Physics - Magnetism Question with Solution | TestHub

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 $AB$ 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 $AB$ 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 $\begin{array}{rl}& {\mathrm{B}}_{\mathrm{R}}={\mathrm{B}}_1+{\mathrm{B}}_2\\ & =-\frac{{\mu }_0{\mathrm{i}}_1\times 270}{4\pi \mathrm{R}}+\frac{{\mu }_0{\mathrm{i}}_2\times 90}{4\pi \mathrm{R}}\text{ [From Eq. (i) and (ii)] }\end{array}$ 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. $\begin{array}{rl}{\mathrm{V}}_1& ={\mathrm{V}}_2\\ {\mathrm{i}}_1{\mathrm{R}}_1& ={\mathrm{i}}_2{\mathrm{R}}_2\end{array}$ where, ${\mathrm{R}}_1$ and ${\mathrm{R}}_2=$ Resistance of respective segments The wire is uniform so $\begin{array}{rl}& \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{{\mathrm{L}}_1}{{\mathrm{L}}_2}=\frac{\mathrm{R}\times 270}{\mathrm{R}\times 90}[\because \text{ length of arc }=\text{ radius }\times \\ & \text{ angle] }\end{array}$ 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 $3i_1=i_2$ From Eq. (iii) and (v), we get $\begin{array}{rl}{\mathrm{B}}_{\mathrm{R}}& =\frac{{\mu }_0}{4\pi \mathrm{R}}\left(-270{\mathrm{i}}_1+90{\mathrm{i}}_2\right)\\ & =\frac{{\mu }_0}{4\pi R}\left[-270{\mathrm{i}}_1+90\left(3{\mathrm{i}}_1\right)\right]\\ & =\frac{{\mu }_0}{4\pi \mathrm{R}}\left(-270{\mathrm{i}}_1+270{\mathrm{i}}_1\right)=0\end{array}$