Motion of charged particles in complex fields > Motion in a homogeneous magnetic field > Problem 10.1.24

Statement

$10.1.24.$ a. An electron moves in a uniform magnetic field around a circle. Any other circle is drawn through it, the $OO'$ axis of which is directed along the magnetic field. Show that the sum of $M +(1/2π)eΦ$, where $Φ$ is the magnetic field flux through this circle, and $M$ is the angular momentum of the electron relative to the $OO'$ axis, does not depend on the position of the electron.

b. Show that the sum of $M +(1/2π)eΦ$ does not change in the case of an electron moving in a homogeneous magnetic field along a helical line

Solution

a)
Consider two circles, one with a radius of $R$, along which an electron rotates, the other on an axis that is located at a distance of $d$ from the center of the first. Consider the moment in time when R forms an angle with d $\alpha$. The angular momentum is the product of the momentum on the shoulder, which is perpendicular from the axis to the straight line on which the velocity vector lies. In our case, for $R < d$:

$$h = dcos\alpha - R$$

And the angular momentum, remembering the expression for momentum in terms of the radius of the orbit:

$$M = m \upsilon (dcos\alpha - R)=BeRdcos\alpha - BeR^2$$

The flow through the side circle is:

$$\Phi = B\pi R^2 = B \pi (R^2 + d^2 - 2Rdcos\alpha)$$

Let's substitute everything into our desired invariant:

$$M + \frac{1}{2\pi} e \Phi = BeRdcos\alpha - BeR^2 + \frac{BeR^2}{2} + \frac{Bed^2}{2} - BeRdcos \alpha = \frac{Be(d^2-R^2)}{2}$$

As you can see, there really is no dependence on the position.

b)

When moving in a spiral, we have two velocity components, however, the component parallel to the axis does not contribute to the projection of the angular momentum onto the axis of the magnetic field, and the addition of the scalar of the flow and the angular momentum module makes sense only in the sense of projection onto the axis perpendicular to the section.

Answer

а) Q.E.D

b) Q.E.D