temp3 REVIEWED

This is the Title PhL 3.26.05 [ OK, on 5/1/14 I tried to make this section fly, but it is just too slippery and I think my existing Appendix P opening section is much better. // But later I came back to this approach combined with a perturbation iterative approach (5/5/14) ] P.1 Eddy Current Analysis There are whole books on the subject of eddy currents, and the web reveals a proliferation of papers and theses on eddy current applications. However, simple analytic examples of eddy current problems are hard to find. The general theory is quite complicated, and we present here a simplified approach which suits are limited purposes, and then we work through two specific sample problems. Consider this figure, where the free-hand lines generically represent E and B field lines, Fig P.1 An external apparatus has current density J1 flowing in some wires and creates magnetic field B1 and there is some associated electric field E1 created as well. The Maxwell curl equations for Fig P.1 are: curl E1 = -jωB1 curl B1 = μaJ1 // region (a) curl E1 = -jωB1 curl B1 = jωμbεbE1 // region (b) (P.1.1) In region (a) we ignore the displacement current inside the wires. In region (b) J1 = 0 and there is then some displacement current jωεE1. We now bring in a Device Under Test (DUT) to obtain a new picture: The Maxwell curl equations are now: curl E2 = -jωB2 curl B2 = μaJ2 // region (a) curl E2 = -jωB2 curl B2 = jωμbεbE2 // region (b) curl E'2 = -jωB'2 curl B'2 = μcJ2' // region (c) (P.1.2) In regions (a) and (c) we ignore the displacement currents (the DUT is a good conductor). The quantity J'2 is the eddy current in the DUT, and E'2 is its associated electric field. For arbitrarily large ω, the solution of the problem of the fields inside the DUT is complicated and one must solve a vector Helmholtz equation. Setting J'2 = σc E'2 we can write for region (c) curl curl E'2 = -jω curl B'2 = - jωμc(σcE'2) (P.1.3) But we know that curl curl E'2 = grad div E'2 - 2 E'2 = - 2 E'2 (P.1.4) since div E'2 = 0 inside the DUT (ρ=0). Combining (P.1.3) and (P.1.4) we get - 2 E'2 = - jωμcσcE'2 or (2 + βc2)E'2 = 0 where βc2 = - jωμcσc and this is the equation encountered in *****. Of course to obtain a full solution, we have to consider the Helmholtz equations in each region and match boundary conditions at all interfaces, a formidable problem. With respect to the DUT, we have both an interior and an exterior problem to worry about. If ω is small, we can take a different approach to solving the region (c) problem. We first define Beddy as the difference between B1 of the first drawing and B'2 of the second drawing, and we rename J'2 and E'2 Beddy ≡ B'2 - B1 Jeddy ≡ J'2 Eeddy ≡ E'2 The region (c) curl equations of (P.1.2) are then curl Eeddy = -jω(B1 + Beddy) curl Jeddy = -jωσc(B1 + Beddy) curl (B1 + Beddy) = μcJeddy // region (c) If ω is small, the middle equation implies that Jeddy is small. If this is the case, we expect that Beddy will also be small. In this case, we expect that the eddy currents don't alter J1 very much when the DUT is added to Fig 1 to get Fig 2, so we then set J2 = J1 in the second drawing. If Beddy is small, we then have roughly B'2 = B1 + Beddy ≈ B1 . The above equations are then curl Jeddy ≈ -jωσB1 curl B1 ≈ μcJeddy // region (c) It is the first of these two equations that is our main interest. The second equation says that B1 has a small curl inside the DUT due to the eddy current there. One should not interpret this second equation as saying that the small eddy current somehow generates the large field B1. After all, in region (b) curl B1 = 0 and we have a large B1 there.