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## 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..!