d
(d) The field at \(O\) due to \(AB\) is \(\frac{{{\mu _0}}}{{4\pi }}.\frac{i}{a}\hat k\) and that due to \(DE\) is also \(\frac{{{\mu _0}}}{{4\pi }}.\frac{i}{a}\hat k\).
However the field due to \(BCD\) is \(\frac{{{\mu _0}}}{{4\pi }}.\frac{i}{a}\left( {\frac{\pi }{2}} \right)\,\hat k\).
Thus the total field at \(O\) is \(\frac{{{\mu _0}}}{{4\pi }}.\frac{i}{a}\left( {2 + \frac{\pi }{2}} \right)\,\hat k\)