The average emf induced in a coil in which current changes from 2 A to 4 A in 0.05 sec is 8 V. The self - inductance of coil is:

0.2 H

The correct answer is Option (2) → 0.2 H

Self induced emf, $ε=-L\frac{ΔI}{Δt}$

Where,

$ε = 8V$ (average emf induced)

L: Self-inductance

$ΔI=4A-2A=2A$ (change in current)

$Δt = 0.05 s$

$L=\frac{ε.Δt}{ΔI}=\frac{8×0.05}{2}$

$L=\frac{0.4}{2}=0.2H$