As an electrical engineer I know that any AC transmission line can induce an AC voltage over a long cable located in a close proximity to the transmission line. I am interested to know how to calculate all parts of device capable of producing 12V to 110VAC with a current of about 1,000MA to 5Amp from a transmission line of 110KV to 220KV at the distance of 50m to 100m. The most important question is the length of the device and its approximate cost. thank you in advance

  • \$\begingroup\$ Is that 1,000 Megamps, or milliAmps? \$\endgroup\$ – John D Apr 25 '14 at 2:55
  • 4
    \$\begingroup\$ I would seriously doubt that enough electromagnetic field exists at 50-100m to generate that much current and voltage. The multiple conductors of the power lines would be effectively very close and cancelling out much of each other fields. \$\endgroup\$ – Joe Apr 25 '14 at 2:58
  • 3
    \$\begingroup\$ This sounds like one of those schemes to steal power from high voltage transmission lines. \$\endgroup\$ – Samuel Apr 25 '14 at 3:17
  • 2
    \$\begingroup\$ The voltage on the transmission line (110 kV etc.) is irrelevant - it's the current in the T-line that induces voltage in another wire. \$\endgroup\$ – Andy aka Apr 25 '14 at 11:22

The magnetic flux density at a distance of 50 meters from a 3 phase transmission line with conductors, say, 10 meters apart, carrying 100 amps is very small, on the order of 1 microtesla. At that distance you need a truly huge loop to extract any appreciable energy from inductive coupling.

I read a paper where in a similar situation to the above they were able to extracy maybe 10 watts of power using a loop of 12 AWG wire nearly a kilometer in total length...so 5 amps at 12 volts might be possible, but even with several turns the optimal size of this secondary will be truly enormous.

Just buy a solar panel.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.