# Quick Help!

1. Feb 7, 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 = ??

Can someone help me out on the answer for this problem? Thanks!

2. Feb 7, 2005

### dextercioby

Pay attention with the units... :grumpy:
HINT:
$$|\vec{j}|=\frac{I}{S}= n |q_{el.}| |\vec{v}_{drift}|$$

Daniel.