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

Solution :
It is given that emf is zero i.e.,
$$\eqalign{ & e = - L\frac{{di}}{{dt}} = 0 \cr & {\text{or}}\,\,L\frac{{di}}{{dt}} = 0 \cr & {\text{or}}\,\,\frac{d}{{dt}}\left( {{t^2}{e^{ - t}}} \right) = 0\,\,\left( {{\text{As,}}\,\,i = {t^2}{e^{ - t}}} \right) \cr & {\text{or}}\,\,2t \times {e^{ - t}} + {t^2} \times \left( { - 1} \right){e^{ - t}} = 0 \cr & {\text{or}}\,\,t{e^{ - t}}\left( {2 - t} \right) = 0\,\,{\text{or}}\,\,t = 2\,s\,\,\,\left( {\because t{e^{ - t}} \ne 0} \right) \cr} $$