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 ?

1 Answer
Oct 25, 2015

Answer:

#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.