• 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)                                            $4\pi$

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, $M=0.005\,H,\,{{I}_{0}}=10\,A,\,\omega =100\pi \,\,rad/s$ $\therefore {{e}_{\max }}=0.005\times 10\times 100\pi =5\pi$