Question

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 magnetic field induction at the centre of the loop $O$ is

• A. zero
• B. $\frac{{\mu }_{0}I\theta }{4\pi r}$
• C. $\frac{{\mu }_0}{2\pi }\frac{I}{r}(\pi -\theta )$
• D. $\frac{{\mu }_0}{2\pi }\frac{I}{r}(2\pi -\theta )$

Answer 1

Answer: (C)

Magnetic field induction at $O$ due to current through $ACB$ is
$B_1=\frac{{\mu }_0}{4\pi }\frac{I\theta }{r}$
It is acting perpendicular to the paper downwards.
Magnetic field induction at $O$ due to current through $ADB$ is
$B_2=\frac{{\mu }_0}{4\pi }\frac{I}{r}(2\pi -\theta )$
It is acting perpendicular to paper upwards.
$\therefore$ Total magnetic field at $O$ due to current loop is
$B=B_2-B_1=\frac{{\mu }_0}{4\pi }\frac{I}{r}(2\pi -\theta )-\frac{{\mu }_0}{4\pi }\frac{I}{r}\theta =\frac{{\mu }_0}{2\pi }\frac{I}{r}(\pi -\theta )$
acting perpendicular to paper upwards.