Answer
See answer.
Work Step by Step
There is no force on the top and bottom segments of wire, because the current runs parallel to, or opposite to, the magnetic field.
The only force is on the left segment, of length L, assumed to be in meters. By the right-hand rule, the force on the left segment is out of the page.
The current is perpendicular to the magnetic field, so use equation 20–2 to find the force.
$$F=I_{wire}L_{wire}B$$
Use equation 20–8 for the magnetic field inside a solenoid.
$$F= I_{wire}L_{wire}(\frac{\mu_o NI_{solenoid}}{L_{solenoid}})$$
$$F = (5.0A)L_{wire} (\frac{4\pi \times10^{-7} (700)(7.0A)}{0.15m})=0.21 L_{wire}\;newtons$$
