A current of 5 A is passing through a metallic wire of cross-sectional area $5 × 10^{-6}\, m^2$. If the density of the charge carriers in the wire is $5 × 10^{26}/m^3$, the drift velocity of the electron is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $1.25 × 10^{-2}\, m/s$
Drift velocity is given by:
$v_d = \frac{I}{n e A}$
Given: $I = 5 \, \text{A}$, $n = 5 \times 10^{26} \, \text{m}^{-3}$, $A = 5 \times 10^{-6} \, \text{m}^2$, $e = 1.6 \times 10^{-19} \, \text{C}$
$v_d = \frac{5}{5 \times 10^{26} \cdot 1.6 \times 10^{-19} \cdot 5 \times 10^{-6}}$
$v_d = \frac{5}{4 \times 10^2} = 0.0125 \, \text{m/s}$
Drift velocity $v_d = 0.0125$ m/s