= Solution
Use $d\mathbf F=I\,d\boldsymbol\ell\times\mathbf B$ for the magnetic <Lorentz force> on each wire segment. Let $+x$ point right, $+y$ upward and $+z$ out of the page, so $\mathbf B=-B\hat{\mathbf z}$. With clockwise <electric current>, the four magnetic <forces> are
$$
\boxed{\mathbf F_{\rm top}=IaB\hat{\mathbf y},\quad \mathbf F_{\rm bottom}=0,\quad \mathbf F_{\rm right}=IBh\hat{\mathbf x},\quad \mathbf F_{\rm left}=-IBh\hat{\mathbf x}.}
$$
The upper side carries <electric current> to the right and experiences an upward <force>, opposing the fall. The bottom side lies outside the <magnetic field> and has no magnetic <force>. On the right side the <electric current> is downward and the <force> is outward to the right; on the left it is upward and the <force> is outward to the left. Each side's in-field vertical length is $h$, so these horizontal <forces> cancel in the net <force> on the loop.
The wire's total <mass> is $M=4a\rho_mA_w$. Uniform wire gives each side a downward <weight> $Mg/4$. Internal mechanical <forces> at the corners transmit the unequal loads and maintain the assumed square shape; they cancel for the loop as a whole. The total external vertical <force> is therefore $Mg-IaB$ downward.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-falling-loop-forces.png]
{title=Current and magnetic forces during exit from the field, with terminal-speed balance}
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