Answer
Please see the work below.
Work Step by Step
(a) We know that the magnetic field is pointed out of the page, which is perpendicular to all four sides of the square loop. Thus, the direction of the net force will act away from the wire.
Now $F_{net}=IL\frac{\mu_{\circ}I^{\prime}}{2\pi}(\frac{1}{r_{close}}-\frac{1}{r_{far}})$
We plug in the known values to obtain:
$F_{net}=\frac{(1.0m)(2.5A)(4\pi\times 10^{-7}\frac{T.m}{A})(14A)}{2\pi}(\frac{1}{0.2m}-\frac{1}{1.2m})=3\times 10^{-5}N$
(b) We know that $F=ILB$. Thus, if we extend the horizontal side with width $2.0m$, the force is doubled and hence the net force is increased by a factor of two.
(c) We know that the net force will increase as there is less force on the bottom side of the loop to cancel the force on the top side.
