The drift velocity of electrons is $2\times 10^{-6} m/s$ in a conductor in which the number of free electrons is $6 \times 10^{28}\, per \, m^3$. If the cross section of the conductor is $5\times 10^{-4}m^2$, then the current flowing through the consuctor is

(Given $e=1.6\times 10^{-19}coul$)

9.6 Amp

The correct answer is option (3) : 9.6 Amp

We know,

$i=neAV_d ⇒i=6\times 10^{28} \times 1.6\times 10^{-19} \times 5 \times 10^{-4} \times 2 \times 10^{-6}$

$⇒i=6 \times 1.6\times 10\times 10^{28}\times 10^{-29}=9.6 A$