Answer
$B=3.6\times10^{-5}T$
$\theta=17^o$ below horizontal
Work Step by Step
$B_{wire}=\frac{\mu_0I}{2\pi r}=\frac{(2\times10^{-7})(24.0A)}{0.200m}=2.4\times10^{-5}T$
$B_y=2.4\times10^{-5}T-(5\times10^{-5}T)\sin(44^o)$$=1.07\times10^5T$ South
$B_x=(5\times10^{-5}T)\cos(44^o)=3.60\times10^{-5}T$
$B=\sqrt{B_x^2+B_y^2}=3.6\times10^{-5}T$
$\theta=\tan^{-1}(\frac{1.07\times10^5T}{3.60\times10^{-5}T})=17^o$ below horizontal
