856_Physics ProblemsTechnical Physics

856_Physics ProblemsTechnical Physics - 196 P30.23 Sources...

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196 Sources of the Magnetic Field P30.23 From Ampere’s law, the magnetic field at point a is given by B I r a a a = µ π 0 2 , where I a is the net current through the area of the circle of radius r a . In this case, I a = 100 . A out of the page (the current in the inner conductor), so B a = ×⋅ × = 41 0 1 0 0 2 1 00 10 200 7 3 TmA A m T toward top of page ej af . . . Similarly at point b : B I r b b b = 0 2 , where I b is the net current through the area of the circle having radius r b . Taking out of the page as positive, I b =−= 300 200 .. . A A A, or I b = . A into the page. Therefore, B b = × = 0 2 0 0 2 3 00 10 133 7 3 A m T toward bottom of page . . . P30.24 (a) In B I r = 0 2 , the field will be one-tenth as large at a ten-times larger distance: 400 cm (b) Bk k =+ 0 1 0 2 22 I r I r ±± so B = F H G I K J = 0 2 1 03985 1 04015 750 7 T m 2.00 A A m m nT a f . (c) Call r the distance from cord center to field point and 2 3 00 d = . mm the distance between conductors. B I rd rd I d rd = + F H G I K J = 00 2 11 2 2 7 50 10 2 00 10 2 00 300 10 225 10 10 7 3 26 .
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