Li for thin strip

1 C.5 The DC internal inductance of a thin strip conductor The strip geometry is as follows, where we assume J z is uniform and total current is I : F i g 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 J z 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 direc tions shown on the right, both inside and outside the surface of the conductor. The current density is given by I/area, J z = I / ( w d ) . ( 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) = J zx = Ix/(wd) H/I = x/(wd) . (C.5.2) To get the internal inductance per unit length L i we compute its contribution from the dashed rectangle, then multiply by (w/s) to get L i for the entire strip. Therefore, Li = (w/s) μ ∫dashed dS (H/I)2 = (w/s) μ∫ dashed (sdx) [x/(wd)]2 2 = (w/s) μ s (wd)-2 ∫-d/2 d/2 dx x2 = (w/s) μ s (wd)-2 2 ∫0 d/2 dx 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).