Answer
0.12 N to the south
Work Step by Step
Use equation 20–8 for the magnetic field inside a solenoid.
This field is perpendicular to the current in the wire. Use equation 20–2 to find the magnitude of the force on the wire segment.
$$F=I_{wire}\mathcal{l}_{wire}B_{solenoid}$$
$$F=I_{wire}\mathcal{l}_{wire}\frac{\mu_oNI_{solenoid}}{\mathcal{l}_{solenoid}}$$
$$=(22A)(0.030m)\frac{(4\pi \times10^{-7} T\cdot m/A)(550)(38A)}{0.15m}$$
$$=0.1156N\approx0.12N$$
The right hand rule tells us the force is to the south.
