Question

Figure shows a small loop carrying current $I$. The curved portion is an arc of a circle of radius $R$ and the straight portion is a chord to the same circle subtending at an angle $\theta$. The magnetic induction at center $O$ is :

• A. zero
• B. always inward irrespective of the value of $\theta$
• C. inward as long as $\theta$ is less than $\pi$
• D. always outward irrespective of the value of $\theta$

Answer 1

The correct option is C inward as long as $\theta$ is less than $\pi$

Magnetic field at O due to curved portion: $B_{curved}=\frac{{\mu }_{0}i}{2R}\times \frac{\theta }{2\pi }$ is out of the plane of the paper
Magnetic field at O due to the straight wire: $B_{wire}=\frac{{\mu }_{0}i}{4\pi d}\times 2\sin \left(\frac{\theta }{2}\right)$ where $d=R\cos \frac{\theta }{2}$

$\Rightarrow B_{wire}=\frac{{\mu }_{0}i}{4\pi R\cos \frac{\theta }{2}}2\sin \left(\frac{\theta }{2}\right)=\frac{{\mu }_{0}i}{2\pi R}\tan \frac{\theta }{2}$ into the plane of the paper.
The net magnetic field $B_{net}=B_{wire}-B_{curved}=\frac{{\mu }_{0}i}{2\pi R}(\tan \frac{\theta }{2}-\frac{\theta }{2})$

For $\theta >\pi$ $B_{net}$, is negative ie. it is out of the plane of the paper.