Problem 21.17

22.17. a)
We are given that the magnetic field away from a segment of wire is given by
$B = \frac{4 \mu_{0} I}{4 \pi a} (\cos \frac{\pi}{4} - \cos \frac{3\pi}{4})$
where, the distance a is given by
$a = \frac{\ell}{2}$
and it represents the distance from any side of the square of wire to the center. From the right hand rule, the magnetic field at the center of the box from all sides of the wire will be in to the paper. Plugging in the constants to the above equation, we find that
$B = \frac{4 \times 10^{-6}}{.2}(\sqrt{2})$
$B = 28.3 \mu T$ into the paper.

b)
For a single turn of wire with circumference $4\ell$, we have
$2 \pi R = 4 \ell$
$B = \frac{\mu_{0}I}{2 R}$
$B = \frac{\mu_{0}\pi I}{4 \ell}$
$B = 24.7 \mu T$ into the paper