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**[SOLVED] Current in a Wire**

## Homework Statement

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 = 8

n = density??

A = (2.06)^2 * 3.14

v = 5.4 x 10^-5

q = 1.6x10^-19

Express your answer numerically in m^-3 to two significant figures.

## Homework Equations

I = ne[tex]v_{d}[/tex]A

A = [tex]\pi r^2[/tex]

## The Attempt at a Solution

A = [tex]\pi r^2[/tex] = [tex]\pi (2.06x10^-3)^2[/tex] = 1.3 x 10^-5

So manipulating the equation, I get n = I / evA

n = [tex]\frac{8}{(1.6x10^-19)(5.4x10^5)(1.3 x 10^-5)}[/tex] = 6.94 x 10^28

and since I need it in millimeters, I multiply it by 1000 (since I did the calculations in SI) and get 6.94 x 10 ^31, but it says it's wrong. Any ideas where I went wrong?

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