Q. There is a current of $1.344\ amp$ in a copper wire whose area of cross-section normal to the length of the wire is $1\ mm^{2}.$ If the number of free electrons per $cm^{3}$ is $8.4 \times 10^{22}$, then the drift velocity would be
A) $1.0\ mm/sec$
B) $1.0\ m/sec$
C) $0.1\ mm/sec$
D) $0.01\ mm/sec$
Explanation
(c) $v_{d} = \frac{i}{nAe} = \frac{1.344}{10^{- 2} \times 1.6 \times 10^{- 19} \times 8.4 \times 10^{22}}$
$= \frac{1.344}{10 \times 1.6 \times 8.4} = 0.01cm/s = 0.1mm/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.