Answer
1.7 A
Work Step by Step
Given: Electric field E= 0.012 V/m,
Area=$ 2.0 mm\times2.0mm= 4.0\times10^{-6}m^{2}$
For aluminium,
Conductivity $\sigma= 3.5\times10^{7}\Omega^{-1}m^{-1}$
Recall: I= JA where J is the current density equal to $\sigma E$. Then, $I= \sigma EA$.
Substitute the values of $\sigma$,E and A:
Current I= $3.5\times10^{7}\Omega^{-1}m^{-1}\times0.012 \frac{V}{m}\times4.0\times10^{-6}m^{2}\approx 1.7 A$
