The magnetic field at O due to current in the infinite wire forming a loop as shown in figure is

$\begin{array}{l}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi d}}\left[\sin {\text{θ}}_1+\sin {\text{θ}}_2\right]\\ \mathrm{But} {\theta }_1={90}^0-{\phi }_1 and {\theta }_2={90}^0 -{\phi }_2\\ \sin {\mathrm{\theta }}_1=\sin \left({90}^0-{\mathrm{\varphi }}_1\right)=\cos {\mathrm{\varphi }}_1\\ \mathrm{Similarly} \sin {\theta }_2=\cos {\mathrm{\varphi }}_2 .\mathrm{Magnetic} \mathrm{field} \mathrm{on} \mathrm{the} \mathrm{axis} \mathrm{of} \mathrm{a} \mathrm{current} \mathrm{carrying} \mathrm{conductor} \mathrm{is} \mathrm{zero}\\ \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi d}}\left(\cos {\mathrm{\varphi }}_1+\cos {\mathrm{\varphi }}_2\right)\end{array}$