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(D) The mutual inductance $M$ is given as $5 \,mH = 5 	imes 10^{-3} \,H$.The current in the primary coil is $i = 50 \sin(500t) \,A$.The induced e.m.f

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$125$

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(D) The mutual inductance $M$ is given as $5 \,mH = 5 	imes 10^{-3} \,H$.The current in the primary coil is $i = 50 \sin(500t) \,A$.The induced e.m.f. $\varepsilon$ in the secondary coil is given by $\varepsilon = -M \frac{di}{dt}$.Taking the magnitude,$\varepsilon = M \left| \frac{di}{dt} ight|$.$\frac{di}{dt} = \frac{d}{dt} [50 \sin(500t)] = 50 	imes 500 \cos(500t) = 25000 \cos(500t)$.Therefore,$\varepsilon = (5 	imes 10^{-3}) 	imes (25000 \cos(500t)) = 125 \cos(500t) \,V$.The peak value of the induced e.m.f. is the coefficient of the cosine term,which is $125 \,V$.

Two coaxial coils are very close to each other and their mutual inductance is $5 \,mH$. If a current $i = 50 \sin(500t) \,A$ is passed in one of the coils,then the peak value of the induced e.m.f. in the secondary coil will be ........... $V$.

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