A long straight horizontal wire carries a steady current I1 = 200 A. A rigid rectangular loop of mass 0.60 g lies in the vertical plane containing…

For levitation the loop must be repelled by the wire, so the current in the lower horizontal side must be opposite to $I_1$; the current in the upper side is then parallel to $I_1$ and is attracted downward. The forces on the two vertical sides are equal and opposite and cancel. The net upward force is $F=\frac{\mu_0 I_1 I_2 \ell}{2\pi}\left(\frac1{d_1}-\frac1{d_2}\right)$ with $\ell=0.50$ m, $d_1=0.02$ m, $d_2=0.08$ m. Here $\frac1{d_1}-\frac1{d_2}=50-12.5=37.5\ \mathrm{m^{-1}}$, so $F=2\times10^{-7}\times200\times0.5\times37.5\,I_2 = 7.5\times10^{-4}I_2$ newton. Setting $F=mg=0.6\times10^{-3}\times10=6\times10^{-3}$ N gives $I_2=8.0$ A. Using only the lower side gives 6.0 A, adding the two contributions gives 4.8 A, and using the loop height instead of $d_2$ gives 9.0 A, all of which are wrong.