Question

A short bar magnet placed with its axis at \(45^\circ\) with a uniform external magnetic field of \(28.3 \times 10^{-3} \, \text{T}\) experiences a torque of magnitude \(3.6 \times 10^{-5} \, \text{J}\). The magnitude of magnetic moment of the magnet is nearly:

• A. \( 1.8 \times 10^{-3} \, \text{J·T}^{-1} \)
• B. \( 1.2 \times 10^{-3} \, \text{J·T}^{-1} \)
• C. \( 2.4 \times 10^{-3} \, \text{J·T}^{-1} \)
• D. \( 1.6 \times 10^{-3} \, \text{J·T}^{-1} \)

Hint:

Use \( \tau = MB \sin \theta \) for magnetic dipoles. Remember to convert angles properly and use accurate trigonometric values.

Answer 1

The Correct Option is A

Torque on a magnetic dipole in a magnetic field is given by: \[ \tau = MB \sin \theta \] Given: - \( \tau = 3.6 \times 10^{-5} \, \text{J} \)
- \( B = 28.3 \times 10^{-3} \, \text{T} \)
- \( \theta = 45^\circ \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \)
\[ M = \frac{\tau}{B \sin \theta} = \frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \cdot \frac{1}{\sqrt{2}}} = \frac{3.6 \times \sqrt{2}}{28.3} \times 10^{-2} \approx 1.8 \times 10^{-3} \, \text{J·T}^{-1} \]