A long straight wire carries a current of 12 A. Another straight wire of length 5 cm placed parallel to it at a distance of 5 cm, carries a current of 2 A. The magnetic force on the smaller wire is

$4.8 × 10^{-6} N$

The correct answer is Option (2) → $4.8 × 10^{-6} N$

Given:

Current in long wire, $I_1 = 12 \ \text{A}$

Current in short wire, $I_2 = 2 \ \text{A}$

Length of short wire, $l = 5 \ \text{cm} = 0.05 \ \text{m}$

Distance between wires, $r = 5 \ \text{cm} = 0.05 \ \text{m}$

Magnetic field due to long wire at distance $r$: $B = \frac{\mu_0 I_1}{2 \pi r}$

Magnetic force on short wire: $F = I_2 l B = I_2 l \frac{\mu_0 I_1}{2 \pi r}$

Substitute values:

$F = 2 \cdot 0.05 \cdot \frac{4 \pi \times 10^{-7} \cdot 12}{2 \pi \cdot 0.05}$

$F = 0.1 \cdot \frac{48 \pi \times 10^{-7}}{0.1 \pi} = 48 \times 10^{-7} = 4.8 \times 10^{-6} \ \text{N}$

Magnetic force on the smaller wire: $F = 4.8 \times 10^{-6} \ \text{N}$