Q 12-03-257JEE MainJEE Main 2019 (9 Jan, Shift 1)Easy
Drift speed of electrons, when $1.5\ \text{A}$ current flows in a copper wire of cross section $5\ \text{mm}^2$, is $v_d$. If the electron density in copper is $9\times10^{28}\ \text{m}^{-3}$ the value of $v_d$ in $\text{mm s}^{-1}$ is close to (Take charge of an electron to be $1.6\times10^{-19}\ \text{C}$)
Answer: (D) $0.02$
$$v_d = \frac{I}{neA} = \frac{1.5}{9\times10^{28}\times1.6\times10^{-19}\times5\times10^{-6}} = \frac{1.5}{7.2\times10^{4}}$$
$$v_d \approx 2.1\times10^{-5}\ \text{m s}^{-1} = 0.02\ \text{mm s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics