• question_answer
                    Two coils have a mutual inductance of 0.005 H. The current changes in the first 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)                 \[2\,\pi \]           

    B)                 \[5\,\pi \]

    C)                                            \[\,\pi \]              

    D)                                            \[4\pi \]

    Correct Answer: B

    Solution :

                              Key Idea: For maximum value of emf in the second coil, the rate of change of current \[\left( \frac{dI}{dt} \right)\]should be maximum.                 The given equation of current changing in the first coil is                 \[I={{I}_{0}}\sin \omega t\]                                         ...(i)                 Differentiating Eq. (i) with respect to time, we have                                 \[\frac{dI}{dt}=\frac{d}{dt}({{I}_{0}}\sin \omega t)\]                 \[or\frac{dI}{dt}={{I}_{0}}\frac{d}{dt}(\sin \omega t)\]                 \[or\frac{dI}{dt}={{I}_{0}}\omega \,\cos \,\omega t\]                 For maximum \[\frac{dI}{dt}\], the value of \[\cos \,\omega t\] should be equal to 1.                 So, \[{{\left( \frac{dI}{dt} \right)}_{\max }}={{I}_{0}}\,\omega \]                 The maximum value of emf is given by \[\therefore {{e}_{\max }}=M{{\left( \frac{dI}{dt} \right)}_{\max }}=M{{I}_{0}}\omega \]                 Here, \[M=0.005\,H,\,{{I}_{0}}=10\,A,\,\omega =100\pi \,\,rad/s\] \[\therefore {{e}_{\max }}=0.005\times 10\times 100\pi =5\pi \]

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