Q. A coil of N =100 turns carries a current I =5 A and creates a magnetic flux $\phi = 10^{- 5}\ {Tm}^{2}$ per turn. The value of its inductance L will be
A) 0.05 mH
B) 0.10 mH
C) 0.15 mH
D) 0.20 mH
Explanation
(d) The inductance of a coil of wire of $N$ turns is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ L = N\frac{\phi}{i}\ $
Where $i$ is current and $\phi$ the magnetic flux.
Given, $N = 100,\ i = 5A,\phi = 10^{- 5}{Tm}^{2}(turn)^{- 1}$
$\therefore\ \ \ \ \ \ \ \ L = 100 \times \frac{10^{- 5}}{5} = 0.20\ mH$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.