Answer
$\Delta V = 2.5\times 10^{-4}~V$
Work Step by Step
We can find the resistance of a 2.0-cm length of the aluminum wire:
$R = \frac{\rho~L}{A}$
$R = \frac{\rho~L}{\pi~(\frac{d}{2})^2}$
$R = \frac{4~\rho~L}{\pi~d^2}$
$R = \frac{(4)~(2.65\times 10^{-8}~\Omega~m)~(0.020~m)}{(\pi)~(0.020~m)^2}$
$R = 1.687\times 10^{-6}~\Omega$
We can find the potential difference $\Delta V$:
$\Delta V = I~R$
$\Delta V = (150~A)(1.687\times 10^{-6}~\Omega)$
$\Delta V = 2.5\times 10^{-4}~V$
