A solenoid of length 0.4 m having 500 turns of wire carries a current of 3 A. A coil of 10 turns and radius 0.01 m carries a current of 0.4 A. What will be the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid?

$5.9 × 10^{-6} Nm$

The correct answer is Option (2) → $5.9 × 10^{-6} Nm$

Given:

Solenoid: $n_s = 500$ turns, $l = 0.4\ \text{m}$, $I_s = 3\ \text{A}$

Coil: $N_c = 10$ turns, radius $r = 0.01\ \text{m}$, current $I_c = 0.4\ \text{A}$

Magnetic field inside a solenoid: $B = \mu_0 \frac{N_s I_s}{l}$

$B = 4\pi \times 10^{-7} \cdot \frac{500 \cdot 3}{0.4} = 4\pi \times 10^{-7} \cdot 3750 = 4.712 \times 10^{-3}\ \text{T}$

Magnetic moment of the coil: $m = N_c I_c A$, where $A = \pi r^2$

$A = \pi (0.01)^2 = 3.1416 \times 10^{-4}\ \text{m}^2$

$m = 10 \cdot 0.4 \cdot 3.1416 \times 10^{-4} = 1.25664 \times 10^{-3}\ \text{A·m}^2$

Torque on the coil when its axis is perpendicular to the field: $\tau = m B \sin 90^\circ = m B$

$\tau = 1.25664 \times 10^{-3} \cdot 4.712 \times 10^{-3} \approx 5.92 \times 10^{-6}\ \text{N·m}$

∴ Torque required = 5.92 × 10⁻⁶ N·m