2)
$ \begin{equation*} \boxed{d\bar{H}=\frac{I(\bar{R}')d\bar{l}'\times(\bar{R}-\bar{R}')}{4\pi\abs{\bar{R}-\bar{R}'}^3}} \end{equation*} $
$ \begin{align*} \text{\underline{along x}:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idx\hat{x}\times(-x\hat{x})}{4\pi x^3}=0\\ & &\bar{R}'&=x\hat{x} &\\ & &\abs{\bar{R}-\bar{R}'}&=x && &\\ \text{\underline{along y}:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idy\hat{y}\times(-y\hat{y})}{4\pi y^3}=0\\ & &\bar{R}'&=y\hat{y} &\\ & &\abs{\bar{R}-\bar{R}'}&=y && & \end{align*} $
$ \begin{equation*} \text{\underline{Superposition}: Total field} \qquad \boxed{\bar{H}=0} \end{equation*} $
