A square loop PQRS carrying a current of 6.0 A is positioned near a...

Let the long straight wire lie along the y-axis, carrying current $I_1 = 10\text{ A}$ in the $+y$ direction.

The magnetic field $\vec{B}$ produced by this wire at a perpendicular distance $x$ is directed into the plane of the paper ($-z$ direction). Its magnitude is given by $B = \dfrac{\mu_0 I_1}{2\pi x}$.

For the segment $PQ$, the current $I_2 = 6.0\text{ A}$ flows away from the wire (in the $+x$ direction). The magnetic force on a small element $dx$ is:

$d\vec{F}_{PQ} = I_2 (dx\,\hat{i}) \times (-B\,\hat{k}) = I_2 B dx\,\hat{j}$

The total force on $PQ$ is directed upwards ($+y$ direction) and its magnitude is:

$F_{PQ} = \int_{1}^{3} I_2 \left(\dfrac{\mu_0 I_1}{2\pi x}\right) dx = \dfrac{\mu_0 I_1 I_2}{2\pi} \ln(3)$

For the segment $RS$, the current $I_2$ flows towards the wire (in the $-x$ direction). The magnetic force on a small element $dx$ is:

$d\vec{F}_{RS} = I_2 (-dx\,\hat{i}) \times (-B\,\hat{k}) = -I_2 B dx\,\hat{j}$

The total force on $RS$ is directed downwards ($-y$ direction) and its magnitude is:

$F_{RS} = \int_{1}^{3} I_2 \left(\dfrac{\mu_0 I_1}{2\pi x}\right) dx = \dfrac{\mu_0 I_1 I_2}{2\pi} \ln(3)$

Thus, the forces on segments $PQ$ and $RS$ are equal in magnitude and opposite in direction.