The electron has two degrees of freedom on the sphere delimited by its orbit. In the direction perpendicular to the orbit shown, the positive nucleus relatively slides along the \(B\) field line and experience no other force. The electron's movement along the orbit however, rotates the \(B\) field around the nucleus. Any movement of the nucleus towards the electron results in a force pointing into the path of the electron further ahead, ie. closer to the electron and any movement away from the electron results in a force that send the nucleus further away from the electron. The system seems not to confine the nucleus!
The nucleus is confined only if it is moving relatively slowly within the bounds of the orbit of the electron. As the nucleus move away from the electron slowly, the electron swings pass and comes ahead of the nucleus. So eventually all movement results in the nucleus and the electron coming closer together. The Lorentz's force pushes the nucleus before the electron each time and the nucleus orbit in the same plane as the electron, in the same direction. A repulsive force develop between the two opposite charges orbiting in parallel orbits in the same direction. This repulsive force keeps the atom from collapsing. The electron does not collide into the nucleus.
At the same time, the nucleus is driven along the path of the \(B\) field lines. The nucleus is also in orbit; a \(B\) orbit.
The net result of these two factors is a torus. The nucleus orbiting around a wider \(B\) orbit and the electron in a tight orbit around the nucleus. The nucleus's movement within the confines of the electron has to be relatively slow (\(\lt\lt c\)) in order that it is always acted upon by a force that send it into the electrons orbit further ahead. The nucleus is also in a spin. The system stabilize when the relative spins of the nucleus and the electron results in no Lorentz's force on the nucleus. Both the nucleus and electron will be spinning in the same direction.
This then is the nucleus plus one electron. A torus. The important point is that the relative spin between the positive nucleus and the electron negates \(F_{LP}\) and the quadratic equation for \(r_e\) becomes,
\( (Av^{ 2 }+\sum \limits_{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ 4\varepsilon _{ o }r^{ 2 }_{ ei } } } )r^{ 2 }_{ e }-m_{ e }v^{ 2 }r_{ e }+\cfrac { 1 }{ 4 } \left\{{4nGm_pm_e} -\cfrac { T_{ e }T_{ n } }{ \pi \tau _{ o } } \right\} =0\)
At last we have a torus. This is highly speculative.
