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 xR x = + µ 0 2 22 32 2 ej If each coil has N turns, the field is just N times larger. BB B NI R Rx R B R Rx x R xx =+= + + −+ L N M M M O Q P P P = + + +− L N M M M O Q P P P 12 0 2 2 2 0 2 2 11 2 a f e j FIG. P30.55 (b) dB dx R xx R R x x R x R =−+ + L N M O Q P −− 0 2 52 2 3 2 2 3 2 2 2 af eje j Substituting x R = 2 and canceling terms, dB dx = 0. dB dx R x x R xR R x x R 2 2 0 2 2 72 2 3 2 2 2 = +−++ + L N M O Q P e j Again substituting x R = 2 and canceling terms, dx 2 2 0 = . P30.56 “Helmholtz pair” separation distance = radius B IR RR IR R I R = + = + = 2 1 140 0 2 2 2 0 2 1 4 3 0 µµ bg . for 1 turn. For N turns in each coil, B NI R == × π 0 7 3 41 01 0 0 1 0 0 1400500 180 10 . . .. . 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 r : Bs Bk r ⋅= = × z dI Br r J B Jr J πµ 0 0 2 00 2 inside ± Now the total field at P inside the saddle coils is the field due to
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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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