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Li for thin strip

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Section C.5 of the Appendix C notes on DC internal inductance, in the Transmission Lines notes by chapter. It assumes uniform current density in a strip with width much greater than thickness, applies Ampere's law to find H(x) inside the conductor, and integrates H squared over the cross section. The result is Li = (1/12) μ (d/w), and a closing remark notes this is half the value in Holloway and Kuester equation (25).

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C.5 The DC internal inductance of a thin strip conductor The strip geometry is as follows, where we assume Jz is uniform and total current is I : Fig C.5 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.6 The uniform current density Jz is out of the plane of paper as shown on the right. Based on the current direction we expect the H field to have the directions shown on the right, both inside and 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 so Jz 2x s = 2H(x) s => H(x) = Jzx = Ix/(wd) H/I = x/(wd) . (C.5.2) To get the internal inductance per unit length Li we compute its contribution from the dashed rectangle, then multiply by (w/s) to get Li for the entire strip. Therefore, Li = (w/s) μ ∫dashed dS (H/I)2 = (w/s) μ∫ dashed (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 (2/3) (1/8) = (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).