Source: High school physics (Chinese)
Problem
A long straight wire carries a current $I_1 = 30$ A. In the same plane as this wire, a rectangular conducting loop of length $L = 30 \text{ cm}$ and width $w = 8 \text{ cm}$ carries current $I_2 = 20$ A. The long sides of the loop are parallel to the wire; the side of the loop nearest the wire is at distance $a = 1 \text{ cm}$ from it, and the current in this near side flows in the same direction as the current in the straight wire. Find the magnitude and direction of the net magnetic force that the straight wire exerts on the loop.
The two short sides of the loop (perpendicular to the wire) carry currents in opposite directions through the same field profile, so the forces on them are equal in magnitude and opposite in direction — they cancel. The net force comes only from the two long sides.
For a long side of length $L$ at distance $x$ from the wire, the field magnitude is $B(x) = \dfrac{\mu_0 I_1}{2\pi x}$, giving force magnitude $F(x) = I_2 L B(x) = \dfrac{\mu_0 I_1 I_2 L}{2\pi x}$.
- Near side ($x = a$, current parallel to $I_1$): attractive force toward the wire.
- Far side ($x = b = a+w$, current antiparallel to $I_1$): repulsive force away from the wire.
Net force (toward the wire since $a With $a = 0.01$ m, $b = 0.09$ m, $L = 0.30$ m, $I_1 = 30$ A, $I_2 = 20$ A: