Mathcad - 04-01

Mathcad - 04-01 - A m = in air B = μ Η μ o 4 π ⋅ 10...

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Problem 4-1: Solution A conductor constitutes one turn, so the MMF is equal to the current F 12.5 A = The magnetic field intensity is the MMF divided by the path length. Considering a circle of radius r surrounding the conductor, the path length is 2 π r r 1.0 m = l 2 π r = l 6.283m = H F l = H 1.989
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Unformatted text preview: A m = in air B = μ ο Η μ o 4 π ⋅ 10 7-⋅ Wb A m ⋅ ⋅ = B μ o H ⋅ = B 2.5 10 6-× T = obviously a very weak magnetic field since H is inversely proportional to the distance from the conductor, doubling the distance will halve the flux density...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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