Q. Every atom makes one free electron in copper. If 1.1 A Current is flowing in the wire of copper having 1 mm diameter, then the drift velocity(approx.) will be (density of copper=$9 \times 10^{3}\ kg\ m^{- 3}$and atomic weight of copper=63)
A) $0.1\ mms^{- 1}$
B) $0.2\ mms^{- 1}$
C) $0.3\ mms^{- 1}$
D) $0.2\ mms^{- 1}$
Explanation
(a) $v_{d} = \frac{i}{n\ Ae};where\ n = N_{\rho}/M$
$= 6.023 \times 10^{26} \times 9 \times 10^{3}/63 = 0.860 \times 10^{29}$
$= 8.6 \times 10^{28}$
and $A = \frac{\pi D^{2}}{4} = \frac{22}{7} \times \frac{\left( 10^{- 3} \right)^{2}}{4m^{2}};$
$= \frac{11}{14} \times 10^{- 6}m^{2}$
$v_{d} = \frac{1.1}{8.6 \times 10^{28} \times \frac{11}{14} \times 10^{- 6} \times 1.6\ \times 10^{- 19}}$
$= \frac{1}{9.6 \times 10^{+ 3}} = \frac{100 \times 10^{- 4}}{96} = 1.0 \times 10^{- 4}m/s$
$= 0.1\ mms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.