Here B = 0.02 T, θ = 30° and t = 0.060 N m
(i) ∴ Magnetic moment of the magnet m = t/Bsinθ = 0.060/0.02 × 0.5000 = 6 A m² .
(ii) For stable equilibrium of the magnet in the magnetic field θ = 0°
Class 12 Physics Chapter 5 Magnetism and Matter Session 2024-2025.
A short bar magnet place with its axis at 30° to a uniform magnetic field of 0.02 T experiences a torque of 0.060 N m. (i) Calculate magnetic moment of the magnet, and (ii) find out what orientation of the magnet corresponds to its stable equilibrium in the magnetic field.
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(i) ∴ Magnetic moment of the magnet
M = τ/Bsinθ = 0.060/0.02×0.5 = 6A\cdotpm².
(ii) For stable equilibrium, the magnet’s axis aligns with the magnetic field (θ=0°).
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