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So this question is really about skin depth. I have been introduced to the skin depth by a simple model (simple equation for electrons in a metal with a damping term) of polarisability for a metal. In this calculation, using the dilute form of the classes most relation the permittivity of the metal was found. This had a complex form, and thus so did the refractive index.

As a result the wave vector of any electromagnetic signal is complex and the complex part attenuates the wave and represents energy loss. I understand this part. However, I was then told that this is the reason that sometimes hollow conductors are preferred (like a hollow copper tube). Because beyond the skin depth the field is rapidly attenuated. However, surely the attenuation only takes place in the direction of the wave vector k? How would a hollow tube cary a signal? If the k vector is along its long axis? Surely it gets attenuated by the time it reaches the end?

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For microwaves the skin depth is too small and it does not conduct efficiently, so you have to design it and use it as a waveguide. For lower frequencies the skin depth causes the conduction to be near the surface. See the answer and comments already given at http://physics.stack-exchange.com/questions/109897/does-electricity-flow-on-the-surface-of-a-wire-or-in-the-interior

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The reason is simply that if only the surface can carry a current, there is no point in having all that other material (heavy, expensive) sitting in your wire not doing anything.

If I can remove 50% of the material of my conductor without affecting its function (conduct current without dissipating too much energy), I probably take away close to half the cost. I could then consider using the same amount of material to make a larger diameter hollow conductor instead - same cost, but lower effective resistance (because while the skin depth is the same, the circumference is now bigger so I get a larger conductive surface).

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