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**Unformatted text preview: **ECE-329 Spring 2009 Homework 3 Solution February 10, 2009 1. Faradays law, C E d l =- d dt S B d S , states that the electromotive force E = C E d l around any closed loop C equals the time rate of change of the magnetic flux = S B d S through any surface S bounded by the loop. a) If B = at all times, then the magnetic flux is zero ( = 0 ), and therefore, according to Faradays law, the electromotive force over any closed loop C is also zero ( E = 0 ). b) If B 6 = but time-independent, then the magnetic flux through a surface bounded by a fixed loop C is also time-independent and therefore d dt = 0 . Because of this, and according to Faradays law, the electromotive force over the loop C will be zero ( E = 0 ). c) Let us define a closed loop C passing through the fixed points P 1 and P 2 (see the next figure). P 2 P 1 d ~ l d ~ l path A path B C Since B is time-independent, the corresponding magnetic flux is also time independent, and therefore, C E d l =- d dt = 0 . Breaking the closed path integral into two parts, we have C E d l = P 1 P 2 path A E d l + P 2 P 1 path B E d l = 0 . Reversing the direction of integration of the second integral, it can be shown that P 1 P 2 path A E d l = P 1 P 2 path B E d l . In consequence, the line integral 12 E d l does not depend on the path taken from P 1 to P 2 . d) If B 6 = but time-independent, it is still possible for the emf E to be non-zero if the path C is disturbed or displaced in such a way that the magnetic flux = S B d S varies in time. 1 ECE-329 Spring 2009 2. Given 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 magneticaround the following closed paths....

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