Answer
Diameter of the wire is 0.5 mm
Work Step by Step
Current density of a material,
$$j=I/A$$
Maximum current density is $500A/cm^{2}$ and current applied is $1A$
Solving for $A$,
$A= \frac{1A}{500Acm^{-2}}= 2 \times 10^{-3}cm^{2}$
Finding diameter of wire,
$A=π(\frac{d}{2})^{2}$
$d=\sqrt \frac{4\times2 \times 10^{-3}cm^{2}}{π} \approx 0.5mm$
