# Physics 11 Electricity Problem

1. Feb 27, 2005

### eno

Copper has approximately 10^29 free electrons per cubic meter. What is the approximate average velocity of electons in a 1.0 mm radius wire carrying 1.0A?

resisivity, p of copper = 1.68 * 10^-8 ohms meter

The answer in the book apparently is 2 * 10^-5 m/s.

2. Feb 27, 2005

### vincentchan

3. Feb 27, 2005

### eno

i have tried it many times and it did not reach tat ans.

4. Feb 28, 2005

### vincentchan

the current $I$ is equals to free electrons density $\rho$ times charge per electrons $e$ times drift velocity $v$ times cross section area $A$
$$I=\rho e v A$$
ring the bell?
since you have everything except v, this is one equation with one unknown... solving it is straight forward....
you must be careful on the unit conversion, it is a little bit tricky....
the area is not given directly, but you have radii, the area shold be piece of cake
charge of electron is on the back of your textbook.... if not, look it up on internet

5. Feb 28, 2005

### eno

Within 1 sec, e drifted for length L

L = v * 1 sec = v

total e drifted within 1 sec is 1 Ampere.

1.0 A = 6.24*10^18 e = 10^29 * v * (π * (0.001)^2)

v = 2 * 10^-6 m/s

therefore, does not match with the ans in the book.

6. Feb 28, 2005

### vincentchan

what the hack are you doing here?

7. Feb 28, 2005

### vincentchan

$$I=\rho e v A$$
$$\rho$$ is the free electronss density
$$e$$ is the charge per electron....
$$A$$ is the cross section area
$$I$$ is the current
plug in all number and solve for v....

8. Mar 1, 2005

### eno

it appears that I have punched in the wrong notation the whole time. ths for ur help.

9. Mar 1, 2005

### Mo

I dont know why, but i just cant seem to get the answer :(

the formula i use is

I = NaQv

so V = I / NAQ

A (cross sectional area) = (0.001)^2 multiplied by Pi
Q = 1.6 * 10^-19
N= 10^29
I = 1

is this right so far?

Regards,
Mo

10. Mar 1, 2005

### dextercioby

Yes,it's identical to what the OP did.You should get the same answer.$$2\times 10^{-5}m \ s^{-1}$$

Daniel.