d
(d) Directions of currents in two parts are different, so directions of magnetic fields due to these currents are opposite. Also applying Ohm’s law across \(AB\)
\({i_1}{R_1} = {i_2}{R_2} \Rightarrow {i_1}{l_2} = {i_2}{l_2}\)     \(\left( {\;R = \rho \frac{l}{A}} \right)\)
Also \({B_1} = \frac{{{\mu _o}}}{{4\pi }} \times \frac{{{i_1}{l_1}}}{{{r^2}}}\) and \({B_2} = \frac{{{\mu _o}}}{{4\pi }} \times \frac{{{i_2}{l_2}}}{{{r^2}}}\)    (\(\;l = r\theta \))
\(\therefore \,\frac{{{B_2}}}{{{B_1}}} = \frac{{{i_1}{l_1}}}{{{i_2}{l_2}}} = 1\)
Hence, two field induction’s are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of \(\theta \).