P30_020 - 20 The two small wire-segments each of length a/4...

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20. The two small wire-segments, each of length a/ 4, shown in Fig. 30-39 nearest to point P , are labeled 1 and 8 in the ±gure below. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . i P 1 8 2 3 4 5 6 7 Let ~e be a unit vector pointing into the page. We use the results of problems 13 and 16 to calculate B P 1 through B P 8 : B P 1 = B P 8 = 2 µ 0 i 8 π ( a/ 4) = 2 µ 0 i 2 πa , B P 4 = B P 5 = 2 µ 0 i 8 π (3 a/ 4) = 2 µ 0 i 6 , B P 2 = B P 7 = µ 0 i 4 π ( a/ 4) · 3 a/ 4 [(3 a/ 4) 2 +( a/ 4) 2 ] 1 / 2 = 3 µ 0 i 10 , and B P 3 = B P 6 = µ 0 i 4 π (3 a/ 4) · a/ 4 [( a/ 4) 2 +(3 a/ 4) 2 ] 1 / 2 = µ 0 i 3 10 . Finally, ~ B P = 8 X n =1 B Pn
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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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