Question

For what electric field strength would the current in a 1.9-mm-diameter nichrome wire be the same as the current in a 0.50-mm-diameter aluminum wire in which the electric field strength is 0.0080 V/m ?

Express your answer to two significant figures and include the appropriate units.

Answer 1

we know,

rho_nichrome = 10^-6 ohm.m
rho_aluminum = 2.65*10^-8
we know, E = J*rho

E ---> electric field
J ---> current density
rho --> resistivity

so,
E = (I/A)*rho

I = E*A/rho

I_nichrome = E_nochrome*A_nichrome/rho_nichrome

and

I_aluminum= E_aluminum*A_aluminum/rho_aluminum

given

I_nichrome = I_aluminum

E_nochrome*A_nichrome/rho_nichrome = E_aluminum*A_aluminum/rho_aluminum

E_nichrome = E_aluminum*(A_aluminum/A_nichrome)*(rho_nichrome/rho_aluminum)

= 0.0080*(0.5/1.9)^2*(10^-6/(2.65*10^-8))

= 0.021 V/m <<<<<<<<<----------------Answer