Old Sections B.2 thru B.5 for archive

Old Sections B.2 thru B.5 for archive PhL retired 10.15.14 B.2 Calculation of H from the current J in a conductor (a) An expression for H in terms of J Start with Maxwell's equation (1.1.1), and we are now working inside a conductor so E = 0 and then curl H = jωεE + J = J (B.2.1) where J is the conduction current. Apply curl to both sides and use curl curl = grad div -2 to get grad div H - 2H = curl J . (B.2.2) But in a uniform medium div H = 0 since div B = 0 so 2H = - curl J . (B.2.3) Now let's assume that we have J = Jz(x,y) and assume that the solution H does not depend on z. In that case we have 22D H(x,y) = - curl J J = Jz(x,y) . (B.2.4) The particular solution to this PDE is shown in (I.1.8) to be H(x,y) = ∫d2x' [ ln(1/R) ] curl' J(x') R = |x-x'| (B.2.5) or H(x,y) = - ∫d2x' ln(R2) curl' J(x') R = |x-x'| (B.2.6) where ln(1/R) is the Poisson 2D free-space propagator. This gives H in terms of J. (b) An alternative derivation using the vector potential Az An alternate derivation of (B.2.5) makes use of the vector potential Az. Start with (1.5.4), (2 + β2)A(x) = - Σi μiJi (x) . all of region R (1.5.4) Apply this to a single conductor and assume A(x) = Az(x,y),  so the above equation becomes -22D Az(x) = μJz(x) . (B.2.7) The particular integral from (I.1.8) is then Az(x) = ∫d2x' [ln(1/R)/2π] [μJz(x') = - (μ/2π) ∫d2x' ln(R) Jz(x') . (B.2.8) Then use B = curl A = (∂yAz - ∂zAy) + (∂zAx - ∂xAz) + (∂xAy - ∂yAx) = (∂yAz) + (- ∂xAz) (B.2.9) so that μH = (∂yAz) + (- ∂xAz) . (B.2.10) Now compute, ∂yAz(x) = ∂y [- (μ/2π) ∫d2x' ln(R) Jz(x') ] = - (μ/2π) ∫d2x' Jz(x') ∂y ln(R) . But ∂y ln(R) = - ∂y' ln(R) since R = |x-x'|. But then do parts integration to move ∂y'to Jz(x') picking up an offsetting minus sign, and the parts vanish on a great circle surrounding the conductor. Thus, ∂yAz(x) = - (μ/2π) ∫d2x' ln(R) ∂y'Jz(x') ∂xAz(x) = - (μ/2π) ∫d2x' ln(R) ∂x'Jz(x') . (B.2.11) Then from (B.2.10) one gets, H = - (1/2π)∫d2x' ln(R) [ ∂y'Jz(x') - ∂x'Jz(x')] . But the coordinates x = (x,y) and x' = (x',y') have the same unit vectors = ' and = ' . Since J = Jz we then end up with H(x) = - (1/2π)∫d2x' ln(R) curl' Jz(x') = + (1/2π)∫d2x' ln(1/R) curl' J(x') which agrees with (B.2.5). (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.2.5) at some point on the conductor surface such as the point shown in Fig 3.3. We have (t = transverse) Ht(x) = ∫d2x' [ ln(1/R) ] [curl' J(x')]t R = |x-x'| , (B.2.12) 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) 22DHt = 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 the above 3D vector Helmholtz equation, 2H = - curl J . (B.2.3) In Cartesian coordinates it is three scalar Helmholtz equations which can be solved as in (H.1.8) to give H(x) = ∫d3x' [ ] curl' J(x') R ≡ |x-x'| (B.2.13) where 1/4πR is the 3D free-space Poisson propagator. Write this in components and define R as shown, 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 end of Appendix A.1), 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) 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 an alternate way to obtain H from J in a 2D problem. 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 the H field at all points in the wire cross-sectional plane section using (B.2.6) or (B.2.24) H(x,y) = - ∫d2x' ln(R2) curl' J(x') R = |x-x'| . (B.2.6) H(x,y) = ∫d2x' J(x') x R R ≡ x - x' (B.2.24) or use the third method of first computing Az, Az(x) = - (μ/2π) ∫d2x' ln(R) Jz(x') (B.2.8) B = curl A = (∂yAz) + (- ∂xAz) H = B/μ (B.2.9) 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 formula (B.2.6). 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.2.5) 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).