Problem 22.51

22.51. a)
Take a small element of the segment, ds. Then ds is a distance r from P. The direction of the magnetic field at P due to ds is out of the page since $d\vec{s} \times \hat{r}$ is out of the page. In fact, all elements ds contribute a component to the magnetic field that is out of the page, so we only need to find the magnitude of the field at P since the direction is determined.
In this case, $d\vec{s} = \hat{i} dx$. So, $\vert d\vec{s} \times \hat{r}\vert = dx \sin \theta$
Where $\theta$ is the angle that is made between $d\vec{s}$ and $\hat{r}$.
Plugging into Biot-Savart, we find
$dB = \frac{\mu_{0} I dx \sin \theta}{4 \pi r^{2}}$
In order to make any use of this equation, we have to eliminate two of the three variables, r, $\theta$, and x. I choose to keep $\theta$. We look at the geometric relations in the problem and realize that
$r = \frac{a}{\sin \theta}$
or, $r = a \csc \theta$
and also we see that
$\tan \theta = \frac{a}{x}$
solving for x, we find that
$x = a \tan \theta$
taking a derivative, we see that
$dx = a \csc^{2} \theta d\theta$
Substituting into our magnetic field equation, we see that
$dB = \frac{mu_{0} I a \csc^{2} \theta \sin \theta d\theta}{4 \pi a^{2} \csc^{2} \theta}$
$dB = \frac{\mu_{0} I}{4 \pi a}\sin \theta d\theta$
Integrating, we find
$B = \frac{\mu_{0} I}{4 \pi a} \int_{\theta_{1}}^{\theta_{2}} \sin \theta d\theta$
$B = \frac{\mu_{0} I}{4 \pi a} (\cos \theta_{1} - \cos \theta_{2})$
Where the angles are given in the problem.

b)
For an infinite wire, $\theta_{1} \rightarrow 0$ and $\theta_{2} \rightarrow -\pi$. Therefore, our result reduces to the expected form for the magnetic field a distance a away from a straight wire,
$B = \frac{\mu_{0}I}{2 \pi a}$