$=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{y}}\left(1+\frac{{x}}{\sqrt{{x}^{2}+{y}^{2}}}\right) \ldots \ldots(1)$

${B}_{\text {due to wire }(2)}=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{x}}\left(\sin 90^{\circ}+\sin \theta_{2}\right)$

$=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{x}}\left(1+\frac{{y}}{\sqrt{{x}^{2}+{y}^{2}}}\right) \ldots \ldots(2)$

Total magnetic field

${B}={B}_{1}+{B}_{2}$

${B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{1}{{y}}+\frac{{x}}{{y} \sqrt{{x}^{2}+{y}^{2}}}+\frac{1}{{x}}+\frac{{y}}{{x} \sqrt{{x}^{2}+{y}^{2}}}\right]$

${B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{{x}+{y}}{{xy}}+\frac{{x}^{2}+{y}^{2}}{{xy} \sqrt{{x}^{2}+{y}^{2}}}\right]$

${B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{{x}+{y}}{{xy}}+\frac{\sqrt{{x}^{2}+{y}^{2}}}{{xy}}\right]$

${B}=\frac{\mu_{0} {I}}{4 \pi {xy}}\left[\sqrt{{x}^{2}+{y}^{2}}+({x}+{y})\right]$