There are 8.7 + 10^18 current-carrying electrons which are effectively drifting along at .065 m/s in a uniform conducting wire of length 4.85 m. What is the current in electrons/sec, and what is the current in amps?
Since electrons drift at .065 m/s, in 1 second all the electrons in .065 meters of the wire will pass a given point.
The distance moved in a second is .065 meters, which is is .065 / 4.85 = .0134 of the length of the wire.
This increment therefore contains .0134 * 8.7 * 10^18 = 116.5 * 10^15 electrons.
So in a second, the rate at which electrons pass the point is 116.5 * 10^15 electrons.
An electron carries a charge of approximately 1.6 * 10^-19 Coulombs, so the charge carried in a second is
If there are nElectrons current-carrying electrons in a length L of wire, and if they drift a distance `dL in one second, then in each second the proportion `dL / L of the nElectrons electrons, or (nElectrons * `dl / L) electrons will pass a given point. The electrons each carry charge 1.6 * 10^-19 Coulombs, so the current is
1.6 * 10 ^ -19 Coulombs / electron * (nElectrons * `dL / L) electrons / sec = 1.6 * 10^-19 * nElectrons * `dL / L amps.
The figure below shows a length L of wire containing nElectrons current-carrying electrons, and a segment of length `dL through which electrons travel in a second.
The resulting current in electrons/second is easily converted to Coulombs/second.
"