# 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 ?

Oct 25, 2015

$6.3 \cdot {10}^{19} {\text{e"^(-)"s}}^{- 1}$

#### 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 $\text{10 A}$ is equivalent to

$\text{10 A" = "10 coulomb"/"s}$

A coulomb is the equivalent of roughly $6.241 \cdot {10}^{18}$ 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 $\text{10 C}$ is passing through the cross-section per second, how many electrons would that be equivalent to?

Well, if $\text{1 C}$ is equivalent to $6.241 \cdot {10}^{18}$ electrons per second, $\text{10 C}$ 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 $1.6 \cdot {10}^{- 19} \text{C}$, which means that you would get $6.25 \cdot {10}^{18}$ 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.