Wire A is subjected to two primary forces: the downward gravitational force per unit length ($w = \lambda g$) and the upward magnetic force per unit length ($f_m$) exerted by wire B due to the current interaction.
The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $h$ is given by the formula: $$f_m = \frac{\mu_0 I_1 I_2}{2\pi h}$$ Substituting $I_1 = I$ and $I_2 = 2I$: $$f_m = \frac{\mu_0 I (2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$$
For wire A to not rise from the floor, the upward magnetic force must not exceed the downward gravitational force. The limiting case (minimum height) occurs when the forces are balanced: $$f_m = \lambda g$$ $$\frac{\mu_0 I^2}{\pi h} = \lambda g$$
Rearranging the equation to solve for $h$: $$h = \frac{\mu_0 I^2}{\pi \lambda g}$$
Final Answer: \frac{\mu_{0}I^{2}}{\pi\lambda g}
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