Classical model of diagamanetism

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

My difficulty is :
If electron with orbital motion is taken as a dipole, then since the magnetic field is uniform, the net force due to magnetic field is 0.
But, in eqn. 6.6, while calculating force acting on it , e is taken as a moving charged particle, not as a dipole?
Why should the two ways of taking e gives two different answer?

 

[haruspex]

the net force due to magnetic field is 0.

There is no force parallel to the field. The force on the orbiting electron is radial, in the plane of the orbit.
If you view it as a point charge in orbit then you will get a rapidly rotating transverse force, but if you view it as smoothed out around the circle of the orbit then there is no net force at all since all the radial forces on the portions of it cancel.

 

[rude man]

Homework Statement

View attachment 209269View attachment 209270

Homework Equations

The Attempt at a Solution

My difficulty is :
If electron with orbital motion is taken as a dipole, then since the magnetic field is uniform, the net force due to magnetic field is 0.
But, in eqn. 6.6, while calculating force acting on it , e is taken as a moving charged particle, not as a dipole?
Why should the two ways of taking e gives two different answer?

Fact is, it's a funny dipole, in the sense that one charge is very massive compared to the other. So, just as we approximate that the Earth revolves around the Sun in a circular orbit, the electron revolves around the much more massive proton which just sits there. So the only moving charge is the electron, which sees the electrostatic attraction of the proton as well as the Lorentz force due to the B field.

(Strictly as an aside, the computational assumption is here made that the radius does not change when the B field is applied. For the classical view of the electron revolving around the nucleus as described here, that is incorrect. And the view of electrons revolving in quantized but constant radii around the nucleus is false to begin with. So the radius in the above picture is acually a function of B. I state this only to stir up another argument with whoever, as usual! And I've done the analysis. You the OP should ignore it completely.