A current of \( 5\,\mathrm{A} \) is passing through a metallic wire of cross-sectional area \( 5\times10^{-6}\,\mathrm{m^2} \). If the density of the charge carriers in the wire is \( 5\times10^{26}/\mathrm{m^3} \), the drift velocity of the electron is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.25\times10^{-2}\,\mathrm{m/s} \)
The current in a conductor is related to the drift velocity of the carriers by \( I = n e A v_d \), where \( n \) is the carrier density, \( e \) the charge of an electron and \( A \) the area of cross-section.
\( \displaystyle v_d = \frac{I}{n\,e\,A} \)
\( \displaystyle n\,e\,A = (5\times10^{26})(1.6\times10^{-19})(5\times10^{-6}) = 400\,\mathrm{C\,m^{-1}} \)
\( \displaystyle v_d = \frac{5}{400} = 1.25\times10^{-2}\,\mathrm{m\,s^{-1}} \)