A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field…

The magnetic field at the centre of an arc subtended at an angle $\theta$θ is given by $\mathrm{B}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} \times \frac{\theta}{2 \pi}$B=2Rμ0​i​×2πθ​ Then, the magnetic field due to larger arc $A B$AB is $\mathrm{B}_1=\frac{\mu_0 \mathrm{i}_1}{2 \mathrm{R}} \times \frac{270}{2 \pi}$B1​=2Rμ0​i1​​×2π270​ which acts in inward direction according to right hand thumb rule. And magnetic field due to smaller arc $A B$AB is $\mathrm{B}_2=\frac{\mu_0 \mathrm{i}_2}{2 \mathrm{R}} \times \frac{90}{2 \pi}$B2​=2Rμ0​i2​​×2π90​ which acts in outward direction. The resultant magnetic field $$\begin{aligned} & \mathrm{B}_{\mathrm{R}}=\mathrm{B}_1+\mathrm{B}_2 \\ & =-\frac{\mu_0 \mathrm{i}_1 \times 270}{4 \pi \mathrm{R}}+\frac{\mu_0 \mathrm{i}_2 \times 90}{4 \pi \mathrm{R}} \text { [From Eq. (i) and (ii)] } \end{aligned}$$​BR​=B1​+B2​=−4πRμ0​i1​×270​+4πRμ0​i2​×90​ [From Eq. (i) and (ii)] ​ which acts in inward direction as $\mathrm{B}_1>\mathrm{B}_2$B1​>B2​. Two arcs can also be seen as the two resistances in parallel combination. So, the potential across them will be same i.e. $$\begin{aligned} \mathrm{V}_1 & =\mathrm{V}_2 \\ \mathrm{i}_1 \mathrm{R}_1 & =\mathrm{i}_2 \mathrm{R}_2 \end{aligned}$$V1​i1​R1​​=V2​=i2​R2​​ where, $\mathrm{R}_1$R1​ and $\mathrm{R}_2=$R2​= Resistance of respective segments The wire is uniform so $$\begin{aligned} & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{R} \times 270}{\mathrm{R} \times 90} [\because \text { length of arc }=\text { radius } \times \\ & \text { angle] } \end{aligned}$$​R2​R1​​= L2​L1​​=R×90R×270​[∵ length of arc = radius × angle] ​ From Eq. (iv), we get $\Rightarrow \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{90}{270}=\frac{1}{3}$⇒i2​i1​​=R1​R2​​=27090​=31​ or $3 i_1=i_2$3i1​=i2​ From Eq. (iii) and (v), we get $$\begin{aligned} \mathrm{B}_{\mathrm{R}} & =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+90 \mathrm{i}_2\right) \\ & =\frac{\mu_0}{4 \pi R}\left[-270 \mathrm{i}_1+90\left(3 \mathrm{i}_1\right)\right] \\ & =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+270 \mathrm{i}_1\right)=0 \end{aligned}$$BR​​=4πRμ0​​(−270i1​+90i2​)=4πRμ0​​[−270i1​+90(3i1​)]=4πRμ0​​(−270i1​+270i1​)=0​