I've tried very hard to find an answer to this question, and every path leads me to an abstract discussion of fundamental forces. Therefore, I will propose two very specific scenarios and see if they yield the result that I am looking for.

Scenario One

Let's say that I have a vertical tube exactly 1 inch in diameter that is completely incapable of holding an electromagnetic charge and has a frictionless surface. Resting inside this tube is a steel ball also exactly 1 inch in diameter. If a cylindrical magnet, also exactly 1 inch in diameter is slowly lowered into the tube, how does one determine the exact point at which the force being applied to the steel ball by the magnet will cause the ball to overcome gravity and rise toward the magnet? Is there even any way to determine this? What further information would I need?

Scenario Two

I have the same tube from above with the cylindrical magnet resting on the bottom of the tube, north pole facing upward. Suspended by a weightless string in the tube is an identical magnet, north pole facing downward. If the bottom magnet is slowly raised, how does one calculate the exact point at which the suspended magnet will begin to move upward? Is this calculation possible? What further information would I need for this calculation?

Extra Question

How are weight capacities on magnets calculated? I.e. if a whitepaper says that a magnet is capable of lifting 25 pounds, how is the correct size magnet calculated?

  • $\begingroup$ I just meant like static electricity. I'm basically just eliminating every variable from the scenario. $\endgroup$ – Nick Anderegg Oct 11 '11 at 18:21
  • $\begingroup$ @Richard Terrett Thanks for the correction. I originally had force, then I changed it because I thought that was wrong. I'm only a freshman engineering student, so physics isn't quite my thing yet. $\endgroup$ – Nick Anderegg Oct 11 '11 at 18:23

If you have a steel ball bearing with high permeability, then the magnetic potential energy of this configuration will be proportional to the square of the B-field at the position of the ball bearing. For a simple dipole magnetic field on-axis, this will be proportional to $r^{-6}$, where $r$ is the distance from the centre of the magnet.

The force associated with this will be the gradient of the potential and will go as $r^{-7}$.

The exact force I think is difficult to calculate because of geometric factors and the finite sizes of the components - e.g. integrals of the dipole field over the volume of the ball bearing.


I am consider Scenario One:

For given ball and magnet the magnitude of the interaction force between them, depending on the distance $x$, follows the formula:

$$F=\frac{const}{x^7}$$ where $const$ depends on the ball's radius, on the magnetic dipole moment of the magnet and on the magnetic permeability(expected to be constant) of the material of the ball.

You can determine $const$ by measuring the distance $x$ under gravity(with known weight of the magnet).
A derivation of the formula for $F$ requires the vector calculus. I guess it is not particularly familiar with you.

  • $\begingroup$ ""and on the magnetic permeability(expected to be constant) of the material of the ball. "" This is very optimistic for a "steel" ball. $\endgroup$ – Georg Oct 12 '11 at 8:43
  • $\begingroup$ Well, I know calculus, just not physics yet. This was very helpful though. $\endgroup$ – Nick Anderegg Oct 12 '11 at 17:38
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    $\begingroup$ @Georg I know. I did things as simple as possible, taking into account Nick's background. $\endgroup$ – Martin Gales Oct 13 '11 at 5:37

I would measure the force using a well calibrated weighing scale. You will see the weight decrease as a function of distance. The difference is the magnetic force. Note that the weighing scale gives kilogram force, 1 kgf being about 9.81 Newton.


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