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(B) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$,where $L$ is the self-inductance and $i$ is the current.Let the i

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$0.25$

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(B) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$,where $L$ is the self-inductance and $i$ is the current.Let the initial current be $i_1$ and the initial energy be $U_1 = \frac{1}{2} L i_1^2$.When the current is halved,the new current $i_2 = \frac{i_1}{2}$.The new energy stored is $U_2 = \frac{1}{2} L i_2^2 = \frac{1}{2} L \left( \frac{i_1}{2} \right)^2 = \frac{1}{2} L \left( \frac{i_1^2}{4} \right) = \frac{1}{4} \left( \frac{1}{2} L i_1^2 \right) = \frac{1}{4} U_1$.Thus,$U_2 = 0.25 U_1$. Therefore,the energy becomes $0.25$ times the previous value.

If the current in a coil is halved,then the energy stored becomes how many times the previous value?

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