What is the current in the wire?

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The electric field in a 1.2mm x 1.2mm aluminum wire is given as 2.0×10^-2 V/m, and the initial calculation for current resulted in 1.087 A. However, the user later discovered an error in the resistivity value used, which should be 2.82×10^-8 ohm-m instead of 2.65×10^-8 ohm-m. A different method for calculating current yielded a result of 1.80679 A. Ultimately, the corrected answer for the current in the wire is 1.02 A. The discussion highlights the importance of using accurate values for resistivity in electrical calculations.
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Homework Statement



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

Homework Equations



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

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!
 
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Looks good to me.
 
TSny said:
Looks good to me.

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

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