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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.
 Fall '08
 SPRUNGER
 Physics

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