Question

At a place the horizontal component of earth’s magnetic field is \( 3 \times 10^{-5} \) T and the magnetic declination is \( 30^\circ \). A compass needle of magnetic moment \( 18 \) A\(m^2\) pointing towards geographic north at this place experiences a torque of:

• A. \( 36 \times 10^{-5} \) Nm
• B. \( 18 \times 10^{-5} \) Nm
• C. \( 54 \times 10^{-5} \) Nm
• D. \( 27 \times 10^{-5} \) Nm

Hint:

The torque on a magnetic dipole in a magnetic field is given by \( \tau = M B \sin\theta \), where \( M \) is the magnetic moment, \( B \) is the magnetic field, and \( \theta \) is the angle between them.

Answer 1

The Correct Option is D

Step 1: Apply Torque Formula The torque experienced by a magnetic dipole in a magnetic field is given by: \[ \tau = M B \sin\theta \] Step 2: Compute Torque \[ \tau = (18) \times (3 \times 10^{-5}) \times \sin 30^\circ \] Since \( \sin 30^\circ = \frac{1}{2} \), we get: \[ \tau = 18 \times 3 \times 10^{-5} \times \frac{1}{2} \] \[ \tau = \frac{54 \times 10^{-5}}{2} \] \[ \tau = 27 \times 10^{-5} { Nm} \] Thus, the correct answer is \( 27 \times 10^{-5} \) Nm.