A wire loop PQRSP is constructed by joining two semi circular coils of radii ${\mathrm{r}}_1 \mathrm{and} {\mathrm{r}}_2$ respectively as shown in the fig. current is flowing in the loop. The magnetic induction at point ${}^{\text{'}}o{ }^{\text{'}}$ will be

Given data:

The figure depicts the construction of a wire loop, which is formed by combining two semi-circular coils having radii $r_1$ of and $r_2$.

Formula used: $B_C=\frac{{\mu }_{0}iN}{2r}$

Detailed Solution:

So, therefore form the question we get magnetic field at C is 

$B=\frac{{\mu }_{0}i\left(0.5\right)}{2R_1}=\frac{{\mu }_{0}I}{2R_1}$ out of plane of paper

The magnetic field at C due to Semicircular coil of radius $R_2$,

$$\begin{gathered} B_2=\frac{{\mu }_{0}I\left(0.5\right)}{2R_2} \\ =\frac{{\mu }_{0}I}{4R_2} \end{gathered}$$

Hence, the net magnetic field at C is 

 $$\begin{gathered} B_{net}=B_1-B_2 \\ =\frac{{\mu }_{0}I}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \end{gathered}$$ 

Therefore, the net magnetic field is $\frac{{\mu }_{0}I}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$.