# AC - impedance for a coaxial cable

Problem

I want to calculate the AC impedance for a coaxial cable. A rough image is presented below: Because we have an AC current, suppose that the skin depth, which we can define as $$\\delta\$$ is much less than $$\a\$$ and $$\c-b\$$ respectively. Also, suppose that the conductivity is given by $$\\sigma\$$ and that the AC current is driven by the angular frequency $$\\omega\$$.

Ideas

Here's my idea: the skin depth is really small, so we might conclude that all the current that passes through the cable is concentrated at the surface of the 'inner' and 'outer' cylinder so to speak. From here, I think that we might be able to approximate the coaxialcable as two infitesimally thin cylindrical sheets, one of inner radius a and one of outer radius $$\b\$$ or $$\c\$$. For the outer shell, I can't really determine how the current will distribute itself, so that's why I'm not sure on how to pick that outer radius.

My guess is that it'll be an outer radius of $$\b\$$. We know that a magnetic field is induced by the current in the inner shell, this magnetic field will exist outside the inner shell, but it weakeans with distance. Inside the outer shell, we might assume that the magnetic field is constant throughout from the current passing through itself. Therefore, the total magnetic field experienced inside the outer shell should be larger on the inside rather on the outside. I think this discrepancy will cause more Eddy currents to form on the inner shell.

Either way, this is just all my reasoning, and the real answer should land at

$$\frac{1}{2\pi \sigma \delta}(1/a+1/b).$$

according to my answer sheet in my textbook. I'm really lost at how to attack this problem. I'd be glad if anyone out there could show me a way of deriving this.

Update

If I continue with my argument above, we might assume the inner and outer shells placed at radii $$\a\$$ and $$\b\$$ respectively, both with thickness $$\\delta\$$. Then we might use the expression for the resistance given by:

$$R = \frac{l}{\sigma A}$$

Where we basically use the effective area instead. Since we might as well find the impedance per length, we can study the expression:

$$R_l = \frac{1}{\sigma A}$$

Now we can add the resistances from the inner and outer shell since they're in series. The area just becomes the effective area, so for the inner shell, $$\2\pi a \delta\$$ and for the outer shell $$\2\pi b \delta\$$. We therefore land at precisely:

$$R_{l,tot} = \frac{1}{2\pi \sigma \delta}(1/a+1/b).$$

I'm not sure if this is the correct way of thinking of it. I hope I can get feedback on my solution.

Thank you.

• You start with square root of L/C but get a different result that includes the log of b/a and not what you have Feb 28 at 20:55
• You've shown a formula but not equated it to anything and also you have not really stated what $\sigma$ equals. Are you talking about characteristic impedance or some other impedance? It doesn't look like a Z0 formula I've ever seen. Feb 28 at 21:13
• It must also include relative dielectric constant, so you have the wrong formula or the wrong target Feb 28 at 21:15
• @Hoagie It doesn't if the relative dielectric constant is just 1, then it vanishes from the expression. Between the inner and outer shell we might assume we're dealing with air. Mar 1 at 15:06
• @Andyaka I'm sorry. It's the AC impedance for the setup given in the figure. I've not been given more information than that unfortunately. I've added some more to my post, I hope that'll clear things up. Mar 1 at 15:07