Problem 3 – Hint 5

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I Note that a spherical permanent magnet of homogeneous magnetisation creates outside itself a magnetic field identical to that of an ideal point dipole. Indeed, electric and magnetic point dipoles produce isomorphic fields (obtainable from each other by replacing 1/\varepsilon_0 with \mu_0 or vice versa). Homogeneous magnetisation means that there is an homogeneous volume density of point dipoles. A sphere homogeneously filled with electric dipoles can be regarded as a superposition of two slightly displaced homogeneously charged spheres, with equal by magnitude and opposite volume densities of charge. Outside of a homogeneously charged sphere, its field is identical to that of a point charge at its centre, so the two spheres can be replaced by two point charges.

II Note that the field of a point dipole is inversely proportional to the cubed distance from it, and can be calculated from the superposition of the fields due to two opposite point charges.

III The magnetisation of a ferromagnetic plate can be expressed directly in terms of the B-field from the relationships B = \mu_0(M + H) and B=\mu_0\mu_r H while taking advantage of the fact that the normal component of the B-field is continuous at the surface, and that the field due to a very thin plate is much smaller than the external field. The field caused by a ferromagnetic sphere cannot be neglected, so it should be taken into account, or alternatively, the magnetisation of a sphere can be obtained from Ampère’s circuital law for a loop extending along the axis of symmetry of the sphere from minus to plus infinity (and closing onto itself infinitely far from the sphere) while making use of the fact that the B-field inside a homogeneously magnetised sphere is homogeneous.

IV Once M has been obtained, the magnetic dipole moment m =MV of a scrap particle can be calculated. At the moment when the particle is lifted, the gravitational force is balanced by the dipole-dipole attraction force between the permanent magnet, and the scrap particle. The latter can be found by taking the derivative of the potential energy U =-\vec B\cdot \vec m.

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