P30_058 - 58(a The contribution to BC from the(infinite straight segment of the wire is BC 1 = µ0 i 2πR BC 2 = µ0 i 2R The contribution from

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Unformatted text preview: 58. (a) The contribution to BC from the (infinite) straight segment of the wire is BC 1 = µ0 i . 2πR BC 2 = µ0 i . 2R The contribution from the circular loop is Thus, BC = BC 1 + BC 2 = µ0 i 2R 1+ 1 π . BC points out of the page. (b) Now BC 1 ⊥ BC 2 so BC = 2 2 BC 1 + BC 2 = µ0 i 2R 1+ 1 , π2 and BC points at an angle (relative to the plane of the paper) equal to tan−1 BC 1 BC 2 = tan−1 1 π = 18◦ . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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