Q. Due to 10 ampere of current flowing in a circular coil of $10\ cm$ radius, the magnetic field produced at its centre is $3.14 \times 10^{- 3}weber/m^{2}$. The number of turns in the coil will be
A) 5000
B) 100
C) 50
D) 25
Explanation
(c) $B = \frac{\mu_{0}}{4\pi}.\frac{2\pi Ni}{r}$
$\Rightarrow 3.14 \times 10^{- 3} = \frac{10^{- 7} \times 2 \times 3.14 \times N \times 10}{(10 \times 10^{- 2})} \Rightarrow N = 50$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.