Problem 22.18

21.18.
For the radial parts of the wire, $d\vec{s}$ is in the same direction as $\vec{r}$, so the magnetic field felt at P due to the radial parts of the wire is zero. We need only consider the arc part of the wire. In that part, $d\vec{s}$ is perpendicular to $\vec{r}$, so
$d\vec{s} \times \hat{r} = \vert ds\vert$
 
Now, from Biot - Savart, we have that
$B = \int dB ~=~ \int k_{m} \frac{I \vert d\vec{s} \times \hat{r}\vert}{r^{2}}$
Since the radius, r is a constant along the arc part of the wire, we can take it out of the integral. We are left with
$B = \frac{I k_{m}}{r^{2}} \int d\vec{s}$
$B = \frac{I k_{m}}{r^{2}} s$
Where the length s is the length of the arc part of the wire. We find that
$s = r \theta ~=~ (.6 m)(\frac{\pi}{6})$
s = .314 m
So, the magnetic field is given by
$B = (10^{-7})\frac{3A}{.6 m^{2}} (.314 m)$
B = 261 nT into the paper