JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Question Bank Self Evaluation Test - Moving Charges and Mangetism

  • question_answer
    A closely wound solenoid of 2000 turns and area of cross-section \[1.5\times {{10}^{-4}}\,{{m}^{2}}\] carries a current of 2.0 A. It suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field \[5\times {{10}^{-2}}\]tesla making an angle of \[30{}^\circ \] with the axis of the solenoid. The torque on the solenoid will be;

    A) \[3\times {{10}^{-2}}N-m\]

    B) \[3\times {{10}^{-3}}N-m\]

    C) \[1.5\times {{10}^{-3}}N-m\]

    D) \[1.5\times {{10}^{-2}}N-m\]

    Correct Answer: D

    Solution :

    [d] Torque on the solenoid is given by \[\tau =MB\,\sin \,\theta \] where \[\theta \] is the angle between the magnetic field and the axis of solenoid. \[M=niA\] \[\therefore \,\,\,\tau =niA\,B\,\sin \,{{30}^{o}}\] \[=2000\times 2\times 1.5\times {{10}^{-4}}\times 5\times {{10}^{-2}}\times \frac{1}{2}\] \[=1.5\times {{10}^{-2}}N-m\]


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