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A short bar magnet placed with its a×is at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10-2 J. What is the magnitude of magnetic moment of the magnet?

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A short bar magnet placed with its a×is at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10-2 J. What is the magnitude of magnetic moment of the magnet?

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  1. Solution:
    Torque x = MB sinθ 4.5 × 10-2 = M (0.25 sin 30°) Magnetic dipole moment, M = 0.36 J T-1.

    Check the complete chapter with solutions.

    NCERT Solutions for 12th Class Physics: Chapter 5-Magnetism and Matter