Question

A current $I$ flows in a long straight wire with cross-section having the form of a thin half-ring of radius $R$ (figure shown above). If the induction of the magnetic field at the point $O$ is given as $B=\frac{x{\mu }_{0}i}{{\pi }^{2}R}$. Find $x$

Answer 1

First of all let us find out the direction of vector $\overrightarrow{B}$ at point $O$. For this purpose, we divide the entire conductor into elementary fragments with current $di$. It is obvious that the sum of any two symmetric fragments gives a resultant along $\overrightarrow{B}$ shown in the figure below and consequently, vector $\overrightarrow{B}$ will also be directed as shown
So, $|\overrightarrow{B}|=\int dB\sin \phi$ (1)
$=\int \frac{{\mu }_0}{2\pi R}di\sin \phi$
$={\int }_0^{\pi }\frac{{\mu }_0}{2{\pi }^{2}R}i\sin \phi d\phi ,(asdi=\frac{i}{\pi }d\phi )$
Hence $B=\frac{{\mu }_{0}i}{{\pi }^{2}R}$