The mutual inductance of a pair of coils,each | vedclass

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(D) The total induced $emf$ $(E)$ in a coil of $N$ turns due to mutual inductance $M$ is given by Faraday's law: $E = M \frac{dI}{dt}$.Given that the

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$\frac{MI}{Nt}$

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(D) The total induced $emf$ $(E)$ in a coil of $N$ turns due to mutual inductance $M$ is given by Faraday's law: $E = M \frac{dI}{dt}$.Given that the current changes from $I$ to $0$ in time $t$,the magnitude of the rate of change of current is $\frac{dI}{dt} = \frac{I}{t}$.Therefore,the total induced $emf$ in the coil is $E = M \frac{I}{t}$.Since the question asks for the $emf$ induced per turn,we divide the total induced $emf$ by the number of turns $N$.However,note that the total flux linkage is $N\phi = MI$. The induced $emf$ in the entire coil is $E = \frac{d(N\phi)}{dt} = M \frac{dI}{dt} = \frac{MI}{t}$.The $emf$ induced per turn is $\frac{E}{N} = \frac{MI}{Nt}$ is incorrect based on standard definitions; rather,the $emf$ induced in the whole coil is $E = \frac{MI}{t}$. If the question asks for $emf$ per turn,it is $\frac{E}{N} = \frac{MI}{Nt}$.Wait,re-evaluating: The total $emf$ induced in the secondary coil is $E = M \frac{dI}{dt} = \frac{MI}{t}$. This $emf$ is distributed across all $N$ turns. Thus,$emf$ per turn is $\frac{E}{N} = \frac{MI}{Nt}$.

The mutual inductance of a pair of coils,each of $N$ turns,is $M$ henry. If a current of $I$ ampere in one of the coils is brought to zero in $t$ seconds,the $emf$ induced per turn in the other coil,in volts,will be

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