Since the magnetic field due to all current elements at P are parallel to the z-direction, the field at P due to a wire, the ends of which make angles and at P is given by a straightforward integration
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and 
at P is given by a straightforward integration ![\begin{eqnarray*} \vec B &=& \frac{\mu_0I}{4\pi a} \int_{\theta_1}^{\theta_2}\si... ... \frac{\mu_0I}{4\pi a} \left[\sin\phi_1+\sin\phi_2\right] \hat k \end{eqnarray*}](../main3_clip_image024_0003.gif)
