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- Thread starter antonantal
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That being the case, if there is ANY electric field whatsoever inside the conductor, the electrons will almost instantly move to cancel it out. Electrons can't, however, escape the physical boundaries of the conductor -> so they can accumulate and move along the outside.

Does that help?

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That being the case, if there is ANY electric field whatsoever inside the conductor, the electrons will almost instantly move to cancel it out. Electrons can't, however, escape the physical boundaries of the conductor -> so they can accumulate and move along the outside.

Does that help?

remember the leftover protons from when the electrons slosh over to the other side, those guys are tied into nuclii that don't move from their lattice positions.

i think the fact is that

now as the frequency increases, you get this skin effect where the current density in the center is less than the current density closer to the conductor edge. at very high frequencies, virtually all of the current is at the cylinderical edge of the conductor. on a linear scale, the current density vs. radius function would look spike-like (perhaps approximated with a dirac-impulse function). that is the extreme end of

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...at very high frequencies, virtually all of the current is at the cylinderical edge of the conductor...

The thing that got me confused was that in an ideal electric conductor the phenomenon occurs at any frequency.

But looking at the formula for the skin depth [tex]\delta = \sqrt{\frac{2}{\sigma\omega\mu}}[/tex] it can be seen that, for infinite conductivity, the skin depth is zero no matter what the frequency is.

I'm still confused about what would be the skin depth at DC in an ideal conductor (superconductor).

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In a theoretical, perfect, ideal conductor (effectively superconductors count) the skin-depth is infinitesimal... effectively zero.

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The thing that got me confused was that in an ideal electric conductor the phenomenon occurs at any frequency.

But looking at the formula for the skin depth [tex]\delta = \sqrt{\frac{2}{\sigma\omega\mu}}[/tex] it can be seen that, for infinite conductivity, the skin depth is zero no matter what the frequency is.

I'm still confused about what would be the skin depth at DC in an ideal conductor (superconductor).

I don't know about the equation you've given. That equation, if my memory is correct, is for a planar sheet, not a cylindrical wire. I'll check my reference book when I get home, but I remember a different equation for wires. In grad school in the late '70's, we studied the math for skin effect. It involved "ber" and "bei" functions, which are derived from Bessel functions of the first kind. I'll look it up and post again. What happens when sigma becomes infinite is a good question, and my curiosity is now piqued. BR.

Claude

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