Q.
Equal current $i$ flows in two segments of a circular loop in the direction shown in figure. Radius of the loop is $r$. The magnitude of magnitude field induction at the centre of the loop is
A) zero
B) $\frac{\mu_{0}}{4\pi}\frac{i\theta}{r}$
C) $\frac{\mu_{0}}{2\pi}\frac{i}{r}\ (\pi - \theta)$
D) $\frac{\mu_{0}}{2\pi}\frac{i}{r}(2\pi - \theta)$
Explanation
(c) Magnetic field induction at $O$ due to current through $ACB\ is\ B_{1} = \frac{\mu_{0}i\theta}{4\pi r}$
It is acting perpendicular to the paper downwards.
Magnetic field induction at $O$ due to current through $ABD\ is\ B_{2} = \frac{\mu_{0}}{4\pi}\ \frac{i(2\pi - \theta)}{r}$
It is acting perpendicular to paper upwards.
$\therefore\ \ $ Total magnetic field at $O$ due to current loop is
$B = B_{2} - B_{1} =$ $\frac{\mu_{0}}{4\pi}\frac{i}{r}(2\pi - \theta)\frac{\mu_{0}}{4\pi}\frac{i}{r}\theta$
$\ = \ \frac{\mu_{0}}{2\pi}\frac{i}{r}\ (\pi - \theta)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.