The current through a coil of self-inductance | vedclass

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(B) The induced $emf$ $(e)$ in a coil is given by the formula $e = -L \frac{dI}{dt}$.For the $emf$ to be zero, the rate of change of current $\frac{dI

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(B) The induced $emf$ $(e)$ in a coil is given by the formula $e = -L \frac{dI}{dt}$.For the $emf$ to be zero, the rate of change of current $\frac{dI}{dt}$ must be zero.Given $I = t^2 e^{-t}$.Using the product rule for differentiation: $\frac{dI}{dt} = \frac{d}{dt}(t^2) \cdot e^{-t} + t^2 \cdot \frac{d}{dt}(e^{-t})$.$\frac{dI}{dt} = 2t e^{-t} + t^2 (-e^{-t}) = e^{-t} (2t - t^2) = e^{-t} t(2 - t)$.Setting $\frac{dI}{dt} = 0$, we get $e^{-t} t(2 - t) = 0$.Since $e^{-t} 
eq 0$ for finite $t$, the solutions are $t = 0$ or $t = 2 \ s$.At $t = 0$, the current is zero, but the $emf$ becomes zero at $t = 2 \ s$ as the current reaches its maximum value.

The current through a coil of self-inductance $L = 2 \ mH$ is given by $I = t^2 e^{-t}$ at time $t$. How long will it take for the induced electromotive force $(emf)$ to become zero (in $s$)?

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