Section B.2 rewrite

Rewrite of Section B.2 PhL 10.15.14 B.2 Calculation of H from the current J in a conductor (a) An expression for H in terms of J First, we want to clarify the connection between H and J in terms of Maxwell's equations. Since J = σE, we can write the curl H equation (1.1.1) in these three equivalent ways: a curl H = jωεE + J = displacement current + conduction current (1.1.1) (B.2.1a) b curl H = jω(ξ/σ) J ξ ≡ ε - jσ/ω = ε + σ/jω (B.2.1b) c curl H = jωξ E . (B.2.1c) Using the identity curl curl = grad div -2 and noting that div H = μ div B = 0 from (1.1.4) we find curl curl H = -2H . (B.2.2) Applying curl to the three forms above, then using (B.2.2) and doing some small algebra as shown below, one obtains these three exactly equivalent equations for H : a (2 + κ2)H = -curl J κ2 = ω2εμ // agrees with (1.5.26) (B.2.3a) b 2H = - jω(ξ/σ) curl J = - [1 + jω(ε/σ) ] curl J ξ ≡ ε + σ/jω (B.2.3b) c (2 + β2 )H = 0 β2 = ω2ξμ // agrees with (1.5.27) (B.2.3c) Forms a and c make use of (1.1.2) which says curl E = -jωB = -jωμH . Algebra: a -2H = curl (jωεE + J) = jωε(-jωμH) + curl J = ω2εμH + curl J = κ2H + curl J b -2H = curl (jω(ξ/σ) J) = jω(ξ/σ) curl J c -2H = curl (jωξ E) = (jωξ)( -jωμH) = ω2ξμ H = β2H We showed below (2.2.2) that for f << 1018 Hz, we have ωε/σ << 1 (copper). Assuming this inequality for any practical ω, we know that ε << σ/ω so ξ = σ/jω and β2 = ω2(σ/jω)μ = - jωμσ which agrees with (1.5.1d). In this situation, since - jω(ξ/σ) = -1, we can write (B.2.3b) above as 2H = -curl J . // copper for f << 1018 Hz (B.2.4) One might wonder how this last equation and (B.2.3a) can both be valid. The reason is that |κ2/β2| = ω2εμ/ (- jωμσ) = |-ωε/σ| = ωε/σ << 1 (B.2.5) Then using this fact and (B.2.3c) we find | κ2H| << | β2 H| = |2 H| so | κ2H| << |2 H| (B.2.6) so the κ2 term in item 1 makes no contribution. Recalling (H.1.8), -2 f(x) = s(x) => f(x) = ∫d3x' [1/4πR] s(x') + homogeneous solutions The Poisson Equation (H.1.8) one can write the Helmholtz particular integral solution of (B.2.4) as H(x) = ∫d3x' [1/4πR] curl' J(x') (B.2.7) where [1/4πR] is the Poisson free-space propagator. (b) An alternative derivation using the vector potential A Consider a conductor with μ,ε,σ,ξ surrounded by a dielectric with μd,εd,σd,ξd . Here are two equations for the vector potential associated with the conductor current J. The first equation is for the King gauge, while the second is for the Lorenz gauge: (2 + βd2)A(x) = - μ J(x) (1.5.4) βd2 = ω2μdξd // King gauge (B.2.8a) (2 + β02)A(x) = - μ J(x) . (1.5.28) β02 = ω2με // Lorenz gauge (B.2.8b) Recalling (H.1.9), - (2+k2) f(x) = s(x) => f(x) = ∫d3x' [e-jkR/4πR] s(x') + homogeneous solutions The Helmholtz Equation (H.1.9) we may write down the Helmholtz particular integral solutions to (B.2.8), A(x) = μ ∫d3x' [e-jβR/4πR] J(x') King gauge R = |x-x'| (B.2.9a) A(x) = μ ∫d3x' [e-jβR/4πR] J(x') Lorenz gauge R = |x-x'| (B.2.9b) where [e-jβR/4πR] is the usual 3D Helmholtz propagator. We know that B = curl A and therefore, using βx to stand for either βd or β0, H(x) = (1/μ) curl A(x) = ∫d3x' curl ( [e-jβR/4πR] J(x') ) (B.2.10) Notice that J(x') is a constant in terms of the unprimed curl operator. A useful vector identity then is x (φQ) = φ x Q + φ ( x Q) = (φ) x Q Q = constant vector . Thus one may write H(x) = ∫d3x' [e-jβR/4πR] x J(x') . Since R = |x-x'| we know acting on any function of R is the same as -' on that function, so H(x) = - ∫d3x' '[e-jβR/4πR] x J(x') or in components (implied sum on r and s) , Hi = - ∫d3x' εirs ∂'r [e-jβR/4πR] Js(x') . Doing parts integration and dropping the parts (see Section A.0 (d)), we get Hi = + ∫d3x' [e-jβR/4πR] εirs [∂'r Js(x')] or H(x) = ∫d3x' [e-jβR/4πR] curl' J(x') . (B.2.11) We can compare this result to (B.2.7) obtained assuming f << 1018 Hz, H(x) = ∫d3x' [1/4πR] curl' J(x') . 3D R = |x-x'| (B.2.7) We conclude that in either gauge, and for this frequency range with copper, the exponential factor e-jβR in (B.2.11) may be ignored. This basically says that the main contribution to the integral comes from the region near R = 0. At low ω and in particular at DC with ω = 0, we may assume that J = J(x,y) with no z dependence. In this case we can do the dz' integration in (B.2.7) as shown in (J.10) which converts [1/4πR] to [-ln(s)/2π ]. Renaming s to be R in 2D, we obtain H(x,y) = ∫d2x' [-ln(R)/2π] curl' J(x',y') 2D R = |x-x'| (B.2.12) where now [-ln(R)/2π] is the 2D Poisson free-space propagator of (I.1.8). In fact, this 2D result follows directly from (I.1.8) if we assume that H = H(x,y) so that 2H = 2D2 H. (c) Boundary conditions Recall from (B.1.1) that the tangential component of H is continuous through the boundary between conductor and dielectric, even if μ1 ≠ μ2. Consider then the transverse component of (B.2.7) at some point on the conductor surface such as the point shown in Fig 3.3. We have (t = transverse) Ht(x) = ∫d2x' [1/4πR] ] [curl' J(x')]t R = |x-x'| . This particular integral is naturally continuous at the boundary between the media, and this agrees with the fact that Ht(x) must have this property. Therefore, no homogeneous solutions of (B.2.4) 2Ht = 0 need be added in, so (B.2.12) is the complete solution for Ht(x). This solution can then be used in (B.1.10) to find the magnetization surface current. (d) The Biot-Savart Law in 3D and 2D Recall from above the 3D vector Helmholtz equation valid for f << 1018 Hz, and its solution 2H = - curl J (B.2.4) H(x) = ∫d3x' [1/4πR] curl' J(x') . R = |x-x'| (B.2.7) (B.2.13) We shall now reverse the steps done in the previous section, but this time with e-jβR = 1 based on the conclusion just drawn above. In components (B.2.7) reads, Hi(x) = ∫d3x' [ ] εijk ∂'jJk(x') , R ≡ x - x' = points to observation point x . (B.2.14) Then move ∂j' from Jk to (1/R) by parts integration (pick up minus sign) and throw out the parts for the usual reasons (see Section A.0 (d)), Hi(x) = - ∫d3x' ∂'j () εijk Jk(x') . (B.2.15) Then note that ∂'jR-1 = -R-2 ∂'jR and ∂'jR = ∂'j = (1/2)(1/R) 2(x'j-xj) = R-1 (x'j-xj) = - R-1 Rj (B.2.16) so that ∂'jR-1 = +R-3Rj. Only the parts minus sign remains, so Hi(x) = - ∫d3x' [ ] εijk Rj Jk(x') (B.2.17) or reversing the cross product order, H(x) = ∫d3x' J(x') x R . R ≡ x - x' (B.2.18) This equation is basically the 3D Biot-Savart Law, see for example Panofsky and Philips p 125 (7.31). For a short piece ds' of thin wire carrying current I, one writes J(x') d3x' = I ds' so the above becomes, H(x) = I ds' x R or dH(x) = I ds' x R . (B.2.19) We can apply the same process to obtain a 2D Biot-Savart Law as follows. Start with 22D H(x,y) = - curl J J = Jz(x,y) (B.2.4) and its solution (B.2.5) obtained from (I.1.8) H(x,y) = ∫d2x' [ ln(1/R) ] curl' J(x') R = |x-x'| (B.2.5) where ln(1/R) is the Poisson 2D free-space propagator. Then, inverting 1/R, Hi(x,y) = - ∫d2x' [ ln(R) ] εijk ∂'jJk(x') . (B.2.20) Doing the same parts integration gives Hi(x,y) = + ∫d2x' [ ∂'j ln(R) ] εijk Jk(x') (B.2.21) and now using result (B.2.16) from above, ∂'j ln(R) = R-1∂'jR = R-1 [- R-1 Rj] = -R-2Rj (B.2.22) we get Hi(x,y) = - ∫d2x' [ R-2Rj ] εijk Jk(x') = - ∫d2x' εijk Rj Jk(x') (B.2.23) or H(x,y) = ∫d2x' J(x') x R R ≡ x - x' (B.2.24) which is the 2D Biot-Savart Law. It provides a way to obtain H from J in a 2D problem. In such a problem, we assume that J = J(x,y) with no z dependence. B.3 General Method for computing the surface current Jm on a wire Here are the steps for a wire of arbitrary cross sectional shape: 1. Compute H from J. Consider these sets of equations obtained above: 3D H(x) = ∫d3x' curl' J(x') . R = |x-x'| (B.2.7) H(x) = ∫d3x' J(x') x R . R ≡ x - x' Biot-Savart (B.2.18) A(x) = μ ∫d3x' J(x') R = |x-x'| H = (1/μ) curl A (B.2.9) 2D H(x,y) = ∫d2x' curl' J(x',y') R = |x-x'| (B.2.12) H(x,y) = ∫d2x' J(x',y') x R R ≡ x - x' Biot-Savart (B.2.24) A(x,y) = μ ∫d2x' J(x',y') R = |x-x'| H = (1/μ) curl A (B.3.1) where the last line for A is the same 3D→2D resulting used for H. For either 3D or 2D, we thus provide three different methods for computing H from J. 2. Evaluate this H field at xb = (xb,yb) for all points xb on the cross section boundary. 3. Compute the component of H which is tangential to the boundary in the cross sectional plane. Call this component Hθ. 4. The surface current density is then given by (B.1.10), Kz = - ( - ) Hθ amps/m (B.1.10) B.4 Surface current on a round wire with uniform J For a round wire of radius a with uniform Jz (as would be the DC case ω = 0), geometric symmetry makes the calculation of H very easy. One need only apply Ampere's Law separately for a point r outside the wire, and for another point r inside the wire. For the outside case one finds 2πr Hθ(r) = I => Hθ(r) = I/(2πr) r ≥ a . (B.4.1) And then for the inside case the "current enclosed" is determined by a simple area fraction. 2πr Hθ(r) = I (πr2/πa2) => Hθ(r) = I r/(2πa2) r ≤ a . (B.4.2) At the boundary the two expressions agree and we have Hθ = I/(2πa) . (B.4.3) If this wire has magnetic permeability μ2 and is embedded in an infinite medium of μ1, then the surface magnetization current induced on the wire is Kz = - ( - ) Hθ = - ( - ) I/(2πa) amp/m (B.4.4) Kz = Kz (B.4.5) and this surface current is in the direction opposite J if μ2 > μ1. If μ1 = μ2, the surface current vanishes. This surface current could be expressed in volume density form as Jm = Kzδ(r-a) amp/m2 . (B.4.6) B.5 Computing Hθ for a round wire using the General Method of B.3 For a wire of some general cross section, symmetry is not available to allow the simple solution for Hθ outlined in the previous section. We then have to use the more general method outlined in Section B.3 above. As a check on the viability of this general method, we shall carry out "step 1" of the method and show how Hθ may be computed from J using the 2D formula (B.2.12). The conduction current density in a round wire with uniform Jz is given by Jz(r) = J0θ(a-r) (B.5.1) where θ is the Heaviside step function. Our first step is to compute curl J, and we do this in cylindrical coordinates by just staring at the cylindrical-coordinates curl formula, curl J = [ r-1∂θJz - ∂zJθ] + [∂zJr - ∂rJz] + [ r-1∂r(rJθ) - r-1∂θJr ] (B.5.2) and finding the only non-zero piece which is this (uniform Jz), curl J = [-∂rJz(r)] . (B.5.3) Inserting Jz(r) from above we find ∂r Jz(r) = J0 ∂rθ(a-r) = - J0 δ(r-a) (B.5.4) => curl J(r) = J0 δ(r-a) . (B.5.5) so we have a "ring source of curl J". For use in our integral for H we then have curl' J(r') = ' J0 δ(r'-a) . (B.5.6) For a current distribution which tapers off smoothly to 0 at the wire edge one would not have this singular contribution, but for a wire with prescribed uniform current, it is present, and curl J vanishes everywhere but on the boundary. The relevant picture is this: Fig B.4 From (B.2.12) the H field at any point x = (x,y) is then given by H(x,y) = - ∫d2x' ln(R2) curl' J(x') = - ∫d2x' ln(R2) ' J0 δ(r'-a) = - !Syntax Error, Idθ' ln(R2)|r'=a ' or H(r,θ) = - !Syntax Error, Idθ' ln [ r2 + a2 - 2ar cos(θ'-θ) ] ' . (B.5.7) The figure shows that ' = cosθ' - sinθ' (B.5.8) so then H(r,θ) = - !Syntax Error, Idθ' ln [ r2 + a2 - 2ar cos(θ'-θ) ] [cosθ' - sinθ' ] . (B.5.9) Next, let x ≡ θ'-θ. Since the ∫dθ' has full range 2π, one can replace !Syntax Error, Idθ' = !Syntax Error, Idx . Then H(r,θ) = - !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] [cos(x+θ) - sin(x+θ) ] . (B.5.10) Now writing H = Hx + Hy , decompose the above into two equations Hx(r,θ) = + !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] sin(x+θ) Hy(r,θ) = - !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cos(x+θ) (B.5.11) or Hx(r,θ) = + !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] [ sinxcosθ+cosxsinθ ] Hy(r,θ) = - !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] [cosxcosθ - sinxsinθ ] . (B.5.12) Since !Syntax Error, Idx is over an even range, throw out odd integrand terms, and then fold the negative range into the positive adding a factor of 2 to get Hx(r,θ) = + sinθ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx ≡ sinθ Q Hy(r,θ) = - cosθ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx = -cosθ Q (B.5.13) where Q ≡ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx . (B.5.14) Before evaluating this integral, we see that H = Hx + Hy = - Q [ cosθ -sinθ ] = - Q = Hθ . (B.5.15) Thus we find that the resulting H is entirely in the direction and Hθ = - Q . (B.5.16) We seek now to evaluate this integral Q, Q ≡ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx = !Syntax Error, Idx ln [ {a2}{ (r/a)2 + 1 - 2(r/a) cos(x)} ] cosx = !Syntax Error, Idx { ln (a2) + ln[(r/a)2 + 1 - 2(r/a) cos(x)] } cosx = ln (a2)[!Syntax Error, Idx cosx ] + !Syntax Error, Idx ln[(r/a)2 + 1 - 2(r/a) cos(x)] cosx = ln (a2)[0] + !Syntax Error, Idx ln[α2 + 1 - 2α cos(x)] cosx where α ≡ r/a = !Syntax Error, Idx ln[α2 + 1 - 2α cos(x)] cosx . // = Q (B.5.17) This integral is the n=1 special case of the following integral from GR7 page 589 4.379.6, Therefore we find Q = (B.5.18) so Hθ = - Q = . (B.5.19) Now the total current in the wire is I = J0πa2 so (aJ0/2) = (I/2πa) and then Hθ = = . (B.5.20) Thus, we finally arrive at the same results for Hθ as obtained in (B.4.2) and (B.4.1).