The magnetic potential energy stored in a cer | vedclass

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(D) The magnetic potential energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$. Given: $U = 25 \ mJ = 25 	imes 10^{-3} \

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$20$

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(D) The magnetic potential energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$. Given: $U = 25 \ mJ = 25 	imes 10^{-3} \ J$ $I = 50 \ mA = 50 	imes 10^{-3} \ A$ Rearranging the formula to solve for inductance $L$: $L = \frac{2U}{I^2}$ Substituting the values: $L = \frac{2 	imes 25 	imes 10^{-3}}{(50 	imes 10^{-3})^2}$ $L = \frac{50 	imes 10^{-3}}{2500 	imes 10^{-6}}$ $L = \frac{50 	imes 10^{-3}}{2.5 	imes 10^{-3}}$ $L = 20 \ H$.

The magnetic potential energy stored in a certain inductor is $25 \ mJ$,when the current in the inductor is $50 \ mA$. What is the inductance (in $H$)?

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