When the current change from + 2A to – 2A in 0.05 second, an e.m.f. of 8 V is induced in a coil. The coefficient of self-induction of the coil is
0.1 H
0.2 H
0.4 H
0.8 H
e=Ldidt⇒8=L×2−−20.05⇒L=0.1H