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Manipal Medical Manipal Medical Solved Paper-2014

  • question_answer
    Two coils have mutual inductance 0.005 H. The current changes in the from coil according to equation, \[I={{I}_{0}}\sin \omega t.\]Where \[{{I}_{0}}=10\,A\]and \[\omega =100\pi \]rad/s. The maximum value of emf in the second coil is

    A) \[12\,\pi \]                                            

    B) \[8\,\pi \]

    C) \[5\,\pi \]                                               

    D) \[2\,\pi \]

    Correct Answer: C

    Solution :

    Mutual inductance between two coils \[M=0.005H\] Peak current \[{{l}_{0}}=10\,A\] Angular frequency \[\omega =100\,\pi \]rad/s                  Current \[l={{l}_{0}}\sin \,\omega \,t\]           \[\frac{d}{dt}=\frac{d}{dt}({{l}_{0}}\sin \omega \,t)\]                                               \[={{l}_{0}}\cos \omega \,t.\omega \]                                               \[=10\times 1\times 100\pi \]                                               \[=1000\,\pi \] Hence, induced emf is given by \[E=M\times \frac{di}{dt}\]  \[=0.005\times 1000\times \pi =5\pi \,V\]


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