When a current I flows through a coil,it stor | vedclass

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(B) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance of the coil.From this for

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$3E_1/4$

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(B) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance of the coil.From this formula,we can see that $E \propto I^2$.Initially,when the current is $I$,the energy is $E_1 = \frac{1}{2} L I^2$.When the current is reduced to $I' = I/2$,the new energy $E_2$ is given by $E_2 = \frac{1}{2} L (I/2)^2 = \frac{1}{2} L (I^2/4) = \frac{1}{4} E_1$.The change in energy is $\Delta E = E_1 - E_2$.Substituting the value of $E_2$,we get $\Delta E = E_1 - \frac{E_1}{4} = \frac{3E_1}{4}$.

When a current $I$ flows through a coil,it stores the energy $E_1$ in it. Now the current is reduced to $I/2$,the energy stored in the coil becomes $E_2$. The change in the energy is

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