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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Self Evaluation Test - Electromagnetic Induction

  • question_answer
    Magnetic flux linked with a stationary loop of resistance R varies with respect to time during the time period T as follows: \[\phi =at(T-t)\] The amount of heat generated in the loop during that time (inductance of the coil is negligible) is

    A) \[\frac{aT}{3R}\]          

    B) \[\frac{{{a}^{2}}{{T}^{2}}}{3R}\]

    C) \[\frac{{{a}^{2}}{{T}^{2}}}{R}\]     

    D) \[\frac{{{a}^{2}}{{T}^{3}}}{3R}\]

    Correct Answer: D

    Solution :

    [d] Given that \[\phi =at(T-t)\] Induced emf, \[E=\frac{d\phi }{dt}=\frac{d}{dt}[at(T-t)]\] = at \[(0-1)+a(T-t)=a(T-2t)\] So, induced emf is also a function of time. \[\therefore \] Heat generated in time T is \[H\int\limits_{0}^{T}{\frac{{{E}^{2}}}{R}}dt=\frac{{{a}^{2}}}{R}\int\limits_{0}^{T}{{{(T-2t)}^{2}}}\]     \[dt=\frac{{{a}^{2}}{{T}^{3}}}{3R}\]


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