A wire carries a current of 2.0 A. The number of electrons crossing through its cross-section in 10 s will be

$1.25 × 10^{20}$

The correct answer is Option (4) → $1.25 × 10^{20}$

Given:

$I = 2.0 \, A$, $t = 10 \, s$

Charge flowing, $Q = I \times t = 2 \times 10 = 20 \, C$

Number of electrons, $n = \frac{Q}{e}$

$n = \frac{20}{1.6 \times 10^{-19}}$

$n = 1.25 \times 10^{20}$

Answer: $1.25 \times 10^{20}$ electrons