another shot at eddy theory REVIEWED

Try Again PhL 4.15.14 [ I wrote this before Cape Cod, I tried to make it fly in temp3.doc, but I could not get happy with it and decided to stick with my existing Section P.1 which opens Appendix P. ] Consider this figure, where the free hand lines generically represent E and B field lines, 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 the above figure are: curl E1 = -jωB1 curl B1 = μJ1 // region (a) curl E1 = -jωB1 curl B1 = jωμεE1 // region (b) 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 = μJ2 // region (a) curl E2 = -jωB2 curl B2 = jωμεE2 // region (b) curl E'2 = -jωB2 curl B2 = μJ2' // region (c) 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 = σ E'2 we can write for region (c) curl curl E'2 = -jω curl B2 = - jωμ(σE'2) But we know that curl curl E'2 = grad div E'2 - 2 E'2 = - 2 E'2 since div E'2 = 0 inside the DUT (ρ=0). Thus we must solve - 2 E'2 = - jωμσE'2 or (2 + β2)E'2 = 0 where β2 = - jωμσ and this is the equation encountered in *****. 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 B2 of the second drawing, and we rename J'2 , Beddy ≡ B2 - B1 Jeddy ≡ J'2 The region (c) curl equations are then curl Jeddy = -jωσ(B1 + Beddy) curl (B1 + Beddy) = μJeddy // region (c) If ω is small, the first 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 B2 = B1 + Beddy ≈ B1 . The above equations are then curl Jeddy ≈ -jωσB1 curl B1 ≈ μJeddy // 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. CONTINUE HERE AFTER CAPE COD! Maybe compare the above threading to earlier attempts. Each attempt seems to fail after I let it cool off for a day, so we need to get this thing stabilized. Then it can be installed at the start of Appendix P.