Two long and parallel straight wires, each carrying current 'I' in the same direction, are separated by a distance of 40 cm. If the force of attraction per metre between them, is $2 × 10^{-6} N/m$, the value of 'I' will be:

(Given: $μ_0 = 4 π × 10^{-7} T\, m\, A^{-1}$)

2 A

The correct answer is Option (3) → 2 A

The force per unit length between two parallel currents is:

$\frac{F}{L} = \frac{\mu_0 I^2}{2 \pi d}$

Given: $\frac{F}{L} = 2 \times 10^{-6} \, \text{N/m}$, $d = 0.4 \, \text{m}$, $\mu_0 = 4 \pi \times 10^{-7} \, \text{T·m/A}$

Substitute values:

$2 \times 10^{-6} = \frac{4 \pi \times 10^{-7} \cdot I^2}{2 \pi \cdot 0.4}$

$2 \times 10^{-6} = \frac{2 \times 10^{-7} I^2}{0.4}$

$2 \times 10^{-6} = 5 \times 10^{-7} I^2$

$I^2 = \frac{2 \times 10^{-6}}{5 \times 10^{-7}} = 4$

$I = 2 \, \text{A}$

Current I = 2 A