Drift speed of electrons,

Question:

Drift speed of electrons, when $1.5 \mathrm{~A}$ of current flows in a copper wire of cross section $5 \mathrm{~mm}^{2}$, is $v$. If the electron density in copper is $9 \times 10^{28} / \mathrm{m}^{3}$ the value of $v$ in $\mathrm{mm} / \mathrm{s}$ close to (Take charge of electron to be $=1.6 \times 10^{-19} \mathrm{C}$ )

1. (1) $0.02$

2. (2) 3

3. (3) 2

4. (4) $0.2$

Correct Option: 1

Solution:

(1)  Using, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$

$\therefore$ Drift speed $\mathrm{v}_{\mathrm{d}}=\frac{1}{\text { neA }}$

$=\frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}}=0.02 \mathrm{~mm} / \mathrm{s}^{-1}$