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I know that electric current is the flow of electrons but electrons have a very slow drift speed (about 2mm/s). How is it that electric current reaches its destination at almost the speed of light?

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The speed at which the signal in a circuit travels is not the speed of drifting free electrons but rather the speed of electromagnetic waves whose velocity factor is usually $0.5$ to $0.99$(50%-99% $c$).

Therefore, the speed of electric signal is the speed of the propagation of the electric field waves of the conductor but not the drift speed of the free electrons drifting through the conductor.

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  • $\begingroup$ And to clarify further, the signal moves continuously ahead, while the electrons in the wire oscillate back and forth (for AC power system.) For an AC dynamo connected to a distant incandescent bulb, the current is a back-and-forth motion over a small distance, with quite low peak velocity, while the signal/energy is an EM wave, and travels at a fraction of c. $\endgroup$
    – wbeaty
    Apr 4 '16 at 7:49
  • $\begingroup$ An easier to understand analogy is that the waves in the sea move along the surface a lot faster that the water does. $\endgroup$
    – Tricky
    Sep 25 '17 at 7:34
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Drift speed is the average speed of the average electron. Electric current reaches the destination because the wires that carry the electrons are already filled with loose electrons (because the wires are metals) so the electrons do not actually travel from source to destination but rather provide a push to the ones at the end of the wire so electrical transmission is instantaneous.

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There are electrons already present in the conductive material. What is being transferred from one electron to another is a virtual photon, the force carrier of the electromagnetic force: the push in particle form. The virtual photon travels at the speed of light, which therefore makes the electric current travel close to/at the speed of light.

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