Answer
$J = 2.15\times 10^6~A/m^2$
Work Step by Step
We can express the diameter in units of $cm$:
$d = (0.0400~in)(\frac{2.5~cm}{1~in}) = 0.100~cm$
We can find the current:
$i = \frac{V}{R}$
$i = \frac{V}{\rho~L/A}$
$i = \frac{V~A}{\rho~L}$
$i = \frac{V~\pi~r^2}{\rho~L}$
$i = \frac{(1.20~V)~(\pi)~(5.0\times 10^{-4}~m)^2}{(1.69\times 10^{-8}~\Omega\cdot m)(33.0~m)}$
$i = 1.69~A$
We can find the magnitude of the current density:
$J = \frac{i}{A}$
$J = \frac{i}{\pi~r^2}$
$J = \frac{1.69~A}{(\pi)~(5.0\times 10^{-4}~m)^2}$
$J = 2.15\times 10^6~A/m^2$
