b
(b) According to question resistance of wire \(ADC\) is twice that of wire \(ABC\). Hence current flows through \(ADC\) is half that of \(ABC\) i.e. \(\frac{{{i_2}}}{{{i_1}}} = \frac{1}{2}\). Also \({i_1} + {i_2} = i\) \(==>\) \({i_1} = \frac{{2i}}{3}\) and \({i_2} = \frac{i}{3}\)

Magnetic field at centre \(O\) due to wire \(AB\) and \(BC\) (part \(1\) and \(2\)) \({B_1} = {B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}\sin {{45}^o}}}{{a/2}} \otimes \)\( = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,{i_1}}}{a} \otimes \)

and magnetic field at centre \(O\) due to wires \(AD\) and \(DC\) (i.e. part \(3\) and \(4\)) \({B_3} = {B_4} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\sqrt 2 \,{i_2}}}{a}\odot\)
Also \(i_1 = 2i_2. \,So \,(B_1 = B_2) > (B_3 = B_4)\)

Hence net magnetic field at centre \(O\)
\({B_{net}} = ({B_1} + {B_2}) - ({B_3} + {B_4})\)
\( = 2 \times \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \, \times \left( {\frac{2}{3}i} \right)}}{a} - \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,\left( {\frac{i}{3}} \right) \times 2}}{a}\)
\( = \frac{{{\mu _0}}}{{4\pi }}.\frac{{4\sqrt 2 \,i}}{{3a}}(2 - 1)\, \otimes = \frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \otimes \)