Q. When current in a coil changes from 2 A to 4 A in 0.05s, an emf of 8 V is induced in the coil. Self inductance of the coil is
A) 0.1 H
B) 0.2 H
C) 0.4 H
D) 0.8 H
Explanation
(b) From $e = LdI/dt$, $L = \frac{edt}{dI} = \frac{8 \times 0.05}{2} = 0.2$H
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.