# How Fast Do Electrons Travel in a Copper Wire?

• fishturtle1
In summary, a current of 1.51 x 106 A flows through a lightbulb with a cross sectional area of 3.30 x 10-6 cm2.
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?

J = nqvd

## 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

vd = J / n|q|

I substituted all my values in:

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

vd = (3.105 x 106) / 13.167 m/s

vd = 2.28 x 105 m/s

the correct answer is vd = 0.111 mm/s

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

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

gneill said:
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

fishturtle1 said:
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.

fishturtle1 said:
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.

gneill said:
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 vd = J / n|q|.
J = 1.51 x 106
n = 8.5 x 1028 free electron / m3
|q| = 1.602 x 10-28 Cvd = (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.

gneill

## 1. What is drift velocity?

Drift velocity is the average velocity of charged particles, such as electrons, when they move through a material in response to an electric field.

## 2. How is drift velocity calculated?

Drift velocity is calculated by dividing the current density by the charge density, multiplied by the material's conductivity.

## 3. What factors affect drift velocity?

Drift velocity is affected by the strength of the electric field, the material's conductivity, and the charge and mass of the particles.

## 4. How does drift velocity relate to electrical conductivity?

Drift velocity is directly proportional to electrical conductivity. As conductivity increases, so does drift velocity.

## 5. Can drift velocity be measured experimentally?

Yes, drift velocity can be measured experimentally using methods such as the Hall effect, where the induced voltage in a conductor is measured to calculate drift velocity.

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