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daudaudaudau
daudaudaudau is offline
#5
Nov20-12, 06:36 AM
P: 296
Quote Quote by Enthalpy View Post
In the second equation, E must be complemented by dA/dt, possibly with some sign if you like.

This is all-important in induction. For instance in a generator, copper wires shall have low loss, meaning E~0, but you get a V at the terminals thanks to the induction dA/dt summed over the conductor path (or d phi / dt if you prefer). Or even, E=0 in a superconductor, which is considered practically for generators and motors, at orientable pods for boats for instance. Though there, it would supposedly be a type II superconductor, which has a resistance.
Thanks for the answers. Where exactly should A be included ? In the first equation, ∇E=-j ω B, I would put [itex]E=-\nabla V-j \omega A[/itex] to give
[tex]
\oint\left(\nabla V+j\omega A\right)\cdot dl=j\omega \Phi
[/tex]

So does this mean that my voltage is
[tex]
V=j\omega\Phi-j\omega\oint A\cdot dl
[/tex]
?