The magnetic flux through a circuit carrying | vedclass

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(A) The magnetic flux $\phi$ is proportional to the current $I$ in the circuit,so $\phi = LI$,where $L$ is the self-inductance.First,we find the self-

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$2$

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(A) The magnetic flux $\phi$ is proportional to the current $I$ in the circuit,so $\phi = LI$,where $L$ is the self-inductance.First,we find the self-inductance $L$:$L = \frac{\phi_1}{I_1} = \frac{0.8\,Wb}{2.0\,A} = 0.4\,H$.Now,calculate the flux $\phi_2$ when the current $I_2 = 1.5\,A$:$\phi_2 = L 	imes I_2 = 0.4\,H 	imes 1.5\,A = 0.6\,Wb$.The change in flux is $\Delta \phi = \phi_1 - \phi_2 = 0.8\,Wb - 0.6\,Wb = 0.2\,Wb$.The induced $emf$ is given by Faraday's law: $|e| = \frac{|\Delta \phi|}{\Delta t}$.$|e| = \frac{0.2\,Wb}{0.1\,s} = 2.0\,V$.

The magnetic flux through a circuit carrying a current of $2.0\,A$ is $0.8\,Wb$. If the current reduces to $1.5\,A$ in $0.1\,s$,the induced $emf$ is......$V$.

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