Q. An air cored coil has a self-inductance of 0.1 H. A soft iron core of relative permeability 100 is 1/10 th. The value of self-inductance now becomes
A) 1 mH
B) 10 mH
C) 0.4 H
D) 0.8 H
Explanation
(c) From $L = \ \frac{\mu_{0}N^{2}A}{l} = \ \frac{\mu_{0}\mu_{r}N^{2}A}{l},$
When $\mu = 1000\ \ and\ N\ \ becomes\frac{1}{10}\ $
$\therefore\ \ \ L\ becomes\ 1000 \times \left( \frac{1}{10} \right)^{2} = 10$ times
$ie,\ \ \ \ L = 10 \times 0.1 = 1$H
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.