Question

A current $I$ flows through a closed loop as shown in figure. The magnetic field induction at the centre $O$ is

• A. $\frac{\mu_{0} I}{4 \pi R} \theta$
• B. $\frac{\mu_{0} I}{4 \pi R}(\theta+\sin \theta)$
• C. $\frac{\mu_{0} I}{4 \pi R}(\pi-\theta+\sin \theta)$
• D. $\frac{\mu_{0} I}{2 \pi R}(\pi-\theta+\tan \theta)$

Answer 1

Answer: (D)

$B_{O}=B_{P Q S}+B_{S P}$
$=\frac{\mu_{0}}{4 \pi} \frac{I}{R}(2 \pi-2 \theta)+\frac{\mu_{0}}{4 \pi} \frac{I}{R \cos \theta}(\sin \theta+\sin \theta)$
$=\frac{\mu_{0}}{2 \pi} \frac{I}{R}[\pi-\theta+\tan \theta]$