865_Physics ProblemsTechnical Physics

865_Physics ProblemsTechnical Physics - 205 Chapter 30 *...

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Chapter 30 205 * P30.58 From example 30.6, the upper sheet creates field Bk = µ 0 2 J s ± above it and 0 2 J s ± k ej below it. Consider a patch of the sheet of width w parallel to the z axis and length d parallel to the x axis. The charge on it σ wd passes a point in time d v , so the current it constitutes is q t wdv d = and the linear current density is J wv w v s == . Then the magnitude of the magnetic field created by the upper sheet is 1 2 0 µσ v . Similarly, the lower sheet in its motion toward the right constitutes current toward the left. It creates magnetic field 1 2 0 v ± k above it and 1 2 0 v ± k below it. + + + + d x y z w FIG. P30.58 (b) Above both sheets and below both, their equal-magnitude fields add to zero . (a) Between the plates, their fields add to 00 vv −= ± k away from you horizontally. (c) The upper plate exerts no force on itself. The field of the lower plate,
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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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