Electric Field Associated with an infinite wire carrying current I REVIEWED

Electric Field Associated with an infinite wire carrying current I PhL 4.15.14 I looked at this again on May 5, 2014, seems OK. My appendix P never talks about fields outside a round wire. Inside the wire things seem to agree with Appendix P and Chapter 2. For very low ω, for a round wire, and for μ1 = μ2 = μ, the conclusion is that there is a small E field Ez both inside and outside the wire which is shown in this summary quoted from the doc end. The E field inside agrees with the low ω limit of Chapter 2. [ reviewed 4/30/14 ] Magnetic field: inside: Bθ(r) = (μI/2πa2)r outside: Bθ(r) = (μI/2π)(1/r) Electric field: inside: Ez(r) = I /(σπa2) + jωμ I (1/8π) (2r2/a2 - 1) outside: Ez(r) = I/(σπa2) + jω (μI/8π) [ 4ln(r/a) + 1] Current: inside: Jz(r) = I /(πa2) + jωσμI (1/8π) (2r2/a2 - 1) outside: Jz(r) = 0 Here σ is always the conductivity inside the wire, and we assume outside is an insulator. I am assuming low ω in this entire doc so Jr ≈ 0, there is no surface charge pumping, wavelength is very long, etc. I also assume that μ applies to the wire and to its surroundings. We know the magnetic field for this problem, it is this: inside: 2πr Hθ = I (r/a)2 => Bθ = (μI/2πa2)r outside: 2πr Hθ = I => Bθ = (μI/2π)(1/r) But what is going on with electric fields here? Well, we know that curl E = -jωB or curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] Assuming no z dependence, because we assume a very long wavelength, because we assume a very low ω, this says curl E = [ r-1∂θEz] + [- ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] We then have these three equations r-1∂θEz = 0 - ∂rEz = -jωBθ r-1∂r(rEθ) - r-1∂θEr = 0 The first says that Ez is independent of θ, so we have just Ez(r). The third I let be satisfied with both Eθ and Er = 0, call it an ansatz. The second equation then says ∂rEz = jωBθ Inside the wire this says ∂rEz = jωBθ = jω (μI/2πa2)r => Ez(r) = jω (μI/2πa2)(1/2)r2 + C1 (*) This seems to be the skin effect in the low ω limit, I will have to check that momentarily. [ Compare this to (P.4.6) with Jext = I/(πa2) so here we would have Ez(r) = jω μJext(1/4)r2 + C1 ] Outside the wire we have instead ∂rEz = jωBθ = jω (μI/2π)(1/r) => Ez(r) = jω (μI/2π) ln(r) + C2 We know that Ez is continuous at the wire surface, so we can match these two at that point jω (μI/2πa2)(1/2)a2 +C1 = jω (μI/2π) ln(a) + C2 or jω (μI/4π) + C1 = jω (μI/4π) 2ln(a) + C2 so that C2 = C1 + jω (μI/4π) (1 - 2 lna) The total current in the wire is given by I = ∫rdrdθ Jz = σ ∫rdrdθ Ez = σ 2π ∫rdr [jω (μI/2πa2)(1/2)r2 +C1] = 2πσ { jω (μI/2πa2)(1/2) ∫rdr r2 + C1 ∫rdr } = 2πσ { jω (μI/2πa2)(1/2) (1/4)a4 + C1 (1/2)a2 } = σπa2 { jω (μI/8π) + C1} where I checked dimensions on the first term. Then we have I/(σπa2) = jω (μI/8π) + C1 => C1 = I/(σπa2) - jω (μI/8π) Then inside the wire we must have Ez(r) = jω (μI/2πa2)(1/2)r2 + C1 = jω (μI/4πa2)r2 + I/(σπa2) - jω (μI/8π) = jω (I μ/4πa2)r2 + I /(σπa2) - jω(μ I /8π) = I /(σπa2) + jω (μ I /8πa2)2r2 - jω(μ I /8π) = I /(σπa2) + jωμ I (1/8π) (2r2/a2 - 1) (*) Notice that Ez(a) = I /(σπa2) + jωμ I (1/8π) For small ω which is our regime, the second term in (*) is a small correction term which is negative near the wire center and positive near the surface, expressing the skin effect. The first term is just the DC current we would expect. So this seems good. What about outside the wire? We find that C2 = C1 + jω (μI/4π) (1 - 2 lna) = I/(σπa2) - jω (μI/8π) + jω (μI/4π) (1 - 2 lna) = I/(σπa2) - jω (μI/8π) + jω (μI/8π) (2 - 4 lna) = I/(σπa2) + jω (μI/8π) (2 - 4 lna - 1) = I/(σπa2) + jω (μI/8π) (1 - 4 lna) Then outside the wire we get Ez(r) = jω (μI/2π) ln(r) + C2 = jω (μI/2π) ln(r) + I/(σπa2) + jω (μI/8π) (1 - 4 lna) = I/(σπa2) + jω (μI/8π) 4ln(r) + jω (μI/8π) - jω (μI/8π) 4 ln(a) = I/(σπa2) + jω (μI/8π) 4ln(r/a) + jω (μI/8π) = I/(σπa2) + jω (μI/8π) [ 4ln(r/a) + 1] Notice that Ez(a) = I/(σπa2) + jω (μI/8π) which matches the inside result shown above. Does my inside result agree with Chapter 2 of lines doc? Ez(r) = I /(σπa2) + jωμ I (1/8π) (2r2/a2 - 1) Chapter 2 says this Ez(r) = (-jω/β) But for small ω we have small β and then J0(x) ≈ 1 - x2/4 J1(x) ≈ (x/2)(1 - x2/8) 1/J1(x) ≈ (2/x) (1 + x2/8) so this gives Ez(r) = ( 1 - β2r2/4 ) (2/βa) (1 + β2a2/8) (-jω/β) = ( 1 - β2r2/4 ) (1 + β2a2/8) (-2jω/β2a) Now β2 = ω2μ ξ = ω2μ (- jσ/ω) = -jσμω so let's install that in three places to get = ( 1 + jσμω r2/4 ) (1-jσμω a2/8) (-2jω/[-jσμω]a) = ( 1 + jσμω r2/4 ) (1-jσμω a2/8) (2/[σμ]a) = [ 1 + (jσμω/8)(2r2-a2) ] (2/[σμ]a) = [ 1 + (jσμω/8)(2r2-a2) ] = + (jωμI/8πa2)(2r2-a2) // Chapter 2 small ω limit The limit I just got in this document was = I /(σπa2) + jωμ I (1/8π) (2r2/a2 - 1) and this does agree exactly, hurray. Summary of Fields Associated with a Wire Low ω only, and μ1 = μ2 = μ only. Magnetic field: inside: Bθ(r) = (μI/2πa2)r outside: Bθ(r) = (μI/2π)(1/r) Electric field: inside: Ez(r) = I /(σπa2) + jωμ I (1/8π) (2r2/a2 - 1) outside: Ez(r) = I/(σπa2) + jω (μI/8π) [ 4ln(r/a) + 1] Current: inside: Jz(r) = I /(πa2) + jωσμI (1/8π) (2r2/a2 - 1) outside: Jz(r) = 0 Here σ is always the conductivity inside the wire, and we assume outside is an insulator. Comment: At low ω, the Ez both inside and "near" the wire is very small. For example, the adder term has this size at 100 Hz : ωμI ≈ 2π 100 Hz * 4π x 10-7 * 1 amp ≈ 0.8 mV/m while the main term for a 1 mm radius wire carrying 1 amp is I/(πa2σ) ≈ 5.5 mV/m What about this curl equation: curl E = -jωB I could apply this (integral form) to a rectangular loop parallel to the wire which captures the flux. I then get line integral of E = dz [ Ez(r+dr)-Ez(r) ] = -jω B dz dr so that ∂rEz = -jωB which we already saw above. If we think of B = large and ω = small,