A battery 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  

Let $I_1$ and $I_2$ be the lengths of two parts $ARB$ and $ASB$ of the ring, and $\rho$ be resistance per unit length of the conductor.

The resistance of the portion $ARB$ will be $R_1=l_1\rho$. The resistance of the portion $ASB$ will be $R_2=l_2\rho$. Let $i_1$ and $i_2$ be the  currents in $ARB$ and $ASB$ respectively.

Since, these parts are in parallel, the potential difference across their ends will be same.

$\therefore i_1{R}_1=i_2{R}_2$

or $i_1{l}_1\rho =i_2{l}_2\rho$

or $i_1{l}_1=i_2{l}_2$

According to Biot Savart's law, the magnetic fields at the centre $O$ due to current through circular ring $ARB$ and $ASB$ will be 

$B_1=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{i_1{l}_1 \sin 90°}{r^2}$

and $B_2=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{i_2{l}_2 \sin 90°}{r^2}$

Since, $i_1{l}_1=i_2{l}_2$

$\therefore B_1=B_2$

According to right hand rule, the directions of $B_1$ and $B_2$ are opposite to each other, hence the resultant magnetic field at $O$ will be zero.