Q. A circular coil of radius $5\ cm$ has 500 turns of a wire. The approximate value of the coefficient of self induction of the coil will be
A) $25\ millihenry$
B) $25 \times 10^{- 3}millihenry$
C) $50 \times 10^{- 3}millihenry$
D) $50 \times 10^{- 3}millihenry$
Explanation
(a) $\phi = Li \Rightarrow NBA = Li$
Since magnetic field at the centre of circular coil carrying current is given by $B = \frac{\mu_{0}}{4\pi}.\frac{2\pi Ni}{r}$
$\therefore N.\frac{\mu_{0}}{4\pi}.\frac{2\pi Ni}{r}.\pi r^{2} = Li \Rightarrow L = \frac{\mu_{0}N^{2}\pi r}{2}$
Hence self inductance of a coil
$= \frac{4\pi \times 10^{- 7} \times 500 \times 500 \times \pi \times 0.05}{2} = 25\ mH$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.