Q. A circular coil carrying a current has a radius $R.$ The ratio of magnetic induction at the centre of the coil and at a distance equal to $\sqrt{3}\ R$ from the centre of the coil on the axis is
A) 1 : 1
B) 1 : 2
C) 2 : 1
D) 8 : 1
Explanation
(d) $\ \ \ \ \ \ \ \ B_{axis} = \frac{\mu_{0}niR^{2}}{2\left( R^{2} + x^{2} \right)^{3/2}}$
$\ \ \ \ B_{centre} = \frac{\mu_{0}ni}{2R}$
At $x = \sqrt{3}R,\ \ \ \ \ \ \ \ B_{axis} = \frac{\mu_{0}niR^{2}}{2\left( R^{2} + {3R}^{2} \right)^{3/2}} = \ \frac{\mu_{0}ni}{16R}$
$\therefore\frac{B_{centre}}{B_{axis}} = \frac{8}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.