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(adsbygoogle = window.adsbygoogle || []).push({}); E = Rj + v×BwithB = const(and orthogonal toj).

There is quite a bit of subtlety involved in how to interpretvin that sort of ad-hoc generalization of Ohm's law, and I'm not 100% sure I got that right. If we defineρbyj := ρvand assume a static background-chargeρ'so that∇E = (ρ' + ρ)/εI 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 = 0given that this PDE appears to have nontrivial solutions forj'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 thusρ > 0! Then it could possibly be shown that the PDE cannot be satisfied. Investigating...

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# ∇E = 0 in an Ohmic, 2d Hall conductor with constant B?

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