CUET · PHYSICS · PYQ PAPER 2025
A short bar magnet placed with its axis at \(30^\circ\) with a uniform external magnetic field of 0.24 T experiences a torque of magnitude 0.048 N m. The magnetic moment of the magnet is
- A \( 0.1 \, \text{A}\cdot\text{m}^2 \)
- B \( 0.2 \, \text{A}\cdot\text{m}^2 \)
- C \( 0.4 \, \text{A}\cdot\text{m}^2 \)
- D \( 0.23 \, \text{A}\cdot\text{m}^2 \)
Answer & Solution
Correct Answer
(C) \( 0.4 \, \text{A}\cdot\text{m}^2 \)
Step-by-step Solution
Detailed explanation
\(\tau = MB\sin\theta\) \(0.048 = M \times 0.24 \times \sin 30^\circ\) \(M = \frac{0.048}{0.24 \times 0.5}\) \(M = 0.4 \, \text{A}\cdot\text{m}^2\)
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