unrolling the thin shell REVIEWED.

This is the Title PhL 3.26.05 I removed this section from Appendix C after I realized that my strip Li derivation was wrong. The basic facts presented here I think are correct in terms of the H plots, but the whole subject of transverse H fields ruins things. In the rolled state, all the B field is "along the strip", but in the unrolled state half the Li energy is in the newly appearing " perp to the strip" Li B field. So the comparison is very weak and not worth writing about. (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 integrating the squared H field, 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).