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What does curl E = const. on Ω say about E on ∂Ω?

by ManDay
Tags: ∂Ω, Ω, curl e const
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ManDay
#1
Apr5-13, 05:27 AM
P: 155
Assume I have a simply connected domain Ω and a twice differentiable vector field E for which I know that "∇×E = const." (1) and "∇E = 0" (2) on Ω - I am interested in solving a BC Problem on ∏ = (Ʃ ⊃ Ω)\Ω, the remainder of Ʃ less Ω.

(1) and (2) imply certain restrictions on the BC on ∏. Question:

Which are the restrictions equivalent to (1) and (2)?

By Stokes' theorem, ∫dr·E = ∫dA·const. along the boundary of ∂Ω, but that alone can't possibly be equivalent, can it? I might pick an E which satisfies a certain curve integral value along ∂Ω and which can't satisfy (1) and (2), I assume.

Context: A conductor is forming a loop the hole in which is pierced by a changing magnetic field - how this can be re-formulated into BCs on the conductor's domain? Can it, at all?
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Jano L.
#2
Apr7-13, 04:32 AM
PF Gold
P: 1,168
Starting with the assumptions 1,2 and Maxwell's equations, I do not see how one could arrive at simple boundary conditions. Probably many different boundary conditions are possible.

On the other hand, for metallic ring in slowly changing magnetic field, the boundary conditions on its surface are known. The electric field E has continuous tangential component, and the normal component has jump proportional to surface charge density. The magnetic field B has continuous normal component, and the tangential component has jump proportional to surface current density. Inside the metallic ring, the current density may be assumed to be proportional to electric field (Ohm's law).


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