A current of 15 A flows in a wire of cross-sectional area $5\, mm^2$ with a drift velocity of $3 × 10^{-3} m/s$. The electron density in the wire will be

$625 × 10^{25} m^{-3}$

The correct answer is Option (2) → $546 × 10^{25} m^{-3}$

Given:

Current: $I = 15 \,\text{A}$

Cross-sectional area: $A = 5 \,\text{mm}^2 = 5 \times 10^{-6} \,\text{m}^2$

Drift velocity: $v_d = 3 \times 10^{-3} \,\text{m/s}$

Charge of electron: $e = 1.6 \times 10^{-19} \,\text{C}$

Formula: $I = n e A v_d$

$n = \frac{I}{e A v_d}$

$n = \frac{15}{(1.6 \times 10^{-19})(5 \times 10^{-6})(3 \times 10^{-3})}$

Denominator = $(1.6 \times 5 \times 3) \times 10^{-19 -6 -3}$

Denominator = $24 \times 10^{-28} = 2.4 \times 10^{-27}$

$n = \frac{15}{2.4 \times 10^{-27}} = 6.25 \times 10^{27} \,\text{m}^{-3}$

Final Answer: $6.25 \times 10^{27} \,\text{m}^{-3}$