Charged particles and magnetic confinement
Why can an invisible magnetic field hold a fuel that would melt or vaporise any solid container?
Prior knowledge
- A plasma is a quasi-neutral mixture of ions and free electrons that reacts strongly to magnetic fieldschapter Plasma: the fourth state of matter
- Uniform circular motion needs a centripetal force $F = mv^2/r$, and kinetic energy is $E_\text{k} = \frac{1}{2}mv^2$ (earlier physics courses)
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 2.
Think about it
A fusion plasma is about $1.5 \cdot 10^{8}\ \text{K}$ hot. No material can touch it: even tungsten, the metal with the highest melting point, melts at about $3400\ ^\circ\text{C}$, and a plasma that touches a wall cools down at once. So the plasma must float inside its vessel without touching anything. In the previous unit you learnt that a plasma is made of free ions and electrons. Charged particles feel magnetic forces.
How could a magnetic field, which you cannot see or touch, keep millions of degrees away from the walls?
Review: plasma and magnetic fields
Predict
Make your predictions before investigating 2 questions
Prediction 1: a stronger field
Prediction 2: a magnetic mirror
Investigate
One charged particle in a magnetic field. Choose the view, the particle, its speed and the field, and watch the orbit. The blue arrow is the velocity, the red one the force.
Simulation
How to use
View: Uniform field shows $\overline{B}$ perpendicular to the page (⊗ into, ⊙ out of it): the particle circles. The checkbox Electric field adds a field $\overline{E}$ in the page: the centre of the circle then drifts sideways, the same way for ions and electrons. Such drifts push the plasma outward in a ring (see the theory). View: Field along the particle shows the field from the side: the pitch angle $\theta$ sets how much of the velocity is along the field; tick Magnetic mirror to make the field stronger at both ends. The canvas says how long one grid square is and how much time is slowed down. Playback changes only the animation speed.
Assumptions of the model
One particle only, with no collisions and no other particles; a real plasma has billions of interacting particles in every cubic millimetre. The particle is non-relativistic (the fastest speed in the simulation, $10^{7}\ \text{m/s}$, is $3.3\ \%$ of the speed of light). The uniform field is exact. In the mirror the field grows from $B_0$ in the middle to $B_{\max} = 4B_0$ at the ends, and the motion is computed with the guiding-centre model, which conserves the magnetic moment $m v_{\perp}^2 / (2B)$. All readings are model values, not measurements.
How the calculation works
$r = mv_{\perp}/(|q|B)$ and $T = 2\pi m/(|q|B)$ with CODATA masses (proton $1.673 \cdot 10^{-27}\ \text{kg}$, deuteron $3.344 \cdot 10^{-27}\ \text{kg}$, alpha particle $6.645 \cdot 10^{-27}\ \text{kg}$ with $q = 2e$, electron $9.109 \cdot 10^{-31}\ \text{kg}$) and $e = 1.602 \cdot 10^{-19}\ \text{C}$. In the side view $v_{\parallel} = v\cos\theta$ and $v_{\perp} = v\sin\theta$. The mirror ratio is $B_{\max} : B_0 = 4$.
Experiment 1: field strength and radius
Test prediction 1. A proton moves at $2.0 \cdot 10^{6}\ \text{m/s}$ in a uniform field. Change only the field $B$, for example $0.5$, $1.0$, $2.0$ and $4.0\ \text{T}$, and record the radius $r$ and the period $T$ each time. Watch the grid scale on the canvas: it changes when the circle becomes very small or large.
Initial state: view: uniform field, particle: proton, speed $v$ = $2\,000\,000$, field $B$ = $1$. Kept constant: view, particle, speed $v$ $(\text{m/s})$. Varied: field $B$ $(\text{T})$.
| field $B$ $(\text{T})$ | $r\;(\text{mm})$ | $T\;(\text{ns})$ |
|---|
Experiment 2: the magnetic mirror
Test prediction 2. The view shows the field from the side; make sure Magnetic mirror is ticked. A proton starts in the middle at $2.0 \cdot 10^{6}\ \text{m/s}$. Change only the pitch angle $\theta$, for example $10^\circ$, $20^\circ$, $40^\circ$, $60^\circ$ and $80^\circ$. Watch each orbit for a few seconds, then record whether the proton stays trapped.
Initial state: view: field along the particle, particle: proton, speed $v$ = $2\,000\,000$, field $B$ = $1$, $\theta$ = $60$. Kept constant: view, particle, speed $v$ $(\text{m/s})$, field $B$ $(\text{T})$. Varied: $\theta$ $(\text{degrees})$.
| $\theta$ $(\text{degrees})$ | $\text{trapped or escapes}$ |
|---|
Optional Optional experiment: which particle?
Keep the speed $2.0 \cdot 10^{6}\ \text{m/s}$ and the field $1.0\ \text{T}$, and change only the particle. The alpha particle is a helium nucleus, $q = 2e$. Which two particles have almost the same circle, and why? Which one turns the other way?
Initial state: view: uniform field, particle: proton, speed $v$ = $2\,000\,000$, field $B$ = $1$. Kept constant: view, speed $v$ $(\text{m/s})$, field $B$ $(\text{T})$. Varied: particle.
| particle | $r\;(\text{mm})$ | $T\;(\text{ns})$ |
|---|
Word bank
magnetic field · magnetic force · radius · period · inversely proportional · pitch angle · along / across the field · trapped · reflected · to escape · loss cone
Sentence starters
- When we doubled the field, the radius …
- The period did not depend on …
- Protons with a pitch angle of … were …, whereas …
- The evidence suggests that a magnetic field can hold particles across …, but not along …
- I agree / disagree, because …
Conclusion: what can a magnetic field do?
Discuss with a partner, then write four or five sentences. Use numbers from your two tables.
How does the radius depend on the field? Which protons does the mirror hold and which escape? What does this tell you about holding a plasma with a magnetic field?
Model answer
When we doubled the field from $1.0\ \text{T}$ to $2.0\ \text{T}$, the radius fell from about $20.9\ \text{mm}$ to $10.4\ \text{mm}$, so the radius is inversely proportional to $B$. The period also halved, from about $65.6\ \text{ns}$ to $32.8\ \text{ns}$. In the mirror, protons with pitch angles of $40^\circ$ or more were trapped, whereas protons at $10^\circ$ and $20^\circ$ escaped through the ends. So a strong field holds particles well across the field lines, on small circles, but motion along the field lines is a problem: a machine needs field lines without ends. These are model values for one particle.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary.
Theory summary
The magnetic force on a moving charge
A magnetic field $\overline{B}$ pushes on a charge $q$ only when the charge moves. The force is called the magnetic part of the Lorentz force. In the formula below, $v$ is the speed and $\alpha$ is the angle between the velocity $\overline{v}$ and the field $\overline{B}$.
$$F = |q|\,v B \sin\alpha$$
A charge that moves along the field feels no force.
The force is always perpendicular both to $\overline{v}$ and to $\overline{B}$. For a positive charge, the right-hand rule gives its direction; a negative charge is pushed the opposite way. Because the force is perpendicular to the motion, it does no work: it changes the direction of the particle, not its speed.
Radius and period of the orbit
When $\overline{v}$ is perpendicular to a uniform $\overline{B}$, the magnetic force is a constant-size force towards a centre, so the particle moves on a circle. The magnetic force is the centripetal force:
$$\begin{aligned} |q|\,v B &= \frac{m v^2}{r} \\[8pt] r &= \frac{m v}{|q| B} \\[8pt] T &= \frac{2 \pi r}{v} = \frac{2 \pi m}{|q| B} \end{aligned}$$
The radius $r$ is called the gyroradius. A stronger field gives a smaller circle. The period $T$ does not depend on the speed: a faster particle runs on a larger circle in the same time.
Along the field: helices and magnetic mirrors
Usually the velocity has a part along the field, $v_{\parallel} = v\cos\theta$, and a part across it, $v_{\perp} = v\sin\theta$. The angle $\theta$ is the pitch angle. The field does not affect $v_{\parallel}$, so the particle circles around a field line and slides along it: its path is a helix. The particle is tied to its field line, but it can run freely along it.
In a magnetic mirror the field is stronger at both ends. As a particle moves into the stronger field, more of its motion turns into circling and $v_{\parallel}$ shrinks. If $v_{\parallel}$ reaches zero, the particle turns back. The condition for this is below: $B_0$ is the field in the middle and $B_{\max}$ the field at the ends.
$$\sin^2\theta \geq \frac{B_0}{B_{\max}}$$
Particles with smaller pitch angles, inside the loss cone, escape through the ends.
The Earth's magnetic field is a natural mirror: it traps charged particles in the radiation belts, where they bounce between the north and the south. Particles that do reach the upper air near the poles make the northern lights.
Why a ring, and why it is still hard
A straight field always has ends, and every particle with $v_{\parallel} \neq 0$ eventually leaks out. Bending the field into a ring, a torus, closes the field lines, so there are no ends.
But in a ring the coils are closer together on the inside, so the field is stronger near the centre and weaker on the outside, and the field lines are curved. A circling particle then drifts slowly across the field: ions one way, electrons the other. The charges separate, an electric field forms, and the whole plasma drifts outward into the wall. The cure is to twist the field lines, which is the topic of the chapter From Early Machines to Modern Reactors.
One particle is not a plasma
The simulation follows one particle in a fixed field. A fusion plasma has about $10^{20}$ ions and as many electrons in every cubic metre. They collide, which lets them jump from one field line to another. Their currents change the magnetic field that holds them. Small waves and swirls, turbulence, carry heat out much faster than single-particle theory predicts. Large-scale instabilities can bend or break up the whole plasma: the kinks and wriggles of early pinch machines such as ZETA in 1958 are an example. Single-particle motion explains why magnetic confinement is possible; the collective behaviour of the plasma explains why it is so difficult.
Describe a process: from a straight field to a twisted ring
Put the steps of the argument in a logical order. Then explain the chain to your partner aloud, one step each.
Back to your predictions
Compare your two predictions with your tables. Then answer the opening question in one or two sentences: how can a magnetic field hold a fuel that would melt any solid container?
Would you change your prediction?
Apply
Worked example
Problem
In a tokamak with a magnetic field of $5.0\ \text{T}$, a deuteron (the nucleus of deuterium) moves at $1.0 \cdot 10^{6}\ \text{m/s}$ perpendicular to the field. Calculate the radius and the period of its circular motion. The mass of a deuteron is $3.344 \cdot 10^{-27}\ \text{kg}$ and its charge is $e = 1.602 \cdot 10^{-19}\ \text{C}$.
Step 1/7
Given: $m = 3.344 \cdot 10^{-27}\ \text{kg}$, $q = e = 1.602 \cdot 10^{-19}\ \text{C}$, $v = 1.0 \cdot 10^{6}\ \text{m/s}$, $B = 5.0\ \text{T}$, $\overline{v}$ perpendicular to $\overline{B}$. Asked: the radius $r$ and the period $T$.
Step 2/7
Principle: The magnetic force $|q|vB$ is perpendicular to the velocity, so it keeps the speed constant and acts as the centripetal force $mv^2/r$ of a circular motion.
Step 3/7
Formula: From $|q|vB = mv^2/r$:
$$\begin{aligned} r &= \frac{m v}{|q| B} \\[8pt] T &= \frac{2 \pi m}{|q| B} \end{aligned}$$
Step 4/7
Unit conversions: All quantities are in SI units. $1\ \text{T} = 1\ \text{kg}/(\text{C} \cdot \text{s})$, so $\text{kg} \cdot (\text{m/s}) : (\text{C} \cdot \text{T}) = \text{m}$.
Step 5/7
Substitution: $$\begin{aligned} r &= \frac{3.344 \cdot 10^{-27}\ \text{kg}}{1.602 \cdot 10^{-19}\ \text{C} \cdot 5.0\ \text{T}} \\ &\quad \cdot 1.0 \cdot 10^{6}\ \text{m/s} \\ &= 4.17 \cdot 10^{-3}\ \text{m} \\[6pt] T &= \frac{2 \pi \cdot 3.344 \cdot 10^{-27}\ \text{kg}}{1.602 \cdot 10^{-19}\ \text{C} \cdot 5.0\ \text{T}} \\ &= 2.62 \cdot 10^{-8}\ \text{s} \end{aligned}$$
Step 6/7
Answer: $r \approx 4.2\ \text{mm}$ and $T \approx 26\ \text{ns}$. Two significant figures, as in the speed and the field.
Step 7/7
Reasonableness: The deuteron circles about 38 million times a second on a circle of only a few millimetres, tiny compared with a plasma about $2\ \text{m}$ across. Check with $T = 2\pi r / v$:
You can test the result in the simulation: deuteron, $v = 1.0 \cdot 10^{6}\ \text{m/s}$, $B = 5.0\ \text{T}$.
Apply: can the plasma hold its alpha particles?
Each deuterium–tritium reaction makes an alpha particle (a helium-4 nucleus) with a kinetic energy of $3.5\ \text{MeV}$. Its mass is $6.645 \cdot 10^{-27}\ \text{kg}$ and its charge $2e$. Calculate the radius of its circle in ITER's field $B = 5.3\ \text{T}$, assuming that it moves perpendicular to the field. Give the answer in centimetres.
Hint 1: a guiding question
First find the speed from $E_\text{k} = \frac{1}{2}mv^2$. Change $3.5\ \text{MeV}$ into joules.
Hint 2: what is the connection
$1\ \text{MeV} = 1.602 \cdot 10^{-13}\ \text{J}$, and $v = \sqrt{2E_\text{k}/m}$ gives about $1.3 \cdot 10^{7}\ \text{m/s}$.
Hint 3: an intermediate step
Then $r = mv/(|q|B)$ with
Show the solution
Given
${E_\text{k} = 3.5\ \text{MeV}}$, ${m = 6.645 \cdot 10^{-27}\ \text{kg}}$, ${|q| = 2e = 2 \cdot 1.602 \cdot 10^{-19}\ \text{C}}$, ${B = 5.3\ \text{T}}$, velocity perpendicular to the field
Asked
the radius $r$ of the circle, in centimetres
Principle
The magnetic force is the centripetal force, so $r = mv/(|q|B)$. The speed comes from the kinetic energy; the speed is far below the speed of light (about $4\ \%$), so $E_\text{k} = \frac{1}{2}mv^2$ is accurate enough.
Formula
$$v = \sqrt{\frac{2E_\text{k}}{m}}, \qquad r = \frac{m v}{|q| B}$$
Unit conversions
$$\begin{aligned} E_\text{k} &= 3.5 \cdot 10^{6} \cdot 1.602 \cdot 10^{-19}\ \text{J} \\ &= 5.607 \cdot 10^{-13}\ \text{J} \end{aligned}$$
Substitution
$$\begin{aligned} v &= \sqrt{\frac{2 \cdot 5.607 \cdot 10^{-13}\ \text{J}}{6.645 \cdot 10^{-27}\ \text{kg}}} \\ &= 1.299 \cdot 10^{7}\ \text{m/s} \\[6pt] r &= \frac{6.645 \cdot 10^{-27}\ \text{kg} \cdot 1.299 \cdot 10^{7}\ \text{m/s}}{3.204 \cdot 10^{-19}\ \text{C} \cdot 5.3\ \text{T}} \\ &= 5.08 \cdot 10^{-2}\ \text{m} \\ &= 5.08\ \text{cm} \end{aligned}$$
Answer
$r \approx 5.1\ \text{cm}$
Two significant figures as in the energy and the field.
Reasonableness
The alpha particle has about $340$ times the energy of the deuteron in the worked example (
), but its circle is still only about $5\ \text{cm}$, small compared with ITER's plasma, which is about $4\ \text{m}$ across. So the field can hold the alpha particles long enough for them to give their energy to the plasma by collisions, which keeps the plasma hot.
Optional Find the error
Common misconception
"The magnetic field pushes the particles towards the centre of the machine." It does not. The force on each particle points to the centre of its own small circle, not to the centre of the plasma. A magnetic field cannot stop motion along the field lines at all, and it holds particles across the field lines only because they circle around them. That is why the shape of the field lines matters so much: closed (a torus), twisted, and without ends.
A second mistake is to think that the magnetic force speeds the particles up or heats them. It does no work, because it is always perpendicular to the velocity. A plasma is heated in other ways, for example by its own current, by beams of fast atoms or by radio waves.
Check your understanding
1. Concept: force and speed
2. Representation: the direction of the force
3. Model and reality
Sentence starters
- Charged particles circle around …, but they can move freely …
- A straight field has … , so …
- That is why fusion machines are shaped like …
- However, in a ring the field is …, which makes …
- One limitation of the one-particle model is …
4. Explain: why a ring?
A younger student asks: “Why are fusion machines shaped like a doughnut and not like a long tube?” Answer in four or five sentences. Mention what a magnetic field does to a charged particle, why a tube does not work, and one problem that remains in the ring.
Model answer
A charged particle in a magnetic field circles around a field line, so the field holds it across the field lines, but it can still slide freely along them. In a long tube the field lines end, so particles that move along them leak out at the ends, even with magnetic mirrors, because of the loss cone. Bending the tube into a ring closes the field lines, so there are no ends. However, in a ring the field is stronger on the inside than on the outside, so ions and electrons drift apart and the plasma is pushed outward. That is why the field lines must also be twisted.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary.
Summary
- The magnetic force on a moving charge is $F = |q| v B \sin\alpha$, perpendicular to $\overline{v}$ and $\overline{B}$; it changes the direction, not the speed.
- Perpendicular to a uniform field the particle moves on a circle: $r = mv/(|q|B)$, $T = 2\pi m/(|q|B)$, independent of the speed.
- With a velocity part along the field, the path is a helix around a field line.
- A magnetic mirror reflects particles with $\sin^2\theta \geq B_0/B_{\max}$; the rest escape through the loss cone.
- A torus closes the field lines, but its curved, uneven field makes ions and electrons drift apart; the field lines must be twisted.
- The single-particle model shows why confinement is possible. Collisions, turbulence and instabilities of the whole plasma make it difficult.
Le coin français
The magnetic field is le champ magnétique, and a fusion machine shaped like a ring is un tore. Many words are almost the same as in English. Listen, repeat, and then say the last sentence to your partner. In the hints, capitals show the stressed syllable, and õ, ã and ẽ are nasal vowels: the air goes through the nose and no n is heard.