Force on an iron ball due to a dipole magnet

In summary, the question asks for the force that a soft iron ball, fixed a distance d above the pole of a rectangular dipole magnet, feels due to the magnetic field. The dimensions of the dipole magnet are a x a x b, where a < b. The equation B= \frac{μ_0}{4π}\frac{m}{d^3} is used, where m is the dipole moment and p is the magnetic dipole strength. The formula F = qv x B cannot be used as the iron ball is fixed, but the magnetic field from the dipole magnet can act as a dipole, resulting in a dipole-dipole force. The force is calculated using F_{mag}
  • #1
cereal9
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Homework Statement


A soft iron ball is fixed a distance d above the pole of a rectangular dipole magnet which is permanently magnetized. What is the force the iron ball feels due to the magnetic field?

The dimensions of the dipole magnet are a x a x b, where a < b

Homework Equations



[tex]B= \frac{μ_0}{4π}\frac{m}{d^3}[/tex]

Dipole Moment:
[tex]m = pl[/tex]
p = magnetic dipole strength (how is this even calculated?)
l = displacement vector between poles

The Attempt at a Solution



I know the following:

[tex]F = qv x B[/tex]

But I don't think I can use that because there's no velocity as the iron ball is fixed.

The ball has to feel some kind of force, though that formula suggests it isn't possible. It seems to me that if I hung a ball from a string, the tension in the string would increase if a magnet was placed below the iron ball. Is there some concept I'm not understanding, here?
 
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  • #2
After thinking about it, I think that with the magnetic field from the dipole magnet pulling charge to one spot in the soft iron, that too would act like a dipole so I'd be looking for dipole-dipole forces.

[tex]F_{mag}=\frac{-3μ_0m_1m_2}{4πr^4}[/tex]

Where m1 and m2 are the masses of the soft iron ball and the aforementioned dipole magnet?
 
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