When the current changes from 2a to 2a in 0.05 sec an emf of 8v is induced in a coil the coefficient of self-inductance of the coil is?

The coefficient of self-inductance ( L ) can be calculated using the formula for induced emf: ( \text{emf} = -L \frac{di}{dt} ). Here, the change in current ( di = 2A - 0A = 2A ) and the time interval ( dt = 0.05s ). Thus, ( \frac{di}{dt} = \frac{2A}{0.05s} = 40 A/s ). Rearranging the formula gives ( L = -\frac{\text{emf}}{\frac{di}{dt}} = -\frac{8V}{40 A/s} = -0.2 H ), so the self-inductance ( L ) is 0.2 H (Henries).