simplify section G.4 REVIEWED

This is about reusing Q(r',r) from Appendix B. All is installed. G.4 Solution for Az using the 2D Helmholtz Integral This method of finding Az is technically more difficult than the first two methods shown in Section G.2 (solving the ODE and adding a homogeneous solution to meet the boundary conditions) and Section G.3 (instant Ampere's Law solution). The method is important because our entire Chapter 4 is based on using Helmholtz integrals to develop the theory of transmission line parameters, and this is one Helmholtz integral that we can actually compute without too much effort. An important result we find is that, when μ1≠μ2, the Helmholtz integral by itself does not supply the complete solution, and one must add in some amount of homogeneous solution of 22D Az(r) = 0 to meet the required boundary conditions at r=a. We really have a Poisson integral since β12 = 0, but the full Helmholtz integral works the same way so we keep referring to it as a Helmholtz integral. Recall from above: 22D Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθ with Bθ = - ∂rAz . (G.1.7) Using the 2D free-space Green's function (propagator) as reviewed in Appendix I equation (I.1.6), g(x|x') = (1/2π) ln(1/R) = - (1/4π) ln(R2) R = R = |x-x'| , (G.4.1) we may write the particular solution to (G.1.7) as the following "Helmholtz" integral [see (I.1.8)] -AzH(r) = ∫dA' [(1/4π) ln(R2) ] [Iμ2/(πa2)] θ(r≤a) = !Syntax Error, Ir' dr'!Syntax Error, Idθ' [(1/4π) ln(r2+ r'2- 2rr'cos(θ-θ'))] [Iμ2/(πa2)] = (1/4π) [Iμ2/(πa2)] !Syntax Error, Ir' dr' [ !Syntax Error, Idθ' ln(r2+ r'2- 2rr'cos(θ-θ')) ] = !Syntax Error, Ir' dr' [ 2 Q(r',r) ] (G.4.2) where Q(r',r) ≡ (1/2) !Syntax Error, Idθ' ln(r2+ r'2- 2rr'cos(θ-θ')) = (1/2) !Syntax Error, Idθ" ln(r2+ r'2- 2rr'cos(θ")) = !Syntax Error, Idx ln(r2+ r'2- 2rr'cosx) . (G.4.3) But we have already computed this AzH(r) in Appendix B where it was called Az(c)(r,θ), see (B.7.3) and (B.7.4). We may therefore borrow the solution (B.7.7) to obtain the results, AzH(r>a) = - { (1/2) a2lnr } = - lnr AzH(r<a) = { (a2-r2)/2 - a2lna } . (G.4.4) The derivatives are ∂rAzH(r>a) = - ∂rAzH(r<a) = - r . (G.4.5) Recall the two boundary conditions, [Az(a)]1 = [Az(a)]2 . (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . (G.2.6) For our particular Helmholtz integral AzH(r) we evaluate these boundary conditions to find - lna = { (a2-a)/2 - a2lna } (1/μ1) [- ] = (1/μ2)[- a] or 1 = 1 (μ2/μ1) = 1 . (G.4.6) Thus, only in the case μ1 = μ2 does the Helmholtz particular solution meet both boundary conditions. If μ1≠ μ2, we must add to the particular solution some amount of homogeneous solution of 22D Az(r,θ) = 0. So we then write generally, Az(r) = AzH(r) + Azhomo(r) (G.4.7) where we know that Azhomo(r) can only have terms α + β ln r. We then write for the two regions Az(r) = - ln(r) + α + β lnr r>a Az(r) = { (a2-r2)/2 - a2lna } + α ' + β' lnr r<a . (G.4.8) As earlier, we choose the zero point for Az(r) by requiring that the large r behavior be K ln(r) without a constant added, which then means α = 0. And for r<a we must have β' = 0 to be finite at r = 0. So Az(r) = - ln(r) + β lnr r>a Az(r) = { (a2-r2)/2 - a2lna } + α ' r<a (G.4.9) -∂rAz(r) = [ - β](1/r) r>a -∂rAz(r) = (2r) r<a . (G.4.10) The boundary conditions are then [Az(a)]1 = [Az(a)]2 (1/μ1) [(-∂rAz)(a)]1 = (1/μ2) [(-∂rAz)(a)]2 (G.2.6) or - ln(a) + β lna = { (a2-a2)/2 - a2lna } + α ' (1/μ1) [ - β](1/a) = (1/μ2) (2a) or [- Iμ2/(2π) + β] lna = - Iμ2/(2π) lna + α ' (1/μ1)[ Iμ2/(2π) - β](1/a) = I/(2πa) // simplify or β lna = α ' Iμ2/(2π) - β = μ1I/(2π) // simplify some more so we find that β = I/(2π) (μ2-μ1) α' = I/(2π) (μ2-μ1) lna . (G.4.11) The full solution is then Az(r) = [- Iμ2/(2π) + β] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' r<a or Az(r) = [- Iμ2/(2π) + {I/(2π) ( μ2-μ1)}] lnr r>a Az(r) = - Iμ2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + { I/(2π) ( μ2-μ1) lna } r<a or Az(r) = I/(2π) [- μ2 + ( μ2-μ1)] lnr r>a Az(r) = - Iμ2/(πa2) (1/4) (r2-a2) - Iμ2/(πa2) (1/2) a2lna + I/(2π) ( μ2-μ1) lna r<a or Az(r) = - (I/(2π) [μ1] lnr r>a Az(r) = - Iμ2/(πa2) (1/4) (r2-a2) - Iμ1/(π) (1/2) lna r<a or Az(r) = - [Iμ1/2π] lnr r>a Az(r) = - [Iμ2/(4πa2)] (r2-a2) - [Iμ1/2π] lna r<a . (G.4.12) This result matches the results (G.2.7) and (G.3.3) of the previous two methods and then results in the B field solution, Bθ = - ∂rAz = [Iμ1/2π]∂rln(r) = [Iμ1/2π](1/r) r > a region 1 Bθ = - ∂rAz = [Iμ2/(2πa2)] r = [Iμ2/2π](r/a2) r < a region 2 . (G.2.8)