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Teltron Tube Experiment and How to Get The Specific Charge of a Particle

Important Formula

Formula: Specific Charge of a particle in a Magnetic Field

$$\frac{q}{\class{brown}{m}} ~=~ \frac{2 \, U_{\text B}}{r^2 \, \class{violet}{B}^2}$$ $$q ~=~ \frac{2 \, \class{brown}{m} \, U_{\text B}}{r^2 \, \class{violet}{B}^2}$$ $$\class{brown}{m} ~=~ \frac{q \, r^2 \, \class{violet}{B}^2}{2 \, U_{\text B}}$$ $$U_{\text B} ~=~ \frac{q \, r^2 \, \class{violet}{B}^2}{2\class{brown}{m}}$$ $$\class{violet}{B} ~=~ \frac{1}{r} \sqrt{\frac{2 \, \class{brown}{m} \, U_{\text B}}{q}}$$ $$r ~=~ \frac{1}{\class{violet}{B}} \, \sqrt{ \frac{2 \, \class{brown}{m} \, U_{\text B}}{q} }$$

Rearrange formula

What do the formula symbols mean?

Electric charge

$$ q $$ Unit $$ \mathrm{C} = \mathrm{As} $$

Electric charge of the particle (e.g. electron) moving on a circular path in a magnetic field.

The ratio of the charge $q$ to the mass $m$ of the particle is called specific charge $ \frac{q}{m} $. For example, the electron has the following specific charge: $ \frac{q}{m} = - 1.758 \cdot 10^{11} \, \frac{\text C}{\text{kg} } $.

Mass

$$ \class{brown}{m} $$ Unit $$ \mathrm{kg} $$

Mass of the particle.

If the particle is a electron, then the mass is $ m = 9.1 \cdot 10^{-31} \, \text{kg} $.
If the particle is a proton, then the mass is $ m = 1.67 \cdot 10^{-27} \, \text{kg} $.

Natürlich kannst du auch andere geladene Teilchen haben...

Acceleration voltage

$$ U_{\text B} $$ Unit $$ \mathrm{V} $$

Accelerating voltage set, for example, in the electron gun in the teltron tube experiment to change the velocity of electrons or other charged particles.

Magnetic flux density (B-field)

$$ \class{violet}{B} $$ Unit $$ \mathrm{T} = \frac{\mathrm{kg}}{\mathrm{A} \, \mathrm{s}^2} $$

Magnetic flux density describes how strong the magnetic field is in which the charged particle moves.

Radius

$$ r $$ Unit $$ \mathrm{m} $$

Radius of the circular path on which the charged particle moves. By changing the magnetic field $\class{violet}{B}$ you can easily make the radius $r$ of the circular path larger or smaller.

Table of contents

In a teltron tube experiment, we can visualize the circular path of charged particles deflected in a magnetic field due to the Lorentz force.

How a teltron tube works...

With a electron gun we generate a straight electron beam. The electron gun consists of a heating spiral which is heated up using a heating voltage $ U_{\text H} $. This causes an electron cloud to form around the heating spiral. By means of an applied accelerating voltage $ U_{\text B} $ between the spiral and a permeable anode, the electrons are accelerated. Behind the anode the particles have the velocity $ \class{blue}{v} $. In this way, a beam of electrons is obtained.

Schematic experimental setup of the teltron tube with shown circular electron path.

The electron beam is in a glass bulb containing a gas (for example molecular hydrogen, $\mathrm{H}_2$). The flying electrons will transfer some of their energy to the $\mathrm{H}_2$ molecules via collisions. The atoms are therefore energetically excited. After a very short time, they emit the obtained energy in the form of visible light. In this way, we can see the electron path with our naked eyes!

The gas pressure in the glass bulb should not be too high, because then electrons are slowed down too much by the gas atoms. A slower speed leads to a larger, distorted circular path radius. Therefore, you choose the pressure in the glass bulb as low as possible, but so that you can still see a visible excitation of the atoms (a glow).

Now we would see a straight electron beam. To force the electrons into a circular path, we place the glass bulb into a Helmholtz coil. It consists of two round coils placed side by side, through which an electric current flows. This current generates a homogeneous magnetic field $\class{violet}{B}$ between the two coils where the glass bulb is placed.

A Helmholtz coil.

The glass bulb is aligned in such a way that the electron movement is exactly perpendicular to the magnetic field. This perpendicular motion makes the equations much simpler.

Moving charged particles (electrons) in the magnetic field experience a Lorentz force $\class{green}{F}$ that forces the electrons into a circular path.

Origin of circular motion due to Lorentz force pointing perpendicular to the velocity of the electron and the magnetic field.

Direction of electron movement

Point your thumb, index finger and middle finger as shown here to find out the deflection direction of the electrons.

You determine the direction of the Lorentz force with the left or right hand rule. For negative charges you use your left hand. For positive charges you use your right hand. Since we are dealing with electrons here, you use the left hand.

Thumb - points in the direction of movement of the electrons.
Index finger - points in the direction of the magnetic field.
Middle finger - points (after alignment of the thumb and index finger) in the direction of the Lorentz force.

If you have applied the three-finger-rule correctly, your middle finger - at each position of the electron - will point into the center of the circle at the circular movement. The Lorentz force always acts perpendicular to the direction of motion of the charge.

Specific charge of an electron

You can use the teltron tube experiment to find out the specific charge $ \frac{q}{\class{brown}{m}} $ of an electron (or other charged particle). The specific charge is the ratio of the charge $q$ to the mass $ \class{brown}{m} $ of the particle. Why is this quantity important? Well, sometimes you don't know neither the mass nor the charge of the particle. In this case, you can at least find out their ratio.

Example: Difference between charge and specific charge

The elementary charge $ e = 1.6 \cdot 10^{-19} \, \mathrm{C} $ of an electron and a proton is the same, but their specific charges $q/\class{brown}{m}$ are not! And this is because the mass of the proton $\class{brown}{m_{\text p}}$ is much larger than the mass of the electron $\class{brown}{m_{\text e}}$.

Since we have aligned the glass bulb in the external magnetic field so that the direction of motion of the electrons and the magnetic field lines are exactly orthogonal to each other, we can use the following simple formula for the Lorentz force $\class{green}{F}$ (magnetic force):

$$ \begin{align} \class{green}{F} ~=~ q \, \class{blue}{v} \, \class{violet}{B} \end{align} $$

The Lorentz force does exactly the same what the centripetal force $F_{\text z}$ does, namely it keeps a body (in our case a charged particle) on a circular path. We can calculate the centripetal force as follows:

$$ \begin{align} F_{ \text z} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{ r } \end{align} $$

Centripetal force keeps the particle on a circular path.

The Lorentz force IS in this case the centripetal force, because both hold the electron on the circular path! We can therefore set the two forces in Eqs. 1 and 2 equal:

$$ \begin{align} q \, \class{blue}{v} \, \class{violet}{B} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{r} \end{align} $$

Rearrange Eq. 3 with respect to the specific charge $q/\class{brown}{m}$:

$$ \begin{align} \frac{q}{\class{brown}{m}} ~=~ \frac{\class{blue}{v}}{r \, \class{violet}{B}} \end{align} $$

Perfect! The strength of the magnetic field $ \class{violet}{B} $ is known to us because we set the magnetic field ourselves. We can determine or estimate the radius $ r $ of the resulting circular path with a ruler. We know the velocity $ \class{blue}{v} $ indirectly as well. For this we just have to express it somehow with the acceleration voltage $ U_{\text B} $. The voltage is set with a voltage source and sets the velocity of the electrons.

How to calculate the velocity of electrons

The electron receives the total kinetic energy $ \frac{1}{2} \class{brown}{m} \, \class{blue}{v}^2 $ after passing through the accelerating voltage $ U_{\text B} $ in the electron gun. This voltage accelerates the particle to the velocity $ \class{blue}{v} $. After the acceleration phase, the electron has completely converted its electric energy $q \, U_{\text B}$ into kinetic energy. So you can equate the two formulas for kinetic and electric energy:

$$ \begin{align} \frac{1}{2} \, \class{brown}{m} \, \class{blue}{v}^2 ~=~ q \, U_{\text B} \end{align} $$

Rearrange Eq. 5 for the specific charge $ \frac{q}{\class{brown}{m}} $:

$$ \begin{align} \frac{q}{\class{brown}{m}} ~=~ \frac{\class{blue}{v}^2}{2\, U_{\text B}} \end{align} $$

Now you can equate the specific charge from Eq. 4 with Eq. 6 and rearrange for the velocity:

$$ \begin{align} \class{blue}{v} ~=~ \frac{2\, U_{\text B}}{r \, \class{violet}{B}} \end{align} $$

Calculate specific charge by means of acceleration voltage

To get the specific charge, which depends only on experimentally accessible, known quantities, we must substitute the derived formula for velocity 7 into Eq. 4:

$$ \begin{align} \frac{q}{\class{brown}{m}} ~=~ \frac{2 \, U_{\text B}}{r^2 \, \class{violet}{B}^2} \end{align} $$

We're done! You have learned how a teltron tube experiment works, and how you can use it to find out the specific charge of a charged particle. In the next lesson, we'll look at what else the Lorentz force has to offer: We will be able to use it to construct a velocity filter (WIEN filter) that will allow us to choose the desired velocity of the charged particles.