Two long and parallel straight wires A and B carrying currents of 8.0…

The correct answer is Option (3) → $2 × 10^{-5}N$, attractive

Current in wire A: $I_1 = 8.0 \,\text{A}$

Current in wire B: $I_2 = 5.0 \,\text{A}$

Distance between wires: $d = 0.04 \,\text{m}$

Length of wire considered: $L = 0.10 \,\text{m}$

Force per unit length between two parallel currents:

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}$

Therefore, $F = \frac{\mu_0 I_1 I_2}{2 \pi d} \cdot L$

Using $\mu_0 = 4 \pi \times 10^{-7}$ H/m:

$F = \frac{4 \pi \times 10^{-7} \times 8 \times 5}{2 \pi \times 0.04} \times 0.10$

$F = \frac{160 \pi \times 10^{-7}}{2 \pi \times 0.04} \times 0.10$

$F = \frac{160 \times 10^{-7}}{0.08} \times 0.10$

$F = \frac{160 \times 10^{-7} \times 0.10}{0.08}$

$F = \frac{16 \times 10^{-7}}{0.08}$

$F = 200 \times 10^{-7}$

$F = 2 \times 10^{-5} \,\text{N}$

Answer: The force on 10 cm length of wire B is $2 \times 10^{-5}$ N (attractive).