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EAMCET Medical EAMCET Medical Solved Paper-1998

  • question_answer
    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:

    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 \[={{90}^{o}}-{{30}^{o}}={{60}^{o}}\]


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