The magnetic potential energy stored in a cer | vedclass

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(C) The energy stored in an inductor is given by the formula: $E = \frac{1}{2} LI^2$.Given: Energy $E = 64 	imes 10^{-3} \ J$ and current $I = 80 \ m

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$20$

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(C) The energy stored in an inductor is given by the formula: $E = \frac{1}{2} LI^2$.Given: Energy $E = 64 	imes 10^{-3} \ J$ and current $I = 80 \ mA = 80 	imes 10^{-3} \ A$.Rearranging the formula to solve for inductance $L$: $L = \frac{2E}{I^2}$.Substituting the values: $L = \frac{2 	imes 64 	imes 10^{-3}}{(80 	imes 10^{-3})^2}$.$L = \frac{128 	imes 10^{-3}}{6400 	imes 10^{-6}} = \frac{128 	imes 10^{-3}}{6.4 	imes 10^{-3}} = \frac{128}{6.4} = 20 \ H$.

The magnetic potential energy stored in a certain inductor is $64 \times 10^{-3} \ J$,when the current in the inductor is $80 \ mA$. The inductance of this inductor is: (in $H$)

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