329sum11hw6sol - ECE 329 Homework 6 Solution Due: July 6,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 5

329sum11hw6sol - ECE 329 Homework 6 Solution Due: July 6,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online