While energy is not conserved here due to ohmic dissipation, according to the Emmy Noether’s theorem, each symmetry gives rise to a conservation law, and there are many symmetries in this problem. The mechanical momentum of a charged particle in magnetic field is not conserved due to the Lorentz force \vec{F}_L=q\vec v\times \vec B, but what about the Newton’s 3rd law? It appears that a moving charge exerts a force equal to -\vec{F}_L on the electromagnetic field which means that -\vec{F}_L equals to the time derivative of its momentum \vec P_f=\frac 1{c\mu_0}\int \vec E\times \vec B\mathrm d^3\vec r. So, the momentum of the electromagnetic field together with the momentum of the charge is conserved. While the integral expression of \vec P_f is very inconvenient, there is a nice simple expression for it in terms of the vector potential. All this might sound very complicated, but actually there is no need to even use vector potential (although you can do that, too): you can just play with the Newton’s 2nd law for the charged particle in magnetic field and see if you can deduce from it a conservation law (i.e. an expresio the time derivative of which is zero).

Apart from the conservation law discussed above, you might want to solve an auxiliary problem of finding the polarizability of the “dumbbell”, i.e. calculating the dipole moment induced on it by an external electric field \vec E as a function of the angle between the field and the axis of the “dumbbell”.

Chinese translation of hint 2 (中文)

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