Question

A 5m length of conducting wire has overall resistance 2`Omega` with current flowing through it. It dissipates 8W of power along its length. Wire contains `8*10^22` free electrons. Calculate the average drift velocity of the electrons?

Charge on electron `1.6*10^-19` C

Answer 1

The drift velocity was `0.00078 m/s`

Explanation:

One of the power formulas is `P = I^2*R`, so `I = sqrt(P/R)`.

So in this case, the current is given by

`I = sqrt((8 W)/(2 Omega)) = 2 A`

Assume for my discussion that the 5 m length of wire is delivering those electrons to a light bulb. That means that each second there are 2 Coulombs of charge flowing out of the wire to the bulb. The question says that the charge per electron, `epsilon`, is

`(1.6*10^-19 C)/epsilon`. (Read `C/epsilon` as Coulombs per electron.)

I need to define a variable to be the number of electrons. Let it be `n_epsilon`. So we can determine how many electrons per second are passing to the bulb as follows:

`n_epsilon = 2 cancel(C) * (1 epsilon)/(1.6*10^-19 cancel(C)) = 1.25 * 10^19 epsilon`

The question says that there are `8*10^22` electrons in that 5 m length of wire. `1.25 * 10^19` of them leave the wire each second. The ratio of that number to the total in the 5 m length of wire is only

`(1.25 * 10^19)/(8*10^22) = 0.156*10^-3 = 0.000156`

Assuming the electrons that left the wire were the ones that had the shortest distance to drift since the end of the previous second, they traveled

`5 m * 0.000156 = 0.00078 m`

So their drift velocity was `(0.00078 m)/(1 s) = 0.00078 m/s`

Actually the term "drift" is descriptive. They wander, bouncing of some things in the structure of the wire. Some electrons get lucky and make comparatively good time. Others that should have easily left the wire in that second took a wrong turn and did not make it in this second. We call that value we just calculated "average drift velocity" because of that.

I hope this helps,
Steve