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.