1
$\begingroup$

According to the skin effect, alternating current in a conductor has the highest current density on the surface, and it drops with the distance from the surface, such that the current is conducted in a layer near the surface.

For a tube, is the inner surface a skin?

I am wondering about it because in cylindrical conductor with a radius larger than the depth of the skin, there is a part of the cylinder not part of the skin, so there should be no current. I can remove the part of the conductor that is not conducting current by creating cylindrical hole. The result is a tube.

If I removed something that was not an active part of the conductor, why does it change the current in the active part?

I can also create a tube in a different way than drilling: Bending a sheet to a tube, which can plausibly create a tube where the inner surface is a surface with conducting skin depth.

I start with a flat sheet, with the thickness more than two times the skin depth. It's length is the same as the length of the cylinder. I bend it to almost form a tube by bending its shorter side incrementally to a section of a circle, until I have a tube with a small open gap parallel to its center. There are two surfaces building the gap. As a conductor, the voltage between the sides of the gap should be the same based on the geometric symmetry of the cross section. If I close the gap to create a proper tube, the geometry changes very little, but the topology changes, the surface gets a hole. The cross section allows for a loop current now, but as the surfaces that came in contact had the same voltage, there should be no current in the loop. So it seems the inner skin should behave like the outer skin.

This means the inner surface has a skin.
The method by creating a hole seems seems to indicate it is somehow different.

Can I change a conductor by removing something not part of it?

$\endgroup$
0
$\begingroup$

For a tube, is the inner surface a skin?

If you're using the tube as a circular waveguide, with EM waves propagating along the tube, then yes, it certainly is.

in cylindrical conductor with a radius larger than the depth of the skin, there is a part of the cylinder not part of the skin, so there should be no current. I can remove the part of the conductor that is not conducting current by creating cylindrical hole. The result is a tube.

Yes, this is commonly done (as I understand it) in power transmission. Instead of making a solid aluminum wire, the center of the wire is replaced with steel, which is a poor conductor but provides greater strength for hanging wires from towers. The skin depth for 60 Hz in aluminum is about 10 mm, so for any wire thicker than 20 or 30 mm the inner part of the wire is not contributing much to the conduction.

I start with a flat sheet, with the thickness more than two times the skin depth. It's length is the same as the length of the cylinder. I bend it to almost form a tube by bending its shorter side incrementally to a section of a circle, until I have a tube with a small open gap parallel to its center.

You can build this.

But not much current will flow on the "inner" surface of this structure. Because this part is shielded from the return conductor (another wire or the actual earth ground) by the outer part of the structure.

If I close the gap to create a proper tube, the geometry changes very little, but the topology changes, the surface gets a hole. ... there should be no current in the loop. So it seems the inner skin should behave like the outer skin.

It doesn't change. In both cases, little current flows on the inner surface, because it is not exposed to the return path.

Above I elided this part of your question,

The cross section allows for a loop current now, but as the surfaces that came in contact had the same voltage,

How will you excite the structure to flow current in one direction on the outer surface and in the other direction on the inner surface? You can do it by treating the inner cavity as a waveguide, like I mentioned above. But that becomes a much different use case than treating it as a wire with return current flowing on a different conductor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.