A long horizontal wire $A B$ rests on the surface of a table and carries a current $I$ . Horizontal wire $C D$ is vertically above wire $A B$ , and is free to slide up and down on the two vertical metal guides $C$ and $D$ (as shown in figure). Wire $C D$ is connected through the sliding contacts to another wire that also carries a current $I$ , opposite in direction to the current $C D$ is $λ$ . To what equilibrium height $h$ will the wire $C D$ rise, assuming that magnetic force on it is wholly due to the current in wire $A B$ ? If h is given by $h = \frac{μ_{0} I^{2}}{x π λ g}$ . Find x.

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Solution

We need the magnetic and gravitational forces to cancel.
$λ L g = \frac{μ_{0} I^{2}}{2 π} \Rightarrow h = \frac{μ_{0} I^{2}}{2 π λ g}$

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