Question

A copper wire of cross-sectional area 2.0 `mm^2` carries a current of 10 A. How many electrons pass through a given cross-section of the wire in one second ?

Answer 1

`6.3 * 10^19"e"^(-)"s"^(-1)`

Explanation:

This is a classic example of a trick question. Sort of.

The last thing you need to worry about is the cross-section of the wire. Here's why.

What is an ampere?

An ampere is the equivalent of one coulomb per second. In your case, a current of `"10 A"` is equivalent to

`"10 A" = "10 coulomb"/"s"`

How about a coulomb?

A coulomb is the equivalent of roughly `6.241 * 10^(18)` elementary charges, or

`"1 C" = 6.241 * 10^(18) xx underbrace(1.60217662 * 10^(-19)"C")_(color(blue)("elementary charge"))`

So if a total charge of `"10 C"` is passing through the cross-section per second, how many electrons would that be equivalent to?

Well, if `"1 C"` is equivalent to `6.241 * 10^(18)` electrons per second, `"10 C"` will be equivalent to

`10 color(red)(cancel(color(black)("C"))) xx (6.241 * 10^(18)e^(-)"s"^(-1))/(1color(red)(cancel(color(black)("C")))) = 6.241 * 10^(18)"e"^(-)"s"^(-1)`

Now, the elemental charge is often given as `1.6 * 10^(-19)"C"`, which means that you would get `6.25 * 10^(18)` electrons in one coulomb.

In that case, the answer will indeed be

`10 color(red)(cancel(color(black)("C"))) xx (6.25 * 10^(18)e^(-)"s"^(-1))/(1color(red)(cancel(color(black)("C")))) = 6.3 * 10^(18)"e"^(-)"s"^(-1)`

Rounded to two sig figs, of course.

So remember, think about the basic concepts and don't get distracted by "additional information", which can sometimes be misleading.