A) \[30{}^\circ \]
B) \[60{}^\circ \]
C) \[40{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: B
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 \[={{90}^{o}}-{{30}^{o}}={{60}^{o}}\]You need to login to perform this action.
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