849_Physics ProblemsTechnical Physics

849_Physics ProblemsTechnical Physics - Chapter 30 P30.3 B=...

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Chapter 30 189 P30.3 (a) B I a =− F H G I K J 4 44 3 4 0 µ π ππ cos cos where a = A 2 is the distance from any side to the center. B = × + F H G I K J = 400 10 0200 2 2 2 2 2 2 10 28 3 6 5 . . . T T i n t o t h e p a p e r (b) For a single circular turn with 4 2 A = R , B I R I == = × = µπ 0 0 27 24 41 0 1 0 0 40400 24 7 A ej af . . . T into the paper FIG. P30.3 P30.4 B I r × 0 7 7 2 01 0 0 21 0 0 200 10 . . . A m T P30.5 For leg 1, d sr ×= ± 0, so there is no contribution to the field from this segment. For leg 2, the wire is only semi-infinite; thus, B I x I x = F H G I K J = 1 22 4 00 into the paper . FIG. P30.5 P30.6 We can think of the total magnetic field as the superposition of the field due to the long straight wire (having magnitude 0 2 I R and directed into the page) and the field due to the circular loop (having
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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