Problem 22.23

22.23.
Each wire is distant from the wire by a distance of
$d = .2 m \cos 45{}^{\circ}$
d = .141 m
Each wire prodcues a magnetic field at point P of equal magnitude.
$B = \frac{\mu_{0}I}{2\pi d}$
$B = \frac{(2 \times 10^{-7} T \cdot m)(5 A)}{(.141 m \cdot A)}$
$B = 7.07 \mu T$
in the direction determined by the right hand rule. Now, each wire creates a magnetic field that has equal components both downward and to either the right or the left. The wires B and C create components of the magnetic field to the right and the wires A and D create magnetic field components to the left. These effects cancel since the wires are all equidistant from P. Therefore, only the four downward components add together. The net magnetic field is given by
$B_{net} = 4 B_{A} \cos 45{}^{\circ}$
$B_{net} = 20.0 \mu T$
or, $\vec{B_{net}} = 20.0 \mu T$ downward