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A long, horizontal wire AB rests on the surface of a table and carries a current I. Horizontal wire CD· is vertically above wire AB, and is free to slide up and down on the two vertical metal guides C and D (as shown in figure). Wire CD is connected through the sliding contacts to another
wire that also carries a current I, opposite in direction to the current in wire AB. The mass per unit length of the wire CD is $λ$ . To what equilibrium height h will wire CD rise, assuming that magnetic force on it is wholly due to the current in wire AB?

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$\frac{3 μ_{0} I^{2}}{2 πλg}$

$\frac{μ_{0} I^{2}}{3 πλg}$

$\frac{5 μ_{0} I^{2}}{2 πλg}$

$\frac{μ_{0} I^{2}}{2 πλg}$

answer is D.

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Detailed Solution

Gravitational force and magnetic force are equal in magnitude and opposite in direction.

$\Rightarrow λLg = \frac{μ_{0} I^{2} L}{2 πh} \Rightarrow h = \frac{μ_{0} I^{2}}{2 πλg}$

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