# Current through two different types of wire of different diameter

• RichardEpic

## Homework Statement

For what electric field strength would the current in a 2.0 mm diameter nichrome wire be the same as the current in a 1.0 mm diameter aluminum wire in which the electric field strength is 0.0080 V/m?

aluminum diameter = 1.0 mm
Same current(I) flowing through both wires

## Homework Equations

J = sigma*E... I/A = sigma*E
so...I = sigma*E*A
sigma = 1/rho
sigma = conductivity; rho = resistivity

## The Attempt at a Solution

Equating the two currents, I solved for the unknown E-field:

E(nichrome) = (E*A*sigma(aluminum))/(A*sigma(nichrome))
rho(nichrome) = 1.1*10^-6 Ωm
rho(aluminum) = 2.82*10^-8 Ωm

I calculated the E-field to be approximately 0.078 V/m, but this was apparently wrong. Help would be appreciated..!

Really? All I did was "Equate the two current equations"...
So here it is:

σEA = σEA

The left side representing Nichrome, and then the right side representing aluminum. Then I solved for the E-field of the unknown wire getting:

E = (σEA)/(σA)

The top part of the fraction being aluminum properties, and the bottom being the properties of Nichrome. Is there anything that I have done wrong?

Your derivation was correct, I got the same result. What is the problem with it?

ehild

Really? All I did was "Equate the two current equations"...
So here it is:

σEA = σEA

The left side representing Nichrome, and then the right side representing aluminum. Then I solved for the E-field of the unknown wire getting:

E = (σEA)/(σA)

The top part of the fraction being aluminum properties, and the bottom being the properties of Nichrome. Is there anything that I have done wrong?

How are the two areas related?

The homework website says my answer of 0.078 V/m is wrong. I'm utterly confused. I'll have to address my professor about this problem, as well as the other problem. I've double checked every single possible error for units, etc. as well.