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:
