Explanation

(b) According to question, resistance of wire $ADC$ is twice that of wire $ABC$. Hence current flowing through $ADC$ is half that of $ABC,\ i.e.,\frac{i_{2}}{i_{1}} = \frac{1}{2}$. Also $i_{1} + i_{2} = i$
$\Rightarrow i_{1} = \frac{2i}{3}$ and $i_{2} = \frac{i}{3}$
Magnetic field at centre $O$ due to wire $AB$ and $BC$ (part 1 and 2) $B_{1} = B_{2} = \frac{\mu_{0}}{4\pi}.\frac{2i_{1}\sin{45{^\circ}}}{a/2}\bigotimes = \frac{\mu_{0}}{4\pi}.\frac{2\sqrt{2}i_{2}}{a}\bigotimes$
And magnetic field at centre $O$ due to wires $AD$ and $DC$
[$i.e.$ part 3 and 4] $B_{3} = B_{4} = \frac{\mu_{0}}{4\pi}\frac{2\sqrt{2}i_{2}}{a}\bigodot$
Also $i_{1} = 2i_{2}$. So $\left( B_{1} = B_{2} \right) > \left( B_{3} = B_{4} \right)$
Hence net magnetic field at centre $O$
$B_{net} = \left( B_{1} + B_{2} \right) - (B_{3} + B_{4})$
$= 2 \times \frac{\mu_{0}}{4\pi}.\frac{2\sqrt{2} \times \left( \frac{2}{3}i \right)}{a} - \frac{\mu_{0}}{4\pi}.\frac{2\sqrt{2}\left( \frac{i}{3} \right) \times 2}{a}$
$= \frac{\mu_{0}}{4\pi}.\frac{4\sqrt{2}i}{3a}(2 - 1) \otimes = \frac{\sqrt{2}\mu_{0}i}{3\pi\ a} \otimes$