Solving Skin Depth for Copper at 1.0 GHz

[purejoker]

hi there, I am stuck on this question can some give me a hint on how to solve the question:

the question is:

Calculate the skin depth for copper at 1.0 GHz . σ = 5.8x10^7 S/m, εr=1, μr=1.

the formula i think that i need to use is

tan δ = σ / (ω)(ε0)(εr)

ω = 2nf = 2 x n x 1 x 10^9
ε0 = 8.85 x 10^-12
er = 1
where do i put μr?


im missing a another step but i don't no what, can some give me a hint please?


the answer to the queation is
tan δ >>1, d = 2.1um

 

[physicsboy]

You've got to understand what skin depth is and you'll get this question. When you solve Maxwell's equations in a conducting material, you get a complex wavenumber (i.e. a complex dielectric constant) and that imaginary part of the wavenumber leads to a decay of the wave in the conductor. Remember from electrostatics that E-fields should be zero inside a conductor? Well, for time-dependent fields in "good conductors" the E-fields damp out pretty darn quickly. Copper is such a conductor. Start with Maxwells equations in vacuum (your dielectric constant and mag. permeability are identically 1).

$$\nabla \times \vec{B} = \frac{4 \pi}{c} \vec{J} + \frac{1}{c} \partial_t \vec{E} \hspace{1cm} \nabla \times \vec{E} = -\frac{1}{c} \partial_t \vec{B}$$

Assume your E,B fields have time dependence of the form
$$e^{-i \omega t}$$

And that for the conductor the current density is proportional to the E-field
$$\vec{J}=\sigma \vec{E}$$

Take the curl of both equations and combine to get a modified wave equation for E. It looks like
$$\nabla^2 \vec{E} + k^2 \vec{E} = 0$$

Where our k is
$$k=\frac{w}{c}\sqrt{1+\frac{i4 \pi \sigma}{\omega}}$$

For "good conductors" (like copper) we have $$\sigma \gg \omega$$ This amounts to killing the displacement current term in the first place. Or here our k becomes

$$k = \frac{\omega}{c}\sqrt{i4 \pi \frac{\sigma}{\omega}} = \frac{1}{c}\sqrt{i4 \pi \sigma \omega} = \frac{(1+i)}{c}\sqrt{4 \pi \sigma \omega}$$

Be sure of how your text/course defines the skin depth. Sometimes it's 1/Im(k) sometimes its 1/[2Im(k)] where Im(k) is the imaginary part of k. Be careful here - I've used Gaussian units throughout so just adapt the technique to your units which are probably SI.