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(D) The induced e.m.f. in the second coil is given by the formula: $e = M \frac{dI_1}{dt} \dots (i)$Given $I_1 = I_0 \sin \omega t$, we differentiate

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$5$

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(D) The induced e.m.f. in the second coil is given by the formula: $e = M \frac{dI_1}{dt} \dots (i)$Given $I_1 = I_0 \sin \omega t$, we differentiate with respect to time $t$:$\frac{dI_1}{dt} = I_0 \omega \cos \omega t$Substituting this into equation $(i)$, we get:$e = M I_0 \omega \cos \omega t$The maximum value of e.m.f. $(e_{max})$ occurs when $\cos \omega t = 1$:$e_{max} = M I_0 \omega$Given values: $M = 5 	imes 10^{-3} 	ext{ H}$, $I_0 = 10 	ext{ A}$, and $\omega = 100 \pi 	ext{ rad/s}$.Substituting these values:$e_{max} = (5 	imes 10^{-3}) 	imes 10 	imes 100 \pi$$e_{max} = 5 	imes 10^{-3} 	imes 10^3 \pi$$e_{max} = 5 \pi 	ext{ V}$

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