I was well into my electrical-engineering career years ago when I first learned that electricity travels far slower than I would ever have imagined. I was as surprised as anyone, but I did not doubt that it was a fact. Ever since, I have wanted to calculate for myself the speed of electrons in a wire, particularly for the AC case, but never got around to it--until now! I am posting the result here for the "electric train" fans who may find this sort of thing interesting.
When we flip on a light switch, the light comes on so quickly that we may imagine the electrons in the circuit traveling between the switch and the light in a fraction of a second--but in fact the electrons hardly move at all! For example, a DC current of 1 ampere takes nearly half a minute to move the tiny distance of 1 millimeter in 14-AWG copper wire. In an AC circuit, the electrons travel even less, merely vibrating in place over a distance of less than a thousandth of a millimeter, and never make any forward progress.
It seems like the electrons are moving fast, because they are already in place throughout the wires before the switch closes the circuit. They do not need to travel from the switch to the light; instead, they simply push each other through the wire, and the electrons at one end move immediately when the electrical voltage pushes on the electrons at the other end.
A bicycle chain works the same way. When the pedals begin to turn the chainwheel at the front, there is no delay for the sprocket at the back to begin turning and no wait for the chain to travel the foot or so from one point tothe other. The chain is already in place and begins to move all at once, no matter how fast or slow the rider begins to pedal.
Let us call the speed of any particular electron past any particular place on the wire, S. In one second, a cylindrical volume of electrons with length S and with a cross-sectional area a passes the point of interest on the wire.
So the rate of the volume of electrons flowing in the wire is
S * a
This has the units of cubic meters per second. I will assume that the wire is 14-AWG, which has an area a of 2.081e-6 m*m. We can convert the units to electrons per second by multiplying by the conduction-electron density in the wire, which I assume is made of copper:
S * a * rho
where rho = 8.491e28 /(m*m*m). Then we can convert electrons per second to coulombs per second by multiplying by the charge q of a single electron, which is -1.602e-19 C:
S * a * rho * q
Coulombs per second is amperes, so we now have current as a function of speed:
I = S * a * rho * q
Solving for S:
S = I / a / rho / q = I * -35.33 um/C
Where u substitutes for the lower-case Greek letter mu. The minus sign reflects the negative charge of the electron and tells us that the electrons are moving backwards, in the direction opposite to whatever direction we assumed for S.
For a DC circuit, the current I is constant, so the speed S is also constant at -35.33 um/s for a current of 1 ampere.
In an AC circuit, the current varies sinusoidally as a function of time, and alternately in opposite directions. Let the instantaneous current be
I(t) = Irms * sqr(2) * sin(120 * pi * t)
where Irms is the root-mean-square current, that is, the square-root of the average of the square of the instantaneous current. The distance D that the electrons move is the definite integral of their speed over a complete positive half-sinewave, like the interval from t = 0 to s/120:
D = integral(S * dt, 0, s/120)
= Irms * sqr(2) / a / rho / q * integral(sin(120 * pi * t) * dt, 0, s/120)
= Irms * sqr(2) / a / rho / q / 60 / pi = -265 nm