Imagine a magnetic conductor with a cylindrical cross section, surrounded by a coil with a time varying current of $$I = I_0\cdot \cos (2\pi f t)$$ The conductor is split into two parts, the first with a conductivity and a relative permeability of $\kappa, \mu$, the second with $4\kappa, \mu$. There is a magnetic field $B$ through the conductor, which is caused by the current and therefore time varying as well: $$B = B_0\cdot \cos (2\pi f t)$$

enter image description here

The change of this magnetic field induces a voltage inside the material and causes a current density $J$. This current density has the value $J_1$ on the surface of the left conductor and $J_2$ on the right side.

The skin depth $\delta$ is defined by the distance from the surface where $J = 0.37 \cdot J_1$, respectively $J = 0.37 \cdot J_2$ , with $0.37 = 1/e$ and also:

$$\delta = \frac{1}{\sqrt{\pi f\kappa\mu}} = \frac{\sqrt{2j}}{\alpha}$$

where a $\alpha$ is the propagation constant. I found out by simulation, that at the boundary between both materials, the blue one, and the orange one, applies:

$$\frac{1}{\delta_{12}} = \frac{1}{2}(\frac{1}{\delta_{1}}+\frac{1}{\delta_{2}})$$

and therefore

$$\alpha_{12} = \frac{1}{2}(\alpha_1 + \alpha_2)$$

But I'm really struggling to prove that. Can someone give me some hints, how I could get these relations analytically?

Here another plot:

The upper one shows the current density at the surface. The second one shows the contour line where the current density decreased about $63\% = skin depth$. At $z=0$ is the boundary between both materials. Though the current density is a step function, the skin depth is continuous and has the value $\delta_{12}=\frac{2}{\frac{1}{\delta_{1}}+\frac{1}{\delta_{2}}}$ at $z=0$.

enter image description here

  • 1
    $\begingroup$ This is a really good question. I don't have an answer right now - but upvoting it in the meantime. $\endgroup$
    – Floris
    Aug 21, 2014 at 23:25
  • $\begingroup$ Can you just clarify what $\delta_{12}$ and $\alpha_{12}$ represent? $\endgroup$
    – ProfRob
    Aug 22, 2014 at 8:10
  • $\begingroup$ @RobJeffries I added another plot, is it clearer now? $\endgroup$ Aug 22, 2014 at 8:53
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    $\begingroup$ Much clearer. I don't have time to think about this one now, but will be following with interest. $\endgroup$
    – ProfRob
    Aug 22, 2014 at 10:59
  • $\begingroup$ This looks similar to those capacitor problems where you have two different dielectrics with two different arrangements. For example, put the two dielectrics on top of each other (i.e., like two capacitors in series) and then find the total capacitance. The result will have a form similar to the mathematical form of your $\delta_{12}$. I think this is similar in that you are dealing with an inductive effect. In any case, look at Chapter 8 [wave guides] in Jackson's E&M book. $\endgroup$ Oct 3, 2014 at 0:05

1 Answer 1


Well, I hope I didn't go too much off track. I'm open for discussion.(possible solution path is on the bottom)

Looking at the formula for $\delta$ I see that it is actually related to the speed of EM waves in the medium. The speed of propagation for a medium with the properties $\kappa_1$ and $\mu_1$ is: $$c_1=\frac{1}{\sqrt{\kappa_1\mu_1}}$$ Therefore we can rewrite it as: $$\delta=\frac{1}{c\sqrt{\pi f}}$$ Looking at it, I would like to have the $\omega$ in it rather than $f$: $$\delta=\frac{\sqrt{2}}{c\sqrt{\omega}}$$ Lets square it: $$\delta^2=\frac{2}{c^2\omega}$$ What stays same, I guess its $\omega$: $$\omega=\frac{2}{c^2\delta^2} $$ The frequency is the same everywhere, so it is: $$\frac{1}{c_1^2\delta_1^2}=\frac{1}{c_2^2\delta_2^2} $$ or: $$\frac{c_2}{c_1}=\frac{\delta_1}{\delta_2} $$ So now we actually reduced it to a problem of finding the speed of propagation of EM-waves at the interface. That will not help directly but it gives me another idea.

I would now suggest to look at it similar to a semiconductor PN-junction problem, and calculate the change of $\kappa$ on the interface due to the differences in charged particle density. The drift currents will do the charge compensation on the interface.

So the diffusion current is: $$I_{diff}=qD\frac{dn}{dx}$$ Using this formula you get some current, this affects the local $\kappa$ and creates a $\kappa'$. Using this value you should get the $\delta$ on the boundary(and even the functional dependence on the distance from the interface).

  • $\begingroup$ It's an interesting approach, I will think about soon. The first part of your question is similar to what I finally wrote in my thesis. In addition I suggested that I need the average propagation speed at a fixed depth. The average speed of two movements with fixed distance is calculated by the harmonic mean. This way I get my desired solution. It's not really a proof, but was satisfying enough for now. $\endgroup$ Mar 6, 2015 at 21:11

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