unused eddy current text

This is the Title PhL 3.26.05 Unused eddy current text, don't really like it much now. Note that page numbering is turned on in this template. curl (Bext + Beddy) = μ(Jext + Jeddy) (1) curl (Jext + Jeddy) = - jωσ(Bext + Beddy) . (2) We assume next that Jeddy in the DUT exists in a region of space which is disjoint from the region where Jext flows in the generating apparatus. In the DUT region, Jext = 0, and in the generating region we have Jeddy = 0, following from our definitions of Jext and Jeddy. Any eddy currents which the generating system induces into itself we assume are included in Jext. Then curl (Bext + Beddy) = μ(Jext) // external (1a) curl (Jext) = - jωσ(Bext + Beddy) // external (2a) curl (Bext + Beddy) = μ(Jeddy) // within DUT (1b) curl (Jeddy) = - jωσ(Bext + Beddy) // within DUT (2b) These equations are exact. We want to solve somehow for curl (Beddy) curl (Beddy) + curl (Bext) = μ(Jeddy) // within DUT (1b) Start over. Assume that the DUT is not present and we just have curl (Bext) = μ(Jext) (1) curl (Jext) = - jωσ(Bext) . (2) and both Bext and Jext are "large". We now cause the DUT to materialize in its prescribed region and this causes a "small" Jeddy to appear inside the DUT. We write this as α Jeddy where α is a small dimensionless parameter. This eddy current in turn creates a new small B field everywhere which we write as α Beddy. We then have curl (Bext + αBeddy) - μ(Jext + αJeddy) = 0 (3) curl (Jext + αJeddy) + jωσ(Bext + αBeddy) = 0 (4) or [curl (Bext) - μ(Jext)] + α [curl (Beddy) - μ(Jeddy)] = 0 (5) [curl (Jext) + jωσ(Bext)] + α [curl (Jeddy) + jωσ(Beddy)] = 0 (6) Subtract (1) and (2) from (5) and (6) to get α [curl (Beddy) - μ(Jeddy)] = 0 (5) α [curl (Jeddy) + jωσ(Beddy)] = 0 (6) Start Over #2 We start with some "external apparatus" in which Jext flows through some wires and this creates field Bext everywhere in space. We identify a region of space close to, but disjoint from, the region of the external apparatus where we shall be placing a conductive Device Under Test or DUT. Assume that the DUT is not yet present and we just have curl (Bext) = μ(Jext) (1) which is valid in all regions of space including where the DUT will be placed. Since the currents Jext don't exist in the region where the DUT will be, we can then write the above as curl (Bext) = 0 // in the DUT region, no DUT present (2) We now cause the DUT to materialize in its prescribed region of space. To simplify our problem, we assume that the DUT is surrounded by air and both the DUT and air have the same magnetic permeability μ0, so the DUT does not then alter Bext in the DUT region. This Bext within the DUT creates eddy currents according to Faraday's Law, so curl (Jeddy) = - jωσ(Bext) // in the DUT region (3) These eddy currents in turn create a new magnetic field Beddy according to Ampere's Law, curl (Beddy) = μ(Jeddy) (4) We then have to update (3) to adjust for this new eddy B field, so curl (Jeddy) = - jωσ(Bext+Beddy) // in the DUT region, updated (3) The field Eext causes an eddy current Jeddy to flow in the DUT according to Ohm's Law Jeddy = σ Eext. This eddy current creates a new field Beddy which exists everywhere. The above equations then become curl (Bext+Beddy) = 0 // in the DUT region (5) curl (Jeddy) = - jωσ(Bext) // in the DUT region (6) and this causes an "induced" Jeddy to appear inside the DUT according to curl (Jeddy) = - jωσ(Bext) . (2) This eddy current in turn creates a new small B field which we write as Beddy according to curl (Beddy) = μ(Jeddy) We then have curl (Bext + Beddy) = μ(Jext + Jeddy) (3) curl (Jext + Jeddy) = - jωσ(Bext + Beddy) . (4) where Bext and Jext are exactly as they were before the DUT appeared. For example, perhaps the external apparatus is driven by a current source which does not allow Jext to be altered by the newly created field Beddy which exists everywhere including in the external apparatus. In practice, Beddy will be very small and such a current source is not needed, it is just a theoretical tool. We then subtract equations (1) and (2) from equations (3) and (4) to get curl (Beddy) = μ(Jeddy) (5) curl (Jeddy) = - jωσ(Beddy) . (6) Oops. I have "lost my signal" because I want Bext on the right of (6)! Beddy << Bext (3) So that curl (Bext) ≈ μ(Jext) // external (1c) curl (Jext) ≈ - jωσ(Bext) // external (2c) curl (Bext) ≈ μ(Jeddy) // within DUT (1d) curl (Jeddy) ≈ - jωσ(Bext) // within DUT (2d) Here is the equation I want to end up with somehow: curl (Beddy) ≈ μ(Jeddy) . (6) Here is one way to get it. Rewrite, curl (Bext + Beddy) = μ(Jext + Jeddy) // everywhere (1) curl (Bext) ≈ μ(Jeddy) // within DUT (1d) Subtract the second from the first and you get (6). Jeddy << Jext . (4) Then (1) becomes curl (Bext) ≈ μ(Jext) (5) Subtracting (5) from (1) gives, curl (Beddy) ≈ μ(Jeddy) . (6) We assume next that Jeddy in the DUT exists in a region of space which is disjoint from the region where Jext flows in the generating apparatus. In the DUT region, Jext = 0, and in the generating region we have Jeddy = 0, following from our definitions of Jext and Jeddy. Any eddy currents which the generating system induces into itself we assume are included in Jext. Then we can write (2) as curl (Jeddy) ≈ - jωσ(Bext) // within the DUT (7) ********************************