Two semicircular ring of same radius $\left(R\right)$ having currents  $i_1$ and  $i_2$ are kept parallel to each other in $y-z$ plane as shown in the figure. Centre of first ring is kept at $x = 0$ and that of second ring is kept at  $x=\frac{7}{12}R$ . The magnetic field due to semi-circular rings at point  $P(x=\frac{4R}{3})$ is found to be along the x-axis and its value is  $\frac{n{\mu }_0{i}_1}{20R}$. Find the value of n.

Ring 1 will produce magnetic field along -x axis and in lower half of y-z plane.

Ring 2 will produce magnetic field along -x axis and in upper half of y-z plane.

Since , it is given that the magnetic field at point P only exist in x direction, we can say that lower and upper half of magnetic field due to ring 1 and ring 2 respectively, will cancel out each other.

$B_{yz Ring1}=B_{yz Ring2}$

$\Rightarrow \frac{{\mu }_0}{4\pi }\frac{i_1\left(\pi R\right)}{\left({\left({\displaystyle \frac{4R}{3}}\right)}^2+R^2\right)}\cos {37}^0=\frac{{\mu }_0}{4\pi }\frac{i_2\left(\pi R\right)}{\left({\left({\displaystyle \frac{3R}{4}}\right)}^2+R^2\right)}\cos {53}^0$

Using, $\sin {37}^0=\frac{R}{\sqrt{{\left(\frac{4R}{3}\right)}^2+R^2}}$ and $\sin {53}^0=\frac{R}{\sqrt{{\left(\frac{3R}{4}\right)}^2+R^2}}$

$\Rightarrow \frac{{\mu }_0}{4R}{i}_1{\sin }^2{37}^0\cos {37}^0=\frac{{\mu }_0}{4R}{i}_2{\sin }^2{53}^0\cos {53}^0$

$\begin{array}{l}{i}_1{\sin }^{2}37\cos 37=i_2{\sin }^{2}53\cos 53\\ i_2=\frac{3}{4}{i}_1\\ B_x=\frac{{\mu }_0{i}_1}{4R}{\sin }^3{37}^0+\frac{{\mu }_0}{4R}\times \frac{3}{4}{i}_1{\sin }^3{53}^{\circ }\\ \frac{{\mu }_0{i}_1}{4R}\left[\frac{27}{125}+\frac{3}{4}\times \frac{64}{125}\right]\\ \frac{{\mu }_0{i}_1}{4R}\times \frac{75}{125}=3\frac{{\mu }_0{i}_1}{20R}\end{array}$