The Way Back Assumption INSTALLED

The Way Back Assumption PhL 6.29.14 This explanation of "the large σd" assumption is now installed. It is referred to just after it used in (1.3.20) and then at the end of that section I give a long explanation showing that it implies a lower limit for ω for which the A damped wave equation is valid. File this under Chapter 1. WAY BACK just above (1.3.20), while deriving the wave equation for A in Region 2, I snuck in the following assumption which I have long kept well swept under the rug: E << ∂tA in magnitude where E and A are measured inside a conductor of a transmission line. When you say the words "perfect conductor", you think |E | ≈ 0 and so you expect this to be true. But to test it, I have to know something about the vector potential inside the wire. Plan A no go Maybe Appendix G has something to say about this. For a round wire I compute that Az(r) = - [Iμ1/2π] ln(r) r > a region 1 Az(r) = - [Iμ2/(4πa2)] (r2-a2) - [Iμ1/2π] ln(a) r < a region 2 . (G.2.7) where Region 2 is inside the wire. But this is all a DC calculation! And there is that DC offset that is arbitrary so how do you deal with that anyway? Plan B a hint Recall that Ez(x1) - Ez(x2) = - ∂z V(x1,x2) - jω W(x1,x2) (4.11.6a)' I argue that the left side, although it varies, is very small and is the difference of the two large terns shown on the right. That at least HINTS that maybe Ez << jωAz somehow. Plan C best If I claim to have a reasonable solution of the transmission line system, I must have some estimate of the vector potential inside the wire in the time varying world. In Chapter 5 I claim that [ t2 + (βd2-k2)] φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.3.10) [ t2 + (βd2-k2)] Azt(x,y) = 0 Azt(C1) = W1 Azt(C2) = W2 W1- W2 = K (5.3.11) where (βd2- k2) = . so at least I am saying something about Azt on the outer surface of the conductors. Maybe it is like φ and is continuous through the surface (possible slope jump). Then we sort of have Azt(x,y) ~ K and then Az(x,y,z) = i(z) Azt(x,y) so we then have Az(x,y,z) ~ i(z) K and lo!, we have some kind of crude estimate for the size of Az on the conductors, and perhaps inside the conductors as well. So in this case I need to show that Ez << ∂tAz Ez << jωAz Ez << ω i(z) K ? Then fiddle some more Jz << ω I σ K ? But roughly I = πa2Jz Jz = I/(πa2) so I then want to show that I/(πa2) << ω I σ K ? 1/(πa2) << ω σ K ? ω >> 1/(πa2) / [ σ K] ? ω >> ? ω >> ? Dimensions check: RHS = m-2 m/henry * ohm-m = 1/henry * ohm = ohm / [ohm-sec] = sec-1 correct Recall that μσ = 73 roughly, so ω >> ≈ = .055 (1/a2) (1/K) Suppose a = 1 cm = 10-2 m. Suppose K = 17.5. Then we need ω >> 104 * .055 / 17.5 = 100 * 5.5 / 17.5 = 31.4 radians/sec f = ω/2π = 5 Hz That is something I can live with. Now do the Belden: .055 (1/a2) (1/K) Let's just guess that maybe K = 5. We have a = 394μ ≈ 400e-6 = 4e-4. Then .055 (1/a2) (1/K) = .055 So this says roughly that we need ω > 10 KHz for the Belden cable, again that is OK.