Q. A current of 5 A is passing through a metallic wire of cross-sectional area $4 \times 10^{- 6}m^{2}$. If the density of charge carries of the wire is $5 \times 10^{26}m^{- 3}$, the drift velocity of the electrons will be
A) $1 \times 10^{2}{ms}^{- 1}$
B) $1.56 \times 10^{- 2}{ms}^{- 1}$
C) $1.56 \times 10^{- 3}{ms}^{- 1}$
D) $1 \times 10^{- 2}{ms}^{- 1}$
Explanation
(b) Drift velocity, $v_{d} = \frac{I}{neA}$
$\Longrightarrow v_{d} = \frac{5}{\left( 5 \times 10^{26} \right) \times \left( 1.6 \times 10^{- 19} \right) \times \left( 4 \times 10^{- 6} \right)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{64} = 1.56 \times 10^{- 2}ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.