Save low freq limit section of Appendix D REVIEWED

Save low freq limit section of Appendix D PhL 12.4.13 Appendix D has grown a lot in the last month, and I think the following section should be removed. The limit itself has no particular interest to me, and nobody uses a transmission line in this limit anyway, so I will just park it here. Other unused App D stuff is also stored here at the end. D.5 The Low Frequency Limit not very interesting, throw out! If we assume that ω is small enough that |xa| = |β'|a << 1, we can greatly simplify the above results. For a good conductor, β' ≈ β = ej3π/4 (/δ) so |β'|a << 1 => (a/δ) << 1 or (a/δ) << 1, which means the skin depth is much larger than the radius of the wire. In terms of frequency ω this condition is a << 1 => ω << . Using our ongoing example, for the Belden 8281 coax center conductor with a = .394 mm, μ = μ0 = 4π x 10-7 henry/m, and σ = 5.81 x 107 mho/m (copper), the limit requires that ω << = 88 KHz . All limits come from the leading small-x term of Jn(x) which is Jm(x) ≈ xm/ (2mm!) m ≥ 0 // Spiegel (24.2) (D.5.1) To avoid errors, we have Maple compute the limits of the various terms appearing in box ***. The left column is for visual check, the right column then provides the limits. Our limit is xa<<1 and since x ≤ xa we also have x << 1, but the ratio (x/xa) = (r/a) is merely ≤ 1 and not << 1. From the column on the right above we read off the limits: A = ≈ 2(m+1)(x/xa)m B = ≈ (1/2m)(x/xa)m xa2 ≈ 0 since xa << 1 C = ≈ (x/xa)m-1 D = ≈ (x/xa)m+1 E = ≈ (x/xa)m+1 xa2 ≈ 0 F = ≈ 2 G = ≈ (x/xa) (D.5.2) In terms of the letters A through G the E(r,m) fields from box *** are given by Ez(r,m) = (1/4) ηm I Rdc [A - B ] Ez(r,0) = (1/2) I Rdc [F] Er(r,m) = (j/4) (aβd) ηm I Rdc [ C + D - E ] Er(r,0) = (j/2) (aβd) I Rdc [G] Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - C + D + E ] Eφ(r,0) = 0 (D.5.3) Installing the limits and setting (x/xa) = (r/a) we find E Fields Inside a Round Wire (low frequency limit) Ez(r,m) = (1/2) ηm I Rdc (m+1)(r/a)m Ez(r,0) = I Rdc Er(r,m) = (j/4) (aβd) ηm I Rdc [(r/a)m-1 + (r/a)m+1 ] Er(r,0) = (j/2) (aβd) I Rdc (r/a) Eφ(r,m) = (1/4) (aβd) ηm I Rdc [- (r/a)m-1 + (r/a)m+1 ] Eφ(r,0) = 0 (D.5.4) The equation Ez(r,0) = I Rdc states that in this low frequency limit the wire acts as a resistor having the expected resistance per unit length Rdc = 1/(σπa2) as in (D.4.2). In the transmission line limit aβd << 1 of interest to us, we see that there is always a small radial current which pumps the changing surface charge. In the m = 0 this radial current is linear in r, peaking at the surface. In fact, if you integrate the radial current over the surface of a long piece of wire of length λ/ 2, taking into account the z dependence, you get 2I. [ I think the above has no basis in fact! At DC, we have ω = 0 and Er(a,0) = (j/2) (aβd) I Rdc. But βd = ω ≈ ω = 0 at DC, so the integral of the radial current is 0. For m = 1, the radial current depends on η1 which we don't know, and then Er(a,1) = (j/4) (aβd) ηm I Rdc(2) and we have no conclusion at all. Yes it is proportional to I. I don't know where this strange claim came from! Throwing in the sinφ angle and integrating over half a wavelength is not going to rescue this weird claim. ] See Section 3.7, Figure 2 for a drawing of the current we have just computed. continue here after reading Chapter 3. **************************** other discarded stuff from Appendix D ***************** The Maxwell curl E equation is not tested since we know it is valid since it was used to compute B. Inserting these coefficients we find [(jω/2σ) Nm ]-1(ω/β')Bz(r,m) = [ + ] Jm(x) [(jω/2σ) Nm ]-1 (ω/jβ')Br(r,m) = ( m[ – ] - 2m ) Jm(x)/x + [ + ] Jm+1(x) [(jω/2σ) Nm ]-1 (ω/β')Bφ(r,m) = - ( m[ – ] - 2m ) Jm(x)/x + [ – ] ( + ) Jm+1(x) Now use (D.2.28) on the left and factor out m in two places to get [(j/4) (aβd) ηm I Rdc]-1(ω/β')Bz(r,m) = [ + ] Jm(x) [(j/4) (aβd) ηm I Rdc ]-1 (ω/jβ')Br(r,m) = m { [ – ] - 2 } Jm(x)/x + [ + ] Jm+1(x) [(j/4) (aβd) ηm I Rdc ]-1 (ω/β')Bφ(r,m) = - m { [ – ] - 2 } Jm(x)/x + [ – ] ( + ) Jm+1(x) Now move all factors to the right side Bz(r,m) = (β'/ω)(j/4) (aβd) ηm I Rdc [ + ] Jm(x) Br(r,m) = (jβ'/ω)(j/4) (aβd) ηm I Rdc * m { [ – ] - 2 } Jm(x)/x + [ + ] Jm+1(x) Bφ(r,m) = (β'/ω)(j/4) (aβd) ηm I Rdc * - m { [ – ] - 2 } Jm(x)/x + [ – ] ( + ) Jm+1(x) We can replace (jω/2σ) Nm = (j/4) (aβd) ηm I Rdc (D.2.28) Therefore we get (ω/β')Bz(r,m) = (jω/2σ) Nm [ + ] Jm(x) (ω/jβ')Br(r,m) = + ( m(jω/2σ) Nm [ – ] - (jω/2σ) 2m Nm ) Jm(x)/x + (jω/2σ) Nm [ + ] (ω/β')Bφ(r,m) = - ( m(jω/2σ) Nm [ – ] - (jω/2σ) 2m Nm ) Jm(x)/x + (jω/2σ) Nm [ – ] ( + ) Jm+1(x) Replacing or (1/β')Bz(r,m) = (j/2σ) Nm [ + ] Jm(x) (ω/jβ')Br(r,m) = + ( m(jω/2σ) Nm [ – ] - (jω/2σ) 2m Nm ) Jm(x)/x + (jω/2σ) Nm [ + ] (ω/β')Bφ(r,m) = - ( m(jω/2σ) Nm [ – ] - (jω/2σ) 2m Nm ) Jm(x)/x + (jω/2σ) Nm [ – ] ( + ) Jm+1(x) Jm+1(x) so then (ω/jβd)Br(r,m) = ( m - am) Jm(x)/x + (βd/β') ( + ) Jm+1(x) Finally for Bφ, (ω/βd)Bφ(r,m) = ( m - am) Jm(x) = ( m - am) Jm(x)/x + (βd/β') ( + ) Jm+1(x) = ( m - am) Jm(x) + (βd/β') ( + ) Jm+1(x) (ω/βd)Bθ(r,m) = - [ (β'2/βd2) m - am ] Jm(x)/x + (1 + β'2/βd2) Jm+1(x) jEφ(r,m) =(j/4) (aβd) ηm I Rdc { - + [ + ] } We summarize these results in a box. The m=0 results are all obtainable from the general expressions with m = 0 using J-1 = - J1 : E Fields Inside a Round Wire x = β'r β'2 = β2- βd2 ≈ β2 β = ω ξ =[ε + σ/(jω)] ≈ σ/(jω) conductor xa = β'a βd = ω ξd =[εd + σd/(jω)] ≈ εd dielectric E(r,φz,t) = ej(ωt-βz) E(r,φ) (D.1.2) E = Er + Eφ + Ez (for any arguments) E(r,φ) = E(r,0) + 2!Syntax Error, I[Re{E(r,m)}cos(mφ) - Im{E(r,m)} sin(mφ)] = real (D.1.4) where: Ez(r,m) = (1/4) ηm I Rdc [ - ] a = radius ηm ≡ Ez(r,0) = (1/2) I Rdc [] // = (2.2.22) for βd = 0 η0 = 1 Er(r,m) = (j/4) (aβd) ηm I Rdc [ + - ] Er(r,0) = (j/2) (aβd) I Rdc [ ] Rdc = Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - + + ] Eφ(r,0) = 0 (D.4.6)