Old section D.11 (d) REVIEWED

Old Section D.11 (d) PhL 10.6.14 Replaced by updates on 10.6.14, due to changes in D.9. (d) Low frequency E fields The fields in (D.2.33) quoted above then become, Ez(r,m) = (1/4) ηm I Rdc [aβ'] fm fm = (r/a)|m| (|m|+1) (2/β'a) f0 = 4/(aβ') Er(r,m) = (j/4) ηm I Rdc (ak) gm gm = (r/a)|m|+1 + (r/a)|m|-1 g0 = 2 (r/a) Eθ(r,m) = (1/4) ηm I Rdc (ak) hm hm = (r/a)|m|+1 - (r/a)|m|-1 h0 = 0 or Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = I Rdc Er(r,0) = (j/2) I Rdc (ak) (r/a) Eθ(r,0) = 0 // low ω E fields If n(θ) is real and even in θ, we know from (D.10.4a) that, Ez(r,θ) = (1/4) I Rdc (aβ') [ f0 + 2 Σm=1∞ fm ηm cos(mθ) ] Er(r,θ) = (j/4) I Rdc (ak) [ g0 + 2 Σm=1∞ gm ηm cos(mθ) ] Eθ(r,θ) = (1/4) I Rdc (ak) [ h0 + 2 Σm=1∞ hm ηm cos(mθ) ] . (D.10.4a) Inserting the expressions (D.11.6) then gives Ez(r,θ) = I Rdc { 1 + Σm=1∞ (r/a)m (m+1) ηm cos(mθ) } Er(r,θ) = (j/2) I Rdc (ak) {(r/a) + Σm=1∞ [(r/a)m+1 + (r/a)m-1] ηm cos(mθ) } Eθ(r,θ) = (1/2) I Rdc (ak) { 0 + Σm=1∞ [(r/a)m+1 - (r/a)m-1] ηm cos(mθ) } . (D.11.8) As for surface impedance, from (D.11.8) we find that, for low ω, Zs(θ) ≡ Ez(a,θ)/I = Rdc { 1 + Σm=1∞ (m+1) ηm cos(mθ) } (D.11.9) <Zs(θ)> = Rdc (D.11.10) and this last result certainly seems reasonable. Notice that, since J = σE, = = . (D.11.11) Comment: The above expressions are for the case G = 0. To generalize to the case G>0, one makes the replacement I → I' = I with an appropriate interpretation of k (see Section D.9 and (D.9.31)). This generalization has no effect on the ratio given in (D.11.11). An anomaly. This last result is supposedly valid for very low ω. We would expect that in the limit ω→0 the above ratio should be exactly 1, at least for a non-conducting dielectric G = 0. This is so because we expect there to be no eddy currents at ω = 0 and these are the cause of Jz non-uniformity as discussed in Appendix P. We attribute our anomalous result (that Jz varies with θ) to the inaccuracy of the model at low ω, as outlined in section (a) above. It happens that in the limit ω→ 0 we also have I→ 0 when G = 0 (since then Z0 → ∞), but that is no justification for the anomalous result. See Chapter 7 for further discussion. We ignore this anomaly and proceed with our task of finding the low frequency E fields in more detail. Recall from (4.11.18) that Z0 = 1/Z0 = . (4.12.18) For low ω we find from these expressions for 1/Z0 and from (D.11.1,2) that, I = V/Z0 ≈ V k ≈ -j G > 0 (D.11.12) I = V/Z0 ≈ V k ≈ ω1/2 e-jπ/4 G = 0 (D.11.13) Inserting (D.11.12) into (D.11.7) gives, for G > 0, Small ω limit of the E field solutions, G>0 : Rdc = (D.11.14) Ez(r,m) = (1/2) ηm V Rdc (r/a)|m| (|m|+1) Z0 = Er(r,m) = a (1/4) ηm V Rdc G [(r/a)|m|+1 + (r/a)|m|-1] Eθ(r,m) = -ja (1/4) ηm V Rdc G [(r/a)|m|+1 - (r/a)|m|-1] k = -j Ez(r,0) = V Rdc Er(r,0) = a (1/2) V Rdc G (r/a) Eθ(r,0) = 0 The fields are all finite and there are non-zero expressions for Er and Eθ which account for the expected non-uniform flow of current into the dielectric through the conductor boundaries. We expect that Er(r,θ) just outside the conductor boundary is a strong function of θ for closely spaced conductors (the capacitor problem), and thus so is Jr(r,θ). But Jr(a,θ) is continuous through the boundary at ω = 0, so we expect to see a strong dependence of Jr(a,θ) on θ inside the round wire, as indicated by Er(r,m) in (D.11.14). Reader Exercise: Does (D.11.14) give the correct solution to the implied magnetostatics problem, or are there anomalies like the one noted above? Notice that Z0 is certainly correct based on the reader exercise given in Appendix K (c). The "wave" decays in z according to e-jkz = exp(-z) which also seems reasonable. Next, inserting (D.11.13) into (D.11.7) gives, for G = 0, this limiting form : Small ω limit of the E field solutions, G=0 : Rdc = (D.11.15) Ez(r,m) = (1/2) ηm V ω1/2 Rdc (r/a)|m| (|m|+1) Er(r,m) = a (j/4) ηm V ω C Rdc [(r/a)|m|+1 + (r/a)|m|-1] Eθ(r,m) = a (1/4) ηm V ω C Rdc [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = V ω1/2 Rdc Er(r,0) = a (j/2) V ω C Rdc (r/a) Eθ(r,0) = 0 As ω → 0, all fields vanish, corresponding to the fact that Z0 = → ∞ so I = V/Z0 → 0. In this situation the network model is just an infinite ladder of series resistors with no conductance cross pieces. Fig D.8