Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A circular coil of radius $2R$ is carrying current $I$. The ratio of magnetic fields at the centre of coil and at a point at a distance $6R$ from centre of coil on axis of coil is

JIPMERJIPMER 2010Moving Charges and Magnetism

Solution:

At the centre of coil ,$B_{1} = \frac{\mu_{0}}{4\pi} \frac{2\pi l}{2R}$
At a distance $6R$ from centre of coil and on the axis of coil, $B_{2} = \frac{\mu_{0}}{4\pi} \frac{2\pi I\left(2R\right)^{2}}{\left[\left(2R\right)^{2}+\left(6R\right)^{2}\right]^{3/2}}$
$B_{2} = \frac{\mu_{0}}{4\pi} \frac{2\pi I \times 4R^{2}}{\left(2\sqrt{10}\right)^{3}R^{3}} $
$\therefore \:\:\frac{B_{1}}{B_{2}} = 10 \sqrt{10}$