Answer
$diameter=0.38mm$
Work Step by Step
Current density, J, is given as
$J=\frac{i}{A}$
or $J=\frac{i}{\pi{r^2}}$
$r=\sqrt\frac{i}{\pi{J}}$
Given that $J=440\frac{A}{cm^{2}}=440\times10^4 \frac{A}{m^2}$
We plug in the known values to obtain:
$r=\sqrt\frac{0.50}{(3.1416)(440\times10^4)}$
$r=1.9\times10^{-4}m$
$diameter=2r=2(1.9\times10^{-4})=3.8\times10^{-4}=0.38mm$
