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Work done or torque in turning a magnetic dipole in a field

This type of question, asked 2 times: 26 Apr 2025 S1, 19 May 2026 S1. All 2 questions below, oldest first.

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  1. Question 1

    26 Apr 2025 S1

    7
    The work done in turning a magnet of magnetic moment 'MM' by an angle of 90∘90^{\circ} from the meridian is 'nn' times the corresponding work done to turn it through an angle of 60∘60^{\circ} where the value of ' nn ' is (cos⁡90∘=0,cos⁡60∘=0.5\cos90^{\circ}= 0,\cos60^{\circ}= 0.5)

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  2. Question 2

    19 May 2026 S1

    131
    A magnetic dipole lying parallel to a magnetic field requires 'WW' units of work to turn it through 30∘30^\circ. The torque required to keep this needle in this position is [sin⁡30∘=cos⁡60∘=0.5,sin⁡60∘=cos⁡30∘=32,sin⁡0∘=cos⁡90∘=1]\left[\sin 30^\circ = \cos 60^\circ = 0.5, \sin 60^\circ = \cos 30^\circ = \dfrac{\sqrt{3}}{2}, \sin 0^\circ = \cos 90^\circ = 1\right]

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