Because of the AC current flow, there is a changing magnetic field around the conductor, that in turn induces a changing electric field and vice versa

But does the EM radiation get "loose" from the conductor and propagate through air like with a dipole antenna for example or is it fixed around the conductor getting weaker with distance?


4 Answers 4


Since the summ of all currents in the distribution network is zero, then the magnetic fields caused by each conductor cancells out. The complete cancellation however occurs at infinite distance. But looking practically they cancel in close vicinity, some hundreds of meters from the powerline.

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Since the poweline is not an antenna it does not irradiate the EM power. Like you have mentioned a dipole, it's a setup with two wires going in opposite direction. It also has to be matched, for example dipole has to be half wavelength, you can do a small calculation how distant it would need to be.


To sumarize your question: The field is arround the conductor, only. It does not propagate through the air as antenna. If you look at Maxwell equations, then this is so called near field region.

  • \$\begingroup\$ If I've understood the scale correctly, the earth's magnetic field would be "25" on the vertical scale; so the effects of these fields at ground level are going to be tiny. \$\endgroup\$
    – pjc50
    Oct 11, 2018 at 11:25
  • \$\begingroup\$ So the electric and magnetic fields are fixed around the wire and cant propagate outwards because the powerline is not an antenna but a wire? \$\endgroup\$ Oct 11, 2018 at 11:41
  • \$\begingroup\$ @katrinsterner Yes. \$\endgroup\$ Oct 11, 2018 at 11:44
  • \$\begingroup\$ I'm not sure "the powerline is not an antenna" is quite accurate. Everything is an antenna, some things just aren't designed to be. Power lines will radiate a small amount of energy, but they're designed so that this radiated energy is as small as possible given the other constraints on the system. \$\endgroup\$
    – Hearth
    Oct 11, 2018 at 13:00

EM radiation will continue to propagate at the speed of light and weakening with the inverse square of distance from the lines, unless it is screened by something.


Suppose I'm using 10 amps average, or 100 amps peak, in an audio system. The 100 amps peak comes from the need to recharge the amplifier energy storage capacitors ---- perhaps 100,000 uF --- at the peaks of the 60Hz sine waves.

The electrician who installed the 10 amp wiring inside the wall of the house was running short of "RETURN" wiring, so he simply ran the "HOT" wire thru the wall behind the audio system, to the power-outlet on the wall, and then used a short piece of RETURN wire back to the circuit-breaker box.

Thus the magnetic field of the HOT wire is not cancelled by the magnetic field of the RETURN wire.

If your vinyl record RIAA preamplifier has a 4" by 4" loop in its STAR ground, and that STAR ground loop is located 8" from the HOT wire in the wall, how much non-hum evil-singing (because of the harmonic-content) is induced into the preamplifier's ground?

Use Vinduce == [MU0 * MUr * Area / (2 * PI * Distance)] / dI/dT

This simplified to 2e-7 * Area/Distance * dI/dT.

Let dI/dT = 100 amps/10 microseconds. What is Vinduce?

Vinduce = 2e-7 * 0.1meter * 0.1 meter/ 0.2 meter * 1e+7 amp/sec

Vinduce = 2e-7 * 0.05 * 1e+7 = 0.1 volt.

All because the fluxes did not cancel.


Key is to calculate Poynting vector S = E x H. It turns out, on a closed current loop from the energy source to the load, the potentials of the +- cables and directions of currents give electric and magnetic fields that give a pulsating Poynting vector, at double the AC frequency, pulsating everywhere between zero and a maximum value and always only pointing from source to load by net value. Directions of both fields are such that Poynting vector has no total average flux radiating outside from the circuit. The so-called quasi-static fields do extend outside, but with no Poynting vector outwards flux.


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