Exterior solution for round wire as Section 2_5 REVIEWED

Exterior Solution for Round Wire (Section 2.5) PhL 11.26.13 Comments 12/5/13. I think at this time 11.26 Section 2.5 might have been about an exterior solution for the round wire! Perhaps here is where I developed that solution (since removed from Ch 2). I have not studied what appears below. I see I was debating which Helmholtz equation to use, which β. Since I am planning a comprehensive solution to this problem for Appendix D (not done yet) I won't worry about details below. I now know that the round wire is really the central conductor of a coaxial cable, that there is surface charge, the Er is going to dominate in the exterior, and other things I did not think of when writing this. Here I even got that H1(1)(βr) for exterior B and maybe H0(1)(βr) for exterior E, but I think all that stuff is completely wrong. I will archive this early attempt nevertheless. In this section the E and B fields are functions only of r. Therefore 2 = 22D but to reduce clutter we shall write the 2D Laplacian as 2. In Section 2.1 we solved Maxwell's equations for the E and B fields inside a round wire with results summarized in box (2.1.26) in terms of Bessel functions J0 and J1. The arguments of these functions have phase ej3π/4 which causes them to be complex, and this led to a restatement of the solutions for E and B in terms of the real Kelvin functions as shown in (2.2.11) and (2.2.12). Here we shall complete the solution to the problem by considering the region exterior to the wire. Question: (it did not take long!) Why do I ignore displacement current in (2.1.5) ? Ah, the reason is that in the conductor we approximate E ≈ 0? Then jωεE ≈ 0. I comment on this elsewhere but not at the place I assume it. This won't work in the dielectric. Maybe I should just use the ODE which I quote in (2.1.12) as saying (2 + β2)E(x,ω) = 0 inside the wire. This would be true also outside the wire since ρ = 0 at both places, not sure about the boundary. How would this go? (2 + β2)E(x,ω) = 0 Suppose you maintain the assumptions of (2.1.4) in terms of field directions, so E = E . Then this thing says (2 + β12)E(r) = 0 as a scalar Laplace equation where now β is for the exterior medium. Solutions are E(r) = C' J0(β1r) + D' Y0(β2r) Backup a moment. I know this Damped field wave equations (1.3.36) : βi2 = ω2μiξi (2+β12)E = (1/ε1) Σi=2N+1grad ρi (2+β12)B = 0 // region 1 (2+β22)E = 0 (2+β22)B = 0 // region 2 (2+β32)E = 0 (2+β32)B = 0 // region 3 (1.5.27) Here 1 = dielectric and 2 = conductor, so inside either material we are going to have a homo equation. Thus we have (2 + β2)B(x,ω) = 0 But B is in the θ direction, unlike E which was in the z direction, so this time we really do have to use cylindrical coordinates for the vector 2 operator. Maybe avoid using this equation? But how am I going to find B in the exterior region?? I need an equation for it. OK, suppose B = B(r) . I just went and added this to my diffop doc, so we then have 2B = [2Br -2 r-2∂θBθ - r-2Br] + [2Bθ +2 r-2∂θBr - r-2Bθ] + 2Bz = [ -2 r-2∂θBθ ] + [2Bθ - r-2Bθ] = [2Bθ - r-2Bθ] Thus, the equation (2 + β2)B(x,ω) = 0 becomes 2Bθ - r-2Bθ + β2 Bθ = 0 I have done this recently elsewhere, but I will just do it here again rather than search all over. Call it B, and then we have r-1∂r(r∂rB) - r-2B + β2 B = 0 ∂r(r∂rB) - r-1B + r β2 B = 0 r B" + B' - r-1B + r β2 B = 0 r2 B" + rB' - B + r2 β2 B = 0 r2 B"(r) + rB'(r) + (r2β2-1)B(r) = 0 This is almost Bessel's with n=1. x = βr f(x) = B(x/β) = B(r) f(x) = B(r) B(r) = f(x) f'(x) = (1/β)B'(x/β) = (1/β)B'(r) B'(r) = β f'(x) f"(x) = (1/β)2B"(x/β) = (1/β)2B"(r) B"(r) = β2f"(x) So equation becomes r2 β2 f"(x) + r β f'(x) + (r2β2-1)f(x) = 0 x2 f"(x) +x f'(x) + (x2-1)f(x) = 0 Solutions are like J1(x) = J1(βr). DONE. So, what we find is that B(r) = B(a) H1(1)(βr)/ H1(1)(βa) r > a Aside: Now I need some argument to say that H0(1)(β1r) will be the solution. The usual argument is the large r behavior. So get NIST going again. But wait, I already comment on this somewhere. In I.3 I am doing the Green's for Helmholtz in 2D. But I never quote the result from NIST. OK, I have added this in Appendix I, and I will refer to it. However, in the dielectric the phase of β is not 3π/4. It is this β12 = με1ω2 - jωμ1σ1 = ω2μ1 ( ε1 - jσ/ω) = ω2μ1 ξ1 ξ1 ≡ ε1 - jσ1/ω . (1.5.1) Ouch. In the dielectric, σ1 is very small and we have roughly β12 = με1ω2 - jε But write this as ξ1 ≡ ε1 - jσ1/ω = ε1+ (- j) σ1/ω = ε1+ ej3π/2 σ1/ω Reξ1 = ε1 Imξ1 = -σ1/ω ouch! Won't this cause H(1) problems??? Although β12 has a negative imaginary part, according to our selected branch of square root function, the quantity β1 has a positive imaginary part as shown in this picture where β2 = A - j B. For more details see Comment and Figure in Appendix I ****. Thus, the H0(1)(β1r) does indeed decay for large r. So here is the proposed exterior solution for E. It has n = 0 because E = E , E(r) = K H0(1)(β1r) Here I rewrite the interior solution using 2 subscripts for the conductor. Interior Solution of a Round Wire (2.1.26) = (μ2σ2[r E(r) ] (2.1.7) = jωB(r) (2.1.9) E(r) = E(a) [J0(β2r) / J0(β2a)] (2.1.22) J(r) = J(a) [J0(β2r) / J0(β2a)] // above times σ2, J = Jz = current density B(r) = - (β2/jω) E(a) [J1(β2r) / J0(β2a)] = + B(a) [J1(β2r) / J1(β2a)] (2.1.25) β2 = ej3π/4 (/δ) = ej3π/4 and β22 = -2j/δ2 (2.1.21), (2.1.14) δ ≡ = skin depth (2.1.20) The B field is in the direction and so is a tangential field, so we call upon Ht2 - Ht1 = Kzfree or (1/μ2)Bt2 - (1/μ1)Bt1 = Kzfree (1.1.44) We assume there are no free longitudinal surface currents on the surface of the round wire (only the longitudinal bulk current), so therefore (1/μ2)Bt2 = (1/μ1)Bt1 Thus, our matching condition is (1/μ2) B(a) [J1(β2a) / J1(β2a)] = (1/μ1) K H0(1)(β1a) or (1/μ2)B(a) = (1/μ1) K H0(1)(β1a) Comment: We still have this unknown in one form or the other, B(a) and K. This matching condition does not determine either one. What about the idea that far away, it looks like a thin wire, and then we can use the Green's function solution from Appendix I. Our problem for a thin wire is this Back up. First, write the two wave equations of interest (2 + β2)E = 0 (2 + β2) B = 0 Now assume that in both regions that E = E(r) => only Ez B = B(r) => only Bθ  Use cylindrical coordinates in which 2E = [2Er -2 r-2∂θEθ - r-2Er] + [2Eθ +2 r-2∂θEr - r-2Eθ] + 2Ez = + 2Ez = + 2E(r) 2B = [2Br -2 r-2∂θBθ - r-2Br] + [2Bθ +2 r-2∂θBr - r-2Bθ] + 2Bz = [ -2 r-2∂θBθ] + [2Bθ - r-2Bθ] = [2Bθ - r-2Bθ] since Bθ= Bθ(r) Then the two wave equations become (2 + β2)E(r) = 0 (2 + β2- r-2)B(r) = 0 Using ψ = ψ(r), 2ψ = r-1∂r(r∂rψ) + r-2∂θ2ψ + ∂z2ψ = r-1[r∂r2ψ + ∂rψ] = ψ" + r-1ψ' we write our two equations then as E"(r) + r-1E'(r) + β2E(r) = 0 B"(r) + r-1B'(r) + (β2 - r-2)B(r) = 0 or r2 E"(r) + rE'(r) + β2r2E(r) = 0 r2 B"(r) + rB'(r) + (r2β2 -1)B(r) = 0 Letting B(r) = b(x) and E(r) = e(x) where x = βr, we find for example for B (and similarly for E) b(x) = B(r) => B(r) = b(x) b'(x) = ∂xb(x) = ∂x B(x/β) = (1/β)B'(x/β) = (1/β) B'(r) => B'(r) = β b'(x) b"(x) = (1/β)2 B"(r) => B"(r) = β2 b"(x) The two equations then become r2 β2 e"(x) + r β e'(x)E'(r) + β2r2 e(x) = 0 r2 β2 b"(x) + r β b'(x)E'(r) + (β2r2-1) b(x) = 0 pr x2 e"(x) + xe'(x) + x2e(r) = 0 solutions like e(x) = J0(x) = J0(βr) = E(r) x2 b"(x) + xb'(x) + (x2 -1)b(x) = 0 solutions like b(x) = J1(x) = J1(βr) = B(r) where we recognize the equations in x as standard Bessel equations as in Spiegel 24.1. Now we have to select which kind of Bessel function to use in each region. For r < a we use E(r) = J0(β2r) B(r) = J1(β2r) r < a region 2 conductor E(r) =H0(1)(β1r) B(r) = H1(1) (β1r) r > a region 1 dielectric Normalizing at the surface we can write E(r) = E(a-) J0(β2r)/ J0(β2a) B(r) = B(a-) J1(β2r)/ J1(β2a) r < a region 2 conductor E(r) = E(a+) H0(1)(β1r)/ H0(1)(β1a) B(r) = B(a+) H1(1)(β1r)/ H1(1)(β1a) r > a region 1 dielectric where a+ means just outside the round wire and a- just inside the wire. We now call up our boundary conditions. The E = E(r) field is tangential so box (1.1.50) tells us that Et1 = Et2 => E(a+) = E(a-) ≡ E(a) The B = B(r) is also tangential, so from box (1.1.50), (1/μ2)Bt2 = (1/μ1)Bt1 => (1/μ2)B(a-) = (1/μ1)B(a+) ≡ (1/μ2)B(a) since we assume there is no free surface current on the surface of our wire (only the bulk volume current). The symbols E(a) and B(a) are those just inside the wire and were used in Section ***. We then have E(r) = E(a) J0(β2r)/ J0(β2a) B(r) = B(a) J1(β2r)/ J1(β2a) r < a region 2 conductor E(r) = E(a) H0(1)(β1r)/ H0(1)(β1a) B(r) = (μ1/μ2)B(a) H1(1)(β1r)/ H1(1)(β1a) r > a region 1 dielectric Comment: so here we are with our still undetermined E(a) and B(a) constants. If we apply Ampere's Law just inside the wire surface we get 2πaH(a) = I => 2πaB(a)/μ2 = I => B(a) = μ2I/(2πa) so we now have one of these two constants. From curl E = - ∂tB C E ds = -∂t[∫S B dS] (1.1.36) we showed in (2.1.8) applied to a certain math loop inside the wire (region 2) that = jωB(r) Using the region 2 equations above this says ∂r[E(a) J0(β2r)/ J0(β2a)] = jω [B(a) J1(β2r)/ J1(β2a)] Now ∂r J0(β2r) = β2 J0'(β2r) = - β2 J1(β2r) using Spiegel 24.7. Thus the above becomes -[E(a) β2 J1(β2r)/ J0(β2a)] = jω [B(a) J1(β2r)/ J1(β2a)] Now evaluate this at r = a to get -[E(a) β2 J1(β2a)/ J0(β2a)] = jω [B(a) J1(β2a)/ J1(β2a)] -[E(a) β2/ J0(β2a)] = jω [B(a) / J1(β2a)] E(a) = { (-jω/β2) [J0(β2a)/ J1(β2a)] } B(a) which is consistent with (2.1.25). Thus we have our final frequency domain solutions: E(r) = E(a) J0(β2r)/ J0(β2a) B(r) = B(a) J1(β2r)/ J1(β2a) r < a region 2 conductor E(r) = E(a) H0(1)(β1r)/ H0(1)(β1a) B(r) = (μ1/μ2)B(a) H1(1)(β1r)/ H1(1)(β1a) r > a region 1 dielectric where B(a) = [μ2I/(2πa)] E(a) = [μ2I/(2πa)]{ (-jω/β2) [J0(β2a)/ J1(β2a)] } = Now rewrite the above results E(r) = = B(r) = region 2 E(r) = region 1 B(r) = (μ1/μ2) [μ2I/(2πa)] = [μ1I/(2πa)] region 1 Restate: B(r) = E(r) = region 2 r < a B(r) = E(r) = region 1 r > a Question: Why can't I do Ampere's law outside at radius r and get 2πrB/μ1 = I so that B(r) = μ1I/2πr region 1 r>a Answer: if you draw your loop, you have to include with the current jωε1E(r) in the dielectric! _ The interior solution agrees with the solution shown in box (2.1.26) but now we know the constants. According to this same box we have β2 = ej3π/4 which has phase factor ej3π/4 which motivated us to replace the Jn Bessel functions with their Kelvin function equivalents Jν(ej3π/4z) = berν(z) + j beiν(z) . (2.2.1) In region 1 which we imagine not to be a conductor, σ1 is relatively small and so β12 = μ1ε1ω2 - jωμ1σ1 = ω2μ1 ( ε1 - jσ1/ω) = ω2μ1 ξ1 ξ1 ≡ ε1 - jσ1/ω (1.5.1) so that β12 ≈ μ1ε1ω2 = real (small imaginary part as per ***) Thus we have no need to write out the exterior H1(1)(β1r) functions in terms of Kelvin functions. Question: What happens regarding the complex nature of these functions. B(a) is real, E(a) has phase shown relative to B. Question: How can you "understand" this solution: B(r) = E(r) = region 2 r < a B(r) = E(r) = region 1 r > a where B(a) = E(a) = Generally everything is complex. The Hankel functions for real argument are complex since the real part is J and the imaginary part is Y. I have now made some abs value plots and get things like this where the B field is red and in gauss, and the E field is in V/m. Are these plots believable? E decays very slowly as r-1/2 from the Hankel function.