$A$ closely wound solenoid of $2000$ turns and area of cross-section $1.6 \times 10^{-4} \;m^{2},$ carrying a current of $4.0 \;A,$ is suspended through its centre allowing it to turn in a horizontal plane.
$(a)$ What is the magnetic moment associated with the solenoid?
$(b)$ What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{-2} \;T$ is set up at an angle of $30^{\circ}$ with the axis of the solenoid?

(A) Number of turns on the solenoid,$n = 2000$.
Area of cross-section of the solenoid,$A = 1.6 \times 10^{-4} \;m^{2}$.
Current in the solenoid,$I = 4.0 \;A$.
$(a)$ The magnetic moment $M$ associated with the solenoid is given by $M = nIA$.
Substituting the values: $M = 2000 \times 4.0 \times 1.6 \times 10^{-4} = 1.28 \;A \cdot m^{2}$.
$(b)$ Magnetic field,$B = 7.5 \times 10^{-2} \;T$.
Angle between the magnetic field and the axis of the solenoid,$\theta = 30^{\circ}$.
Since the magnetic field is uniform,the net force on the solenoid is zero.
The torque $\tau$ on the solenoid is given by $\tau = MB \sin \theta$.
Substituting the values: $\tau = 1.28 \times 7.5 \times 10^{-2} \times \sin 30^{\circ} = 1.28 \times 7.5 \times 10^{-2} \times 0.5 = 4.8 \times 10^{-2} \;N \cdot m$.