Answer
$5.32\times 10^{5} A/m^2 $
Work Step by Step
It is given that:
$(R/L) $=$0.150\ \frac{\Omega}{km}$
Resistance of the wire is given by;
$R = \rho \frac{L}{A} $
$A=\frac{\rho L}{R}$
$A=\frac{\rho}{(R/L)}$
Current density of copper is given by $J=\frac{i}{A}$
$J=\frac{iR}{\rho L}$
$J=\frac{(60\ A)(0.150 \Omega.m)}{(1.69\times 10^{-8}\Omega/km)}=5.32\times 10^{5} A/m^2 $
