# EM waves in conductors

On a recent test in my E&M class, we derived what happens to an EM wave propogating in a conductor of conductivity $\sigma$, but I'm having trouble understanding the results.

We started from the wave equation:

$$\triangledown^2 \vec E - \mu\epsilon \frac{\partial^2 \vec E}{\partial t^2} = \frac{\triangledown \rho_{free}}{\epsilon} + \mu \frac{\partial \vec J}{\partial t}$$

Assuming $\rho_{free} = 0$ and $\vec J = \vec E \sigma$,

$$\triangledown^2 \vec E - \mu\epsilon \frac{\partial^2 \vec E}{\partial t^2} = \mu\sigma \frac{\partial \vec E}{\partial t}$$

Assuming a sinusoidal plane wave in the $z$ axis, $\vec E = \vec E_0 e^{j(kz-wt)}.$ (I don't know how common this is, but we're using $j=\sqrt{-1}$)

$$-k^2\vec E - (-\mu\epsilon\omega^2\vec E) = -j\mu\sigma\omega\vec E$$ $$k^2 = j\mu\sigma\omega + \mu\epsilon\omega^2$$ $$k^2 = \mu\omega(j\sigma + \epsilon\omega)$$

Assuming $\sigma \gg \omega\epsilon,$

$$k^2 \approx j\mu\omega\sigma$$ $$k = (1+j)\sqrt{\mu\omega\sigma/2}$$

Substituting this wavenumber back into our plane wave solution,

$$\vec E = \vec E_0 e^{j(zk-t\omega)}$$ $$\vec E = \vec E_0 e^{j(z[(1+j)\sqrt{\mu\omega\sigma/2}]-t\omega)}$$ $$\vec E = \vec E_0 e^{z(j-1)\sqrt{\mu\omega\sigma/2}-jt\omega}$$ $$\vec E = \vec E_0 e^{j(z\sqrt{\mu\omega\sigma/2}-t\omega) - z\sqrt{\mu\omega\sigma/2}}$$ $$\vec E = \vec E_0 e^{-z\sqrt{\mu\omega\sigma/2}}e^{j(z\sqrt{\mu\omega\sigma/2}-t\omega)}$$

Looking at this new middle factor, we see the amplitude degrades exponentially with z. If we apply this to a power line made of copper, we get $\mu \approx \mu_0 = 1.257 \mu H/m$, $\omega = (2\pi) (60 Hz) \approx 377 Hz,$ and $\sigma = 5.9 \centerdot 10^7 S/m.$ Putting that all together, we get $\sqrt{\mu\omega\sigma/2} = 118 m^{-1},$ or $\frac {1}{8.475 mm}$. Within just a few multiples of that, the amplitude will be almost nothing.

That being said, power lines are clearly capable of carrying a 60 Hz sinusoidal current for quite a bit further than 8 mm. What's up with that? What is the difference between carrying an AC current and carrying a 60 Hz plane wave? I've seen people talking about the "skin depth effect," but they're talking about "penetration," not carrying the current along the cable.

• The EM wave in a wire is not propagating inside the conductor but in the space around the conductor, and, yes, the skin depth of 60Hz is somewhere around 8mm, so making solid electrical power lines much thicker than 8mm would be a total waste of material. – CuriousOne Jun 1 '15 at 19:29
• @CuriousOne if it's propagating in the space around the conductor, then why do we need the conductor in the first place? – raptortech97 Jun 1 '15 at 19:32
• To keep the field propagating around the conductor instead of into all of space. The conductor, if you will, keeps the field in place by being a boundary condition of the wave equation. Take that boundary condition away and the field will dissipate quickly. Did you study Poynting vectors en.wikipedia.org/wiki/Poynting_vector, yet? – CuriousOne Jun 1 '15 at 19:35
• @CuriousOne We've talked about them, but not in detail, and only in the context of calculating intensity and power. – raptortech97 Jun 1 '15 at 19:40
• The Poynting vector is perpendicular to the electric and magnetic field. In a wire at low frequency you can treat both the electric and magnetic fields as quasi-static, so the electric field will be perpendicular to the wire and the magnetic field lines will be circular around it. The Poynting vector is therefor parallel to the wire, and that's the direction of the energy flow. – CuriousOne Jun 1 '15 at 19:52

Think about how an AC current might be induced in the cable. You subject it to an alternating electric field directed along the axis of the cable (because $\vec{J}=\sigma \vec{E}$), which must then be propagating in a direction perpendicular to the axis of the cable.
Note that the vector $\vec{E_0}$ in your example above must be perpendicular to the z-axis propagation direction.