Jackson's 6.4 - Uniformly magnetized conducting sphere

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SUMMARY

The discussion centers on the analysis of a uniformly magnetized conducting sphere with a total magnetic moment of m = 4πMR³/3, rotating about its magnetization axis at angular speed ω. It is established that the motion induces an electric field and a uniform volume charge density, given by ρ = -mω/πc²R³, despite the sphere being electrically neutral. The lowest possible electric multipolarity outside the sphere is determined to be quadrupole, with non-vanishing components of the quadrupole moment tensor specified as Q_{33} = -4mωR²/3c² and Q_{11} = Q_{22} = -Q_{33}/2. The solution involves using Ohm's law and Maxwell's equations, emphasizing the complexity of transforming fields in a rotating frame.

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  • Concept of magnetic scalar potential
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Homework Statement


A uniformly magnetized and conducting sphere of radius R and total magnetic moment m = 4\pi MR^3/3 rotates about its magnetization axis with angular speed \omega. In the steady state no current flows in the conductor. The motion is nonrelativistic; the sphere has no excess charge on it.

(a) By considering Ohm's law in the moving conductor, show that the motion induces an electric field and a uniform volume charge density in the conductor, \rho = -m\omega /\pi c^2 R^3.

(b) Because the sphere is electrically neutral, there is no monopole electric field outside. Use symmetry arguments to show that the lowest possible electric multipolarity is quadrupole. Show that only a quadrupole field exists outside and that the quadrupole moment tensor has nonvanishing components, Q_{33} = -4m\omega R^2/3c^2, Q_{11}=Q_{22}=-Q_{33}/2.

Homework Equations



Ohms Law \mathbf{J}=\sigma \mathbf{E} and Maxwell's equations. If I understood, we are considering a steady current, so that \nabla \cdot \mathbf{J} = 0, which reduces to magnetostatics.

The Attempt at a Solution



The first thing I believe is to find the magnetic field due to the magnetization. This is known to be \mathbf{B}=2\mu_0 \mathbf{M}/3 on the inside of the sphere. The only thing I don't feel fine with this is that this result was derived for the case of \mathbf{J} = 0, using the magnetic scalar potential method. The issue is that this is not the case here. The sphere is rotating, so we have a motion of charges, which would make a nonzero current.

Now, when trying to solve the problem, I must use that \mathbf{J}=\sigma \mathbf{E} to find \mathbf{E} as the problem asks. In that case I must find \mathbf{J}. My thought was that since \nabla\times \mathbf{B}=\mu_0\mathbf{J} we could compute just the curl. But the magnetic field I've found is constant, which gives zero current. Actually, this is not inconsistent, since the result above was derived assuming zero current.

Searching online I've seem people changing frames to the rotating frame of the sphere. In that frame there really is no current, since the points on the sphere are at rest. But the formula for transforming the fields to the rotating frame appear to be non trivial to derive, some of then recurring to special relativity, while at this point in the book, the reader isn't expected yet to use relativity theory.

Other approaches for changing frame seems handwavying to me. One merely says that in the rotating frame there is no magnetic field and that the Lorentz force must match in the two frames, but without much explanation of the details.

In summary, I didn't even find how to get started. What should I do?
 
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I found a solution on-line, perhaps the same one that you found, and I've asked for permission from the PF staff to provide you the "link". (Normally under PF rules, homework helpers are not supposed to furnish complete solutions). ## \\ ## That on-line solution uses a relativistic transformation of the fields in a rotating frame. The solution appears to be a good one, and uses Gauss' law to compute the (uniform) charge distribution from the electric field in the laboratory frame, and also computes the surface charge distribution, but the problem and its solution really exceeds the limits of my understanding of E&M. ## \\ ##Perhaps you have already found the "link" that I mentioned above. In any case, I don't see a solution to this without using special relativity transformations. The non-zero electric field is the result of a transformation to get the electric field in the laboratory frame. The electric field is zero in the frame that is rotating with the sphere. ## \\ ## Once he gets the bulk charge density, he computes the surface charge distribution with a detailed multipole analysis that is beyond anything I worked through when we used J.D. Jackson's textbook for a two course E&M sequence in graduate school in 1979. ## \\ ## One thing that is also a puzzle in his solution is, if I understand it correctly, in the laboratory frame, this charge density is stationary even though the object is rotating. The static charge is a result of the static/constant electric field, and is needed to get the resulting electric field (that is non-zero in the laboratory frame) inside the sphere.
 
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