Integration check on (D.10.15) REVIEWED

Integration check on (D.10.15) PhL 9.6.14 Here I verify that if I integrate Jz = σEz with Ez in its large ω form below, I do in fact obtain the correct total current I in the wire! I have to assume that (a/δ) is large. From (D.10.15), Ez(r,θ) = - (jω/σ) e(1+j)(r-a)/δ n(θ) (β/k) Jz(r,θ) = - jωe(1+j)(r-a)/δ n(θ) (β/k) Let's do a little check on this. The integral of Jz should give I, but how is that going to work? I = !Syntax Error, Idθ !Syntax Error, Ir dr Jz(r,θ) = - jω (β/k) !Syntax Error, Ir dr e(1+j)(r-a)/δ !Syntax Error, Idθ n(θ) = - jω (β/k) * 2πN0!Syntax Error, Ir dr e(1+j)(r-a)/δ Now perhaps let x ≡ r/a so dx = dr/a and xdx = r/a dr/a = rdr/a2 and (r-a)/δ = (r/δ) - (a/δ) = (a/δ)(r/a - 1) = (a/δ)(x - 1) The above is then I = - jω (β/k) * 2πN0 !Syntax Error, I a2 xdx / * e(1+j)(a/δ)(x-1) = - jω (β/k) * 2πN0 a2 e-(1+j)(a/δ) !Syntax Error, I dx e(1+j)(a/δ)x The integral is this, where α = (1+j)(a/δ), which is the first time in lines doc I have ever seen the erf appear. If α is very large if a/δ >> 1 then erf(x) = 1 x → ∞ AS page 298 In this case the last blue expression above becomes (1/2) [ 2 eα - ] / [ α] But again if α is large, the first term wins and we have (1/2) [ 2eα ] / [ α] = [ eα] / [ α] = eα/(α) = e(1+j)(a/δ) / [(1+j)(a/δ)] and we then find that I = - jω (β/k) * 2πN0 a2 e-(1+j)(a/δ) * e(1+j)(a/δ) / [(1+j)(a/δ)] = - jω (β/k) * 2πN0 a2 / [(1+j)(a/δ)] Now I claim this above (D.10.15), N0 = (k/2πωa) I . (D.2.31) so my check integral then says I = - jω (β/k) * 2π(k/2πωa) I a2 / [(1+j)(a/δ)] = - jωβ * 2π(1/2πωa) I a2 / [(1+j)(a/δ)] = - jωβ * (1/ωa) I a2 / [(1+j)(a/δ)] = - jωβ * (1/ω) I / [(1+j)(1/δ)] = - jβ I / [(1+j)(1/δ)] But this requires that 1 = - jβ/ [(1+j)(1/δ)] (1+j)(1/δ) = -jβ (1+j) = -jδβ But from above D.2.10 I know that δβ = (j-1) so the check is then (1+j) = -j(j-1) = 1+j and I am duly amazed!