply Faraday’s 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 Faraday’s law as follows E = ˛ C E · d l =- d B dt · ˆ S d S =- d B dt · ˆ n · Area , where ˆ n is the unit vector normal to the surface, and d B dt = B ((cos( ωt )- ωt sin( ωt )) ˆ x + ω cos( ωt ) ˆ z ) . a) For the rectangular path shown in the figure below, x y 1m 1m 1m 1m ˆ n = ˆ z z the area of the enclosed surface S is 1 m 2 and the unit vector normal to S is ˆ n = ˆ z. Therefore, the electromotive force is E =- d B dt · ˆ z =- B ω cos( ωt ) V . b) If we consider the same rectangle of part (a), but the direction of the path is reversed, we have that ˆ n =- ˆ z. x y 1m 1m 1m 1m ˆ n =- ˆ z z 1 As a result, the electromotive force is E = d B dt · ˆ z = B ω cos( ωt ) V ....
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- Summer '08
- Electromotive Force, Magnetic Field, dt, Faraday's law of induction