Question

If a bar magnet of moment \( 10^{-4} \) Am\(^2\) is kept in a uniform magnetic field of \( 12 \times 10^{-3} \) T such that it makes an angle of \( 30^\circ \) with the direction of the magnetic field, then the torque acting on the magnet is:

• A. \( 6 \times 10^{-7} \) Nm
• B. \( 6 \times 10^{-5} \) Nm
• C. \( 12 \times 10^{-7} \) Nm
• D. \( 12 \times 10^{-5} \) Nm

Hint:

The torque on a magnetic dipole is given by \( \tau = MB \sin\theta \).
- Maximum torque occurs when \( \theta = 90^\circ \), and zero torque occurs when \( \theta = 0^\circ \).

Answer 1

The Correct Option is A

The torque acting on a magnetic dipole in a uniform magnetic field is given by: \[ \tau = MB \sin\theta \] where: - \( M = 10^{-4} \) Am\(^2\) (magnetic moment), - \( B = 12 \times 10^{-3} \) T (magnetic field), - \( \theta = 30^\circ \). 1. Substituting values: \[ \tau = (10^{-4}) \times (12 \times 10^{-3}) \times \sin 30^\circ \] \[ = (10^{-4} \times 12 \times 10^{-3}) \times \frac{1}{2} \] \[ = (12 \times 10^{-7}) \times \frac{1}{2} \] \[ = 6 \times 10^{-7} \text{ Nm} \] Thus, the correct answer is \(\boxed{6 \times 10^{-7} \text{ Nm}}\).