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329sp09hw3sol - ECE-329 Spring 2009 Homework 3 Solution 1...

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ECE-329 Spring 2009 Homework 3 — Solution February 19, 2009 1. Faraday’s law, ˛ C E · d l = - d dt ˆ 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 Ψ = ´ S B · d S through any surface S bounded by the loop. a) If B = 0 at all times, then the magnetic flux is zero ( Ψ = 0 ), and therefore, according to Faraday’s law, the electromotive force over any closed loop C is also zero ( E = 0 ). b) If B 6 = 0 but time-independent, then the magnetic flux Ψ through a surface bounded by a fixed loop C is also time-independent and therefore d dt Ψ = 0 . Because of this, and according to Faraday’s 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 dl dl path A path B C Since B is time-independent, the corresponding magnetic flux Ψ is also time independent, and therefore, ˛ C E · d l = - d Ψ dt = 0 . Breaking the closed path integral into two parts, we have ˛ C E · d l = ˆ P 1 P 2 path A E · d l + ˆ P 2 P 1 path B E · d l = 0 . Reversing the direction of integration of the second integral, it can be shown that ˆ P 1 P 2 path A E · d l = ˆ 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 = 0 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 Ψ = ´ S B · d S varies in time. 1

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ECE-329 Spring 2009 2. Given B = B 0 ( 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
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329sp09hw3sol - ECE-329 Spring 2009 Homework 3 Solution 1...

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