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App C_5 REVIEWED

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Reviewed Word draft dated 3.26.05 from Phil's transmission line notes, Appendix C on DC internal inductance. It treats a wide thin strip (w much greater than d) with uniform current, applies Ampere's law to get H(x), and integrates the stored magnetic energy to get Li = (1/12) μi (d/w). A note says this ignores the transverse H field and gives half the Kuester result; the correct coefficient is 1/6.

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This is the Title PhL 3.26.05 This is my original strip calculation where I ignore the transverse H field and I end up getting a result which is only half the Kuester result. This led to my email with him and to my learning my error. 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 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) dLi = μi ∫in dS (H/I)2 = μi (wd)-2 s !Syntax Error, I dx x2 = μi (wd)-2 2s (1/3)(d/2)3 = (1/12) μi (sd/w2) // dS = sdx Ignoring end effects, we then set s = w to get the DC internal inductance of the full strip, Li = (1/12) μi (d/w) = (μi/8π) [ 2π/3 (d/w) ] // strip w>>d (C.5.3) [ the correct solution has 1/6 instead of 1/12 ]