Motion of charged particles in complex fields > Motion in a homogeneous magnetic field > Problem 10.1.25

Statement

$10.1.25.$ Regions $I$ and $II$ of two homogeneous unidirectional magnetic fields with induction $B_1$ and $B_2$ have an axisymmetric thin junction $AA'$, in which the magnetic field has a large radial component. The electron in region I moves along the magnetic field at a distance $R$ from the axis of symmetry of the transition. What angular momentum relative to the axis of symmetry does an electron acquire during the transition from region $I$ to region $II$? Does the sum $M+(1/2π)eΦ$ remain constant during the motion of this particle (see the notation in problem $10.1.24$)?

Solution

To begin with, we will show the invariance of the sum in this case as well. The force acting on the electron inside the transition zone :

$$F = -B_r e \upsilon_z$$

Where $B_r$ is the radial component of induction, $\upsilon_z$ is the component of velocity along the axis of symmetry.

$$\frac{dM}{dt} = -B_r e r \upsilon_z $$

$$dM = - B_r e r dz$$

Now let's consider the flow change through the cross section of a ring of arbitrary radius R:

$$d\Phi = 2 \pi R B_r dz$$

From here:

$$dM = -\frac{1}{2\pi}ed\Phi$$

$$M+\frac{1}{2\pi} e \Phi = const$$

Well, now, it's very easy to find the desired moment, at first the moment of momentum is zero, because the electron does not rotate:

$$\frac{1}{2}eB_1 R^2 = M + \frac{1}{2}eB_2 R^2$$

$$M = \frac{(B_2 - B_1)R^2e}{2}$$

Answer

$$M = \frac{(B_2 - B_1)R^2e}{2}$$