Q. A uniform copper wire of length 1 m and cross-section area $5 \times 10^{- 7}m^{2}$ carries a current of 1 A. Assuming that there are $8 \times 10^{28}$ free electron $m^{- 3}$in copper, how long will an electron take to drift from one end of the wire to the other?
A) $0.8 \times 10^{3}s$
B) $1.6 \times 10^{3}s$
C) $3.2 \times 10^{3}s$
D) $6.4 \times 10^{3}s$
Explanation
(d) Current in wire $i = Anev_{d}$
Here, $i = 1A,\ n = 8 \times 10^{28}\ \ electron/m^{- 3}$
$\ \ \ \ \ \ \ \ \ \ A = 5 \times 10^{- 7}m^{2}$
$\therefore 1 = 8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7} \times v_{d}$
$or\ \ \ v_{d} = \frac{1}{8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7}}$
Now,
$t = \frac{1}{v_{d}} = 8 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 7}$
$\ \ \ = 64 \times 10^{2} = 6.4 \times 10^{3}s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.