Question

Current $I$ flows in an infinitely long straight wire with cross-section in the form of thin uniform semicircular ring of radius $R$ as shown in the diagram. The magnetic field at the centre $O$ will be -

• A. $\frac{{\mu }_{o}I}{{\pi }^{2}R}$
• B. $\frac{{\mu }_{o}I}{4{\pi }^{2}R}$
• C. $\frac{{\mu }_{o}I}{2{\pi }^{2}R}$
• D. $\frac{{\mu }_{o}I}{8{\pi }^{2}R}$

Answer 1

The correct option is A $\frac{{\mu }_{o}I}{{\pi }^{2}R}$

Divide the entire wire into elementary segments, which are infinitely long, each carrying a current $dI$.


The magnetic field due to an elementary segment at centre $O$ is given by

$dB=\frac{{\mu }_{o}dI}{2\pi R}$

Now, the total current $I$ is distributed over an angle $\pi \text{rad}.$

Hence, we can write, $\frac{I}{\pi }=\frac{dI}{d\theta }$

$\Rightarrow dI=\frac{I}{\pi }d\theta$

From symmetry, $dB\sin \theta$ components will get cancelled.

So, the total magnetic field at $O$ will be,

$B=\int dB\sin \theta ={\int }_0^{\pi }\frac{{\mu }_o}{2\pi R}\frac{I}{\pi }\sin \theta d\theta$


$=\frac{{\mu }_{o}I}{2{\pi }^{2}R}[-\cos \theta {]}_0^{\pi }=\frac{{\mu }_{o}I}{{\pi }^{2}R}$


Hence, option $\left(A\right)$ is the correct answer.