When current in a coil changes from 2 A to 5 | vedclass

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(B) The formula for the induced emf in a coil due to self-inductance is given by $\varepsilon = -L \frac{di}{dt}$.Taking the magnitude,we have $|\vare

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$4 \ mH$

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(B) The formula for the induced emf in a coil due to self-inductance is given by $\varepsilon = -L \frac{di}{dt}$.Taking the magnitude,we have $|\varepsilon| = L \frac{|\Delta i|}{\Delta t}$.Given:Initial current $i_1 = 2 \ A$Final current $i_2 = 5 \ A$Change in current $\Delta i = i_2 - i_1 = 5 \ A - 2 \ A = 3 \ A$.Time interval $\Delta t = 0.3 \ s$.Induced emf $\varepsilon = 40 \ mV = 40 	imes 10^{-3} \ V$.Substituting the values into the formula:$40 	imes 10^{-3} = L 	imes \frac{3}{0.3}$.$40 	imes 10^{-3} = L 	imes 10$.$L = \frac{40 	imes 10^{-3}}{10} = 4 	imes 10^{-3} \ H$.$L = 4 \ mH$.

When current in a coil changes from $2 \ A$ to $5 \ A$ in a time of $0.3 \ s$,if the emf induced in the coil is $40 \ mV$,then the self inductance of the coil is

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