P30_072 - y < r B possibility due to wire B carrying the...

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72. (a) We designate the wire along y = r A =0 . 100 m wire A and the wire along y = r B =0 . 050 m wire B . Using Eq. 30-6, we have ~ B net = ~ B A + ~ B B = µ 0 i A 2 πr A ˆ k µ 0 i B 2 πr B ˆ k which yields ~ B net =52 . 0 × 10 6 ˆ kT . (b) This will occur for some value r B <y<r A such that µ 0 i A 2 π ( r A y ) = µ 0 i B 2 π ( y r B ) . Solving, we Fnd y =13 / 160 0 . 081 m.
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Unformatted text preview: y < r B possibility due to wire B carrying the larger current. We expect a solution in the region y > r A where µ i A 2 π ( y − r A ) = µ i B 2 π ( y − r B ) . Solving, we Fnd y = 7 / 40 ≈ . 018 m....
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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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