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
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..!