Apr30-13, 07:46 AM
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 ad-hoc 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 background-charge 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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