# Skin depth challenges?

1. Jul 17, 2008

### deep_tought_e

hi all
i run in to some contradictions while studing electromagnetics.
you know skin depth decreases when frequency increases.It ans that waves of higher frequencies have lower skin depth.
ok
and every one of us knows that X-ray have higher frequencies than other waves for example radio frequencies.
so why X-ray used in hospitals to take photoes from insde of body?
i meant according to skin depth formula X-ray have to have less skin depth in comparison with radio frequencies
why radio frequencies can not be used?

2. Jul 17, 2008

### Staff: Mentor

Re: skin depth challenges????????????

Skin depth in the context of E&M refers to how deep the EM wave penetrates into conductive materials, not actual human skin. X-rays do not pass through conductors.

3. Jul 17, 2008

### ZapperZ

Staff Emeritus
Re: skin depth challenges????????????

The "skin" in skin depth has nothing to do with human skin!

The skin depth is often associated with EM wave penetration on conductors, which has conduction electrons. The penetration of EM wave in insulators (such as human skin, etc), are governed by different mechanisms (read the FAQ in this sub-forum as your starting point).

Zz.

4. Jul 17, 2008

### deep_tought_e

Re: skin depth challenges????????????

hey you are really expert
thanks to answer me in such a short time

5. Jul 19, 2008

### deep_tought_e

Re: skin depth challenges????????????

hi again
i tought about that last day.so i did not gain any result that solve my problem.
when we solve maxwell's equation we do not pay any attention to this subject that we are solving that equation for what kind of material.
for example we solve maxwell's equation for generall lossy medium.and maybe our material for example human body be some material that have some conductivities.so this sentence that skin depth defined for good materials is not true
thanks any way

6. Jul 19, 2008

### ZapperZ

Staff Emeritus
Re: skin depth challenges????????????

Then your solution is incorrect for real-world applications. You ALWAYS have to solve for the boundary conditions correctly based on the nature of your boundary condition.

Now, it is another matter if you are simplifying it to be able to actually find a "good enough" answer. When we solve for the field geometry in waveguides, we always have to make a simplifying assumptions for the nature of the wall of the cavities. If not, the problem is intractable. However, you simply can't use the same boundary condition if you instead have a dielectric structure (as what I am using). That will be a totally different boundary condition and you can't simply use the result from a metallic structure.

So yes, to some extent, you DO have to care about the nature of the material.

Zz.

7. Jul 20, 2008

### uart

That’s actually a very good question DT and I’m not sure that the responses so far have really addressed exactly what you’re asking. So let me try to. First off though I must point out that there is more than one physical reason for the observation you are making. I’ll just deal with the relatively easy case first and provide a link at the end to a second and more complicated phenomenon that also relates to your question.

Let’s start by briefly outlining the theory behind EM attenuation in a conductive medium. As you know the EM wave can be described as the real part of a complex exponential where the x variation is of the form $e^{-\gamma x}$, where gamma is called the propagation constant. If gamma is purely imaginary then we get only cosine and sine terms hence no attenuation. If gamma has a real component then this gives rise to exponential attenuation with distance. Typically we write $\gamma = \alpha + i \beta$ where the real component (alpha) is called the attenuation constant.

The propagation function $\gamma[/tex] of an EM wave is given by, $$\gamma^2 = (\sigma+i \omega \epsilon) (i \omega \mu)$$ For a non conductor (that is a good dielectric) [itex]\sigma = 0$ and so $\gamma = i \omega \sqrt{\mu \epsilon}$, a completely imaginary constant which corresponds to lossless propagation.

Now for a conductor on the other hand we may have $\sigma >> \omega \epsilon$ in which case $\gamma^2 \simeq i \sigma \omega \mu$ resulting in a complex propagation constant and therefore attenuation.

In particular, $\gamma \simeq \frac{1+i}{ \sqrt{2} } \sqrt{\sigma \omega \mu}$ which has a real component (attenuation constant) of $\alpha = \sqrt{\frac{\sigma \omega \mu}{2}}$.

That is, as the wave travels it's amplitude is attenuated by a factor of $e^{-\alpha x}$, so the wave is attenuated by one Neper for each x corresponding to $\sqrt{\frac{\sigma \omega \mu}{2}} x = 1$, giving a skin_depth of $\delta = \sqrt{\frac {2}{\sigma \omega \mu}}$

Clearly as you have noted, the attenuation constant increases with frequency and therefore the skin_depth decreases with frequency.

Now for your observation that this rule does not seem to apply at very high frequencies where the penetration depth of EM waves actually does the opposite and starts to increase! Lets look at why. Note that in derivation of skin depth we used the approximation that $\sigma >> \omega \epsilon$ so that we could make the simplification that $\sigma + i \omega \epsilon$ is approx equal to $\sigma$ when finding the propagation constant. But that inequality surely cannot hold regardless of frequency, after all the RHS of the inequality is proportional to frequency!. So what may happen is that a material may be a “good conductor” ($\sigma >> \omega \epsilon$) at low frequencies but then become a good dielectric ($\sigma << \omega \epsilon$ ) at much higher frequencies. Really it all makes perfect sense when you think about it.

Sea water is actually a good example of this phenomenon. At radio frequencies sea water is a terribly lossy medium and it gets worse with higher frequencies. Communication with a submarine for example is possible only with extremely low frequency EM waves, anything in the “normal” AM/FM radio spectrum and you can just forget about it. However if you increase the frequency high enough then the EM wave starts to penetrate once more. This is clearly demonstrated by the fact that visible light can easily penetrate clear sea water!

The bottom line is that the observations and insights you enquired about are absolutely correct. Well done. :)

8. Jul 20, 2008

### nucleus

Maybe an example of the uses for Eddy Currents can help the OP. They are used for several things such as NDT.
http://asmcommunity.asminternational.org/static/Static%20Files/IP/Magazine/AMP/V165/I12/amp16512p033.pdf?authtoken=87957e332c633e73e9078f937f093ee01e3989a7 [Broken]

Last edited by a moderator: May 3, 2017
9. Jul 20, 2008

### uart

As for the second phenomenon that I mentioned above. Conduction losses are not the only reason why EM waves can be attenuated in a medium. Dielectric losses also cause attenuation and these are normally modelled by having a complex dielectric constant, $\epsilon = \epsilon^{'} + i \epsilon^{''}$, where the imaginary component of epsilon gives rise to dielectric loss and attenuation (like with conductivity but of a different physical origin). The real and imaginary components of the complex dielectric constant (aka "dielectric function") are aften complicated functions of frequency, having resonance peaks and so on, so that some frequencies are propagated well and others are highly attenuated. You can read more about complex dielectric constants here : http://en.wikipedia.org/wiki/Permittivity

Last edited: Jul 20, 2008