In an inductor of self-inductance L = 2 , mH, | vedclass

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(C) Given: Self-inductance $L = 2 \, mH$ and current $i = t^2 e^{-t}$.The induced $emf$ $(E)$ in an inductor is given by the formula $E = -L \frac{di}

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$2$

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(C) Given: Self-inductance $L = 2 \, mH$ and current $i = t^2 e^{-t}$.The induced $emf$ $(E)$ in an inductor is given by the formula $E = -L \frac{di}{dt}$.First,calculate the derivative of current with respect to time:$\frac{di}{dt} = \frac{d}{dt} (t^2 e^{-t}) = (2t) e^{-t} + t^2 (-e^{-t}) = e^{-t} (2t - t^2)$.Now,substitute this into the $emf$ equation:$E = -L [e^{-t} (2t - t^2)]$.For the $emf$ to be zero $(E = 0)$,the term inside the bracket must be zero:$e^{-t} (2t - t^2) = 0$.Since $e^{-t}$ is never zero for finite $t$,we have:$2t - t^2 = 0$$t(2 - t) = 0$.This gives $t = 0 \, s$ or $t = 2 \, s$. Excluding the initial state at $t = 0$,the $emf$ is zero at $t = 2 \, s$.

In an inductor of self-inductance $L = 2 \, mH$,current changes with time according to the relation $i = t^2 e^{-t}$. At what time is the $emf$ zero? (in seconds)

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