Answer
The potential difference is $~~3.8\times 10^{-4}~V$
Work Step by Step
We can find the resistance in a 4.0 cm length of the wire:
$R = \frac{\rho~L}{A}$
$R = \frac{\rho~L}{\pi~r^2}$
$R = \frac{(1.69\times 10^{-8}~\Omega\cdot m)(0.040~m)}{(\pi)~(2.6\times 10^{-3}~m)^2}$
$R = 3.18\times 10^{-5}~\Omega$
We can find the potential difference:
$V = i~R$
$V = (12~V)(3.18\times 10^{-5}~\Omega)$
$V = 3.8\times 10^{-4}~V$
The potential difference is $~~3.8\times 10^{-4}~V$
