When the current changes from 2A to 4A in 0.05 second, an e.m.f. of 8 volt is induced in a coil. The coefficient of self induction of the coil is

0.2H 0.3H 0.4H 0.5H

0.2H

$L = \frac{\xi}{\frac{\Delta I}{\Delta t}} = \frac{\xi \Delta t}{\Delta I} = \frac{8\times 0.05}{2} = 0.2H$