Magnet's force on a metal ball

[sergiokapone]

As one know, the force on mangnetic dipole in magnetic field is
$$ \vec F = (\vec p_m\cdot \nabla) \vec B_0$$,
where B_0 -- external field.
Let consider a some magnetic matherial with permeability \mu. The magnetization of matherial is $$M = (\mu - 1) H$$ (in SI units) and by deffinition $$M = \frac{\sum p_m}{V},$$
where H -- field inside matherial, H = \frac{B}{\mu\mu_0}, B-- field inside matherial.

Let suppose, the material is magnetized so that $$\vec p_m \uparrow \uparrow \vec B_0$$, so the force on magnetic matherial is
$$ F = (\mu - 1) \int H \frac{dB_0}{dr} dV.$$

For the ball, we know the dependence between external and internal field:
$$B = \frac{3\mu}{\mu+2} B_0,$$
and
$$H = \frac{3}{\mu_0(\mu+2)} B_0.$$

Thus, the force on metall ball in a magnetic field:
$$ F = \frac{3(\mu - 1)}{\mu_0(\mu + 2)} \int B_0 \frac{dB_0}{dr} dV = \frac{3(\mu - 1)}{2\mu_0(\mu + 2)} \int \frac{dB_0^2}{dr} dV.$$

If the ball is made of feromagnetic matherial, then \mu \gg 1, then the force does not depend on \mu:
$$ F = \frac{3}{2\mu_0} \int \frac{dB_0^2}{dr} dV.$$

Is the formula correct? Or, can be, I somewhere essentially made a mistake?

 

[kuruman]

... so the force on magnetic matherial is $$
F = (\mu - 1) \int H \frac{dB_0}{dr} dV.$$

I don't see how that follows from $\vec F = (\vec p_m\cdot \vec{\nabla}) \vec B_0.$
It's not clear whether you are using spherical or cylindrical coordinates, but even if $\vec B_0$ has azimuthal symmetry, there should be a $\theta$ dependence in spherical coordinates and a $z$ dependence in cylindrical coordinates.