Problem 4.2.2: Three parallel wires

Each pair of wires acts with ${\mu }_0{I}_i{I}_j/(2\pi d_{ij})$ per unit length, attractive when $I_i{I}_j>0$. Counting forces positive in the $+a_y$ direction and using $d_{12}=a$, $d_{23}=b$, $d_{13}=a+b$:

(Wire 1 is pulled towards $+y$ by both of its neighbours when all currents are parallel, wire 3 towards $-y$, and wire 2 is pulled both ways.)

$F_2=0$ requires $I_1/a=I_3/b$. $F_1=0$ requires $I_2/a=-I_3/(a+b)$. Then $F_3=0$ follows automatically, because the three forces are internal and add to zero. All three vanish together if and only if

For $a=b$ the ratio is $2:-1:2$: the middle wire carries half the outer current, reversed (Figure 4.2.17). With all three currents in the same direction, as drawn, the outer pair always attracts and no balance exists. The sum $F_1+F_2+F_3$ vanishes identically for any currents, as it must for internal forces, and that is why only two of the three equations are independent.