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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    Two coils have mutual inductance 0.005 H. The current changes in the first coil according to equation\[I={{I}_{0}}\] sin cot. Where \[{{I}_{0}}=10\] amp and \[\omega =10\,\pi \] rad/sec. 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 :

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


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