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Unformatted text preview: ECE329 Spring 2009 Homework 3 â€” Solution February 10, 2009 1. Faradayâ€™s law, Ë› C E Â· d l = d dt Ë† S B Â· d S , states that the electromotive force E = Â¸ C E Â· d l around any closed loop C equals the time rate of change of the magnetic flux Î¨ = Â´ S B Â· d S through any surface S bounded by the loop. a) If B = at all times, then the magnetic flux is zero ( Î¨ = 0 ), and therefore, according to Faradayâ€™s law, the electromotive force over any closed loop C is also zero ( E = 0 ). b) If B 6 = but timeindependent, then the magnetic flux Î¨ through a surface bounded by a fixed loop C is also timeindependent and therefore d dt Î¨ = 0 . Because of this, and according to Faradayâ€™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 d ~ l d ~ l path A path B C Since B is timeindependent, the corresponding magnetic flux Î¨ is also time independent, and therefore, Ë› C E Â· d l = d Î¨ dt = 0 . Breaking the closed path integral into two parts, we have Ë› C E Â· d l = Ë† P 1 â†’ P 2 path A E Â· d l + Ë† P 2 â†’ P 1 path B E Â· d l = 0 . Reversing the direction of integration of the second integral, it can be shown that Ë† P 1 â†’ P 2 path A E Â· d l = Ë† P 1 â†’ P 2 path B E Â· d l . In consequence, the line integral Â´ 12 E Â· d l does not depend on the path taken from P 1 to P 2 . d) If B 6 = but timeindependent, it is still possible for the emf E to be nonzero if the path C is disturbed or displaced in such a way that the magnetic flux Î¨ = Â´ S B Â· d S varies in time. 1 ECE329 Spring 2009 2. Given B = B ( t cos( Ï‰t ) Ë† x + sin( Ï‰t ) Ë† z ) Wb / m 2 , we can apply Faradayâ€™s 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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 Fall '08
 Kim
 Electromagnet, Electromotive Force, Magnetic Field, Faraday's law of induction

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