Electricity and Magnetism Qualitative Problems

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10) The vector potential and the magnetic field inside and outside of a diamagnetic sphere in a uniform magnetic field

References: I do not find really good references. There are these related web pages:
http://farside.ph.utexas.edu/teaching/jk1/lectures/node51.html
http://farside.ph.utexas.edu/teaching/jk1/lectures/node40.html

There is also an explanation for a dielectric ellipsoid in Landau, Lifshitz, and Pitaevski,
But there may be better presentations in other books that I do not have, like Jackson or Percell you mentioned you have.

The essence is that
(a) The field inside the sphere is homogeneous and can be regarded as an "inner dipole" field because it has the same dependence on the direction as the "outer dipole" field of a point dipole: Bin = 3 μ / (μ + 2 μo) Be, with vector potential Ain = Br/2 where Be is the applied external homogeneous field.
(b) The field outside the sphere is the sum of the applied homogeneous field (like "inner dipole") and "outer dipole" field,
B = Be + (μo - μ) / (μ + 2 μo) (a3/r3) [Be -3(Ber) r/r2], with vector potential
Ain = He r/2 - (μo - μ) / (μ + 2 μo) [Be r/r3]

I assume that you can find the explanations on how this solution is obtained in your available books. If not and you want the explanation, I can write it down for you separately.

10) The vector potential and the magnetic field inside and outside of a diamagnetic sphere in a uniform magnetic field - DERIVATION

The basic idea is as follows:
Since it is static and there are no free currents, the 2nd and 4th Maxwell's equations - equations (7.55) of Griffiths' book - are
div B = 0 (2)
and
curl H = 0. (4)
From (4), we see that we can use scalar potential,
H = -grad U, (10.1)
and, since B = μH inside the sphere and B = μoH outside, we obtain inside the sphere
0 = div B = div (μH) = μdiv (-grad U) = -μ ΔU , so that
ΔU = 0 (10.2)
And the same holds outside the sphere, that is U satisfies Laplace's equation.

Suppose now the externally imposed uniform magnetic field is He. Since the diamagnetic material is linear, our solution must be linear with respect to He and also satisfy the Laplace's equation. There are only two such solutions, both based on the first Legendre polynomial: c1 Her and c2 Her/r3, where r is counted from the origin taken at the center of the sphere.
Since there should not be any singularity, only c1Hr is acceptable inside the sphere.
Outside the sphere, both solutions are acceptable, and since the second solution vanishes at infinity, the coefficient of the first one is one. That ...