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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    A circular coil of 20 turns and radius 10 cm is placed in uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in coil is 5A, then the torque acting on the coil will be:

    A)  \[31.4\] N-m                 

    B)  \[3.14\] N-m.

    C)  \[0.314\] N-m               

    D)  zero

    Correct Answer: D

    Solution :

    Torque (T) acting on a loop placed in a magnetic field B is given by  \[\tau \,=nBIA\,\sin \theta \]where, A is area of loop, I the current through it, n the number of turns, and \[\theta \] the angle which axis of loop makes with magnetic field B. Since, magnetic field  is parallel to the axis of the coil hence \[\theta =0{}^\circ \] and s\[\sin {{0}^{0}}=0\]\[\tau \,=0\]           


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