# What is the current in the wire?

1. Nov 12, 2012

### Bigworldjust

1. The problem statement, all variables and given/known data

The electric field in a 1.2mm x 1.2mm square aluminum wire is 2.0×10^-2 V/m What is the current in the wire? I = ?A

2. Relevant equations

E = V/length
R = ρL / A
I = V/R = V*A/ ρ*L

3. The attempt at a solution

Electrical field strength E = V / length = 2 x 10^-2 V/m
Resistance of wire R = ρL / A; ρ for aluminium = 2.65 x 10^-8 ohm-m
A = 1.2*1.2 = 1.44 mm^2 = 1.44 x 10^-6 m^2
Current I = V / R = V*A / ρ*L = (V/L)*(A/ ρ) = 2 x 10^-2 (1.44 x 10^-6 / 2.65 x 10^-8 ) = 1.087 A

Alright that was my final answer, but I am just asking here to double check if it is right or not because I only have one more try on this question, and I don't want to get it wrong, lol. Did I make any mistakes? Thank you!

2. Nov 12, 2012

### TSny

Looks good to me.

3. Nov 12, 2012

### Bigworldjust

Damn, it was wrong, lmao. I still have one more try though luckily. You think there is somewhere I went wrong after rechecking it?

4. Nov 12, 2012

### liblenovo

I check it by another method. Your answer is right. My method is:

I=j*S=(sigma*E)*A=(1/ρ)*E*A=1.80679 A

Where you get the value of ρ? In Wikipedia, it is 2.82×10^−8

5. Nov 12, 2012

### Bigworldjust

Ah, that was my error. So the answer was 1.02. Thanks for the help!

Last edited: Nov 12, 2012
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