A for a round wire REVIEWED

What is A and φ for the Round Wire of Chapter 2 ? PhL 10.19.13 Comments 12/5/13. This seems to be the first time I ever tried this calculation, notice early date. Even now this subject is up in the air. I have something in Appendix B, but have not "gotten to it" yet. Remember that Chapter 2 is E and B field oriented, there is no potential. There ought to be a way to get that same result using potential theory, you would think. I later tried this in my "illogical" uniform J Helmholtz treatment. The King method requires a prescribed current, and that just doesn't work well with this problem where one of the "outputs" of the calculation is the current J and its skin effect. This is something I never computed because that section was done using fields only. Remember that we always have gauge freedom for the potentials. In Chapter 2 I assume that E = E = E(r) B =B = B(r) Now let's look at our defining relations: B = curl A E = - grad φ - ∂tA . (1.3.1) We than have B(r) = [curl A] = ∂zAr - ∂rAz E(r) = -∂zφ(r) - ∂tAz(r) = -∂zφ(r) -jω Az(r) We are allowed to pick any gauge we want, so perhaps div A = 0 = ∂zAz + (1/r)∂r(rAr) + (1/r)∂θAθ In addition to the above, we also have [curl A] = 0 => (1/r) ∂θAz - ∂zAθ = 0 [curl A] = 0 => (1/r)∂r(rAθ) - (1/r)∂θAr = 0 [- grad φ - ∂tA] = 0 => - ∂rφ - ∂tAr = 0 => - ∂rφ -jωAr = 0 [- grad φ - ∂tA] = 0 => (1/r)∂θφ - ∂tAθ = 0 => (1/r)∂θφ - jωAθ = 0 So here is a summary of all the equations in the ω domain (1/r) ∂θAz = ∂zAθ 1 ∂r(rAθ) = ∂θAr 2 ∂rφ = -jωAr 3 ∂θφ = rjωAθ 4 ∂zAr - ∂rAz = B(r) 5 ∂zφ(r) +jω Az(r) = - E(r) 6 ∂zAz + (1/r)∂r(rAr) + (1/r)∂θAθ = any function you want 7 One starting point is to require that ∂θφ = 0 as a symmetry condition, which from 4 says Aθ= 0. The equations then simplify to this smaller set (1/r) ∂θAz = 0 1 ∂θAr = 0 2 ∂rφ = -jωAr 3 ∂θφ = 0 4 ∂zAr - ∂rAz = B(r) 5 ∂zφ +jω Az = - E(r) 6 ∂zAz + (1/r)∂r(rAr) = any function you want 7 Now 1 says that Az = Az(r,z). Now 2 says that Ar = Ar(r,z) and this is true for all functions. Question: In item 7, why can't I select "any function" to be (1/r)∂r(rAr). Then the gauge condition says ∂zAz= 0 => Az = Az(r) only The equations of interest are then ∂rφ(r,z) = -jωAr(r,z) 3 ∂zAr(r,z) - ∂rAz(r) = B(r) 5 ∂zφ(r,z) +jω Az(r) = - E(r) 6 Suppose I use 3 to eliminate Ar. then ∂z[-∂rφ(r,z)/jω] - ∂rAz(r) = B(r) 5 ∂zφ(r,z) +jω Az(r) = - E(r) 6 or (1/jω)[∂z∂rφ(r,z)] - ∂rAz(r) = B(r) 5 ∂zφ(r,z) +jω Az(r) = - E(r) 6 Now use 6 to eliminate ∂zφ to get (1/jω)[∂r{ - E(r) - jω Az(r)}] - ∂rAz(r) = B(r) 5 or (1/jω)[∂r{ E(r) + jω Az(r)}] + ∂rAz(r) = - B(r) or [∂r{ E(r) + jω Az(r)}] + jω∂rAz(r) = - jωB(r) or ∂rE(r) + jω ∂rAz(r) + jω∂rAz(r) = - jωB(r) or ∂rE(r) + 2jω ∂rAz(r) = - jωB(r) or 2jω ∂rAz(r) = - jωB(r) - ∂rE(r) ≡ F(r) But in my round wire solution (2.1.9) I found that ∂rE(r) = jωB(r) . (2.1.9) so therefore 2jω ∂rAz(r) = - ∂rE(r) - ∂rE(r) ≡ F(r) = -2 ∂rE(r) This tells me that jω ∂rAz(r) = -∂rE(r) jωAz(r) = - E(r) + constant Now look again at B = curl A E = - grad φ - ∂tA . (1.3.1) You can always shift A by a constant and not affect the E and B fields. This is in addition to the gauge choice you might make which involves only derivatives of A. So write jωAz(0) = - E(0) + constant = 0 So we set the constant to E(0) = E(a)/ J0(βa). Then we have jωAz(r) = - E(r) + E(0) Az(0) = 0. So there is my Big Solution for Az(r). Az(r) = -(1/jω) [E(r) - E(0) ] = (j/ω) [E(r) - E(0)] STOP. I want to see someone else do this problem!