A positive charge with velocity \(v\) will experience a force upwards, so a negative charge will experience a force downwards. A similar force acts on the moving electron below due the \(B\) field established by the moving charge on top. And the two electrons are attracted to one another!
Note: To determine the direction of \(F_L\) look down the direction of \(v\), if the \(E\) field fan outwards then the force on a positive charge is on the left of \(v\). If the \(E\) field is inwards than the force is on the right.
This attraction was observed in current carrying wires. There was no clear indication whether this is the case for electron beams on the World Wide Web, but one has suggested that the beams merged, others, theoretical explanation of what might happen considering both \(E\) and \(B\) fields.
This post and the post "Magnetic Field In General, HuYaa" did not assume a conductor. If a moving charge, in establishing a \(B\) field, loses its electrostatic nature and the \(E\) field around it has zero resultant (from the post "Lorentz Without q"), then the electrostatic force term,
\(F_e=\cfrac{q^2}{4\pi\varepsilon_o r_e^2}\)
is not valid in all the calculations for \(r_e\). In the analysis above, a positive charge is pushed away from the moving electron. The thermal repulsive force that might exist between two hot particles also pushes the electrons and proton apart. We have a new force equation for an electron in orbit around a nucleus,
\(\cfrac{m_ev^2}{r_e}=Av^2-F_{Lp}-\cfrac{T_eT_n}{4\pi\tau_or^2_e}\) --- (*)
where \(A\) is the drag factor at light speed and \(F_{Lp}\) the Lorentz force between the positive nucleus and the moving electron. The drag force then becomes much more significant. It is the sole component that holds the atom together, a role previously played by the electrostatic force.
Does the electrostatic force disappear completely? Or do we have a case of superposition where the magnetic force is added to the electrostatic force? The last situation however is not consistent with the fact that the \(E\) field disappeared outside of a current carrying conductor.
\(F_L\) is strongly dependent on the direction of the \(B\) field. The \(B\) field in turn is orientated by the direction of the charge velocity. The \(B\) field so established is circular around the path of the moving charge. In the path of the charge, there is no \(B\) field and so, no \(F_L\). A negative charge and a positive charge is still attracted to each other head-on.
The electrostatic force does not disappear in the head-on direction of the moving electron. Electrons and protons are still attracted to each other to form atoms. On achieving terminal velocity, the electron begins its path along a helix with the positive charge on its longitudinal axis. Over the nucleus, the electron swings around, its forward velocity curve around the nucleus and it performs circular motion with the nucleus as center. The drag factor provides the centripetal force needed for this circular motion. The electron is captured and is in orbit around the nucleus. The expression (*) above applies.
In addition, electrons on the inner shell attracts the coming electrons and add to the centripetal force.
If \(A\) is the only attraction component, that means \(r_e\) is easily effected by manipulating \(A\) and the related space density, \(d_s\).
A strong magnetic force presence, together with a minimum total energy requirement would result in electron orbits being firmly orientated with respect to each other. Removing \(A\) in the presence of a magnetic field (to serve as reference), would then release the electrons into paths from fixed geometry. If we are to record such paths on photographic films we would have a display of various geometrical shapes for each unique electronic configuration.
The electrostatic force comes into play immediately when the electron stop orbiting around the nucleus and move directly away from the nucleus.
Furthermore, electrons in orbital shells have a zero net magnetic effect on electron above and below them when their orbits are perpendicular to one another. It is possible that, all orbits will align onto one flat orbital plane with maximum attraction between them. Equation (*) then becomes,
\(\cfrac{m_ev^2}{r_e}=Av^2+F_{Le}-F_{Lp}-\cfrac{T_eT_n}{4\pi\tau_or^2_e}\)
where the additional term \(F_{Le}\) is the Lorentz's force attraction between moving electrons in orbits. This force is stronger than just the electrostatic force. It is \(\pi\) times \(qE\). When nucleus gain more electrons (negative charges), it will have a stronger Lorentz's attraction force for the next electron.
Till the next post.
