CUET · PHYSICS · PYQ PAPER 2023
A bar magnet when suspended horizontally and perpendicular to earth's field experiences a torque of \( 3 \times 10^{-4} \text{ N m} \). The Magnetic moment of the magnet would be: (take horizontal component of Earth's magnetic field at the place \( 0.4 \times 10^{-4} \text{ T} \))
- A \( 7.5 \text{ J T}^{-1} \)
- B \( 5.7 \text{ J T}^{-1} \)
- C \( 0.75 \text{ J T}^{-1} \)
- D \( 0.57 \text{ J T}^{-1} \)
Answer & Solution
Correct Answer
(A) \( 7.5 \text{ J T}^{-1} \)
Step-by-step Solution
Detailed explanation
\( M = \frac{\tau}{B_H \sin \theta} \) \( M = \frac{3 \times 10^{-4} \text{ N m}}{0.4 \times 10^{-4} \text{ T} \times \sin 90^\circ} = 7.5 \text{ J T}^{-1} \)
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