Tricky parallel wires and magnetic fields?

Question

You have three parallel conducting rods. Two of them are very long, and the third is 10.0 m long, with a weight of 36.0 N. You wish to conduct the following levitation demonstration: The two long rods will be placed in a fixed horizontal orientation at the same height, 10.0 cm apart. The third rod is to float above and midway between them, 10.0 cm from each one (from an end view, the three rods will form the vertices of an equilateral triangle). You will arrange to pass the same current I through all three rods, in the same direction through the two "supporting" rods and in the opposite direction through the "levitating" rod. What is the magnitude I of the current that will maintain this astounding configuration?

Answer 1

Net Force = 36.0 N

F is force due to one long rod.

Vertical force due to one long rod = F(√3)/2

Vertical force due to two long rods = F(√3).

(Force between two parallel rods carrying current, I),

F = 2x10^(-7)I^2L/d

F(√3) = 36.0

2x10^(-7)I^2L/d(√3) = 36.0

I^2 = 36.0*d(√3)/(L*2x10^(-7))

I = √{36.0*d(√3)/(L*2x10^(-7))}

=√{36.0*0.1(√3)/(10*2x10^(-7))}

=6√{0.1(√3)/(2x10^(-6))}

=(6/10^(-3))√{0.1(√3)/(2)}

=(6x10^(3))√{0.0866}

= 1765.69857

=1770 A 3 significant figures