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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Jun 1, 2015 at 19:29
  • $\begingroup$ @CuriousOne if it's propagating in the space around the conductor, then why do we need the conductor in the first place? $\endgroup$ Jun 1, 2015 at 19:32
  • $\begingroup$ 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? $\endgroup$
    – CuriousOne
    Jun 1, 2015 at 19:35
  • $\begingroup$ @CuriousOne We've talked about them, but not in detail, and only in the context of calculating intensity and power. $\endgroup$ Jun 1, 2015 at 19:40
  • $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Jun 1, 2015 at 19:52

2 Answers 2


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.

If the radius of curvature of the wire were large enough, then we can treat this like EM waves propagating into a plane conductor interface and the skin depth you calculated applies.

Thus the oscillating E-field exists outside the wire and then decays rapidly with skin depth as you go into the cable.


@raportech97....What you have effectively calculated is "skin depth", the impedance along the length of the cable is another matter. The impedance will be dominated by the line spacing, I expect in the order of metres. The loss in the cable will be dominated by series resistance. Your skin depth is handy here, to calculate the "effective cross sectional area" of the conductor, then you use the resistivity of the copper per unit length for that area, you will find that 60 HZ currents will flow with acceptable loss for hundreds of kilometres. The EM fields do not take any significant part in the transmission. the wavelength at 60Hz is 5*10^6 metres. the entire system is incapable of any significant radiation into the far field. However there will be significant energy in the extreme near field usually referred to as the induction field. These fields will be reactive and non radiating and will in part determine the characteristic impedance of the effective transmission line the approximate impedance will be in the order of 600 Ohm's and cannot be used effectively as a true transmission line because of the gross mismatches at generators and substations (in the order of milliohm's). The conclusion is the fields and impedance's play no significant part in the efficiency of the system. Skin depth can be addressed by spacing several 10 mm (or greater) diameter wires hundreds of millimetres apart thereby reducing the inductance and field intensity around the conductors. this reduces the impedance and more importantly increases the effective skin depth, thereby increasing the effective cross sectional area of the conductors.


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