857_Physics ProblemsTechnical Physics

857_Physics ProblemsTechnical Physics - Chapter 30 P30.26...

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Chapter 30 197 P30.26 (a) B NI r inner TmA A m T == ×⋅ × = µ π 0 73 2 41 0 9 0 0 1 4 0 1 0 20 7 0 0 360 ej af . . . (b) B NI r outer A m T × = 0 2 21 0 9 0 01 4 0 130 194 . . . *P30.27 We assume the current is vertically upward. (a) Consider a circle of radius r slightly less than R . It encloses no current so from Bs ⋅= z dI 0i n s i d e Br bg = we conclude that the magnetic field is zero . (b) Now let the r be barely larger than R . Ampere’s law becomes BR I 2 0 πµ = , so B I R = 0 2 . FIG. P30.27(a) The field’s direction is tangent to the wall of the cylinder in a counterclockwise sense . (c) Consider a strip of the wall of width dx and length A . Its width is so small compared to 2 R that the field at its location would be essentially unchanged if the current in the strip were turned off. The current it carries is I Idx R s = 2 up. The force on it is FB u p = F H G I K J ×= I Idx R I R Id x R s AA A 22 4 00 2 intopage radiallyinward . FIG. P30.27(c) The pressure on the strip and everywhere on the cylinder is P F A x Rd x I R = 0 2 0 2 2 4 2 A A inward . The pinch effect makes an effective demonstration when an aluminum can crushes itself as
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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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