Some Skin Effect Notes

Kevin Schmidt, W9CF

Abstract:

I give an animated plot of the radio frequency current and magnetic field in a conductor. For completeness, I include the derivation of the result, however similar descriptions can be found in most electromagnetism texts.

1 Skin Effect

The skin effect is the description given to the phenomenon where electromagnetic fields (and therefore the current) decay rapidly with depth inside a good conductor. In fig. 1 I show the result, derived below, for the magnetic field and current density in a conductor with surface radius of curvature much larger than the skin depth i.e. the surface is reasonably flat on the skin depth length scale.

Figure 1. Here is the current density and magnetic field as a function of depth d in a conductor.


If I look at Maxwell's equations in Gaussian units, at a single frequency, and use the convention for linear media that the time dependence of all fields and currents is given by exp(- i$\omega$t) and the real part is taken, the curl equation are
 

$$\vec{\nabla} \vec{E}_{c}^{} {\frac{i \omega \mu}{c}} \vec{H}_{c}^{}$$ =

$$\vec{\nabla} \vec{H}_{c}^{} {\frac{4 \pi}{c}} \vec{J} {\frac{i \omega \epsilon}{c}} \vec{E}_{c}^{}$$ = (1)

If I am looking for solutions inside a metal, I can evaluate the conduction current $\vec{J}$ using Ohm's law. The second equation becomes

$$\vec{\nabla} \vec{H}_{c}^{} {\frac{4 \pi \sigma }{c}} \vec{E}_{c}^{} {\frac{i \omega \epsilon}{c}} \vec{E}_{c}^{}$$ = (2)

Since the conductivity of copper is 5.2 x 10¹⁷ second ^{- 1}, and $\epsilon$ is of order unity, the term with $\epsilon$$\omega$ is neglible for all radio frequencies, and can be dropped. Taking the curl of the first equation and substituting the second, I get

$$\vec{\nabla} \vec{\nabla} \vec{E}_{c}^{} {\frac{i 4 \pi \sigma \omega \mu}{c}} \vec{E}_{c}^{}$$ (3)

Notice that taking the divergence of the second equation shows that $\vec{\nabla}$ $\cdot$ $\vec{E}_{c}^{}$ = 0 , so that using the identity

$$\vec{\nabla} \vec{\nabla} \vec{E}_{c}^{} \vec{\nabla} \vec{\nabla} \cdot \vec{E}_{c}^{} \nabla^{2}_{} \vec{E}_{c}^{} \nabla^{2}_{} \vec{E}_{c}^{}$$ (4)

waves in the metal must satisfy the equation

$$\nabla^{2}_{} \vec{E}_{c}^{} {\frac{i 4 \pi \sigma \omega \mu}{c}} \vec{E}_{c}^{}$$ (5)

The skin depth $\delta$ is defined to be

$$\delta {\frac{c}{\sqrt{2\pi \sigma \omega \mu}}}$$ (6)

so that the equation above becomes

$$\nabla^{2}_{} \vec{E}_{c}^{} {\frac{2i}{\delta^2}} \vec{E}_{c}^{}$$ (7)

Repeating the calculation for the magnetic field shows that it satisfies the same differential equation

$$\nabla^{2}_{} \vec{H}_{c}^{} {\frac{2i}{\delta^2}} \vec{H}_{c}^{}$$ (8)

The solutions for the two fields are still related by Maxwell's equations.

Outside the conductors, the same calculation shows that the fields satisfy the wave equations

$$\nabla^{2}_{} \vec{E} {\frac{\omega^2}{c^2}} \vec{E}$$

$$\nabla^{2}_{} \vec{H} {\frac{\omega^2}{c^2}} \vec{H}$$ . (9)

Notice that $\omega$/c = 2$\pi$/$\lambda$ where $\lambda$ is the free space wave length. Except for the various factors of 2i and 2$\pi$ , the skin depth is essentially the wave length in the metal. The main thing to notice are that the free space wave length is enormously larger than the skin depth. The other length scales that enter the problem are the size of the conductors and the distances over which the conductor is not essentially flat.

The skin depth in copper is about .005 centimeters for the 160 meter band, .002 centimeters for the 20 meter band, .001 centimeters for the 6 meter band, and .0003 centimers for the 70 centimeter band. It is easy to see that the spatial variation of the fields in vacuum is much smaller than the spatial variation in the metal. Therefore, for the purposes of evaluating the fields in the conductor, the spatial variation from the wave length outside the conductor can be ignored.

The main effect on the form of the fields inside the conductor is governed by the shape of the conductor. For the usual case where the surface of the conductor is flat on the length scale of the skin depth, and the conductor is much thicker than the skin depth, the solution is straightforward. Note that flat on the length scale of the skin depth means that the radii of curvature of the surface should be much larger than a skin depth. Since the external fields are changing on the scale of either the curvature of the conductor or the wave length, and the tangential components of $\vec{E}$ and $\vec{H}$ must be continous across the boundary (this is normally shown by applying Stokes theorem to Maxwell's curl equations above) the tangential components inside also change tangentially in the same way. Similarly the normal component of $\vec{B}$ is continuous, so the normal component of $\mu$$\vec{H}$ is continuous. The result is that the tangential dependence of the magnetic field just inside the material is identical to the dependence outside. The second derivatives in the tangential direction will be of order one over the length scale squared. Since the length scales outside are so much larger than those inside, these second derivatives cannot give a contribution anywhere near the value of 2i/$\delta^{2}_{}$$\vec{H}$ needed to solve Maxwell's equations inside. To an excellent approximation, the dependence of the fields normal to the surface must give this entire contribution and

 

$${\frac{\partial^2}{\partial d^2}} \vec{H}_{c}^{} {\frac{2i}{\delta^2}} \vec{H}_{c}^{}$$ (10)

where d is the distance perpendicular to the conductor surface. The solution to these equations with the boundary condition that there is no energy source deep in the conductor is

$$\vec{H}_{c}^{} \vec{H}_{0}^{} \delta$$ (11)

where $\vec{H}_{0}^{}$ is the component of the magnetic field just outside the surface.

If we take the divergence of the curl of $\vec{E}$ Maxwell equation, we get the divergence of $\vec{H}$ is zero. Taking the divergence of this, the largest terms are from the normal component. Again, since the skin depth is so much smaller than the other length scales, the only way to obtain a zero divergence result is to have a negligible component of $\vec{H}$ normal to the surface. Plugging back in to Maxwell's equations, the magnetic and electric fields in the conductor are

$$\vec{H}_{c}^{} \vec{H}_{\parallel}^{} \delta$$ =

$$\vec{E}_{c}^{} {\frac{(i-1)\mu\omega \delta}{2c}} \hat{n} \vec{H}_{\parallel}^{} \delta$$ = (12)

where $\hat{n}$ is the normal to the surface and $\vec{H}_{\parallel}^{}$ is the tangential magnetic field just outside the surface. This shows that the electric field in the conductor is smaller than the magnetic field by roughly the ratio of the skin depth to the free space wave length.

Notice that the skin depth being so small dominates Eq. 10, so for conductors many skin depths thick, you will always get this exponential decay as you move away from the surface. In particular you will still get this behavior even for surfaces with large radii of curvature once you are many skin depths from the surface.

The current density is given by Ohm's law $\vec{J}$ = $\sigma$$\vec{E}$ , or

$$\vec{J} {\frac{(i-1)c}{4\pi \delta}} \hat{n} \vec{H}_{\parallel}^{} \delta$$ (13)

Plots of the magnetic field and the current density are shown in figure 1. It is amusing to see that the current is not all flowing in the same direction in the conductor. From the figure, we see that there is a wave traveling and decaying into the conductor. When the current density is a maximum at the surface, the current deeper than about 1.5 skin depths is flowing in the opposite direction. The magnetic and electric fields have the same behavior.

To verify the usual results, integrate the current density to get the equivalent surface current,

$$\vec{K}_{\rm eff}^{} \int_{0}^{\infty} \vec{J} {\frac{c}{4\pi}} \hat{n} \vec{H}_{\parallel}^{}$$ (14)

which agrees with the surface current of a perfect conductor. The power dissipated per unit area in the conductor resistance is

$${\textstyle\frac{1}{2}} \int_{0}^{\infty} \vec{E} \cdot \vec{J}^{*}_{} {\textstyle\frac{1}{2 \sigma \delta}} \vec{K}_{\rm eff}^{}$$ (15)

These are the usual results that agree with modeling the current as being constant to a depth of $\delta$ . However, notice that the physical current is quite different.