# Charge carrier density i=navq

1. Mar 6, 2005

### Mo

An end of chapter question that i dont really know how to do..

Question:
A copper connecting wire has a cross-sectional area of 3mm^2 and is 30cm long. (A) Estimate the amount of free charge in this wire. (B) How long will it take for all this free charge to pass through the wire when a current of 3 A flows? (C) Hence find the average speed at which this free charge moves.

Well i know the formula i have to use is; I=naqv where

I = Current
N = Charge carrier density
Q = Charge
V = Drift velocity.

It tells me in the chapter that the atoms of copper have a diameter of 0.25 nm

I need to work out the amount of atoms in a volume of (3mm^2 x 30cm) which equals: 3mm^2 x 300mm = 900mm^3 . = 0.9 m

so now to work out the number of atoms its 0.9 / .25nm = 3.6 x 10^9
then cube that = 4.6656 x 10^28 atoms per 0.9m cubed

actually i dont think ill go further just yet, but are things looking good upto this point??

Help REALLY! appreciated

Regards,
Mo

2. Mar 6, 2005

### dextercioby

The volume of the wire is indeed $900mm^{3}$.However,you've mistaken the conversion into cubic meters...
$$1mm^{3}=(1mm)^{3}=(10^{-3}m)^{3}=10^{-9}m^{3}$$

Daniel.

3. Mar 6, 2005

### Mo

Thanks for the reply.

So 900 mm cubed is actually 0.729m cubed.Is this correct?

4. Mar 6, 2005

### dextercioby

Nope.I told you:
1mm cubed------------------------>10^{-9}m cubed
900mm cubed---------------------->"x" m cubed...

Find "x"...

Daniel.

5. Mar 6, 2005

### Mo

oh lol.ok. 9 x 10^-7

Last edited: Mar 6, 2005
6. Mar 6, 2005

### dextercioby

Perfect,now move on to the next step:the # of free charges (electrons) in the copper wire.

Daniel.

7. Mar 6, 2005

### Mo

Ok thanks. But how am i going to do that now then?

I have a volume.

i know the diameter of 1 copper atom. do i need to find the volume of one copper atom, and then divide the total volume by the volume of 1 copper atom, to find out how many copper atoms i have?

after that (well in the book anway) it says "assume 1 free electron per atom".

??

8. Mar 6, 2005

### dextercioby

You needn't know the diameter of the copper atom...Just the density.

1atom Cu------------------------->64*1.66*10^{-27}kg
"y" atoms Cu----------------------> V_{wire} ([m^{3}])*rho_{wire}([kg/m^{3}])

Find "y"...Then the # of free electrons.

Daniel.

9. Mar 6, 2005

### Mo

Thanks for the reply.I dont mean to sound rude, however i do not think this is what they expect me to do.

Im sure its also correct, however the method in the book (and the one i need to know for the exams) achieves this without needing to know the density.

In the example in the workbook they find the charge carrier density for 1 m.They do so why dividing this 1 m by 0.25 nm and then cubing that, then they assume 1 electron per atom.

I think it would be best if i asked my teacher cause he knows what method the exam board wants.

Thanks for all your help anyway, and sorry for wasting your time!

Regards,
Mo

10. Mar 6, 2005

### dextercioby

I don't know.It looks much more simple & fair through my method.And besides,the density of copper is much more familiar than the average atomic radius...

Daniel.