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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 16

  • question_answer
    A conducting circular loop of radius a is placed in a uniform magnetic field B which is perpendicular to loop. A rod OA touches the loop as shown. A resistor R is also connected between 0 and circumference of loop (loop and rod are resistance less). Rod is rotated with constant angular velocity \[\omega \] in anticlockwise direction. Torque of the external force needed to keep the rod rotating with constant angular velocity \[\omega \] is

    A) \[\frac{{{B}^{2}}{{a}^{2}}\omega }{2R}\]    

    B)        \[\frac{{{B}^{2}}\omega {{a}^{4}}}{8R}\]

    C) \[\frac{{{B}^{2}}a{{\omega }^{2}}}{4R}\]    

    D)        \[\frac{{{B}^{2}}\omega {{a}^{4}}}{4R}\]

    Correct Answer: D

    Solution :

    [d] Emf induced between O and A \[E=\frac{1}{2}B\omega {{a}^{2}}\] so \[i\,\,(in\,\,loop)=\frac{E}{R}=\frac{B\omega {{a}^{2}}}{2R}\] \[\tau =(Bia)\frac{a}{2}=B\left( \frac{B\omega {{a}^{2}}}{2R} \right)\frac{{{a}^{2}}}{2}=\frac{{{B}^{2}}\omega {{a}^{4}}}{4R}\]


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