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12th Class Physics Magnetism Question Bank MCQ - Magnetism and Matter

  • question_answer
    A uniform horizontal magnetic field of \[7.5\times {{10}^{-2}}T\] is set up at an angle of \[30{}^\circ \] with the axis of an solenoid and the magnetic moment associated with it is \[1.28\text{ }J{{T}^{-1}}\]. Then the torque on it is:

    A) \[4.8\times {{10}^{-2}}Nm\]    

    B)             \[1.6\times {{10}^{-2}}Nm\]

    C)             \[1.2\times {{10}^{-2}}Nm\]    

    D)             \[4.8\times {{10}^{-4}}Nm\]

    Correct Answer: A

    Solution :

    (a) \[4.8\times {{10}^{-2}}Nm\] Torque, \[\tau =MB\,sin\theta \] Here   \[M=1.28\text{ }J{{T}^{-1}},\text{ }B=7.5\times {{10}^{-2}}\text{ }T,\text{ }\theta =30{}^\circ \] \[\therefore \tau =1.28\times 7.5\times {{10}^{-2}}sin\,30{}^\circ \]                    \[=1.28\times 7.5\times {{10}^{-2}}\times \frac{1}{2}\]             \[=0.64\times 7.5\times {{10}^{-2}}=4.8\times {{10}^{-2}}Nm\]


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