How to approximate the force on a magnet below a coil  o            x
 o            x   Coil with 4 turns
 o            x
 o            x
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   | |            Neodymium magnet
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I need to know the force acting on a neodymium magnet which is placed below a coil. This simple looking problem is actually very complicated, and no data is known about the system, so it can be simplified as necessary.
If the coil has a diameter of about 1 meter, height 150 mm and has 4 turns, and the magnet is 150mm by  50mm, and is placed bellow the coil, close to the edge (see ASCII drawing above).
What is the force on the magnet as a function of current?
 A: The field from a bar magnet is approximately a dipole. The field from a coil is approximately a dipole. The force between two dipoles will contain both a torque term, and a attraction / repulsion term - both of which will be proportional to current.
Equations for this can be found at http://en.wikipedia.org/wiki/Magnetic_dipole#Forces_between_two_magnetic_dipoles
$$F = \nabla (m_2 \cdot B_1)\\
\Gamma = m_2 \times B_1$$
Note that when you get close to the coil, its field is no longer strictly a dipole; instead you might want to think about the field due to just the current closest to the magnet (since the field due to the conductor on the other side will have minimal effect on the force). Thus to estimate this I would probably use the magnetic moment of the bar magnet, and the current due to four linear wire segments (length equal to the diameter) closest to the magnet. The result will be close (within a factor 2 or so).
If you are on axis, the problem is much simplified. Now you can get $B_1$ from an expression for the field of a circular coil - if you have 4 turns, you evaluate this for 4 distances from the center of the coil (z = 0, 50, 100, 150 mm):
$$B_z = \frac{\mu_0}{4\pi}\frac{2\pi R^2 I}{(z^2 + R^2)^{3/2}}$$
The value of $m_1$ is something you have to get from the parameters of the magnet.
