2. Questions Bank
  3. 12th Class
  4. Physics
  5. Magnetism

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

  • question_answer
    A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines make an angle of \[30{}^\circ \] with the normal of the coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is:

    A) \[33\,Nm\]                     

    B) \[3.3\,Nm\]

    C) \[3.3\times {{10}^{-2}}Nm\]    

    D) \[3.3\times {{10}^{-4}}Nm\]

    Correct Answer: B

    Solution :

    (b) \[33\,Nm\] Given,               \[N=70,\text{ }r=5\text{ }cm\] \[=5\times {{10}^{-2}}m,\,l=8A\] \[B=1.5\,T,\theta =30{}^\circ \] The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil, \[\therefore \tau =NlAB\,\sin \theta =Nl\pi {{r}^{2}}B\sin \,30{}^\circ \]             \[=70\times 8\times 3.14\times {{(5\times {{10}^{-2}})}^{2}}\times 1.5\times \frac{1}{2}\]             \[=3.297\,N\,m=3.3\,Nm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner