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I am a little confused with the theory behind the attenuation of electromagnetic wave on conductors:

$$\nabla ^{2} \vec E - \mu \epsilon \ddot{\vec E} - \mu \sigma \dot{\vec E} = 0$$

Until here, everything is ok. We can find a solution where the amplitude varies as $$E_{0}e^{-k_{r}z}$$

For good conductor, we have that $$\sigma >> \omega \epsilon$$ And so, the penetration depth is given by $z$ such that the amplitude decays to $e^{-1}$ of its value.

$$\delta = k_{r}^{-1} = (\frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}})^{-1}$$

Ok. Now my question is, how do we go from this equation to $$\delta = (\sqrt{\frac{2}{\omega \sigma \epsilon}})$$$$?$$

It seems, to me, that it was made the substituion $\sigma = \omega \epsilon$. But how can that be true? Wasn't our assumption that $\sigma>>\omega \epsilon$?

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  • $\begingroup$ How did you arrive at your first expression for the skin depth? It should already contain omega. $\endgroup$ Jul 16, 2022 at 20:24
  • $\begingroup$ @jensenpaull the omega will comes together with an $k_{0}$, which can be get rid of noticing that $$\sqrt{1/(\epsilon \mu)} = \omega / k_{0}$$ $\endgroup$
    – LSS
    Jul 16, 2022 at 20:34
  • $\begingroup$ this relation isn't correct for em waves in conductors. Unless [I think] you include the conducting electrons in the permitivity constant , which you havent $\endgroup$ Jul 16, 2022 at 20:54
  • $\begingroup$ Where did you derive this formula for the wave speed? Maxwells equation using $\vec{J} = \sigma \vec{E}$ assume that the conducting electrons are not included In the permitivity constant and thus any change to the permitivity and permeability would be the presence of dipoles and/or magnetisation\polarisation current . Which wouldn't be the effect of any free electrons. This relation in the normal homogenous equation is derived using the dispersion relation. The relation for this method of solving for em wave in conductors is far more complex(litterally) $\endgroup$ Jul 16, 2022 at 21:27

1 Answer 1

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The dispersion relation for an EM wave in a conductor, following ohms law [which is actually derived using static fields so the validity of this solution can be challenged greatly]

Way 1:

The way that I prefer to do is that the full dispersion relation in complex form is:

$$|\vec{k}|^2= \mu_0 \omega [\epsilon_0 \omega + \sigma i]$$

$\sigma >> \epsilon_0 \omega$

The left hand side of the bracket is insignificant in comparison to the right, so is set to zero.

We then have:

$$|\vec{k}|= \sqrt{\mu_0 \omega \sigma i}$$

Now if you've actually gone through the derivation before, you know that we want to find

$$|\vec{\alpha}| = re(|\vec{k}|)$$

[**Because the vectors are parrellel the real of the magnitude is the magnitude of $\alpha$]

We need to turn it into a better form:

$$|\vec{k}|= \sqrt{\mu_0 \omega \sigma i}$$

$$|\vec{k}|= i^{\frac{1}{2}} \sqrt{\mu_0 \omega \sigma }$$

$$|\vec{k}|= (e^{i\frac{\pi}{2}})^{\frac{1}{2}} \sqrt{\mu_0 \omega \sigma }$$

$$|\vec{k}|= (e^{i\frac{\pi}{4}}) \sqrt{\mu_0 \omega \sigma }$$

Applying Eulers formula we know that the real part of this expression is

$$\alpha = cos(\frac{\pi}{4}) \sqrt{\mu_0 \omega \sigma } $$

$$\alpha = \sqrt{\frac{\mu_0 \omega \sigma }{2}} $$

$$\frac{1}{\alpha} = \sqrt{\frac{2}{\mu_0 \sigma \omega}} $$

Way 2:

$$-\beta^2 + \alpha^2=-\mu_{0}\epsilon_{0}\omega^2$$

$$2\beta \alpha = \mu_0 \sigma \omega$$

Re arranging these to

$$-\beta^2 + \alpha^2=-\mu_{0}\epsilon_{0}\omega [\omega]$$

$$\frac{2\beta \alpha}{\mu_0 \sigma} = \omega$$

Substituting in $\omega$

$$-\beta^2 + \alpha^2=-\mu_{0}\epsilon_{0}\omega [\frac{2\beta \alpha}{\mu_0 \sigma}]$$

$$-\beta^2 + \alpha^2=- 2\beta \alpha \frac{\epsilon_{0}\omega}{\sigma}$$

When $\sigma >> \epsilon_0 \omega$ the right size is zero

Thus we find that in this limit

$$\alpha = \beta$$

Now, we want to find $$\frac{1}{\alpha}$$

Substituting this new relation in our second dispersion relation we find that

$$2\alpha^2 = \mu_0 \sigma \omega$$

Re arranging we get that

$$\frac{1}{\alpha} = \sqrt{\frac{2}{\mu_0 \sigma \omega}}$$

You have incorrectly stated the skin depth.

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