Current in a Wire: Density of Free Electrons

[cse63146]

[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$$v_{d}$$A

A = $$\pi r^2$$

The Attempt at a Solution

A = $$\pi r^2$$ = $$\pi (2.06x10^-3)^2$$ = 1.3 x 10^-5

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

n = $$\frac{8}{(1.6x10^-19)(5.4x10^5)(1.3 x 10^-5)}$$ = 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?

 

[Kurdt]

The question you've posted says express the answer in m^-3. Does the answer need to be in meters cubed or millimeters cubed?

 

[cse63146]

it's to the power of -3 so m^-3 is millimetres

 

[mukundpa]

it is not 10^-3 m it is 1/(m^3) where m^3 is the unit of volume (cubic meter)

 

[cse63146]

so the answer would be 1 / 6.94 x 10^28 ?

 

[Kurdt]

The variable n is a number density or the number of electrons per unit volume so it is already expressed in terms of m^-3

 

[cse63146]

Thank You.