App M discarded tail REVIEWED

Appendix M Discarded Tail PhL 1.15.14 This seems related to the exterior solution for Az , notice the H(1) below. File Appendix G. The remaining while for the r>a region, AzH(r) = [Iμ2/(πa2)](j/4)2π (1/β1) {H0(1)(β1r) aJ1(β1a) } r > a Since we assumed a uniform current density in (M.5.2), we expect the result to be valid only for small ω and this small β12. One could then at this point use small-argument approximations It is possible to compute this double integral in the order shown, but this general ω result is not really meaningful because we have assumed a uniform J in the wire, but we know from Chapter 2 that the skin effect is going to occur and J is not uniform. The solution here is meaningful only for small ω where there is no skin effect. As will be shown below, the leading small-x term in H0(1)(x) is this H0(1)(x) ≈ (2j/π)ln(x) and when this is inserted into (M.5.2) we get AzH(r) ≈ [Iμ2/(πa2)] (j/4) (2j/π) !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[β1R] = [Iμ2/(πa2)] (-1/2π) (1/2) !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[β12R2] = - (1/4π) [Iμ2/(πa2)] !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[β12R2] = - (1/4π) [Iμ2/(πa2)] !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[R2] + constant which then replicates our Poisson result (M.4.2) for ω = 0. So we need more terms to get something interesting for low frequency behavior. We need to know more terms of small x limit of H0(1)(x) in order to use it in our Az expression above. First, we know that GR7 p 911 that which says K0(z) = (jπ/2)H0(1)(jz) // same as Jackson p 116 (3.101) => K0(-jx) = (jπ/2)H0(1)(x) z = -jx so H0(1)(x) = (2/jπ) K0(-jx) . Near z = 0 we have this expansion for K0(z) from Spiegel 24.40 K0(z) = - [ln(z/2) + γ ]I0(z) + Ax2 + Bx4 + .... // Spiegel 24.40 Since ln(z/2) is very large near z = 0, we neglect all the non-log terms to get K0(z) ≈ - ln(z/2)I0(z) ≈ - ln(z/2) [ 1 + (1/4)z2 + (1/64)z4 + ...] // Spiegel 24.35 Then for small x we find then that H0(1)(x) ≈ - (2/jπ) ln(-jx/2) [ 1 + (1/4)(-jx)2 + (1/64)(-jx)4 + ...] ≈ + (2j/π) ln(x) [ 1 - (1/4)x2 + (1/64)x4 + ... ] where the ln(-j/2) terms do not get large near x = 0 and so we throw them out as well. We have seen already how the leading term replicates the ω= 0 results. The next term will be this ΔAzH(r) = [Iμ2/(πa2)] (j/4) (2j/π) !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[β1R] { - (1/4)(β1R)2 } = - (β12/4) [Iμ2/(πa2)] (j/4) (2j/π) !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[β1R] R2 = - (β12/4) [Iμ2/(πa2)] (j/4) (2j/π)(1/2) !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[β12R2] R2 = - (β12/4) [Iμ2/(πa2)] (j/4) (2j/π)(1/2) !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln[R2] R2 + constant Comment: This next term isn't very simple! My general solution almost seems simpler! The enterprising reader can find some way to do this double integral and thereby obtain AzH(r), perhaps assume small β1 . As in the ω = 0 case, we expect that some homogeneous solutions of (22D + β12) Az(r) = 0 will be required to meet the boundary conditions at least when μ1 ≠ μ2. The form of this homogeneous Helmholtz equation is (1/r)∂r(r∂rf) + β12f = 0 ∂r(r∂rf) + rβ12f = 0 rf" + f' + rβ12f = 0 f" + (1/r)f' + β12f = 0 which unsurprisingly is the same equation we encountered in (2.1.14) which has solutions of the type J0(β1r), Y0(β1r) with β1 = ej3π/4 . These are the Kelvin functions. 4. An alternative is to regard (M.5.1) as a 3D Helmholtz equation (2 + β12) Az(r,z) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (M.5.1) and then the 3D Helmholtz integral becomes (see Appendix H) AzH(r,z) = ∫dV' [ ] [Iμ2/(πa2)] θ(r≤a) = [Iμ2/(πa2)] !Syntax Error, Ir' dr' !Syntax Error, Idθ' !Syntax Error, Idz' R2 = r2+ r'2-2rr'cos(θ-θ') + (z-z')2 From (M.5.2) we expect the dz' integral here will be 4π (j/4) H0(1)(β1R) with R2 = r2+ r'2-2rr'cos(θ-θ').