Solution

The wire carries a current $$I$$ in the negative $$z-direction$$.
We have to consider the magnetic vector field $$\vec {B}$$ at $$(x, y)$$ in the $$z = 0$$ plane.
Magnetic field $$\vec {B}$$ is perpendicular to $$OP$$.
$$\therefore \vec {B} = B\sin \theta \hat {i} - B\cos \theta \hat {j}$$
$$\sin \theta = \dfrac {y}{r}, \cos \theta = \dfrac {x}{r} B = \dfrac {\mu_{0}I}{2\pi r}$$
$$\therefore \vec {B} = \dfrac {\mu_{0}I}{2\pi r^{2}} (y\hat {i} - x\hat {j})$$
or $$\vec {B} = \dfrac {\mu_{0}I (y\hat {i} - x\hat {j})}{2\pi (x^{2} + y^{2})}$$.