IPhO 1987, Теория и експеримент — Задача 2. Electrons in a magnetic field

• 1. 0.0037 T

  Printed in the solution (its part 2) as B1 = 0.37·10⁻² Vs/m².

• 2. 0.0148 T

  Printed in the solution (its part 1) as B = 1.48·10⁻² Vs/m².

• 3. The radial deviation is finite (R ≤ r ≤ r1); the drift is in the direction of the negative z-axis.

1. Determination of B:

The vector of the velocity of any electron is divided into components parallel with and perpendicular to the magnetic field $\vec{B}$:

The Lorentz force $\vec{F} = -e \cdot (\vec{v} \times \vec{B})$ influences only the perpendicular component, it acts as a radial force:

Hence the radius of the circular path that has been travelled is

and the period of rotation which is independent of $v_{\perp}$ is

The parallel component of the velocity does not vary. Because of $\alpha_0 \ll 1$ it is approximately equal for all electrons:

Hence the distance b between the focusing points, using eq. (5), is

From the law of conservation of energy follows the relation between the acceleration voltage $V_0$ and the velocity $v_0$:

Using eq. (7) and eq. (4) one obtains from eq. (6)

and because of $b = \frac{2 \cdot \pi \cdot R}{4}$ one obtains

2. Determination of $B_1$:

Analogous to eq. (2)

must hold.

From eq. (7) follows

3. Finiteness of $r_1$ and direction of the drift velocity

In the magnetic field the lines of force are circles with their centres on the symmetry axis (z-axis) of the toroid.

In accordance with the symmetry of the problem, polar coordinates r and $\varphi$ are introduced into the plane perpendicular to the z-axis (see figure below) and the occurring vector quantities (velocity, magnetic field $\vec{B}$, Lorentz force) are divided into the corresponding components.

Since the angle of aperture of the beam can be neglected examine a single electron injected tangentially into the toroid with velocity $v_0$ on radius R.

In a static magnetic field the kinetic energy is conserved, thus

The radial points of inversion of the electron are defined by the condition

Using eq. (12) one obtains

Such an inversion point is obviously given by

To find further inversion points and thus the maximum radial deviation of the electron the components of velocity $v_{\varphi}$ and $v_z$ in eq. (13) have to be expressed by the radius.

$v_{\varphi}$ will be determined by the law of conservation of angular momentum. The Lorentz force obviously has no component in the $\varphi$ - direction (parallel to the magnetic field). Therefore it cannot produce a torque around the z-axis. From this follows that the z-component of the angular momentum is a constant, i.e. $L_z = m \cdot v_{\varphi} \cdot r = m \cdot v_0 \cdot R$ and therefore

$v_z$ will be determined from the equation of motion in the z-direction. The z-component of the Lorentz force is $F_z = -\, e \cdot B \cdot v_r$. Thus the acceleration in the z-direction is

That means, since B is assumed to be constant, a change of $v_z$ is related to a change of r as follows:

Because of $\Delta r = r - R$ and $\Delta v_z = v_z$ one finds

Using eq. (14) and eq. (15) one obtains for eq. (13)

where $A = \frac{e}{m} \cdot B \cdot \frac{R}{v_0}$

Discussion of the curve of the right side of eq. (17) gives the qualitative result shown in the following diagram:

Hence $r_1$ is finite. Since $R \le r \le r_1$ eq. (16) yields $v_z < 0$. Hence the drift is in the direction of the negative z-axis.

Grading Scheme