Unformatted text preview: a/ 2 R = a √ a 2 + 4 x 2 multiplies the expression of the Feld given by the result of problem 11 (for each side of length L = a ). Since there are four sides, we Fnd B ( x ) = 4 µ µ i 2 πR ¶µ a √ a 2 + 4 R 2 ¶µ a √ a 2 + 4 x 2 ¶ = 4 µ i a 2 2 π ( 1 2 )( √ a 2 + 4 x 2 ) 2 p a 2 + 4( a/ 2) 2 + 4 x 2 which simpliFes to the desired result. It is straightforward to set x = 0 and see that this reduces to the expression found in problem 12 (noting that 4 √ 2 = 2 √ 2)....
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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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