Q. Magnetic fields at two points on the axis of a circular coil at a distance of $0.05\ m$ and 0.2$m$ from the centre are in the ratio 8:1. The radius of the coil is
A) 1.0 $m$
B) 0.1 $m$
C) 0.15 $m$
D) 0.2 $m$
Explanation
(b) $B = \frac{\mu_{0}}{4\pi} \times \frac{2\pi NiR^{2}}{\left( R^{2} + x^{2} \right)^{3/2}} \Rightarrow B \propto \frac{1}{\left( r^{2} + x^{2} \right)^{3/2}}$
$\Rightarrow \frac{8}{1} = \frac{\left( R^{2} + x_{2}^{2} \right)^{3/2}}{\left( R^{2} + x_{1}^{2} \right)^{3/2}} \Rightarrow \left( \frac{8}{1} \right)^{2/3} = \frac{R^{2} + 0.04}{R^{2} + 0.0025}$
$\Rightarrow \frac{4}{1} = \frac{R^{2} + 0.04}{R^{2} + 0.0025}$. On solving $R = 0.1\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.