Q. A vertical circular coil of radius 0.1 m and having 10 turns carries a steady current. When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. If $B_{H} = 0.314 \times 10^{- 4}$ the current in the coil is
A) 0.5 A
B) 0.25 A
C) 2 A
D) 1 A
Explanation
(a) Magnetic field at the centre of circular coil
$\ \ \ B_{H} = \frac{\mu_{0}}{4\pi}\frac{2\pi nI}{r}$
$I$ and $r$ being the current and radius of circular coil respectively.
or $I = \frac{4\pi}{\mu_{0}} = \frac{rB_{H}}{2\pi n}$
$\ \ \ \ \ \ \ \ \ \ = \frac{10^{7} \times 0.1 \times 0.314 \times 10^{- 4}}{2 \times 3.14 \times 10} = 0.5\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.