App C_6 REVIEWED

This is the Title PhL 3.26.05 Perhaps this is my original version of this section where I still have the "unrolling the shell" section included. I later had to drop this section. Only later did I add my discovery of external verification from an old book I found on line. C.6 The DC internal inductance of a hollow pipe The pipe geometry is as follows, where we assume Jz is uniform and total current is I : Fig C.7 As with the round wire case, we can avoid using (C.4.1) to compute H due to symmetry. The total current enclosed within the red circle is this (area = πr2 for any disk), Ienc(r) = = I valid for a ≤ r ≤ b . (C.6.1) For r < a Ienc(r) = 0, and for r > b Ienc(r) = I. Ampere's Law says 2πr Hθ(r) = Ienc(r) . (C.6.2) In passing, note that Hθ(r) = 0 inside the pipe, so this region makes no contribution to Le or Li. Inside the annulus, Hθ(r) = Ienc(r)/ (2πr) = I . a ≤ r ≤b (C.6.3) Recalling that Li = μi ∫in dS (H/I)2 (C.3.7) we conclude that Li = μi ∫i dS (r2 - a2)2 and then using dS = 2πrdr we find Li = μi !Syntax Error, Idr (r2 - a2)2 = μi J . (C.6.4) Maple computes the integral J as follows which we restate as J = (1/4)b4 + (3/4)a4 - a2b2 + a4 ln(b/a) (C.6.5) and then the final result is Li = μi [(1/4)b4 + (3/4)a4 - a2b2 + a4 ln(b/a)] . (C.6.6) (a) limit as a→ 0: should be round wire of radius b We can read off this limit from (C.6.6) Li = μi [(1/4)b4 + (3/4)a4 - a2b2 + a4 ln(b/a)] = μi [(1/4)b4 + 0 - 0 + 0 ln(b/a)] = μi (C.6.7) and we recover the Li of a round wire as found in (C.3.10). (b) limit as b-a → 0: thin shell radius a and thickness d In this limit, the hollow pipe is a thin cylindrical shell of inner radius a and thickness d. Continuing the Maple code, we find Maple then expands this Li function about d = 0, Of course our only interest is in the first term, so in this limit we have found that Li = μi (d/a) = [ (4/3)(d/a) ] thin shell, valid for d << a (C.6.8) where again (μi/8π) is Li for a round conductor of any radius. (c) Unrolling the thin shell? Suppose we were to unroll the thin shell of the previous section to arrive at a wire whose cross section was a strip of with w = 2πa and thickness d, Fig C.8 If this unrolling could be done without affecting Li, we would then have an expression for Li for the strip, which would be Li = [ (4/3)(d/a) ] = = [ (4/3)(2πd/w) ] = [ (8π/3)(d/w) ] // "if ..." Although the current density Jz is the same for rolled and unrolled, the H field pattern is in fact not the same in the two situations: Jz = I/(wd) w = 2πa Jz = I/(wd) Fig C.9 In the above figures we plot the H field in red, giving it linear form inside the conductor. This linear form is justified for the thin shell from (C.3.6), Hθ(r) = Ienc(r)/ (2πr) = I ≈ I ≈ I = I (C.6.9) For the wide strip, we found the linear form H(x) = Ix/(wd) in (C.5.2). Ignoring these specific expressions and just using the fact that H is linear inside the conductor in both cases, and using the Jx expressions above, we will now show that the thin shell has four times more Li than the equivalent strip. For each of the blue loops in Fig * apply Ampere's Law: shell: H1s = Jz sd => H1 = Jzd = I/2πa = I/w => Hshell(x) = H1 ( + ) strip: H2s = Jz s(d/2) => H2 = Jzd/2 = I/(2w) => Hstrip(x) = H2 (2 ) (C.6.10) Then, f1 = H12 !Syntax Error, Idx ( + )2 = H12 (d/3) = (I/w)2 (d/3) = (1/3) I (d/w2) // shell (C.6.11) f2 = H22 !Syntax Error, Idx (2 )2 = H22 (d/3) = [I/(2w)]2 (d/3) = (1/12) I (d/w2) // strip Thus f1 = 4f2. Since Li = μi w (fi/I2) we conclude that L1 = 4L2. This is consistent with our results obtained above, Li = μi (d/a) = [ (4/3)(d/a) ] thin shell, valid for d << a (C.6.8) Li = [ 2π/3 (d/w) ] = [ (1/3) (d/a) ] // strip w= 2πa >>d (C.5.3) As noted earlier, this strip result is half the value quoted in (C.4.11).