How Fast Do Electrons Travel in a Copper Wire?

[fishturtle1]

Homework Statement

A 5.00-A current runs through a 12 gauge copper wire (diameter 2.05mm) and through a light bulb. Copper has 8.5 x 10²⁸ free electrons per cubic meter.

c) At what see does a typical electron pass by any given point in the wire?

Homework Equations

J = nqv_d

The Attempt at a Solution

The previous parts of the question I solved for current density.

J = 3.105 x 10⁶ A / m²

n = 8.5 x 10²⁸ electrons / m³

v_d = J / n|q|

I substituted all my values in:

v_d = (3.105 x 10⁶ ) / (8.5 x 10²⁸)(1.602 x 10^-28)

v_d = (3.105 x 10⁶) / 13.167 m/s

v_d = 2.28 x 10⁵ m/s

the correct answer is v_d = 0.111 mm/s

I'm sorry this is probably an arithmetic error but i can't see what I did wrong

 

[gneill]

Can you show us how you arrived at your current density?

 

[fishturtle1]

Can you show us how you arrived at your current density?

b) What is the current density?

cross section area = A = πr² = πd² / 4 = π(2.05x10⁶) / 4 = 1.61 x 10^-6

I = 5 A

Then current density = J = I / A = 5 / 1.61 x 10^-6 = 3.105 x 10⁶

..
Also the very first part:
a) How many electrons pass through the lightbulb each second?

I = 5A = 5 C/s

5 C/s * 1s = 5C

5C / (-1.602 x 10^-28) = -3.121 x 10²⁸ electrons

 

[gneill]

b) What is the current density?

cross section area = A = πr² = πd² / 4 = π(2.05x10⁶) / 4 = 1.61 x 10^-6

You didn't square your diameter when you calculated the cross sectional area.

a) How many electrons pass through the lightbulb each second?

I = 5A = 5 C/s

5 C/s * 1s = 5C

5C / (-1.602 x 10^-28) = -3.121 x 10²⁸ electrons

Use the magnitude of the charge on the electron. The sign of the charge is unimportant if you're counting quantity, which should be a positive number.

 

[fishturtle1]

You didn't square your diameter when you calculated the cross sectional area.Use the magnitude of the charge on the electron. The sign of the charge is unimportant if you're counting quantity, which should be a positive number.

Ok I see what you mean .. so the correct cross sectional area is:

A = π((2.05)² x 10^-3*2) / 4 = 3.30 x 10^-6
J = 5 / 3.30 x 10^-6 = 1.51 x 10⁶ A / m²

So I know my J and |q| are correct..I also forgot to mention that the problem gave "Copper has 8.5 x 10²⁸ free electrons per cubic meter" which I use as my n value.

So v_d = J / n|q|.
J = 1.51 x 10⁶
n = 8.5 x 10²⁸ free electron / m³
|q| = 1.602 x 10^-28 Cv_d = (1.51 x 10⁶) / (8.5)(1.602) = (1.51 x 10⁶) / 13.617 = 110,890 m/s

never mind I just realized |q| = 1.602 x 10^-19.. so 110,890 * 10^-9 = .111 x 10^-3 which is the answer in the book.

Thanks for helping me on this.