An average induced emf of 0.5 ,V appears in a | vedclass

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(B) The formula for induced emf in a coil due to self-inductance is given by $e = L \left| \frac{dI}{dt} \right|$.Given:Induced emf $e = 0.5 \,V$Initi

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$20$

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(B) The formula for induced emf in a coil due to self-inductance is given by $e = L \left| \frac{dI}{dt} ight|$.Given:Induced emf $e = 0.5 \,V$Initial current $I_1 = 10 \,A$Final current $I_2 = -10 \,A$ (opposite direction)Change in current $\Delta I = I_2 - I_1 = -10 \,A - 10 \,A = -20 \,A$Time interval $\Delta t = 0.8 \,s$Magnitude of rate of change of current $|\frac{\Delta I}{\Delta t}| = \frac{|-20 \,A|}{0.8 \,s} = \frac{20}{0.8} \,A/s = 25 \,A/s$.Substituting these values into the formula:$0.5 = L 	imes 25$$L = \frac{0.5}{25} \,H = \frac{1}{50} \,H = 0.02 \,H$.Converting to millihenry: $0.02 \,H = 0.02 	imes 1000 \,mH = 20 \,mH$.

An average induced emf of $0.5 \,V$ appears in a coil when the current in it is changed from $10 \,A$ in one direction to $10 \,A$ in the opposite direction in $0.8 \,s$. The self-inductance of the coil is: (in $\,mH$)

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