App C_5 retired 2_2_14 REVIEWED

App C_5 retired on 2/2/14 at 8:30 PM PhL 3.26.05 This is another copy of my original wrong section, this is probably the newest version. C.5 The DC internal inductance of a thin ribbon wire The ribbon or strip geometry is as follows, where we assume Jz is uniform and total current is I : Fig C.6 We also assume w >> d so we can ignore end effects. The figure below is an end view of the flat conductor rotated to a vertical position, but only a central segment of the conductor is shown. Fig C.7 The uniform current density Jz is out of the plane of paper as shown on the right. Based on Fig C.2 we expect the H field to have the directions shown on the right, both inside and not far away outside the surface of the conductor. The current density is given by I/area, Jz = I/(wd) . (C.5.1) Applying Ampere's Law to the red loop one finds ∫S J dS = C H ds (1.1.37) so Jz 2x s = 2H(x) s => H(x) = Jzx = Ix/(wd) H/I = x/(wd) (C.5.2) To get the internal inductance we compute its contribution from the dotted rectangle, then multiply by (w/s) to get Li for the entire strip. So using (C.3.6) Li = (w/s) μ ∫dotted dS (H/I)2 = (w/s) μ∫dotted (sdx) [x/(wd)]2 = (w/s) μ s (wd)-2 !Syntax Error, Idx x2 = (w/s) μ s (wd)-2 2 !Syntax Error, Idx x2 = (w/s) μ s (wd)-2 2 (1/3) (d/2)3 = w-1 μ d-2 (2/3) d3/8 = (1/12) μ (d/w) = (μi/8π) [ 2π/3 (d/w) ] // strip w>>d (C.5.3) For some reason, this is 1/2 the value which appears in Holloway and Kuester equation (25).