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Magnetic repulsion formulas

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    We are building a maglev train for an engineering physics project. We have a prototype that is going to work for us but we are struggling with the mathematics. We are using ceramic magnets that are 3/16in thick and 1/2 inch in diameter. One of our project partners has an equation (below) but I am not so sure this explains what we need it to. We are wanting to figure out how much weight our magnets will levitate. Obviously we have like poles facing each other and so we have a force in the Y direction from each magnet but still not sure of the math. Propulsion is from force of gravity. Any help would be greatly appreciated!

    2. Relevant equations

    F(x) = [u(naught)pi]/4 [M^2R^4] [1/x^2 + 1/(x+2h)^2 - 2/(x+h)^2

    M= 2B(naught)/u(naught)

    M=magnetization of the magnets
    h= height/thickness of the magnets
    R= radius of magnets
    x= distance between magnets (vertically)
    B(naught)= magnetic flux density

    3. The attempt at a solution
  2. jcsd
  3. Apr 23, 2016 #2

    Simon Bridge

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    You mean:
    $$F(x)=\frac{\mu_0\pi}{4}M^2R^4\left[\frac{1}{x^2}+\frac{1}{(x+2h)^2} - \frac{2}{(x+h)^2}\right]$$ $$M=\frac{2B_0}{\mu_0}$$
    ... note: "weight" = "force of gravity". The force needed to levitate a car mass m is mg. The first equation tells you the force at different separations ... what is the problem?
    Since you already have the magnets, why not measure the force?
  4. Apr 23, 2016 #3
    I have:
    b(naught)= .000104 T
    M= 165.521
    h= .0047625
    r= .0047625
    u(naught)= 1.2567 x 10^-6
    x= .008128

    plugging in I get F(x) = (.00688N)j

    mg= .423*9.81= 4.1454N

    Doesn't F(x) need to be greater that mg?
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