What is the current in the wire?

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Homework Help Overview

The problem involves calculating the current in a square aluminum wire given the electric field strength and dimensions of the wire. The subject area pertains to electricity and circuits, specifically focusing on the relationships between electric field, voltage, resistance, and current.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply formulas related to electric fields and resistance to find the current. Some participants confirm the calculations, while others question the accuracy of the resistivity value used and suggest alternative methods for verification.

Discussion Status

Contextual Notes

Participants discuss the resistivity value, noting discrepancies in sources, which may affect the calculations. The original poster expresses concern about having limited attempts to submit the answer correctly.

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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!
 
Last edited:

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