old Section B.1 retired 1_16_14 REVIEWED

Old Section B.1 retired on 1.16.14 PhL 1.16.14 B.1 Relationship between surface current K and the field H at a conductor boundary First, consider this blow up of a piece of the boundary between a conductor (medium 2) and a dielectric (medium 1). Both media extend uniformly in the z direction, so we are looking at a piece of the cross section of a transmission line at a particular point on the surface of one of the conductors. Fig B.1 We shall assume that the conduction current is positive in the z direction, so J = Jz with Jz > 0. Since the lower medium is the conductor in the drawing, the B and H field at the boundary are in the - direction, that is to say, they point to the left due to the right hand rule relating J and B or H . At the boundary, we know from (1.1.26) that Hx2 = Hx1 (1/μ1)Bx1 = (1/μ2)Bx2 . (B.1.1) Assuming μ2 ≥ μ1 (which would be the case if μ1 = μ0), the right equation implies |Bx2| ≥ |Bx1| so the B field is larger inside the conductor. But in our picture, both Bx2 and Bx1 are negative, so -Bx2 ≥ - Bx1 which then says Bx2 ≤ Bx1 and finally (Bx2 - Bx1) ≤ 0. Also, Hx2 = Hx1 ≤ 0. For the red loop shown in the figure one then has, B ds ≈ Bx2L - Bx1L = (Bx2 - Bx1)L ≤ 0 . (B.1.2) Now consider (1.1.24) which says ( in the ω domain), B ds = ∫S [ jωεE + μJc + μJm] dA (B.1.3) where dA = dA . Since E dA involves only Ez (parallel to surface), and since by (1.1.25) such Ez is continuous at the boundary, and since Ez ≈ 0 inside the conductor, the εjωE term makes no contribution, giving then B ds = ∫S [μJc + μJm] dA . (B.1.4) Since Jc flows in the + direction, the surface current Jm flows in the - direction, so write Jm = Kz δ(y) (B.1.5) where Kz ≤ 0 is the magnitude of the surface current. The area integral on the right side of (B.1.4) is written as two terms, one for the area above the boundary, and one below: ∫S μ [Jc + Jm] dA = ∫ μ1 [Jc1 + Jm1] dA + ∫ μ2 [Jc2 + Jm2] dA . (B.1.6) The surface current lies on the outside of the conductor surface, so it is region 1. Then ∫S μ [Jc + Jm] dA ≈ μ1Jc1z (s/2)L + μ2Jc2z (s/2)L + μ1 ∫dxdy [Kzδ(y)] dA = [μ1Jc1z + μ2Jc2z ] (s/2)L + μ1Kz L . Then (B.1.4) with (B.1.2) says (Bx2 - Bx1)L = [μ1Jc1z + μ2Jc2z ] (s/2)L + μ1Kz L (B.1.7) Since the volume currents are not infinite at the boundary, as s→ 0 we find μ1Kz = (Bx2 - Bx1) (B.1.8) The signs are consistent with Kz ≤ 0 and (Bx2 - Bx1) ≤ 0 as noted above. Then from (B.1.1) we find μ1Kz = (μ2Hx2 - μ1Hx1) = (μ2-μ1) Hx2 and finally Kz = (μ2/μ1 - 1) Hx2 (B.1.9) As noted earlier, Hx2 < 0 so Kz ≤ 0 is consistent with μ2 ≥ μ1. We now rewrite this result in terms of a different picture: Fig B.2 This shows the cross section of the entire conductor in gray, and Jc is still directed toward the viewer. In this picture a point on the surface is associated with a local coordinate system for which = is normal to the surface and = - is tangent to the surface (so Hθ = -Hx). We are thinking of (r,θ,z) as local cylindrical coordinates at the point shown on the conductor surface, where x = , and the x,y,z directions of the figure match those of the previous figure where as usual x = . Then (B.1.9) says Kz = - (μ2/μ1 - 1) Hθ (B.1.10) where Hθ > 0 and Kz ≤ 0. Conclusion: At the surface of a conductor, if the conduction current is in the + z direction, the surface current is in the - z direction and has a magnitude given by Kz = - (μ2/μ1 - 1) Hθ where Hθ is the tangential H field at the surface of the conductor (either inside or outside since both are the same). If μ1 = μ2 then there is no surface current since there is no imbalance of magnetization current at the boundary. Example: For a round wire of radius a carrying an axially symmetric current distribution, we know that 2πaHθ = I so Hθ = I/(2πa) at the surface. Then Kz = - (μ2/μ1 - 1)[ I/(2πa)] round wire of radius a and μ2, dielectric μ1 (B.1.11) Physical mechanism of the surface current. As a reminder, a surface magnetization current arises at a boundary between media with different μ values just the way surface polarization charge arises at a boundary between media with different ε. In the μ case, here is a suggestive picture : Fig B.3 On the left we look at a round wire end on, while the right shows a top view where the wire has been tilted down. Here μ1= μ0 so there is only vacuum outside the wire. The B field lines up the little magnetic dipoles (or creates them) which we represent schematically as little atoms with orbiting electrons. The atoms in the interior always have cancelling current arrows, but there is an imbalance on the outer surface which is the surface magnetization current. The picture shows why it is that the surface current is directed opposite to the current J which creates it, a sort of magnetic Lenz's Law. The surface current is not seen by H, but it is seen by B. In the case that the outer medium has some μ1 > μ0, both media have surface currents at the boundary, and then when μ1 ≠ μ2 there is a surface current imbalance resulting in a net surface current. If it happens that μ1 < μ2, then the directions shown above are correct, but if μ1 > μ2, the surface current runs in the opposite direction to that shown.