Q. What is the self inductance of an air core solenoid 1 m long, diameter 0.05m, if it has 500 turns? Take $\pi^{2} = 10.$
A) $3.15\ \times 10^{- 4}\ $H
B) $4.8\ \times 10^{- 4}\ $ H
C) $5\ \times 10^{- 4}\ $H
D) $6.25\ \times 10^{- 4}\ $H
Explanation
(d) $L = \ \frac{\mu_{0}N^{2}A}{l} = \frac{\mu_{0}N^{2}\left( \pi r^{2} \right)}{l}$
$= \ \frac{4\pi \times 10^{- 7} \times (500)^{2} \times \pi(0.025)^{2}}{1}$
$= 4 \times 10\ \times 10^{7} \times (500)^{2} \times (0.025)^{2}$
= 6.25 $\times 10^{- 4}$H
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.