Two long parallel wires $A$ and $B$, a distance $d$ apart, carry currents $I_1$ and $I_2$ in opposite directions. A third wire $C$ carrying a current $I$ is to be kept parallel to them at a distance $x$ from $A$ such that the net force acting on it is zero. The possible values of $x$ are:
Answer: (A) $x = \pm\dfrac{I_1d}{I_1-I_2}$
The currents in $A$ and $B$ are opposite, so between the wires the two forces on $C$ point the same way and cannot cancel. $C$ must lie outside, on the side of the weaker current, where the magnitudes balance:
$$\frac{\mu_0I_1I}{2\pi|x|} = \frac{\mu_0I_2I}{2\pi|x-d|}$$
With $x$ and $x - d$ of the same sign: $I_1(x-d) = I_2x$, so
$$x = \frac{I_1d}{I_1-I_2}$$
This is beyond $B$ if $I_1 > I_2$ and on the far side of $A$ (negative $x$) if $I_2 > I_1$, which is option (1).
Solution by Sreeraj P, M.Sc Physics