And Mochi Sticks Too

All the preceding posts just to answer why the positive charges at the nucleus stick together.

They stick, because they are revolving in parallel \(B\) orbits. The attractive force between two orbiting charges in parallel orbits is,

\(F_L=\pi q|v|B=-i\pi qE\)

where \(q\) is the charge in one orbit, \(v\) the velocity of the charge, \(B\) the magnetic field established by the other orbiting charge and \(E\) the electrostatic field at distance \(r\) where \(B\) is, due to the other moving charge, and \(r\) is also the distance between the two orbits.  \(-i\) rotates the force \(F_L\) in the plane perpendicular to \(B\) in the anticlockwise direction.

\(F_L\) is the result of two interacting \(B\) fields.  The maximum interaction occurs when the \(B\) field lines are parallel.   The velocity \(v\) of the charge placed in the established \(B\) field was made perpendicular to the already established \(B\) field lines such that the \(B\) field set up by this moving charge is parallel to the first.

In the derivation of \(F_L\) in the post "Lorentz's without q", a moving negative charge was used that established a \(E\) field downwards.  A moving positive charge was then added to the situation to obtain \(F_L\).

If \(v\) in rotated, \(v\rightarrow -iv\)  to the same direction as the moving negative charge then \(F_L\) is along  \(-E\) as the expression shows.  This is consistent with the observed fact that parallel wires with opposite current repulse each other.

There is no relative velocity concerns here, as long as one charge is moving, it establishes a \(B\) field that will interact with another moving charge.

Have a nice day.