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Magnetic field at a distance - epxression

  1. Aug 30, 2007 #1
    Hello,

    I have a cylindrical Neodymium permanent magnet. I need to find the pull force of the magnet at various distances from the magnet. Is there an expression relating the pull force(or attractive force) to the magnetic flux density B and the distance from the cylindrical magnet? What is the expression that tells us how the B varies with distance?

    The parameters I have for the magnet are here

    http://www.kjmagnetics.com/proddetail.asp?prod=D2C .

    Thanks,

    rsr_life
     
  2. jcsd
  3. Aug 30, 2007 #2

    berkeman

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    Staff: Mentor

  4. Aug 30, 2007 #3
    The wiki doesn't seem to have an expression listed for calculating magnetic flux density at a distance for a cylindrical permanent magnet. There's no current involved, so I just want to be able to simulate and get field values at different points from the magnet.

    Seems there should be a straightforward expression to do this, maybe involving partial derivatives.

    Thanks.
     
  5. Aug 31, 2007 #4
    Far enough away, most magnets can be approximated as a magnetic dipole moment m, which is a vector. For a magnetic dipole m at the origin and oriented in the z-direction, you get

    [tex] \vec{B} = {\mu_o m \over {4\pi r^3}} (2\cos\theta \hat{r} + \sin\theta\hat{\theta})[/tex]

    The trick for your problem will be to find out a good guess at m. As usual, Wikipedia has some good information.
     
  6. Aug 31, 2007 #5
    Thanks for that. What about points closer to the magnet - near the surface, for example, where the gradient is steeper?
     
  7. Aug 31, 2007 #6

    dlgoff

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    Science Advisor
    Gold Member

    Maybe you can find an equation by experiment. Could you use some type of spring device (to messure weight) with a iron-mass on it and find the force at different points around the magnet? Say; leave the spring hanging and moving the magnet at different positions under the mass?
     
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