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  4. A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it from above position, the angle by which it is to be rotated is:

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Q. A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it from above position, the angle by which it is to be rotated is:

EAMCETEAMCET 1998

Solution:

Since, couple acting on the magnet is halved. Hence, $ \tau =MB\sin \theta $ [Since, magnet is same here. Mare constant B is same] $ \frac{1}{2}=MB\sin \theta $ $ \sin \theta =\frac{1}{2}=\sin {{30}^{o}} $ $ \theta ={{30}^{o}} $ $ \theta $ is angle between magnet and direction of field. Hence, the magnet is to be rotated by $ ={{90}^{o}}-{{30}^{o}}={{60}^{o}} $