[Craig Sawyers]

Suppose we take a sine wave current. The electrons move in one direction during the first half cycle. The area under the first current half cycle of a sine wave is the charge that has been moved in that time

Q = I(root 2)/(pi x f)

The root 2 gets from RMS current to peak, which is what you need for the equation.

Lets take 10A flowing at 50Hz. The charge transferred in the first half cycle is 9 x 10^-2 C from the equation. The charge on the electron is 1.6 x 10^-19 C and so the number of electrons for this charge is 9 x 10-2/1.6 x 10^-19 = 5.6 x 10^17.

That sounds like a lot, but there are 8.5 x 10^22 electrons per cc in a copper conductor!

Lets take a 1mm^2 cross section wire (that would be under rated for 10A, but this is a though experiment which will scale for different values). 1mm^2 = 0.01cm^2.

So there are 8.5 x 10^22 x 0.01 = 8.5 x 10^20 conduction electrons per cm of 1mm^2 section wire.

So in the first half cycle of the mains waveform, the electrons move 5.6 x 10^17 / 8.5 x 10^20 = 6.6 x 10^-4 cm = 66 microns.

Then in the next half cycle they move back again. So the electrons shuffle back and forth by 66um peak to peak - or about the thickness of a human hair.

Now let's scale it. Let's take a typical audio mains cable of 12AWG, which is 3.31 mm^2. And the current might typically be 1A for a really chunky class A amplifier (so 240W of standing dissipation).

That means that the peak to peak electron movement in the wire is 1/10 x 1/3.31 x 66 = 2 microns. So about four times the wavelength of green light.

For lower currents of more typical power amps, and certainly preamps the current will be <<1A, so the amplitude of electron motion is down in the wavelength of light territory.

Of course the same calculation holds for signal cables and loudspeaker cables.

SO - if for the moment one accepts that there is an audible difference between cables, the mechanism is definitely not related to the tiny AC amplitude of electron motion.

Craig