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(B) The induced electromotive force $(EMF)$ in an inductor is given by the formula: $V = L \left| \frac{di}{dt} \right|$.Here,the change in current $\

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(B) The induced electromotive force $(EMF)$ in an inductor is given by the formula: $V = L \left| \frac{di}{dt} ight|$.Here,the change in current $\Delta i = 0.25 \; A - 0 \; A = 0.25 \; A$.The time interval $\Delta t = 0.025 \; ms = 0.025 	imes 10^{-3} \; s$.The induced voltage $V = 100 \; V$.Rearranging the formula to solve for self-inductance $L$: $L = \frac{V}{|\Delta i / \Delta t|} = \frac{V \cdot \Delta t}{\Delta i}$.Substituting the values: $L = \frac{100 	imes 0.025 	imes 10^{-3}}{0.25}$.$L = \frac{100 	imes 0.025}{0.25} 	imes 10^{-3} = 100 	imes 0.1 	imes 10^{-3} = 10 	imes 10^{-3} \; H$.Since $1 \; H = 1000 \; mH$,we have $L = 10 \; mH$.

In a fluorescent lamp choke (a small transformer),$100 \; V$ of reverse voltage is produced when the choke current changes uniformly from $0.25 \; A$ to $0 \; A$ in a duration of $0.025 \; ms$. The self-inductance of the choke (in $mH$) is estimated to be:

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