Q 12-03-149JEE MainJEE Main 2023 (8 Apr, Shift 2)Easy
The number density of free electrons in copper is nearly $8\times10^{28}\ \text{m}^{-3}$. A copper wire has its area of cross-section $=2\times10^{-6}\ \text{m}^2$ and is carrying a current of $3.2$ A. The drift speed of the electrons is ______ $\times10^{-6}\ \text{m s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 125
$v_d=\dfrac{I}{neA}=\dfrac{3.2}{8\times10^{28}\times1.6\times10^{-19}\times2\times10^{-6}}=1.25\times10^{-4}=125\times10^{-6}\ \text{m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics