Correct Option: (C) 60°

Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]

The torque ($\tau$) acting on a bar magnet of magnetic moment $\vec{M}$ placed in a uniform magnetic field $\vec{B}$ is given by the cross product:

$$\tau = MB \sin \theta$$

Where $\theta$ is the angle between the magnetic moment vector and the magnetic field vector.

• Initial angle: $\theta_1 = 90^\circ$
• Initial torque: $\tau_1 = MB \sin 90^\circ = MB$
• Final torque: $\tau_2 = \frac{1}{2} \tau_1 = \frac{MB}{2}$

Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]

Using the expression for the final torque:

$$\tau_2 = MB \sin \theta_2$$

Substitute the given condition $\tau_2 = \frac{MB}{2}$:

$$\frac{MB}{2} = MB \sin \theta_2$$ $$\sin \theta_2 = \frac{1}{2}$$

Therefore, the final angle is:

$$\theta_2 = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$$

The question asks for the angle through which the magnet is rotated ($\Delta \theta$):

$$\Delta \theta = \theta_1 - \theta_2 = 90^\circ - 30^\circ = 60^\circ$$

Step 3 (Final Concluded Answer): [1/3 Mark]

The bar magnet must be rotated through an angle of 60°. Thus, the correct option is (C).

💡 EL10 Concept Insight: Think of the torque as the "twisting force" that tries to align the magnet with the field; it is strongest when the magnet is sideways (90°) and zero when it is perfectly aligned (0°). Since the torque is proportional to the sine of the angle, cutting the torque in half means the sine of the angle must also be cut in half, moving the magnet from 90° to 30°.