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

  • question_answer
    A circular coil is 100 turns, radius 10 cm carries a current of 5A. It is suspended vertically in a uniform horizontal magnetic field of 0.5 T and the field lines make an angle of \[60{}^\circ \] with the plane of the coil. The magnitude of the torque that must be applied on it to prevent it from turning is:

    A) 2.93 Nm

    B) 3.43 Nm

    C)             3.93 Nm                       

    D)             4.93 Nm

    Correct Answer: C

    Solution :

    (c) 3.93 Nm Here, \[N=100,\text{ }r=10\text{ }cm=0.10\text{ }m,l=5\text{ }A,B=0.5\text{ }T\] \[\theta =90{}^\circ -60{}^\circ =30{}^\circ \] Area of the coil, \[A=\pi {{r}^{2}}=3.14\times {{(0.1)}^{2}}\] \[\tau =NlBA\,sin\theta \] \[=100\times 5\times 0.5\times 3.14\times {{(0.1)}^{2}}\times sin30{}^\circ \] \[=3.931\text{ }Nm\]


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