A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle $\theta$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is

Directions of currents in two parts are different, so directions of magnetic fields due to these currents are opposite. Also, applying Ohm's law across AB,

${\mathrm{i}}_1{\mathrm{R}}_1={\mathrm{i}}_2{\mathrm{R}}_2\Rightarrow {\mathrm{i}}_1{\mathrm{l}}_2={\mathrm{i}}_2{\mathrm{l}}_2$

$\begin{array}{c}\left(\because \mathrm{R}=\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}\right)\\ \text{ Also, }{\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{{\mathrm{i}}_1{\mathrm{l}}_1}{{\mathrm{r}}^2}\text{ and }{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{{\mathrm{i}}_2{\mathrm{l}}_2}{{\mathrm{r}}^2}(\because \mathrm{l}=\mathrm{r\theta })\\ \therefore \frac{{\mathrm{B}}_2}{{\mathrm{B}}_1}=\frac{{\mathrm{i}}_1{\mathrm{l}}_1}{{\mathrm{i}}_2{\mathrm{l}}_2}=1\end{array}$

Hence, two field induction's are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of $\theta$.