The drift velocity of electrons is \( 2 \times 10^{-6}\,\mathrm{m/s} \) in a conductor in which the number of free electrons is \( 6 \times 10^{28} \) per \( \mathrm{m^3} \). If the cross section of the conductor is \( 5 \times 10^{-4}\,\mathrm{m^2} \), then the current flowing through the conductor is (Given \( e = 1.6 \times 10^{-19} \) coul)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9.6 Amp
The current in a conductor is related to the drift velocity of its free electrons by \( I = n e A v_d \), where \( n \) is the number of free electrons per unit volume and \( A \) the area of cross-section.
\( n = 6\times10^{28}\,\mathrm{m^{-3}} \)
\( A = 5\times10^{-4}\,\mathrm{m^2} \)
\( v_d = 2\times10^{-6}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle I = 6\times10^{28} \times 1.6\times10^{-19} \times 5\times10^{-4} \times 2\times10^{-6}\,\mathrm{A} \)
\( \displaystyle I = (6 \times 1.6 \times 5 \times 2)\times10^{28 - 19 - 4 - 6}\,\mathrm{A} = 96\times10^{-1}\,\mathrm{A} = 9.6\,\mathrm{A} \)