Finding magnitude of current of 3 rods, so 1 can levitate.

1. The problem statement, all variables and given/known data
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?

F= 36 N
L = 10m
distance btw= .1m

2. Relevant equations
F= 4pie-7 * I * I* L / (2Pi*R)

3. The attempt at a solution
The 2 rods at the bottom exert same forces on 3rd on an angle, and their sum has to equal 36N
36 N = sqaureroot ( F^2 + F^2)
F(force of 1 rod on top rod) = 25.4558 N

25.4558 = μ* I^2* 10 / ( 2Pi* .1)
I = 1128.18 A
is this correct?

i saw another solution on the internet

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 = 2×10^(-7)I^2L/d

F(√3) = 36.0
2×10^(-7)I^2L/d(√3) = 36.0
I^2 = 36.0*d(√3)/(L*2×10^(-7))
I = √{36.0*d(√3)/(L*2×10^(-7))}
=√{36.0*0.1(√3)/(10*2×10^(-7))}
=6√{0.1(√3)/(2×10^(-6))}
=(6/10^(-3))√{0.1(√3)/(2)}
=(6×10^(3))√{0.0866}
= 1765.69857
=1770 A 3 significant figures

I think he is using similar method to mine, but different answers. Can someone help me with correct answer?

http://ift.tt/1fv45pR

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