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Calculate pull force of magnet?

  1. Mar 23, 2007 #1

    I have a permanent magnet (actually, an assembly of permanent magnets). I can calculate magnetic flux density induced by the magnets (when no other objects presents).

    Now, I have very small iron particles suspend in water inside a cup. The magnets are then placed on the outside wall of the cup. The iron particles start to move toward the magnets. Ultimately, I would like to calculate the time needed for the particles to reach the cup wall. For the first step, how can I calculate the "pull force" on the iron particles induced by the magnets? Of course, the pull force increases as the iron particles move closer to the magnet. Is there any simplified equation for this?


  2. jcsd
  3. Mar 24, 2007 #2
    No, there is no easy way to do it. The force exerted on each particle is the product of de magnetic field gradient times the magnetic moment of the particle. But the magnetic moment of the particle depends on how the magnetic field polarizes it. That is how the magnetic field transforms the particle in a little magnet (magnetic dipole). Then you must know the magnetic field and its gradient in all places.

    Then there is the magnetic moment of particles. It depends on the magnetic field, the size of particles and on the magnetic properties of ferromagnetic particles witch are highly non linear.
    Then there are the viscosity forces of the liquid against the displacement of particles. For big sized particles the force is known. It is given by Bernouilli and Navier-Stokes equations. These equations do not apply for small particles.

    For smaller particles two new complications arise: the Brownian motion and the electrical charge of particles which avoids colloidal suspensions to settle.

    Are you sure that you want to calculate all this?
    Last edited: Mar 24, 2007
  4. Mar 24, 2007 #3
    Dear lpfr,

    Thank you very much for the reply!

    I have a computer code to calculate the force due to drag (viscous force). For the first attempt to attack this problem, I will ignore the Brownian motion and electrical chage for now. And, I will consider one 1 particle so that there is not interaction among particles. So, the only force left to calculate is magnetic force.

    I found a web site: www.ucs.louisiana.edu/~khh6430/magnet.html. In Eq. 1, it gives: Force = (susceptibility) * volume * [ gradient (H * H / 2)]

    I do not know the source of this equation. It is not clear to me whether H is calculated without the particles. Have you seen this equation?

  5. Mar 25, 2007 #4
    Well, a program code is as good as the method coded. I suppose that you must enter the particle size and that there is a range of valid sizes. You can test this giving a particle size of the order of an atom and see what happens. If it does not complain or crashes, you have better find another one.

    I do have a very old souvenir of this formula. I do not remember what book it was, but it was written in a physicist way: H*grad(H) to separate the two things: the created dipole which is proportional to H (see below) and the gradient. Of course, mathematically the two writings are similar, but not physically.

    The magnetic field is the one without the contribution of particles. The reason is that the particles magnetic field does not exert forces upon themselves.

    The formula is written in the old obsolete emu system. I strongly recommend that you find the SI (Sisteme International) version.

    Last, as the paper says, this formula is valid for dia- and paramagnetic particles. As the paper do not says, the formula is not valid for ferromagnetic particles.
    The reason is that in this version the magnetization of the particle material is proportional to H. This is not true for ferromagnetic material and is still less true for small particles.

    Ferromagnetic material is formed by magnetic domains magnetized in different directions. When you apply a magnetic field, the magnetic domains grow or shrink or change the direction of magnetization. You can find more about all this looking for hysteresis cycles.
    But when particles are small, they have only a unique magnetic domain. They are already magnetized and behave as a small magnet. An external magnetic field just orients them, but do not change the magnetic moment of the particle. In this case, the force is proportional just to the gradient of the field, and your formula does not apply.

    Hope that helps.

    Last edited: Mar 25, 2007
  6. Mar 26, 2007 #5
    Dear Mr. LPFR:

    Thanks a lot for the answer.

    Could you please post the source of the paper (which gave force ~ H * grad(H)) if you happen to find it again?

  7. Mar 27, 2007 #6
    I though where you can find the formula and I remembered the Stern-Gerlach experiment. It involves magnetic forces acting on magnetic dipoles. There is a description – and the formula – in the Feynman (The Feynman Lectures on Physics) in chapter 35-2.

    Good look.
  8. Mar 27, 2007 #7
    Thanks a lot Mr. LPFR!

    You have been a great help!

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