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12th Class Physics Magnetism Question Bank MCQ - Moving Charges and Magnetism

  • question_answer
    The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and carries a current 10 A suspended vertically and the normal to the plane of coil makes an angle of \[50{}^\circ \] with the direction of a uniform horizontal magnetic field of magnitude 0.9 T is:

    A) 1.6 Nm                         

    B) 1.2 Nm                    

    C) 1.4Nm             

    D) 1.8 Mm                    

    Correct Answer: A

    Solution :

    (a) 1.6 Nm.                                                                  \[\tau =NlAB\,\sin \theta \] Here\[N=25,\,l=10A,\,B=0.9\,\,T,\theta =30{}^\circ ~~~~~~~~~~~~~~:\]\[A={{a}^{2}}=12\times {{10}^{-2}}\times 12\times {{10}^{-2}}=144\times {{10}^{-4}}{{m}^{2}}\] \[\therefore \,\,\tau =25\times 10\times 144\times {{10}^{-4}}\times 0.9\times \sin 30{}^\circ =16Nm\]


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