Question
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.]
Options
$\ \frac{\mu_{0}I^{2}}{\pi\lambda g}$
20.
Solution
Sol. [Key Concept: Magnetic force between parallel current-carrying wires]
Force per unit length on wire $A$ by wire $B$
$${F = \frac{\mu_{0}}{4\pi}\frac{2l \cdot 2l}{h} = \frac{\mu_{0}}{4\pi}\frac{4l^{2}}{h}l }{ If\ F_{\text{net}\text{~}} \leq mg}$$, weight of wire $A$, then wire does not rise from the floor.
So, $F \leq mg$,
$$\begin{matrix} & \frac{\mu_{0}}{4\pi}\frac{4I^{2}}{h} \leq \lambda g \\ & \ \therefore h \geq \frac{\mu_{0}I^{2}}{\pi\lambda g} \end{matrix}$$
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