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problem29_74

problem29_74 - c A 10 37 7 1.90 s T 0350 m 20 4 2 2 = = = =...

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29.74: a) For clarity, figure is rotated so B comes out of the page. b) To work out the amount of the electric field that is in the direction of the loop at a general position, we will use the geometry shown in the diagram below. a θ a r E E E π ε θ 2 cos ) cos ( 2 2 but cos loop = = = = dt dB a dt dB r dt dB A dt d a E B 2 2 2 2 loop cos but 2 cos = = Φ = = , 2 2 2 loop dt dB a dt dB a a E = = which is exactly the value for a ring, obtained in Exercise 29.29, and has no dependence on the part of the loop we pick.

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Unformatted text preview: c) A. 10 37 . 7 1.90 ) s T 0350 . ( m) 20 . ( 4 2 2- = = = = = dt dB R L dt dB R A R I d) . V 10 75 . 1 8 ) s T 0350 . ( m) 20 . ( 8 1 8 1 4 2 2- = = = = dt dB L ab But there is potential drop , V 10 75 . 1 4--= = IR V so the potential difference is zero....
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problem29_74 - c A 10 37 7 1.90 s T 0350 m 20 4 2 2 = = = =...

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