A straight conductor carrying current i splits into two parts as...

The magnetic field at the centre of an arc subtended at an angle $\theta$ is given by $\mathrm{B}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} \times \frac{\theta}{2 \pi}$ Then, the magnetic field due to larger arc $A B$ is $\mathrm{B}_1=\frac{\mu_0 \mathrm{i}_1}{2 \mathrm{R}} \times \frac{270}{2 \pi}$ which acts in inward direction according to right hand thumb rule. And magnetic field due to smaller arc $A B$ is $\mathrm{B}_2=\frac{\mu_0 \mathrm{i}_2}{2 \mathrm{R}} \times \frac{90}{2 \pi}$ which acts in outward direction. The resultant magnetic field $$\begin{aligned} & \mathrm{B}_{\mathrm{R}}=\mathrm{B}_1+\mathrm{B}_2 \\ & =-\frac{\mu_0 \mathrm{i}_1 \times 270}{4 \pi \mathrm{R}}+\frac{\mu_0 \mathrm{i}_2 \times 90}{4 \pi \mathrm{R}} \text{ [From Eq. (i) and (ii)] } \end{aligned}$$ which acts in inward direction as $\mathrm{B}_1>\mathrm{B}_2$. Two arcs can also be seen as the two resistances in parallel combination. So, the potential across them will be same i.e. $$\begin{aligned} \mathrm{V}_1 & =\mathrm{V}_2 \\ \mathrm{i}_1 \mathrm{R}_1 & =\mathrm{i}_2 \mathrm{R}_2 \end{aligned}$$ where, $\mathrm{R}_1$ and $\mathrm{R}_2=$ Resistance of respective segments The wire is uniform so $$\begin{aligned} & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{R} \times 270}{\mathrm{R} \times 90} [\because \text{ length\text{ of } arc }=\text{ radius } \times \\ & \text{ angle] } \end{aligned}$$ From Eq. (iv), we get $\Rightarrow \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{90}{270}=\frac{1}{3}$ or 3 i_1=i_2 From Eq. (iii) and (v), we get $$\begin{aligned} \mathrm{B}_{\mathrm{R}} & =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+90 \mathrm{i}_2\right) \\ & =\frac{\mu_0}{4 \pi R}\left[-270 \mathrm{i}_1+90\left(3 \mathrm{i}_1\right)\right] \\ & =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+270 \mathrm{i}_1\right)=0 \end{aligned}$$