P30_039 - current element just to the left of the nearest...

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39. The “current per unit x -length” may be viewed as current density multiplied by the thickness ∆ y of the sheet; thus, λ = J y . Ampere’s law may be (and often is) expressed in terms of the current density vector as follows: I ~ B · d~s = µ 0 Z ~ J · d ~ A where the area integral is over the region enclosed by the path relevant to the line integral (and ~ J is in the + z direction, out of the paper). With J uniform throughout the sheet, then it clear that the right-hand side of this version of Ampere’s law should reduce, in this problem, to µ 0 JA = µ 0 J y x = µ 0 λ x. (a) Figure 30-52 certainly has the horizontal components of ~ B drawn correctly at points P and P 0 (as reference to Fig. 30-4 will con±rm [consider the current elements nearest each of those points] ), so the question becomes: is it possible for ~ B to have vertical components in the ±gure? Our focus is on point P . Fig. 30-4 suggests that the current element just to the right of the nearest one (the one directly under point P ) will contribute a downward component, but by the same reasoning the
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Unformatted text preview: current element just to the left of the nearest one should contribute an upward component to the eld at P . The current elements are all equivalent, as is reected in the horizontal-translational symmetry built into this problem; therefore, all vertical components should cancel in pairs. The eld at P must be purely horizontal, as drawn. (b) The path used in evaluating H ~ B d~s is rectangular, of horizontal length x (the horizontal sides passing through points P and P respectively) and vertical size y > y . The vertical sides have no contribution to the integral since ~ B is purely horizontal (so the scalar dot product produces zero for those sides), and the horizontal sides contribute two equal terms, as shown below. Amperes law yields 2 B x = x = B = 1 2 ....
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