# Question 2

1. Feb 4, 2005

### kyang002

A metallic wire has a diameter of 4.12 mm. When the current in the wire is 8 A, the drift velocity is 5.4*10^-5 m/s.

What is the density of free electrons in the metal?

I have considered a wire of cross-sectional area A, with current I. V is the drift velocity of the free charges in the conductor and n is the number of charged particles per unit volume. Then the current is given by I = nqva, where q is the magnitude of the charge on a charge carrier.

I = 8
n = density??
A = (2.06)^2 * 3.14
v = 5.4 x 10^-5
q = ??

I am unsure of how to finish the problem. We are looking for n. But what is q? Are all my other numbers correct?

2. Feb 4, 2005

### dextercioby

All u need to know (and apparently u don't) is that
$$j=n_{el}e v$$

Daniel.

3. Feb 4, 2005

### kyang002

I am unsure of what the equation you gave was. In my book, I was given I = nqva.

Care to explain on how to do the problem?

4. Feb 4, 2005

### dextercioby

Yes,the two formulas are obviously equivalent.It's better to use yours as it has already built in the numbers/variables u need to plug in.
What is "n" (or as i denoted it,n_{el.}) & what is its SI unit...?

Daniel.

5. Feb 4, 2005

### kyang002

The SI unit is m^-3.

How do I find q? Or do I even need q in this equation?