 Solution

Use $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$ 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 areThe 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 $M=4a{\rho }_m{A}_w$. 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.