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864_Physics ProblemsTechnical Physics

864_Physics ProblemsTechnical Physics - 204 P30.55 Sources...

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204 Sources of the Magnetic Field P30.55 (a) Use equation 30.7 twice: B IR x R x = + µ 0 2 2 2 3 2 2 e j If each coil has N turns, the field is just N times larger. B B B N IR x R R x R B N IR x R R x xR x x = + = + + + L N M M M O Q P P P = + + + L N M M M O Q P P P 1 2 0 2 2 2 3 2 2 2 3 2 0 2 2 2 3 2 2 2 3 2 2 1 1 2 1 1 2 2 µ µ e j a f e j e j FIG. P30.55 (b) dB dx N IR x x R R x xR x R = + + L N M O Q P µ 0 2 2 2 5 2 2 2 5 2 2 3 2 2 3 2 2 2 2 2 a f e j e j a f Substituting x R = 2 and canceling terms, dB dx = 0 . d B dx N IR x R x x R R x xR x R R x xR 2 2 0 2 2 2 5 2 2 2 2 7 2 2 2 5 2 2 2 2 7 2 3 2 5 2 2 5 2 2 = + + + + L N M + O Q P µ e j e j e j a f e j Again substituting x R = 2 and canceling terms, d B dx 2 2 0 = . P30.56 “Helmholtz pair” separation distance = radius B IR R R IR R I R = + = + = 2 2 2 1 1 40 0 2 2 2 3 2 0 2 1 4 3 2 3 0 µ µ µ b g . for 1 turn. For N turns in each coil, B NI R = = × = × µ π 0 7 3 1 40 4 10 100 10 0 1 40 0 500 1 80 10 . . . . . e j a f a f T . *P30.57 Consider first a solid cylindrical rod of radius R carrying current toward you, uniformly distributed over its cross- sectional area. To find the field at distance r from its center we consider a circular loop of radius
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