Q. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field $B$ at the centre of the coil?
A) $\pi \times 10^{- 3}\ T$
B) $2\pi \times 10^{- 4}\ T$
C) $\pi \times 10^{- 4}\ T$
D) Zero
Explanation
(c) Given, $N = 100,\ r = 8\ cm = 0.08\ m$
$I = 0.4\ A$
$B = ?$
$B = \frac{\mu_{0}NI}{2r} = \frac{4\pi \times 10^{- 7} \times 100 \times 0.4}{2 \times 0.08} = \pi \times 10^{- 4}\ T.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.