Electric field in a platinum wire

AI Thread Summary
To calculate the electric field in a platinum wire carrying 0.3 amperes of current, the relevant equations involve conductivity and the relationship between current, charge density, and electric field. The initial calculation yielded an incorrect value of 3e-5 due to unit conversion issues. By adjusting the units correctly, the accurate electric field strength was determined to be 0.03. The discussion highlights the importance of careful unit management in physics calculations. Ultimately, the correct application of formulas leads to the right answer.
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Homework Statement



Consider a platinum wire (σ = 1.0e+7) with a cross-sectional area of 1 mm^2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

Homework Equations



σ=qnu
I=qnAuE
Therefore I=σAE
where phi is the conductivity A the cross sectional area and E the electric field

The Attempt at a Solution


This seemed easy but i got an answer of 3e-5 and that is incorrect?
 
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Never mind i figured it out
 
How did you do it? I am where you were in the first post.
 
Hey I figured it out.
The only problem was dealing with units.
I just multiplied the 3e-5 by 1000 and got .03 as the right answer.
 
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