Two infinetely long, parallel wires are lying on the ground at distance `a=1cm` apart as shown in Fig. A third wire, of length L=10m and mass 400g, carries a current of `I_1=100A` and is levitated above the first two wires, at a horizontal position midway between them. The infinitely long wires carry equal currents `I_1` in the same directions, but in the direction opposite that in the leviated wire. What current must the infinately long wires carry so that the three wires form an equilateral triangle?

Explanation

Since the current in the short wire is opposite to those in
the long wires, the short wire is repelled from both of the others.
Imaging the currents in the long wires in Fig are increased.
The repulsive force becomes stronger, and the levitated wire rises to the point at which the wire is once again levitated in
equilibrium at a higher position. Figure shows the desired
situation with the three wires forming an equilateral triangle.
Because the levitated wire is subjected to forces but does not
The horizontal components of the magnetic forces on the levitated
wire cancel. The vertical components are both positive and added
Fig. and in the plane of the page.
The total magnetic force in the upward direction on the levitated
wire:
`vecF_B=2((mu_0I_1I_2)/(2pia))cos theta hatk=((mu_0I_1I_2)/(2pia))l cos theta hatk`
The gravitational force on the levitaed wire :
`vecF_g=-mg hatk`
Apply the particle in equilibrium model by adding the forces and
setting the net force equal to zero:
`Sigma vec F =vecF_B+F_g= ((mu_0I_1I_2)/(pia))l cos theta hatk-mg hatk=0`
Solve for the current in the wires on the ground:
`I_2=(mgpia)/(mu_0I_1l cos theta)`
Substituting numetical values:
`I_2=((0.400kg)(10ms^-2)pi(0.0100m))/((4pixx10^-7T. mA^-2)(100A)(10m) cos 30^@)=113A`