Where does the sym Jz solution lie REVIEWED

Where does the sym Jz solution lie? PhL 9.17.14 Here I try to understand why I don't get the right DC solution from Appendix D. Much of the stuff below ends up in Chapter 7. I run into the infinite B problem perhaps for the first time (perhaps not). See red comments in each section. Reviewed 10.8.14. 1. Smythian form for 2 round wires with I > 0 ? 1 2. How does the DC solution slip away? 2 3. Superposition and Jz Asym. 7 4. Digression on Atoms in Spherical Coords 8 5. Try starting in App D with ω = 0. 8 6. Precursor of Section 7.3 on getting the DC solution. 9 7. On the infinite B problem 14 1. Smythian form for 2 round wires with I > 0 ? Here I consider the Smythian superposition solution of Chapter 7.2 and conclude that you need a δ(k) coefficient and then you have Jz asym, and the two parallel wires carrying I > 0 just becomes the same old App D TL problem for single k. This is the superposition paradox I have never resolved. Of course those two wires must be infinitely long, then it really is an infinite TL etc etc. Thinking here of the two parallel round wires carrying large I and -I. In one of these wires, the E fields it would seem are determined by Appendix D even though this is not a transmission line. That is to say, the solution to the E Helm equation and the two BC's must have this form for Ez : [ atomic forms argument ] Ez(r,m,z)(3) = ∫dk e-jkz (1/4) ηm 2πω (a/k) N0 rk Rdc (aβ') fm(β'r) = ηm 2πω a2 N0 Rdc∫dk e-jkz (ω/k) rk β' fm(β'r) where rk is the "arbitrary weighting function". This result is based on a "most general possible form for Ez" argument [ see doc "show that the linear combination..." ] along with the two BC's. If you wanted a situation where n(θ,z) = n(θ)r(z) is constant in z, then you need r(z) = constant, which means rk = δ(k). If k is then very small, and we are also interested in ω being small, then β' = will be small in magnitude, and we should be able to use fm = (r/a)|m| (|m|+1) (2/β'a) f0 = 4/(aβ') This says then that Ez(r,m,z)(3) = ηm 2π a N0 Rdc∫dk e-jkz (ω/k) rk (r/a)|m| (|m|+1) (2) m≠ 0 Ez(r,0,z)(3) = 2π a N0 Rdc∫dk e-jkz (ω/k) rk (4) m= 0 Now somehow we are interested in a double limit where k→0 to give uniform r(z), and ω→ 0 which we think is associated with this same uniform r(z), that is, a DC situation where nothing moves. So what are we going to do with the ratio (ω/k)? If we were to make our network model for the two wires as in Fig K.5, which looks just at a differential piece dz of our two conductor system. we obtain the TL equations with k2 = - zy . But this then is our usual k(ω) and we know from App Q that for small ω and G = 0, Re(k) ≈ + Im(k) ≈ - k = (1-j) = e-jπ/2 (ω/k) = e+jπ/2 (1/) . Using k = k(ω) based on the network model, we then at least come up with (ω/k) not blowing up. If we now take our limit rk = δ(k) we find Ez(r,m,z)(3) = ηm 2π a N0 Rdc e+jπ/2 (1/) (r/a)|m| (|m|+1) (2) m≠ 0 Ez(r,0,z)(3) = 2π a N0 Rdc e+jπ/2 (1/) (4) m= 0 But now as ω→ 0, both of these vanish and we get Ez ≡ 0. The DC solution we are hoping to find has somehow slipped out of our hands! Somehow it got excluded. But we know that our elusive DC solution does satisfy the E Helm equation, div E = 0, and both BC's! So where did that solution go??? We thought we had a "most general solution" in terms of the harmonics of the scalar Ez Helm equation. Let's go back at the harmonics we use to construct Er(r,θ,z) : ejmθ e-jkz Jm(r) If you were looking for a constant solution, you would take m = 0 and k = 0 so there is then no variation in either θ or z. The harmonic then becomes ejmθ e-jkz Jm(r) → J0(βr) But alas, this is NOT a constant solution for Ez. However, if we now take ω→0. we get ejmθ e-jkz Jm(r) → J0(0) = 1 and there is your hoped-for constant solution. Note that for β2 ≠ 0, a constant is NOT a solution of the Helm equation! Only for the Laplace when β2 = 0. 2. How does the DC solution slip away? Here I try to figure out how the constant DC solution "slips away". This led me later to write 7.3. So why does this constant solution "get lost" in Appendix D? For β = 0 the solution Ez = constant does satisfy the Helm equation (really Laplace if β2 = 0) , and it also satisfies both BC's since ω = 0 goes with Er = 0 and Eθ ≡ 0 makes the other BC just fine. It also satisfies divE = 0. This missing solution has to be in the m = 0 sector of App D, since we expect no θ asym. That sector has this solution set: Ez(r,0) = (ω/σ) (β'/k) N0 Er(r,0) = j (ω/σ) N0 Eθ(r,0) = 0 or Ez(r,0) = (1/σ) (ω/k) N0 Er(r,0) = j (1/σ) ω N0 Eθ(r,0) = 0 As ω → 0 we know that → jk so we then have Ez(r,0) = (1/σ) jk (ω/k) N0 = (j/σ) ω N0 Er(r,0) = j (1/σ) ω N0 Eθ(r,0) = 0 Now without doing detail, let's assume that k→ 0 in some manner as ω → 0. Then we can use small args in the Bessel functions and say = = = (r/a) We then have Ez(r,0) = (j/σ) ω N0 = (j/σ) ω N0 * = 2 (1/aσ) N0 (ω/k) Er(r,0) = (j/σ) ω N0 = (j/σ) ω N0 (r/a) Eθ(r,0) = 0 Taking the limit now we get Ez(r,0) = 2 (1/aσ) N0 (ω/k) Er(r,0) = 0 Eθ(r,0) = 0 and now we are very close to our hoped-for DC solution, but we have to ponder the ratio (ω/k). We have already assumed that as ω→0 we get k→ 0, but have not specified details of how this happens. \ Question: How do I WANT Ez to come out? Answer: Ez = IRdc This would require that IRdc = (2/aσ) N0 (ω/k) = (2/aσ) q/(2πa) (ω/k) = (1/σ) q/(πa2) (ω/k) = q (ω/k) Rdc which in turn would require that I = q (ω/k) But why would the total conductor current I be proportional to surface charge q? Status: I am searching for the missing DC solution. I look in the m = 0 sector, and I am pretty close with the above field set. but I don't know what to say about (ω/k) . Scenario. Imagine two parallel conductors which have Rdc = 0. In this case, V = 0 can drive any current I that you want. You only need a current source. In this case, assuming also that Gdc = 0, the usual k(ω) formula says that k = -j = ω ≈ ω/vd . Maybe this is a viable model for the two parallel cylinders. In this case we have exactly (ω/k) = vd and then Ez(r,0) = 2 (1/aσ) N0vd = (2/aσ) * q/(2πa) * vd = q * 1/(σπa2) vd = q vd Rdc Er(r,0) = 0 Eθ(r,0) = 0 In this case we have Jz(r,0) = q * 1/(πa2) vd and then I = q vd just as I found above. But if V = 0, then there would be no surface charge q. Well. perhaps we just assume that Rdc ≈ 0 relative to ωL, though this gets harder as ω→ 0. Could write q = CV and then have Ez(r,0) = C vd Rdc V I = CV vd. V = I [ 1/(Cvd)] This last makes it appear that the conductors have R = 1/(Cvd). The Z0 formula however gives Z0 = = = = (1/C) = (1/C) (1/vd) = 1/(Cvd) in agreement with R just computed. So I seem to have concocted an infinite transmission line which has now a finite current I if you apply a source voltage V and you do this at DC, namely, I = (Cvd) V. Of course I have so far only looked at the m = 0 sector. For m ≠ 0 we know that Ez(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (aβ') fm fm = [ - ] So as done above, let's write β' = → jk in the limit ω→0 and so again we use the small arg limit of fm which is fm = (r/a)|m| (|m|+1) (2/β'a). We then have Ez(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (aβ') fm = (1/4) ηm 2πω (a/k) N0 Rdc (aβ') (r/a)|m| (|m|+1) (2/β'a) = (1/4) ηm 2πω (a/k) N0 Rdc (r/a)|m| (|m|+1) (2) = (1/4) ηm 2πa (ω/k) N0 Rdc (r/a)|m| (|m|+1) (2) = ηm πa (ω/k) N0 Rdc (r/a)|m| (|m|+1) But we then have the same (ω/k) factor which we have to set to vd and then Ez(r,m) = ηm πa vd N0 Rdc (r/a)|m| (|m|+1) = ηm πa vd * q/(2πa) * Rdc (r/a)|m| (|m|+1) = (1/2) ηm vd *q * Rdc (r/a)|m| (|m|+1) = (1/2) ηm vd CV Rdc (r/a)|m| (|m|+1) = {(1/2) ηm vd C Rdc (r/a)|m| (|m|+1)} V As usual, we end up with Jz asymmetry. Recall from above Ez(r,0) = C vd Rdc V And then Ez(r,m)/ Ez(r,0) = (1/2) ηm (r/a)|m| (|m|+1) and this is the usual result, just appearing now in our special scenario. [ so I got nowhere ] Below is another little effort that goes nowhere. If I assume the usual k(ω) with G > 0, then Re(k) ≈ (ω/2) (Rdc + ωdLdc) + O(ω2) ω < ωd = (σd/εd) Im(k) ≈ - [ 1 + (tanL/2) (ω/ωd)] + O(ω2) Gdc = Cωd or Re(k) ≈ (ω/2) (Rdc + ωdLdc) Im(k) ≈ - k = (ω/2) (Rdc + ωdLdc) - j Then (ω/k) = ω / [(ω/2) (Rdc + ωdLdc) - j ] → ω/[ - j ] → 0 and I then get Ez = 0. If I assume G = 0, I get instead k = (1-j) and then (ω/k) = ω /[)(1-j)] = / [ )(1-j)] → 0 so in either case I get Ez = 0 and the looked-for solution is lost. Note 1: Something seems wrong with the G > 0 case. There should be some current I for small z, and then it decays away due to the leakage across the dielectric. So I expect to see Ez(r,0,z) be some finite value at z = 0 and then decay away over distance related to 1/ . This is all at ω = 0. But instead I am seeing Ez ≡ 0 at all z due to the (ω/k) factor. [ but in 7.2 3 I show I = V/Z0 = V which is a fine finite current.] [ so I got nowhere again ] ______________________________________________________________________________ 3. Superposition and Jz Asym. So somehow the above expression must have in it somewhere the solution of the two wires with its symmetric Jz. But regardless of the choice of rk, we can see that = = ηm In order to achieve a uniform r(z) [ which we expect for the two long parallel wires near DC ] , we know that we must approach rk = δ(k) since that turns off the z exponential and since r(z) and rk are Fourier transforms. If k is small, then β' = is uncertain, because we are also interested in ω → 0 which causes β→0. In any event, it does seem that |β'| → 0 in this combination of limits, and then we should be justified in using the small arg limits of the f functions, which from D.11 are fm = (r/a)|m| (|m|+1) (2/β'a) f0 = 4/(aβ') and then the above ratio becomes = ηm = ηm where we are thinking low ω and low-skewed rk. Interestingly perhaps, if we set rk = δ(k) [nothing new here! ] _________________________________________________________________________________ 4. Digression on Atoms in Spherical Coords : Look at the spherical harmonics instead. For the Laplace equation, the harmonics are these (MSp26) [ rl, r-l-1], Plm(cosθ), ejmφ // ignore r-l-1 if origin is in region of interest If we want to find a constant in these atoms, we set m = 0 and l = 0 and we get rl Plm(cosθ), ejmφ → 1 For the scalar Helm equation, however, we get instead, [ jl(kr)] Plm(cosθ) ejmφ So in spherical coordinates, in looking for a constant solution, we would have m = 0 and l = 0 and then we get jl(kr)Plm(cosθ) ejmφ → j0(kr) and again, this is NOT a constant solution. But you don't expect (2 + k2)u= 0 to have a constant solution! That is only for Laplace. There is something general Sturm-Liouville going on here. Recall the you have two oscillatory and one expo atom. For the spherical coordinates case, it seems that jl(kr) is the expo factor! But the plots of these functions look oscillatory. The large r behavior is this, and if k is complex, as in our example above, we have jl(kr) ~ e|Im(k)|r /(kr)2 = blows up exponentially, just as rl blows up. _________________________________________________________________________________ 5. Try starting in App D with ω = 0. Plan A. Go through the development of Appendix D at ω = 0 from the start. This means β = 0, and that means that you have a vector Laplace equation instead of a vector Helmholtz equation. Go ahead and work with some particular value of k. The Ez equation will have a solution I guess with partial wave amplitude Ez = constant * rm . Then as before, use divE = 0 and do all the motions. The goal here is to see if you can come up with Jz asymmetry directly at ω = 0 without having to take a limit ω→0. In the end you can take k→0 I guess. I have done all this with ω ≠ 0 and get Jm(β'r) functions, I am just saying "let's try this at ω = 0 from the start". I am of course looking for some mechanism that creates the asym Jz. Now with Rdc= 0 and Gdc= 0 I have a scenario that has finite current, and I will get Jz asym right in that scenario! Do this tomorrow. The details of all these appear in "show that linear....". Maybe I will find out that at DC things really are asym and this is a feature of the vector Laplace equation. 9/19/14. I will now carry out Plan A, but I doubt it will reveal anything. If I set ω = 0 at the start, that just means that β' = = = jk (assuming that sign). Everything in App D then goes through exactly the same and we end up with First summary of the E field solutions (D.2.21) Ez(r,m) = - j (β'/k) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = jk (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) We then come to the boundary conditions which are now these: Er(r=a,m) = 0 // since ω = 0 Eθ(r=a,m) = 0 We can use equations (3) and (4) below (D.2.27) which now say Jm+1(xa) + ( + ) Jm+1(xa) = 0 (3) 2 am xa-1 Jm(xa) - Jm+1(xa) = 0. (4) The second equation says am = 0 for some general value of xa = jka and general m. Then the first equation says that Km = 0 as well. Very interesting. The conclusion is then Ez(r,m) = 0 Er(r,m) = 0 Eθ(r,m) = 0 so we get no solution at all by assuming ω=0 and k ≠ 0. Great. [ same old problem that coeffs vanish ] 6. Precursor of Section 7.3 on getting the DC solution. The following is what let to my "proof" of the true DC solution in 7.3. Suppose we assume both ω and k are 0 at the start, which might be more accurate. In this case, β' = 0 and our equation set is this: [2E]z +0 Ez = 0 : [r2∂r2 + r ∂r - m2 ] Ez(r,m) = 0 (D.1.15) [2E]r + 0 Er = 0 : [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eθ(r,m) = 0 (D.1.17) [2E]θ + 0 Eθ = 0 : [r2∂r2 + r∂r - (m2+1)] Eθ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r [r Er(r,m)] + jmEθ(r,m) = 0 (D.1.19) The acceptable solution to the first equation is this, where for the moment I assume m ≥ 0 : Ez(r,m) = Azm rm r2m(m-1)rm-2 + r mrm-1 - m2rm = rm[ m2- m + m - m2] = 0 As usual, we solve the last for jmEθ and insert that into the second equation to get [r2∂r2 + r∂r - (m2+1)] Er(r,m) + 2∂r [r Er(r,m)] = 0 or [r2∂r2 + r∂r - (m2+1)] Er + 2 { r ∂rEr + Er} = 0 [r2∂r2 + 3r∂r - (m2-1)] Er = 0 Maple shows that the solutions here are rm-1 and r-m-1 : This seems then to say that, for m ≥ 1 Er(r,m) = Ar rm-1 For m = 0 our second equation is really [r2∂r2 + r∂r - 1] Er(r,0) = 0 But the two solutions here according to Maple are (r+1/r) and (r-1/r), but both these diverge at r = 0. This would seem to imply that Er(r,0) = 0! What did I get in App D? Er(r,m) = am x-1 Jm(x) + Jm+1(x) Er(r,0) = a0 x-1 J0(x) + J1(x) = J1(x) ≈ J1(0) = 0 so at least things are consistent. So far then I have Ez(r,m) = Azm rm m ≥ 0 Er(r,m) = Arm rm-1 m ≥ 1 Er(r,0) = 0 m = 0 For m ≥ 1 the div E says, [ no z term since k = 0 ] jmEθ(r,m) = - ∂r [r Er(r,m)] = - ∂r [r Arm rm-1] = - Arm ∂r [ rm] = -m Arm rm-1 which tells us that jEθ(r,m) = - Arm rm-1 . For m = 0 the third equation for Eθ is the same as the second equation for Er when m = 0, but for this latter we had Er = 0. So here then is our summary Ez(r,m) = Azm rm m ≥ 0 Er(r,m) = Arm rm-1 m ≥ 1 Er(r,0) = 0 m = 0 jEθ(r,m) = - Arm rm-1 m ≥ 1 Eθ(r,0) = 0 m = 0 Now we come to the BC's. They are Er(r=a,m) = 0 // since ω = 0 Eθ(r=a,m) = 0 For m = 0, since Er and Eθ are already 0, these BC's are met. For m ≥ 1 we get Arm am-1 = 0 => Arm = 0 Arm am-1 = 0 => Arm = 0 Our solution then seems to be Ez(r,m) = Azm rm m ≥ 0 Er(r,m) = 0 m ≥ 0 Eθ(r,m) = 0 m ≥ 0 I think this is a completely new result!. I guess we could integrate to get for m = 0 we get ∫dA Jz(r,0) = σ Az0 ∫dA = σ Az0 πa2 So we have I = σ Az0 πa2 => Az0 = I/(σπa2) = I Rdc The solution is then Ez(r,m) = Azm rm m > 0 Ez(r,0) = I Rdc m = 0 Er(r,m) = 0 m ≥ 0 Eθ(r,m) = 0 m ≥ 0 There is nothing to determine the coefficients Azm for m ≥1 !!! Let's just make sure the 3rd Helm equation is OK. [r2∂r2 + r∂r - (m2+1)] Eθ(r,m) + 2jmEr(r,m) = 0 For m > 0 we get [r2∂r2 + r∂r - (m2+1)] 0 + 2jm 0= 0 OK For m = 0 we get [r2∂r2 + r∂r - (02+1)] 0 + 2j0 0 = 0 OK Somehow, it seems that when you solve the problem directly at DC, the problem is underspecified because nothing determines the Azm for m > 0. They could be any numbers you want, and your solution still solves the 3 Helm equations, the div E = 0 equation, and the two BC's. So this equation set seems to be incomplete! So let's go back to the Maxwell equations. We need this to be respected: curl E = -∂tB = 0 Start with (taken from "the curl E" doc, curl E(r,θ,z) = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] curl E(r,m,z) = [ r-1jmEz+ jkEθ] + [-jkEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] Now set k = 0 to get curl E(r,m,z) = [ r-1jmEz] + [- ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] But only the Ez field is non-zero so we get curl E(r,m,z) = [ r-1jmEz] + [- ∂rEz] So we have two equations of interest here r-1jmEz = 0 - ∂rEz = 0 Inserting our Ez we get r-1jm Azm rm = 0 - ∂r Azm rm = 0 or m Azmrm = 0 -mAzmrm-1 = 0 These both say the same thing m Azm = 0 and THIS then sets all the other coefficients to 0 !!! Then the problem is determined! Here I discover that you have to then use curlE = 0 to nail down your DC solution, and this is duly carried out now in Section 7.3. Fact: The set of equations Helm + div + BC's is incomplete at ω = 0 ! You need to add curl E = 0 to that set. When you do this, you obtain the classical result that Jz = constant on the cross section, so there is in fact no Jz asymmetry! For ω > 0, the curl E equation is used to find B ! Hypothesis: For ω > 0, the condition corresponding to curl E = 0 for ω = 0 is that B be finite! 7. On the infinite B problem So now I am off on the issue of understanding how I should be getting curl E = 0 as ω→ 0. I know about the little "theorem", but I did not know much about the B fields at this time. Now, for ω > 0 and k > 0 I suspect the problem is over-specified! Ignoring the curl E equation, we obtain these fields if we assume that both ω and k are very small at the same time, Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) m>0 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] We then find (see curl E doc) that [curl E(r,m,z)]r = r-1jmEz+ jkEθ = j (1/4) ηm Rdc { 2 I r-1m (r/a)|m| (|m|+1) + a k2I [(r/a)|m|+1 - (r/a)|m|-1] } where I = 2πω (a/k) N0 = 2πa (ω/k) N0 and N0 = q/(2πa) = CV/(2πa). The second term is never a problem, we can just set it to zero based on k→0, or more carefully for G = 0 k2 ~ ω and I ~ . It is the first term that is the problem because Br(r,m) = +(j/ω) [curl E]r = (j/ω) j (1/4) ηm Rdc { 2 I r-1m (r/a)|m| (|m|+1) Since I = 2πω (a/k) N0 = 2πaN0 e+jπ/4/ we then get Br → ∞ . Escape Hatch: The notion that k ~ must be wrong! My Theorem 1 says that if [curl E]i → ωs [stuff]i , then we must have s ≥ 1 to keep B finite. My problem above is that I have s = 1/2. The only way out is if this is true: (ω/k) → ωs with s ≥ 1 Now suppose k → ωr. Then we have s = 1-r so we need 1-r ≥ 1 which means -r ≥ 0 or r ≤ 0. But cannot have r < 0 since then k blows up which seems completely wrong since we expect no z dependence. Thus we must have r = 0. Fact: if k → small constant as ω→0, then Br will remain finite. [ but see Ch 7 ] I think that the TL equations are wrong as ω→0, but I don't know how to exactly correct them. But whatever that correction is, it must result in k → small constant as ω → 0. But this in turn says there is some residual z dependence, and that seems wrong! Fact 4: The small ω limit for k(ω), assuming ωd = 0, is given by (Q.4.9) Re(k) ≈ + ( 1 - tanL/2) + O(ω3/2) Im(k) ≈ - ( 1 + tanL/2) + O(ω3/2) Fact: If you say k → 0, that in itself is a contradiction because it implies no decay since Im(k) = 0. But that can only happen in a completely lossless line. So yes, there is z dependence at ω = 0 for a real TL. Perhaps the existing imaginary k causes some small real k phase action. How then can the above limit be right? If G = 0, then I = 0 and there is no loss, correct. The G > 0 model seems to work better in this sense. We have Fact 3: The small ω limit for k(ω), assuming ωd > 0, is given by (Q.4.6) Re(k) ≈ (ω/2) (Rdc + ωdLdc) + O(ω2) ω < ωd = (σd/εd) Im(k) ≈ - [ 1 + (tanL/2) (ω/ωd)] + O(ω2) Gdc = Cωd Then as ω→ 0 this says Re(k) = 0 => no wave action Im(k) = - => there is loss going down the line Also, we end up with Br = finite! Let's write k = -j in the limit ω→ 0. k2 = -RdcGdc Then the above says I = 2πωaN0/ [-j] Then go back to [curl E(r,m,z)]r = r-1jmEz+ jkEθ = j (1/4) ηm Rdc { 2 I r-1m (r/a)|m| (|m|+1) + a k2I [(r/a)|m|+1 - (r/a)|m|-1] } = j (1/4) ηm Rdc 2πωaN0/ [-j]{ 2 r-1m (r/a)|m| (|m|+1) - a RdcGdc [(r/a)|m|+1 - (r/a)|m|-1] } Then divide by jω to get Br Br = (1/4) ηm Rdc 2πaN0/ [-j]{ 2 r-1m (r/a)|m| (|m|+1) - a RdcGdc [(r/a)|m|+1 - (r/a)|m|-1] } which is valid only for m ≠ 0. If m = 0 we have instead Ez(r,0) = I Rdc Er(r,0) = (j/2) I Rdc (ak) (r/a) Eθ(r,0) = 0 // low ω E fields We then find (see curl E doc) that [curl E(r,0,z)]r = r-1j0Ez+ jkEθ = 0 + 0 = 0 => Br(m=0) = 0. Other components? [curl E(r,0,z)]θ = -jkEr - ∂rEz = -jk (j/2) I Rdc (ak) (r/a) - ∂r I Rdc = -jk12 (j/2) I Rdc r = -jk12 (j/2) 2πa (ω/k) N0 Rdc r = -jω k1 (j/2) 2πa N0 Rdc r Then get Bθ = - k1 (j/2) 2πa N0 Rdc r = - [-j] (j/2) 2πa N0 Rdc r = [j] (j/2) 2πa N0 Rdc r = - []πa N0 Rdc r = - []πa q/2πa *Rdc r = - [](q/2)*Rdc r = - [](q/2) r But Z0 = at ω = 0, so the above says Bθ = - [1/Z0 ] (1/2) CV r Bθ = - Idc(1/2) C r Needs lost of work! Well, I guess the time has come to do all the B field limits and check them, probably with Maple. I think this last one above is OK and what you expect inside a wire, perhaps get Z0 and q = CV etc. At this time, I had not really computed the B field components in useful detail, but all that later got added to lines doc, and we now how the infinite B section in Chapter 4.