Simple beginner question in Electric Circuits Course (Is the solution wrong?)

[s3a]

The reason why I am suspecting the solution is wrong is because the final step is dividing something with units of electrons/min by something with units of electrons/meter. Am I right about the solution being wrong? If I am wrong, then what am I not seeing? If I am right that the book is wrong, then what is the correct final answer and what must I correct in my book's solution?

The question is:
"An AWG#12 copper wire, a size in common use in residential wiring, contains approximately 2.77 x 10^23 free electrons per meter length, assuming one free conduction electron per atom. What percentage of these electrons will pass a fixed cross section if the conductor carries a constantcurrent of 25.0 A?"

The solution is:
"(25.0 C/s)/(1.602 x 10^(-19) C/electron) = 1.56 x 10^20 electron/s

(1.56 x 10^20 electron/s)(60s/min) = 9.36 x 10^21 electrons/min

(9.36 x 10^21)/(2.77 x 10^23)(100%) = 3.38%"

Any input would be greatly appreciated!
Thanks in advance!

 

[ragamuffin_8]

Formula relating charge to number of electrons:

q = Ne

q - charge
N - number of electrons
e - charge of 1 electron i.e 1.6x10^-19

Try using this to get the number of electrons in 25 A.

 

[s3a]

Unfortunately, I do not see how that relates. Could you please show me more?

 

[NascentOxygen]

The question is:
"An AWG#12 copper wire, a size in common use in residential wiring, contains approximately 2.77 x 10^23 free electrons per meter length, assuming one free conduction electron per atom. What percentage of these electrons will pass a fixed cross section if the conductor carries a constantcurrent of 25.0 A?"

The solution is:
"(25.0 C/s)/(1.602 x 10^(-19) C/electron) = 1.56 x 10^20 electron/s

(1.56 x 10^20 electron/s)(60s/min) = 9.36 x 10^21 electrons/min

(9.36 x 10^21)/(2.77 x 10^23)(100%) = 3.38%"

I see no reason for you to bring minutes into this. The equations are designed for time having units of seconds. Time not being stated, it might be safe to assume 1 second.

Pay close attention to units, and then there will be no conflict.

In one second, how many electrons pass by any point? 1.56x10^20 electrons.
In one metre of wire, there are 2.77x10^23 electrons.
So percentage = ...

It's even simpler than you thought.

Does that answer agree with the answer given in the textbook?

 

[s3a]

Actually, the solution I quoted was what the book said verbatim.

I get an answer of 3.38% metre/minute but the book gives a final answer of 3.38% (without the metre/minute part) so is it safe to assume that the question did not give enough information and that it should have given values of 1 metre and 1 minute such that I get an answer of 3.38% metre/minute * 1 minute/meter = 3.38%?

 

[NascentOxygen]

"An AWG#12 copper wire, a size in common use in residential wiring, contains approximately 2.77 x 10^23 free electrons per meter length, assuming one free conduction electron per atom. What percentage of these electrons will pass a fixed cross section

Reading this afresh, I think we are reading the words "these electrons" inappropriately. I followed your lead and took "these" to mean those in 1 metre of the conductor. I now think that is not what the examiner intended.

The problem would conform to the usual way this type of question is asked if "these" were to mean those in the 12 AWG copper wire, not specifically in just 1 metre of it. This means that time disappears from your answer. Consider any general length, l, and think of electrons flowing linearly along the wire at a velocity v, then rework it.