Register to reply

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

Share this thread:
ManDay
#1
Apr30-13, 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 + vB 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/ρ(Bj + B/(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...
Phys.Org News Partner Physics news on Phys.org
Physicists discuss quantum pigeonhole principle
First in-situ images of void collapse in explosives
The first supercomputer simulations of 'spin?orbit' forces between neutrons and protons in an atomic nucleus

Register to reply

Related Discussions
Hall constant & Electron concentration. Atomic, Solid State, Comp. Physics 0
Ohmic conductor, when R tends to zero I=V/R General Physics 15
Ohmic heating in a conductor Advanced Physics Homework 5
(u.∇)u = ∇(1/2u^2)+w∧u Advanced Physics Homework 13
Hall resistance and fine structre constant General Physics 8