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Skin depth; EM wave and AC

Physics Asked by Aravindh Vasu on August 16, 2021

When I google skin depth, I get the following definition,

Skin effect is the tendency of an alternating electric current (AC) to
become distributed within a conductor such that the current density is
largest near the surface of the conductor, and decreases with greater
depths in the conductor. The electric current flows mainly at the
“skin” of the conductor, between the outer surface and a level called
the skin depth. The skin effect causes the effective resistance of the
conductor to increase at higher frequencies where the skin depth is
smaller, thus reducing the effective cross-section of the conductor.
The skin depth, δ, is defined as the depth where the current density
is just $1/e$ (about $37 % $) of the value at the surface; it depends on the
frequency of the current and the electrical and magnetic properties of
the conductor.

But when I search for skin-depth “EM waves”,

Skin depth defines the distance a wave must travel before its
amplitude has decayed by a factor of 1/e. The skin depth is the
reciprocal of the decay constant β

The second definition, tells us, how deep, an EM wave penetrates into a medium before it almost dies out whereas the first one tells us something about the distribution of current density. Are these two different? If not, How are these two definitions related?

P.S: I found a relevant question but I couldn’t understand the answer.

2 Answers

These definitions seem to imply two different geometries: in the first case, AC current flows in a conductor (which may be, e.g., cylindrical), in the second case, an electromagnetic wave falls on a surface of a conductor (which surface may be, e.g., flat). So how are these definitions related? Remember that there is the following relation between the electric field $E$ and current density $j$ in a conductor: $j=sigma E$, where $sigma$ is the conductivity of the conductor. Therefore, if the electric field penetrates the conductor to the skin depth, the current also penetrates the conductor to the same depth. Specific depths at which the field or the current density is $1/e$ of their magnitudes on the surface of the conductor can be somewhat different in the two different geometries, but they are of the same order of magnitude, so these definitions are pretty consistent.

Answered by akhmeteli on August 16, 2021

The definition of skin depth is mainly applied to conductive media and, as you stated, is the distance that the wave must travel inside a medium in order to experience a decay by $ 1/e $. This decay is due to the medium to be lossy.

The fact that this medium is lossy implies that the propagation constant of the wave becomes complex and this loss is related to the value of the conductivity. The process to obtain the relation is explained at Chapter 1 in the book "Microwave Engineering" by David. M. Pozar.:

First, we get the wave equation, or Helmholtz equation, from Maxwell's equations assuming an harmonic (cosine and sine dependence) time dependence:

Faraday's Law of Induction: $nabla$x $ vec{E} = -jomegamuvec{H} spacespacespacespacespacespacespacespacespace(Eq.1)$

Ampere's Law: $nabla$x $ vec{H} = jomegaepsilonvec{E} +sigmavec{E}$ $spacespacespacespacespacespacespacespacespace(Eq.2)$

being this last term corresponding to $vec{J}=sigmavec{E}$, the surface current density.

Here, j is the imaginary number $sqrt{-1}$ and $omega$ corresponds to the angular frequency $(2pi f)$. $mu$ and $epsilon$ correspond to the magnetic permeabilityand dielectric permitivity of the media.

These equations, in their most basic and simple meaning, state that a variation in time of a magnetic field generates an electric field and that a variation in time of an electric field and/or the presence of an electric current surface generate a magnetic field.

By applying some vector mathematical identities, we get the wave equation for E field:

$ nabla^2 vec{E} + omega^2muepsilon(1-jsigma/(omegaepsilon))vec{E} = 0 spacespacespacespacespacespacespacespacespace(Eq.3) $

If we check the definition of the Helmholtz equation por a generic vector wave $vec{A}$,

$ nabla^2 vec{A} + k^2vec{A} = 0 spacespacespacespacespacespacespacespacespace(Eq.4) $

we see that our expressions are much alike, and that we could define a propagation constant named $gamma$ (corresponding to k in the expression of Helmholtz equation), which will have a real part $alpha$, called the attenuation constant and an imaginary part $beta$, the phase constant.

$gamma=alpha+jbetaspace = omegasqrt{muepsilon}sqrt{(1-jsigma/(omegaepsilon))} spacespacespacespacespacespacespacespacespace(Eq.5)$

In the definition of skin depth we were talking about a power decay, i.e, attenuation. In fact, we can easily check that the skin depth (which from now on we will name $delta_s$) corresponds to the inverse of the attenuation constant

$delta_s=1/alpha spacespacespacespacespacespacespacespacespace(Eq.6)$

Finally, given that $alpha$ is the real part of the propagation constant $gamma$, we can extract the expression of the skin depth by taking the real part of Eq. 5 and also assuming that, since the medium is a conductor, $sigma>>omegaepsilon$.

$ delta_s = sqrt{2/(omegamusigma)} $, which is, by the way, a distance measured in meters (although the very low values at high frequencies might persuade you to use millimeter of even microns)

Here you can see that this parameter is related to the conductivity of the media. By definition, this parameter gives you the distance at which the decay of the electric field is 1/e ~37%. This electric field is related to the surface current, as seen in Eq.2. Since it is a linear proportional relation, we can state that this 37% decay in the electric field will imply a decay of 37% in the surface currents as well.

I saw that you asked in a comment whether this electric field is due to a external wave. As @akhmeteli told you, not necessarily. If you check the equations E1. 1 and Eq.2, you see that the presence of a current is related to the excitations of electromagnetic waves due to that current.

I hope this helps

Answered by Terahertz on August 16, 2021

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