Skin depth: Same for current as for incoming EM wave?

[Wminus]

For a good conductor, an incoming plane electromagnetic wave will be attenuated exponentially as it penetrates a distance $z$ into the conductor, $|\vec{E}(z)| = |\vec{E_0}|e^{-z/ \delta}$. $\delta$ is called the "skin depth". The current generated by this incoming electromagnetic wave is given by ohm's law and thus will be attenuated in the same fashion $\vec{J_E} = \sigma \vec{E} \Rightarrow |\vec{J_E(z)}| = |\vec{J_{E0}}|e^{-z/ \delta}$.

OK, this is all fair enough. It falls out from maxwell's equations.

However, if you have a cylindrical conductor and you connect it to a high-frequency AC power source, it turns out that that the resulting electrical current $\vec{J_{AC}}$ will have the exact same distribution as $\vec{J_E}$. Namely, $\vec{J_{AC}} = \vec{J_{AC0}} e^{-z/ \delta}$, with the exactly same attenuation factor (delta) as before.

Why is this so? Does this mean an AC power source generates a potential difference in the conductor in the same way an incoming electromagnetic wave does?

 

[jasonRF]

For a good conductor, an incoming plane electromagnetic wave will be attenuated exponentially as it penetrates a distance $z$ into the conductor, $|\vec{E}(z)| = |\vec{E_0}|e^{-z/ \delta}$. $\delta$ is called the "skin depth". The current generated by this incoming electromagnetic wave is given by ohm's law and thus will be attenuated in the same fashion $\vec{J_E} = \sigma \vec{E} \Rightarrow |\vec{J_E(z)}| = |\vec{J_{E0}}|e^{-z/ \delta}$.

This is true for a plane wave incident on a plane conductor that occupies the half-space $z>0$.

However, if you have a cylindrical conductor and you connect it to a high-frequency AC power source, it turns out that that the resulting electrical current $\vec{J_{AC}}$ will have the exact same distribution as $\vec{J_E}$. Namely, $\vec{J_{AC}} = \vec{J_{AC0}} e^{-z/ \delta}$, with the exactly same attenuation factor (delta) as before.

I am wondering how you came up with this result, as I think it isn't correct. Qualitatively, would you expect that an AC current in a wire is not a function of $r$? (Here I am assuming you are using cylindrical coordinates with your conductor along the $z$ axis - it helps us understand if you define your coordinates!).

jason

 

[Wminus]

You're right, sorry about the unclear coordinates. $z$ here is the depth, the distance from the surface to a point radially inside the cylinder, so it's parallel to (cylindrical coordinate) r. It's not the length along the z-axis.

Basically I should've called the "depth" $d$ instead of $z$ to avoid confusion.

 

[jasonRF]

Okay - so I understood your coordinates. I am also assuming you are using $e^{i \omega t}$ time dependence for the rest of this post.

While your formula in the second case is approximately true when the radius of the cylinder is much much larger than the skin depth, it is not true in general. Just like in the planar case, inside your good conductor (which is assumed to satisfy Ohm's law, and for which the displacement current is negligible compared to the conduction current) we have
$$\nabla^2 \mathbf{J} = i \mu \omega \sigma \mathbf{J}$$
If you solve this in the planar case with the appropriate boundary condition you get your first answer. For the simple cylindrical case, $\mathbf{J} = \mathbf{\hat{z}}J_z(r)$, so the equation becomes
$$\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial J_z}{\partial r} \right) = i \mu \omega \sigma J_z$$
Solve this with the appropriate boundary conditions and you will get the answer you are looking for. (EDIT: hint: this is a modified Bessel's equation)

jason