Q. A part of a long wire carrying a current $i$ is bent into a circle of radius $r$ as shown in figure. The net magnetic field at the centre $O$ of the circular loop is
A) $\frac{\mu_{0}\ i}{4r}$
B) $\frac{\mu_{0}\ i}{2r}$
C) $\frac{\mu_{0}\ i}{2\pi\ r}(\pi + 1)$
D) $\frac{\mu_{0}\ i}{2\ \pi\ r}(\pi - 1)$
Explanation
(c) The magnitude of the magnetic field at point $O$ due to straight part of wire is
$B_{1} = \frac{\mu_{0}i}{2\pi r}$
$B_{1}$ is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
The field at the centre $O$ due to the current loop of radius $r$ is
$B_{2} = \frac{\mu_{0}i}{2r}$
$B_{2}$ is also perpendicular to the page, directed upwards (right hand screw rule).
$B_{1} + B_{2} = \frac{\mu_{0}i}{2r}\left( \frac{1}{\pi} + 1 \right)$
$= \frac{\mu_{0}i}{2\pi r}\ (\pi + 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.