the nearest one should contribute an upward component to the field at P. The current elements are all equivalent, as is reflected in the horizontal-translational symmetry built into this problem; therefore, all vertical components should cancel in pairs. The field at Pmust be purely horizontal, as drawn. (b) The path used in evaluating GGBdsz⋅is rectangular, of horizontal length ∆x(the horizontal sides passing through points Pand P'respectively) and vertical size δy > ∆y. The vertical sides have no contribution to the integral since GBis purely horizontal (so the scalar dot product produces zero for those sides), and the horizontal sides contribute two equal terms, as shown next. Ampere’s law yields 0012.2BxxBµλµλ∆=∆ ⇒=79. The “current per unit x-length” may be viewed as current density multiplied by the thickness ∆yof the sheet; thus, λ= J∆y. Ampere’s law may be (and often is) expressed in terms of the current density vector as follows
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.