Solution
= Solution
The negative-$z$ segment integrates to the stated azimuthal field. The negative-$y$ segment contributes
$$
\frac{\mu_0I}{4\pi}\frac{x\hat z-z\hat x}{x^2+z^2}\left(1-\frac y{\sqrt{x^2+y^2+z^2}}\right).
$$
Adding this to the first field is the complete answer. On $x=0,y=1$, the field tends to $0$ as $z\to+\infty$ and to $-\mu_0I\hat x/(2\pi)$ as $z\to-\infty$.