Two infinitely long parallel conducting wires A and B carry currents $I$ and $2I$, respectively, in the same direction. The wire A has uniform mass per unit length $\lambda$ and lies on an insulated floor. The wire B is kept fixed at a height $h$ above the floor. The minimum magnitude of $h$ so that the wire A does not rise from the floor is :
[$g$ is the acceleration due to gravity and $\mu_0$ is the permeability of free space.]
Answer: (B) $\dfrac{\mu_0 I^2}{\pi\lambda g}$
Currents in the same direction attract, so wire A is pulled upward towards B. Force per unit length on A:
$$\frac{F}{l} = \frac{\mu_0 (I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$$
A does not rise as long as this does not exceed its weight per unit length $\lambda g$:
$$\frac{\mu_0 I^2}{\pi h} \le \lambda g \;\Rightarrow\; h \ge \frac{\mu_0 I^2}{\pi\lambda g}$$
Minimum $h = \dfrac{\mu_0 I^2}{\pi\lambda g}$.
Solution by Sreeraj P, M.Sc Physics