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JEE Main & Advanced Physics Magnetism Question Bank Self Evaluation Test - Magnetism and Matter

  • question_answer
    A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through\[60{}^\circ \]. The torque required to maintain the needle in this position will be

    A) \[\sqrt{3}W\]

    B) W    

    C) \[\frac{\sqrt{3}}{2}W\]

    D)  2W

    Correct Answer: A

    Solution :

    [a] \[W=MB\,\cos \,{{60}^{o}}\] and    \[\tau =MB\,\sin \,{{60}^{o}}\] \[\therefore \,\,\,\,\,\tau =\sqrt{3}W\].


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