Particle accelerators

What causes the northern lights?

The aurora borealis is shown in the night sky over a coniferous forest with lights of violets, pinks, greens, and yellows.

How did old Television sets work?An older style television set with multicoloured static on the screen, and control dials and switches on the top-right of the set. An inbuild speaker is on the bottom right of the set

Did you know that Melbourne is home to the most powerful synchrotron?

In fact you can search where all accelerators globally are on the International Atomic Energy Agency's webiste.

All of these involve the accelation of particles by magnetic and electric fields. To begin to understand particle acceleration, first review key concepts of electric and magnetic fields with the questions below.

Particle Accelerators

A particle accelerator is a device that accelerates particles through electromagnetic fields to extremely high speeds, often close to the speed of light.

Often it is useful to examine the cathode ray tube when learning about particle accelerators. Cathode ray tubes were used in old televesion sets to create the picture, and are a relatively simple type of particle accelerator. This is due to the configuration of a cathode ray tube having two terminals, negative and positive. The negative terminal (cathode) is heated and as a result will produce free electrons which will then accelerate towards the positive terminal (anode).

An electron gun is the source of the electrons. In an electron gun electrons are ‘boiled’ off a heated wire element acting as the cathode. The electrons are then accelerated from rest across a chamber emptied of air towards a positively-charged plate acting as the anode (the positively-charged electrode) by an electric field created between the charged plates.

A diagram of a cathode ray tube. Electrons are accelerated, then pass through a magnetic field generated by charged plates into a spherical capture area

Check your understanding by filling in the blanks.

Note: ignore relativistic effects.

Calculating the work done and speed of a particle through an electric field

Work is the transfer of energy that occurs when an external force acts on an object over a distance. A particle experiences an acceleration is due to the force and hence there is work done on the particle as it moves through an electric field. To determine the amount of work done, and the velocity of the particle, we need to consider several equations that are commonly used to analyse field concepts.

Consider the electric field strength between two charged plates (cathode and anode):

\[E=\frac{V}{d}\]

Where

  • \(V\) is the accelerating voltage or potential difference across the two plates \((V)\)
  • \(d\) is the distance between the two plates (\(m\))
  • \(E\) is the field strength \((NC^{-1}\) or \(Vm^{-1})

For a uniform electric field the field strength \(E\) is constant at all points between the plates. You might recall from your studies of motion that the work done is given by

\[W= Fs\]

Where

  • \(W\) is work done \((J)\)
  • \(F\) is force \((N)\)
  • \(s\) is displacement \((m)\)

We also know that

\[F=qE\]

Where \(q\) is the charge of the particle \((C)\)

As \(E=\frac{V}{d}\) then \(F=q\frac{V}{d}\). We can also write this as \(Fd = qV\)

Through this algebraic manipulation, we can now show that

\[W = qEd = qV\]

Work is the change in energy of an object due to a force on the object. As a particle is accelerated through a potential difference it gains kinetic energy.

\[qV=\frac{1}{2}mv^{2}\]

This equation can then be used to calculate how fast a charged particle will move when accellerated by an electric field.

Worked Example

Determine the final speed of a single electron, charge \(1.6\times10^{-19}C\) and a mass \(9.1\times10^{-31}kg\), when it is accelerated from rest across a potential difference of 6kV.

1. Set up equation \(qV=\frac{1}{2}mv^{2}\)

2. Substitute know values \(1.6\times10^{-19}\times6000=0.5\times9.1\times10^{-31}v^{2}\)
Note that potential difference is measured in V so a conversion from kV to V is required.

3. Rearrange making v (speed) the subject \(v=\sqrt\frac{1.6\times10^{-19}\times6000}{0.5\times9.1\times10^{-31}}\)

4. State the final speed \(v=4.6\times10^{7}ms^{-1}\)

Calculating the effect of a moving particle through a magnetic field

After a particle has been accelerated through a potential difference, particle accelerators use a series of magnets to change the direction of movement of the particle.

The magnetic force on an electric charge is always perpendicular to the motion of the charge. This results in the charge moving in a circle, as seen in the diagram below:

A diagram shows the motion of a charged particle (-q) in a magnetic field, B-sub-in, directed into the page. The particle moves in a circular path due to the magnetic force, F, which is perpendicular to its velocity, v, at every point, maintaining the circular trajectory

By equating the equations for magnetic force (\(F=qvB\)) and force due to circular motion \((F=\frac{mv^{2}}{r})\) we get:

\[qvB=\frac{mv^{2}}{r}\rightarrow r=\frac{mv}{qB}\]

and the radius of the circular motion can be found.

Worked Example

An electron gun releases electrons from its cathode which are then accelerated across a potential difference of 16 kV, over a distance of 15 cm between a pair of charged parallel plates.

Mass of Electron9.1 x 10-31 kg
Magnitude of Charge of Electron1.60 x 10-19 C

A) Calculate the speed of the electron as it exits the electric field.

  1. Set up equation \(qV=\frac{1}{2}mv^{2}\)
  2. Substitute know values \(1.6\times10^{-19}\times16000=0.5\times9.1\times10^{-31}v^{2}\)

Note that potential difference is measured in V so a conversion from kV to V is required

Rearrange making v (speed) the subject \(v=\sqrt\frac{1.6\times10^{-19}\times16000}{0.5\times9.1\times10^{-31}}\)

\[v=7.5\times10^{7}ms^{-1}\]

B) The electrons then travel through a uniform magnetic field perpendicular to their motion. Given that this field is of strength 0.5 T, calculate the expected radius of the path of the electron beam.

  1. Set up equation \(r=\frac{mv}{qB}\)
  2. Substitute know values  \(r=\frac{9.1\times10^{-31}\times7.5\times10^{7}}{1.6\times10^{-19}\times0.5}\)

\[r=8.5\times10^{-4}m\]

C) The direction of electrons through the magnetic field is given in the diagram:

A particle with velocity, v, travels, left to right, through a magnetic field directed into the page. A vertical plate is right of and facing the field. Top section is labelled X, bottom is labelled Z. Inline of the particle’s current velocity is labelled Y.

Which point, X, Y or Z would the electrons be expected to move towards?

1. Thinking: Based on right hand slap rule

  • The thumb points in the direction of the positive charge
  • The fingers point on the direction of the magnetic field
  • The palm points in the direction of the force experienced by the charge in the magnetic field.

An open-palm left hand shows the application of the left-hand open-palm rule. An arrow pointing to the left from the thumb shows direction of electron movement. An arrow pointing upwards, parallel to the fingers shows direction of magnetic field. An arrow pointing out-of-page away from the palm shows direction of force on electron.

2. As right hand slap rule is for a positive charge. When dealing with a negative charge either: Apply the right-hand palm rule and reverse the direction of the force.

Solution:

For a positive charge, velocity to right of page, magnetic field into page, resultant force would be up the page towards X. As it is a negative charge, reverse the direction of applied force, electron will travel downwards towards Z.