You have three parallel conducting rods. Two of them are very long,and the third

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 toconduct the following levitation demonstration: The two long rodswill be placed in a fixed horizontal orientation at the sameheight, 10.0 cm apart. The third rod is to float above and midwaybetween them, 10.0 cm from each one (from an end view, the threerods will form the vertices of an equilateral triangle). You willarrange to pass the same current I through all three rods, in thesame direction through the two "supporting" rods and in theopposite direction through the "levitating" rod. What is themagnitude I of the current that will maintain this astoundingconfiguration?
I am trouble trying to figure out how to approach thisproblem. I know that Force = (mu*Current1*Current2*Length) /2*pi*R,
and Im supposed toequal the weight of the 3rd rod to magnitude of the two bottom rodsto achieve the "levitating" concept, but I don't know how to putthis into math terms. Please Help!!!
I am trouble trying to figure out how to approach thisproblem. I know that Force = (mu*Current1*Current2*Length) /2*pi*R,
and Im supposed toequal the weight of the 3rd rod to magnitude of the two bottom rodsto achieve the "levitating" concept, but I don't know how to putthis into math terms. Please Help!!!

Explanation / Answer

The force, F1 of repulsion on the levitatingrod by one fixed rod is given by F1 =(1.26*10^-6)*(i^2)*10/(2*pi*0.10). Let i^2 = x. The force ofrepulsion by second fixed rod also has same magnitude and the twoare inclined at angle of 60 degrees> their resultant must beequal to the weight of the levitating wire = 36 N. Now F1=x*[(1.26*10)/(2*0.10*pi)]10^-6 = [2.0*10^-5]x. So we have
x[sq rt (2^2+2^2+2*(cos 60)2^2]*10^-5 = (2*10^-5)*x*sq rt (3) = 36or
x = [36/(2*1.732)]*10^5 = 1.039*10^6 = i^2 or i = 1.02*10^3 A

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