Answer
$5.1\times10^{-6}N$ up the page, toward the wire.
Work Step by Step
The total force on the left and right segments of the wire loop is zero. Use the right-hand rule. The leftward force on a small piece of the left segment of wire is balanced exactly by the rightward force on a similar piece of the right segment of wire.
The magnetic field at the loop’s location due to the current in the long wire is into the page. The magnitude of the field is given by Eq. 20–6.
The force on the upper segment of the loop is up the page, since currents running in the same direction attract each other. The force on the lower segment of the loop, farthest from the wire, is down the page, because the currents are oppositely directed.
Use equation 20–7 to find the force on a current-carrying wire segment. Let upward be the positive direction.
$$F_{net}=F_{upper}-F_{lower}=\frac{\mu_o}{2\pi}\frac{I_1I_2}{d_{upper}}L_{upper}-\frac{\mu_o}{2\pi}\frac{I_1I_2}{d_{lower}}L_{lower}$$
$$F_{net} =\frac{\mu_o}{2\pi} I_1I_2L(\frac{1 }{d_{upper}}-\frac{1}{d_{lower}})$$
$$ =\frac{4\pi\times10^{-7}}{2\pi} (3.5A)^2(0.100m)(\frac{1 }{0.030m}-\frac{1}{0.080})$$
$$ =5.1\times10^{-6}N $$
