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∇E = 0 in an Ohmic, 2d Hall conductor with constant B? 
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#1
Apr3013, 06:46 AM

P: 154

I've been trying to figure this out for long now but unfortunally, I'm not able to prove that ∇E = 0 in an Ohmic, 2d Hall conductor E = Rj + v×B with B = const (and orthogonal to j).
There is quite a bit of subtlety involved in how to interpret v in that sort of adhoc generalization of Ohm's law, and I'm not 100% sure I got that right. If we define ρ by j := ρv and assume a static backgroundcharge ρ' so that ∇E = (ρ' + ρ)/ε I end up at ∇E = 1/ρ²(B×j + B²/(ρR)j)·∇ρ from Ohm's relation. However, I don't see how that would prove that ∇E = 0 given that this PDE appears to have nontrivial solutions for j's which satisfy ∇j = 0. Any ideas would be greatly appreciated. Edit: I just thought of something: Perhaps this depends on assumptions concerning the carriers of charge. In particular that the mobile charges are strictly of one sign and thus ρ > 0! Then it could possibly be shown that the PDE cannot be satisfied. Investigating... 


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