Two infinitely long parallel conducting wires A and B carry...

The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by:

$f = \dfrac{\mu_0 I_1 I_2}{2\pi h}$

Given $I_1 = I$ and $I_2 = 2I$, the upward attractive force per unit length on wire A due to wire B is:

$f = \dfrac{\mu_0 (I)(2I)}{2\pi h} = \dfrac{\mu_0 I^2}{\pi h}$

The downward gravitational force per unit length on wire A is its weight per unit length:

$W = \lambda g$

For wire A not to rise from the floor, the upward magnetic force must be less than or equal to the downward gravitational force:

$f \le W$

$\dfrac{\mu_0 I^2}{\pi h} \le \lambda g$

$h \ge \dfrac{\mu_0 I^2}{\pi \lambda g}$

Therefore, the minimum magnitude of $h$ is $\dfrac{\mu_0 I^2}{\pi \lambda g}$.

Answer: $\dfrac{\mu_0\text{I}^2}{\pi\lambda g}$