# ass22 - SOLUTIONS ASSIGNMENT 22 28.66 r r 0 Idl r...

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SOLUTIONS, ASSIGNMENT 22 28.66. I DENTIFY : Apply 0 2 ˆ 4 μ Id d π r × l r B = r r . S ET U P : The two straight segments produce zero field at P . The field at the center of a circular loop of radius R is 0 2 I B R μ = , so the field at the center of curvature of a semicircular loop is 0 4 I B R = . E XECUTE : The semicircular loop of radius a produces field out of the page at P and the semicircular loop of radius b produces field into the page. Therefore, 0 0 1 1 1 1 2 2 4 a b μ I I a B B B a b a b   = - = - = -     , out of page. E VALUATE : If a b = , 0 B = . 28.69. I DENTIFY : Apply 0 2 ˆ 4 μ Id d π r × l r B = r r . S ET U P : The contribution from the straight segments is zero since 0. d × = l r r r The magnetic field from the curved wire is just one quarter of a full loop. E XECUTE : 0 0 1 4 2 8 μ I μ I B R R = = and is directed out of the page. E

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## This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.

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ass22 - SOLUTIONS ASSIGNMENT 22 28.66 r r 0 Idl r...

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