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Magnet force on an object separated by small space

  1. Nov 28, 2011 #1
    I'm trying to approximate forces on a system.

    One of my Forces is a 6x8x1" Ceramic Grade 8 magnet on an ferromagnetic object, but they are separated by a small space (about 3/16").

    I know the Pull of the magnet, but is there a way to calculate the pull of the magnet relative to the distance away it is from the object?
  2. jcsd
  3. Nov 29, 2011 #2
    The actual magnetic field and hence force is typically complex and depends on the magnets shapes, magnetization, orientation, and the ferromagnet's shape, permeability, history, location, and orientation. To a good approximation though, most single permanent magnets create dipole fields, and dipole fields die off as 1/r3. So if the maximum force you measure when there is no separation is F0, then the force F as a function separation r would be F = F0/r3 to a crude approximation.
  4. Nov 29, 2011 #3
    Shouldn’t that be F=Fo x d^3/r^3 ? where d is the length from one pole to the middle of the magnet and r is also to be taken from the middle of the magnet. (ofcourse also to a crude approximation).
  5. Nov 30, 2011 #4
    Actually, it would be more like F = F0/(r+d)3 if r is the distance from the pole's surface, and d is the distance from the pole's surface to the center of the magnet. Sorry for error
  6. Nov 30, 2011 #5
    If you have a little experience with programming, vectors, and numerical methods, it's actually not that hard to write a little code that solves for the three-dimensional static magnetic field due to any shaped magnet. Just solve each component of [itex]\nabla[/itex]2 A = μ0 Jmag using the relaxation method with sources. Here Jmag is the effective magnetization current describing the magnet. Then the magnetic field is B = [itex]\nabla\times[/itex] A.
  7. Nov 30, 2011 #6
    chrisbaird, the problem I’m referring to is that you stated that Fo is a force, which is fine, but then it follows that Fo/r^3 cannot possibly be a force as well. It would give a very strange answer. Therefore I suggested that F can be worked out by Fo times the ratio d^3 and r^3. This way at least the answer is in Newton and you still have an 1/r^3 dependency. This way if both distances are taken from the centre then when r=d, F=Fo, which is correct.
    I haven’t tried that formula and I wish I had more time to find out. It would be nice if mmartelli could tell us.
  8. Dec 1, 2011 #7
    You are right, in my effort to simplify things for the OP, I keep making the equation unphysical. I should stop digging this hole. :blushing:

    Here is the full form for a magnetic dipole:
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