Magnetic Force on Moving Charges

Beginner

Lay a wire across the gap of a horseshoe magnet and switch on a current: the wire jumps sideways, out of the gap or deeper into it. Reverse the current and it jumps the other way. Turn the magnet round as well and it goes back to the first way.

Why does this happen? A current is nothing but electric charges on the move. A magnet pushes on every one of those moving charges, and since they cannot leave the wire, the wire is dragged along with them. A charge sitting still beside a magnet feels nothing at all: the push appears only once the charge moves.

The direction of the push is the surprising part. It is never along the motion: it acts sideways, at right angles both to the way the charge is going and to the magnetic field. A sideways push cannot make anything go faster or slower; it can only turn it, like a string whirling a ball round your head. So a charge flying freely through a magnet's field keeps turning, and its path bends into a circle.

That is what the animation shows. The grey × marks are a magnetic field pointing into the screen. Follow the red charge: the blue arrow shows where it is heading, the red arrow shows the push. The push always points to the middle of the circle, and the charge never speeds up or slows down.

The same push guides particles from the Sun down towards Earth's poles, where they make the sky glow as auroras, and huge rings of magnets keep particles racing round accelerators.

Check yourself: an electron sits motionless between the poles of a strong magnet. What magnetic force acts on it? Answer: none, since only moving charges feel one.

Intermediate

Charge, velocity and field

A magnet, however strong, exerts no magnetic force on a charge sitting still beside it. Three things set the magnetic force on a particle: its charge $q$, in coulombs and with its sign ($+1.60\times 10^{-19}\mathrm{C}$ for a proton); its velocity $\vec{v}$, direction included; and the magnetic field $\vec{B}$ where the particle is, measured in tesla (T). A tesla is a strong field: Earth's is 25–65 µT, a fridge magnet gives a few millitesla and an MRI scanner 1.5–3 T. The animation uses $B=0.10\mathrm{T}$, about the field between the poles of a good horseshoe magnet.

On flat diagrams a cross × is the tail of an arrow flying away from you, so the grey crosses in the animation mean a field into the screen; a dot • is an arrow tip coming towards you, a field out of the screen. The angle $\theta$ is measured between $\vec{v}$ and $\vec{B}$; the proton moves straight across the field, so $\theta =90^{\circ }$ all the time.

The size of the force

Experiments with particle beams show that the force is proportional to the charge, the speed and the field, and that it depends on the angle between motion and field:
$F=qvB\sin \theta$

Only the part of the velocity that crosses the field lines, $v\sin \theta$, counts. A charge moving along the field lines feels nothing, one crossing them at right angles feels the most, $F=qvB$, and at 30° it feels half of that. For a negative charge use the size of $q$; the sign only changes the direction.

A charge at rest feels no magnetic force at all, however strong the magnet: that is the sharpest difference from an electric field. The rule also defines the tesla: 1 T pushes with 1 N on 1 C crossing it at 1 m/s, so $1\mathrm{T}=1\mathrm{N}\cdot s/(\mathrm{C}\cdot \mathrm{m})=1N/(\mathrm{A}\cdot \mathrm{m})$.

Reading the diagram: the right-hand rule

The force is at right angles both to the velocity and to the field. Its size and its direction fit into one vector equation:
$\vec{F}=q\vec{v}\times \vec{B}$

The vector product $\vec{v}\times \vec{B}$ has the size $vB\sin \theta$, and its direction follows the right-hand rule: point the fingers of your right hand along $\vec{v}$, curl them towards $\vec{B}$, and your thumb shows the force on a positive charge. The order matters, since $\vec{B}\times \vec{v}$ points the opposite way.

Try it on the animation. At the bottom of the circle the proton moves to the right and the field goes into the screen, so the thumb points up, towards the centre, just like the red $F$ arrow: the proton turns anticlockwise. Reversing the field or the velocity reverses the force; reversing both leaves it unchanged.

No work, so the path is a circle

Because $\vec{F}$ is always at right angles to $\vec{v}$, it does no work. The kinetic energy cannot change, so neither can the speed: the $|v|$ readout stays at $1.0\times 10^{6}m/s$. A force of fixed size that only turns the velocity is exactly what uniform circular motion needs, so the magnetic force acts as the centripetal force of Circular Motion:
$qvB=\frac{m{v}^2}{r}$

Solving for the radius, and using $T=2\pi r/v$ for the period and $\omega =2\pi /T$ for the angular frequency, gives three working formulas:

• $r=\frac{mv}{qB}$
• $T=\frac{2\pi m}{qB}$
• $\omega =\frac{qB}{m}$

A faster or heavier particle runs a wider circle; a stronger field or a larger charge, a tighter one. The surprise is in the second line: the speed has cancelled, so a faster particle runs a bigger circle in exactly the same time.

Worked example: the proton in the animation

A proton has $q=1.60\times 10^{-19}\mathrm{C}$ and $m=1.67\times 10^{-27}\mathrm{kg}$. In the animation it moves at $v=1.0\times 10^{6}m/s$ straight across a field of $B=0.10\mathrm{T}$, so $\sin \theta =1$:
$F=qvB=1.60\times 10^{-19}\cdot 1.0\times 10^6\cdot 0.10=1.6\times 10^{-14}\mathrm{N}$
$r=\frac{mv}{qB}=\frac{1.67\times 10^{-27}\cdot 1.0\times 10^6}{1.60\times 10^{-19}\cdot 0.10}\approx 0.104\mathrm{m}$
$T=\frac{2\pi m}{qB}=\frac{2\pi \cdot 1.67\times 10^{-27}}{1.60\times 10^{-19}\cdot 0.10}\approx 6.6\times 10^{-7}\mathrm{s}$

These are the values on the readouts: $F=1.6\times 10^{-14}\mathrm{N}$, $r=10.4\mathrm{cm}$ and a constant $|v|=1.0\times 10^{6}m/s$. The real proton goes round in 0.66 µs, about 1.5 million times a second; the animation, at 4 s per turn, is slowed down some six million times.

The force looks tiny, but the mass is far tinier: the acceleration $F/m\approx 9.6\times 10^{12}m/{\mathrm{s}}^2$ is about $10^{12}$ times $g$, so gravity plays no part. The blue $v$ arrow stays tangent to the dashed circle with a fixed length, while the red $F$ arrow lies along the grey radius $r$ and points at the centre: always at 90° to $v$, it does no work.

Worked example: sorting ions by mass

A mass spectrometer weighs atoms by the size of their circles. Singly charged neon ions, with $q=1.60\times 10^{-19}\mathrm{C}$, leave the source with a spread of speeds, so they first cross a velocity selector, where an electric force $qE$ and a magnetic force $qvB$ push in opposite directions. Only ions with $qE=qvB$ go straight on through a slit. With 250 V across plates 1 cm apart, $E=2.5\times 10^{4}V/m$, and with $B=0.25\mathrm{T}$,
$v=\frac{E}{B}=\frac{2.5\times 10^4}{0.25}=1.0\times 10^{5}m/s$

In the same field, without the plates, the ions then turn through half a circle onto a detector. Neon-20 has $m=3.32\times 10^{-26}\mathrm{kg}$:
$r_{20}=\frac{mv}{qB}=\frac{3.32\times 10^{-26}\cdot 1.0\times 10^5}{1.60\times 10^{-19}\cdot 0.25}\approx 0.083\mathrm{m}$

Neon-22 is 10% heavier, and at a fixed speed $r$ is proportional to $m$, so $r_{22}\approx 0.091\mathrm{m}$. Each ion lands one diameter from the slit, at 16.6 cm and 18.3 cm, so the two isotopes arrive 1.7 cm apart. Without the selector they would blur: a neon-20 ion just 1% too fast lands 1.7 mm further out.

Common mistakes

• The force is not along the field lines, as an electric force would be, but at right angles to them: the proton's $1.6\times 10^{-14}\mathrm{N}$ push points towards the centre of its circle, never into the screen.
• A magnetic field cannot change a charge's speed. The proton keeps its $1.0\times 10^{6}m/s$ and its 5.2 keV of kinetic energy turn after turn; accelerators add energy with electric fields and only steer with magnetic ones.
• A slanting motion gets less force. At 30° to the field instead of 90°, the proton would feel only $8.0\times 10^{-15}\mathrm{N}$, half as much.
• The sign of the charge matters. An electron at the same speed is pushed the other way and circles clockwise, on a radius of about 0.057 mm, 1836 times smaller, like its mass.

Where you meet it

Auroras are the most beautiful case: charged particles from the Sun are steered along Earth's field lines into the polar regions, where they make the upper atmosphere glow. Hospital cyclotrons accelerate protons to make the isotopes used in PET scans, mass spectrometers date ancient carbon and test athletes for doping, and particle detectors read momenta from the curvature of tracks.

The same force acts on the charges drifting along a wire. If the moving charge $Q$ in a length $L$ of wire drifts at a speed $v$, the current is $I=Qv/L$, and the forces add up to $QvB=BIL$ for a wire across the field: 0.025 N for 5.0 A in 5.0 cm of wire across 0.10 T. This push makes a wire jump between the poles of a horseshoe magnet and turns every electric motor, in Motors, Generators and Transformers. Moving a wire through a field pushes its charges along it instead: the start of Electromagnetic Induction.

Advanced

The Lorentz force defines the fields

The whole lesson rests on one law, and its most important consequence is that a magnetic field can steer a charge but never speed it up. A point charge $q$ moving with velocity $\vec{v}$ through an electric field $\vec{E}$ and a magnetic field $\vec{B}$ feels the Lorentz force
$\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})$

The law can be read backwards, as a recipe for measuring both fields with a test charge. Held still, the charge feels only $q\vec{E}$, and that gives $\vec{E}$. Set moving, it feels an extra force that grows with its speed and always points at right angles to its velocity. This extra force is zero for motion along one particular line, and that line is the direction of $\vec{B}$. For any other velocity it fixes the part of $\vec{B}$ across $\vec{v}$; a second, non-parallel velocity supplies the missing part along the first one. Two measurements therefore fix all three components of $\vec{B}$.

The direction of the push comes from the rule for the vector product. With unit vectors $\hat{x}$, $\hat{y}$, $\hat{z}$ along the axes, $\hat{x}\times \hat{y}=\hat{z}$, and the same holds for the cyclic versions $\hat{y}\times \hat{z}=\hat{x}$ and $\hat{z}\times \hat{x}=\hat{y}$. So a positive charge moving along $+y$ in a field along $+z$ is pushed along $\hat{y}\times \hat{z}=+\hat{x}$.

Now the energy. A force changes the kinetic energy $K$ at the rate $\vec{F}\cdot \vec{v}$, the power it delivers. With the Lorentz force,
$\frac{dK}{dt}=q\vec{E}\cdot \vec{v}+q\vec{v}\cdot (\vec{v}\times \vec{B})$
The last term is zero, because the vector product $\vec{v}\times \vec{B}$ is perpendicular to $\vec{v}$. So $dK/dt=q\vec{E}\cdot \vec{v}$: only the electric field changes a particle's energy, however the magnetic field is shaped. A magnetic force can turn the velocity but never lengthen or shorten it. When a changing magnetic field seems to do work, as in a transformer, the work is done by the electric field that the change induces.

Circular motion: two coupled oscillators

Legend

$C$	Centre of the circle, the point $(x_c,y_c)$.
$r$	Radius of the circle, $r=mv/(qB)$, drawn from $C$ to the starting point.	$\mathrm{m}$
$\omega t$	Angle turned since the start; $\omega =qB/m$ is the rate of turning.	$\mathrm{rad}$
$q$	Charge of the particle, here a proton; the open dot is where it started.	$\mathrm{C}$
$\vec{v}$	Velocity: tangent to the circle, its length the constant speed $v$.	$m/s$
$\vec{F}$	Magnetic force $q\vec{v}\times \vec{B}$: at right angles to $\vec{v}$, towards $C$.	$\mathrm{N}$
$\vec{B}$	Magnetic field, uniform and pointing into the page (the crosses).	$\mathrm{T}$
$v_x$	Component of $\vec{v}$ along $x$, $v\cos \omega t$; negative at the moment drawn.	$m/s$
$v_y$	Component of $\vec{v}$ along $y$, $v\sin \omega t$.	$m/s$

The motion in a uniform magnetic field can be solved exactly. The method splits the velocity into components that drive each other, and it is worth learning for its own sake. The result, a circle run at a fixed rate, underlies everything later in the lesson.

Figure 1 sets out the geometry of the animation. The field points into the page (the crosses). With $x$ to the right, $y$ up and $z$ out of the page, that is $\vec{B}=-B\hat{z}$. The force is at right angles to $\vec{B}$, so it never gives the proton a velocity along $z$, and the motion stays in the $xy$-plane. On the left, follow the proton of charge $q$: it started at the open dot at the bottom of the circle, moving along $+x$, and has since turned through the angle $\omega t$ about the centre $C$. Its velocity $\vec{v}$ is tangent to the path, and the force $\vec{F}$ points along the radius towards $C$.

On the right, the same $\vec{v}$ is drawn again from a fixed point. Its shadows on the axes, the horizontal and vertical bars, are its components $v_x$ and $v_y$. The arrow has turned through the same angle $\omega t$ from the $v_x$ axis, because the velocity stays at right angles to the radius and so turns exactly as fast as the radius does.

Now the equations. Newton's second law is $md\vec{v}/dt=q\vec{v}\times \vec{B}$, with $m$ the proton's mass. Working out the vector product with $\vec{B}=-B\hat{z}$ gives $\vec{v}\times \vec{B}=B(-v_y,v_x,0)$. Dividing by $m$ and writing $\omega =qB/m$, the two components of the law read
${\dot{v}}_x=-\omega v_y,{\dot{v}}_y=\omega v_x$
where a dot means the rate of change in time.

Each component drives the other: $v_y$ sets how fast $v_x$ changes, and $v_x$ sets how fast $v_y$ changes. To separate them, differentiate the first equation once more and replace ${\dot{v}}_y$ by the second:
${\ddot{v}}_x=-\omega {\dot{v}}_y=-{\omega }^2{v}_x$
This is the equation of simple harmonic motion with angular frequency $\omega$, and the same steps give it for $v_y$. The starting conditions choose the solution. At $t=0$ the proton moves along $+x$ with its full speed $v$, so $v_x=v$ and $v_y=0$, hence
$v_x=v\cos \omega t,v_y=v\sin \omega t$
These are the two bars on the right of Figure 1. One more integration gives the position:
$x=x_c+\frac{v}{\omega }\sin \omega t,y=y_c-\frac{v}{\omega }\cos \omega t$
The constants $(x_c,y_c)$ are the point $C$. At $t=0$ these formulas put the proton at $y=y_c-v/\omega$, directly below $C$, as in the figure. The path is a circle of radius
$r=\frac{v}{\omega }=\frac{mv}{qB}$
run anticlockwise when $q>0$. For a negative charge $\omega$ is negative, and the same formulas describe a clockwise circle of radius $|mv/(qB)|$.

Complex numbers make the solution shorter. Combine the two components into one complex number, $w=v_x+i{v}_y$. The pair of equations becomes the single equation $\dot{w}=i\omega w$, whose solution is $w=v{e}^{i\omega t}$: an arrow of fixed length $v$ turning at the steady rate $\omega$, exactly the arrow on the right of Figure 1. In vector language the same statement is $\dot{\vec{v}}=\vec{\omega }\times \vec{v}$ with $\vec{\omega }=-q\vec{B}/m$: the velocity turns (precesses) about the field direction at the rate $qB/m$. For $q>0$ the rotation vector $\vec{\omega }$ points against $\vec{B}$, here out of the page, and a rotation whose vector points at you looks anticlockwise. That is why the proton circles anticlockwise in a field pointing away from you.

Reading the velocity graphs

Legend

$v$	Amplitude of both curves: the constant speed.	$m/s$
$t$	Time, measured from the start.	$\mathrm{s}$
$T/4$	A quarter of the period: $v_y$ peaks while $v_x$ crosses zero.	$\mathrm{s}$
$T$	Period of one full turn, $T=2\pi m/(qB)$.	$\mathrm{s}$
$v_x$	Velocity component along $x$, $v\cos \omega t$.	$m/s$
$v_y$	Velocity component along $y$, $v\sin \omega t$.	$m/s$
$\omega v$	Slope of $v_y$ at $t=0$, its steepest: equal to the acceleration $F/m$.	$m/{\mathrm{s}}^2$

The graphs of $v_x$ and $v_y$ against time hold the whole motion: the constant speed, the period and the force can all be read from them. Figure 2 draws the graphs of the animation's Advanced layer over one period $T$, the time of one full turn.

Start with the shapes. The blue curve $v_x$ is a cosine and the green curve $v_y$ a sine, both of amplitude $v=1.0\times 10^{6}m/s$, the proton's speed. They are a quarter period $T/4$ apart: 1 s in the slowed-down animation, 0.16 µs in reality. Wherever one curve crosses zero the other is at a peak or a trough, as at the dashed line at $T/4$. At every instant $v_x^2+v_y^2=10^{12}{\mathrm{m}}^2/{\mathrm{s}}^2$, because ${\cos }^2\omega t+{\sin }^2\omega t=1$. Both components change all the time; the speed never does.

Then the slopes. The slope of a velocity graph is an acceleration, and by Newton's second law it is the force divided by the mass. At $t=0$, $v_x$ is at its peak, where its graph is flat, while $v_y$ crosses zero with its steepest slope, the red line in Figure 2. That slope is
$\omega v\approx 9.6\times 10^{12}m/{\mathrm{s}}^2$
with $\omega =qB/m\approx 9.6\times 10^{6}rad/s$ for a proton in the animation's 0.10 T. Since $v_x$ is not changing at that moment, this is the whole acceleration $F/m$. It points along $+y$, straight at the centre, which lies above the starting point. A quarter turn later the roles are swapped: $v_x$ is steepest and $v_y$ flat, because the acceleration has turned with the radius and now points along $-x$.

The period does not depend on the speed. The animation's purple proton starts from the same point at $1.5v$ and runs a circle of radius 15.7 cm instead of 10.4 cm. Its components are the same sinusoids with 1.5 times the amplitude and the same period, so it comes back to the start at the same moment as the red proton. This is the content of
$T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$
in which the speed has cancelled. In 0.10 T every proton moving well below the speed of light takes 0.66 µs per turn, whatever the size of its orbit.

Helices and magnetic mirrors

Legend

$\theta$	Angle between the starting velocity and $\vec{B}$; $60^{\circ }$ here.	$\mathrm{rad}$
$\vec{v}$	Velocity at the start; its length is the constant speed $v$.	$m/s$
$v_{\perp }$	Part of $\vec{v}$ across $\vec{B}$, $v\sin \theta$: it turns round the circle.	$m/s$
$v_{\parallel }$	Part of $\vec{v}$ along $\vec{B}$, $v\cos \theta$: it feels no force and never changes.	$m/s$
$r$	Radius of the circle, $r=m{v}_{\perp }/(qB)$.	$\mathrm{m}$
$h$	Pitch of the helix: the distance advanced along $\vec{B}$ in one period $T$, $h=v_{\parallel }T$.	$\mathrm{m}$
$\vec{B}$	Magnetic field, pointing up here.	$\mathrm{T}$

So far the velocity lay across the field. The general case is a velocity at any angle to $\vec{B}$, and it is how charged particles actually arrive in Earth's magnetic field or move in a plasma.

Figure 3 shows the result. At the left, the starting velocity $\vec{v}$ makes the angle $\theta$ with $\vec{B}$, here $60^{\circ }$, and is split into two parts: $v_{\parallel }=v\cos \theta$ along the field and $v_{\perp }=v\sin \theta$ across it. The two parts behave quite differently. The force $q\vec{v}\times \vec{B}$ is always at right angles to $\vec{B}$, so it has no component along the field and $v_{\parallel }$ never changes. And since the vector product of parallel vectors is zero, $v_{\parallel }$ adds nothing to the force: the force depends on $v_{\perp }$ alone. So $v_{\perp }$ behaves exactly as in the previous sections. It turns round a circle of radius $r=m{v}_{\perp }/(qB)$, drawn at the foot of the helix, in the same period $T$.

Put the two together: the circle is carried steadily along the field, and the path is a helix. Follow it up the drawing. Each turn takes the period $T$, and in that time the proton advances along $\vec{B}$ by the pitch
$h=v_{\parallel }T=2\pi r\cot \theta$
The second form follows from $T=2\pi r/v_{\perp }$ and $v_{\parallel }/v_{\perp }=\cot \theta$. At $60^{\circ }$ the pitch is about 3.6 times the radius. For the animation's proton, launched at $1.0\times 10^{6}m/s$ at 60° to the 0.10 T field, $v_{\perp }=8.7\times 10^{5}m/s$ gives a circle of radius 9.0 cm, and $v_{\parallel }=5.0\times 10^{5}m/s$ carries it 33 cm along the field per turn.

Real fields are rarely uniform. Suppose $B$ grows slowly along the field lines, changing little during one turn. Then the quantity $m{v}_{\perp }^2/(2B)$, the kinetic energy of the circling motion divided by $B$, stays nearly constant. As the proton moves into a stronger field, $v_{\perp }^2$ must therefore grow in proportion to $B$. The total speed cannot change, so $v_{\parallel }$ shrinks to make room: the helix tightens, and where $v_{\perp }$ would have to exceed $v$, the proton stops advancing and turns back. A region of strong field acts as a mirror. Earth's field lines crowd together near the poles, so protons and electrons bounce back and forth between the hemispheres, trapped in Earth's radiation belts.

Momentum at any speed

The relations so far were derived for slow particles. Which of them survive at high speed? Because the magnetic force never changes the speed, it only turns the momentum $\vec{p}$, and Newton's law in the form $d\vec{p}/dt=q\vec{v}\times \vec{B}$ holds at any speed. For motion across the field it gives a circle with
$p=qBr$
where $p=\gamma mv$ is the relativistic momentum, $\gamma =1/\sqrt{1-v^2/c^2}$ is the Lorentz factor and $c$ is the speed of light. Since the speed is constant, so is $\gamma$, and the steps of the previous sections go through unchanged with $m$ replaced by $\gamma m$.

In practical units, a track of radius 1 m in a 1 T field belongs to a singly charged particle of momentum 0.30 GeV/c. What does change at high energy is the rate of turning, $\omega =qB/(\gamma m)$, which falls as the energy rises. The cyclotron below has to deal with exactly this.

Crossed fields: the drift

Add a uniform electric field $\vec{E}$ at right angles to $\vec{B}$, and the circles no longer close: they glide sideways. This case matters because it shows how a change of frame can remove an electric field, and it is the principle of the velocity selector.

The trick is to watch from a frame moving at the constant velocity
${\vec{v}}_d=\frac{\vec{E}\times \vec{B}}{B^2}$
At low speed, an observer moving at ${\vec{v}}_d$ measures the electric field ${\vec{E}}^{\prime }=\vec{E}+{\vec{v}}_d\times \vec{B}$. When $\vec{E}$ is at right angles to $\vec{B}$, $(\vec{E}\times \vec{B})\times \vec{B}=-B^2\vec{E}$, so the second term is exactly $-\vec{E}$ and ${\vec{E}}^{\prime }=0$. In that frame only $\vec{B}$ remains, and the particle runs an ordinary circle. Back in the laboratory, that circle is carried along at the drift speed $v_d=E/B$. The drift contains neither $q$ nor $m$, so every particle, of either sign and any mass, drifts the same way at the same speed. The exact relativistic treatment gives the same result, provided $E<cB$, so that $v_d$ stays below the speed of light.

With $E=1.0\times 10^{4}V/m$ in the animation's 0.10 T, the drift speed is $1.0\times 10^{5}m/s$, and the proton's loops advance $v_{d}T=6.6\mathrm{cm}$ per turn. A velocity selector uses the same balance: a particle launched at exactly ${\vec{v}}_d$ feels electric and magnetic forces that cancel, so it goes straight through, while faster and slower particles are deflected.

Worked problem: a small cyclotron

Legend

$R$	Radius of the electrodes: the largest orbit, where the protons leave.	$\mathrm{m}$
$\vec{B}$	Magnetic field, pointing into the page through both electrodes.	$\mathrm{T}$
$V$	Alternating voltage between the two electrodes, across the gap.	$\mathrm{V}$
$q$	Charge of the proton, shown at its starting point.	$\mathrm{C}$

A cyclotron puts the speed-independence of the period to work. Because every proton in a given field takes the same time per turn, a voltage alternating at one fixed frequency can push it again and again, whatever its energy.

Figure 4 shows the machine from above. Two hollow D-shaped electrodes of radius $R$ face each other across a narrow gap. The field $\vec{B}$ points into the page through both of them, and a source of alternating voltage $V$ is connected across them. The proton $q$ starts at the centre. Inside a hollow electrode there is no electric field, only $\vec{B}$, so the proton runs a semicircle and comes back to the gap. If the voltage has reversed by then, it pushes the proton across the gap and adds the energy $qV$. A faster proton runs a larger semicircle, so the path spirals outwards until it reaches the rim at $R$, where the protons leave. The drawing shows only four turns; the machine below needs 30.

The problem. A small cyclotron uses the animation's field, $B=0.10\mathrm{T}$, with electrodes of radius $R=0.50\mathrm{m}$ and an alternating voltage of amplitude 2.0 kV across the gap. Protons start near the centre, almost at rest. At what frequency must the voltage alternate, what energy do the protons reach, and in how many turns? Use $q=1.60\times 10^{-19}\mathrm{C}$ and $m=1.67\times 10^{-27}\mathrm{kg}$.

Step 1, the frequency. Each proton crosses the gap twice per turn and must find the voltage reversed every time. So the voltage must go through one full cycle per turn of the proton: it alternates at the cyclotron frequency
$f=\frac{qB}{2\pi m}=\frac{1.60\times 10^{-19}\cdot 0.10}{2\pi \cdot 1.67\times 10^{-27}}\approx 1.52\mathrm{MHz}$

Step 2, the final energy. The last orbit has the radius $R$, so by $p=qBR$ the protons leave with momentum
$p=qBR=1.60\times 10^{-19}\cdot 0.10\cdot 0.50=8.0\times 10^{-21}\mathrm{kg}\cdot m/s$
and kinetic energy
$K=\frac{p^2}{2m}=\frac{(8.0\times 10^{-21}{)}^2}{2\cdot 1.67\times 10^{-27}}\approx 1.92\times 10^{-14}\mathrm{J}\approx 120\mathrm{keV}$
using $1\mathrm{eV}=1.60\times 10^{-19}\mathrm{J}$. That is a speed $p/m=4.8\times 10^{6}m/s$, 1.6% of the speed of light. The voltage does not appear in this result: the final energy is set by $B$ and $R$ alone, and the voltage only decides how many turns it takes to get there.

Step 3, the number of turns. A proton that crosses the gap at the peak of the voltage gains 2.0 keV per crossing, so 4.0 keV per turn. It needs $120/4.0=30$ turns, which take 30 periods, about 20 µs. Since $K$ grows by equal steps and $K\propto r^2$, the radius after $n$ turns is $R\sqrt{n/30}$: 9.1 cm after one turn, 18.3 cm after four, 27.4 cm after nine. The animation's 10.4 cm circle, at 5.2 keV, lies between the first turn and the second.

Where the design breaks down. The machine relies on every half-turn taking the same 0.33 µs. Relativity spoils this slowly, because the true frequency of revolution is $qB/(2\pi \gamma m)$, and for a kinetic energy $K$, $\gamma =1+K/(m{c}^2)$, where the proton's rest energy $m{c}^2$ is 938 MeV. At 120 keV, $\gamma \approx 1.00013$, and the frequency falls by a harmless 0.013%. At 20 MeV it would be 2.1% low. A proton circling at that energy would fall half a cycle behind the voltage within about 24 turns, and would then reach the gap just when the voltage slows it down. Cyclotrons for higher energies therefore either lower the frequency as the protons gain energy, or let the field grow towards the rim so that $\gamma m/B$, and with it the period, stays constant.

Electric or magnetic: a matter of frame

Whether a force counts as electric or magnetic depends on who is watching. This section shows why, using the animation's proton.

In the laboratory the proton feels only a magnetic force. Now look from its momentary rest frame, a frame moving with it at that instant. There the proton has no velocity, so the force on it cannot be magnetic. Yet the force is real, and at low speed it is the same in both frames, so that observer must see an electric field,
${\vec{E}}^{\prime }\approx \vec{v}\times \vec{B}(v\ll c)$
For the animation's numbers its strength is $vB=1.0\times 10^6\cdot 0.10=1.0\times 10^{5}V/m$, and $q{E}^{\prime }=1.6\times 10^{-14}\mathrm{N}$, exactly the magnetic force $qvB$ measured in the laboratory. A pure magnetic field in one frame is a mixture of electric and magnetic fields in another. Neither field on its own is the same for every observer; only the electromagnetic field as a whole is.

Special relativity makes this exact. Under a change of frame the two fields transform into each other, but two combinations of them take the same values for every inertial observer:
$E^2-c^2{B}^2\text{and}\vec{E}\cdot \vec{B}$
These invariants settle what a change of frame can and cannot do. A purely magnetic field has $E^2-c^2{B}^2<0$, and since that sign is the same in every frame, no observer can see the field as purely electric. Crossed fields with $E<cB$ also have $E^2-c^2{B}^2<0$, and $\vec{E}\cdot \vec{B}=0$, so they can always be reduced to a pure magnetic field: that is the drift frame of the section on crossed fields.

Limits of validity

Every result above treats the proton as a classical point charge that moves in a vacuum and loses no energy. This section checks each of those assumptions with numbers.

Radiation. An accelerated charge radiates, and a charge going round a circle is always accelerated. At low speed the radiated power is given by Larmor's formula,
$P=\frac{q^2{a}^2}{6\pi {\epsilon }_0{c}^3}$
where $a$ is the acceleration and ${\epsilon }_0$ the permittivity of free space. For the animation's proton, with $a\approx 9.6\times 10^{12}m/{\mathrm{s}}^2$, this is about $5\times 10^{-28}\mathrm{W}$. Over one turn it loses $4\times 10^{-19}$ of its energy, so its circle is exact for all practical purposes.

The loss is not always small. For particles of the same energy on circles of the same radius, the loss per turn grows as the inverse fourth power of the mass, so light particles radiate far more. Electrons in a 3 GeV storage ring, bent on a 7 m radius, lose about 1 MeV per turn as synchrotron radiation, and such rings are built to produce it as a powerful source of X-rays. A proton is 1836 times heavier than an electron, and $1836^4\approx 10^{13}$, so protons lose little even in the Large Hadron Collider, where 7 TeV protons in 8.3 T magnets bend on a radius $p/(qB)\approx 2.8\mathrm{km}$.

Relativity. At $1.0\times 10^{6}m/s$, $\gamma -1\approx v^2/(2c^2)=5.6\times 10^{-6}$, so the non-relativistic formulas are excellent. In general, replace $mv$ by $p$ and $m$ by $\gamma m$, as in the section on momentum.

Quantum mechanics. In quantum theory the circling motion can only have certain energies, the Landau levels, spaced by $\hslash \omega$, where $\hslash$ is the reduced Planck constant. For the animation's proton the spacing is about $6\times 10^{-9}\mathrm{eV}$, $10^{-12}$ of its 5.2 keV, far too fine to notice. For an electron in 10 T it is 1.2 meV, more than ten times the typical thermal energy at 1 K, and there the quantisation shows up, most strikingly in the quantum Hall effect.

The vacuum. Finally, all of this assumes a vacuum. In air, collisions with molecules would stop a 5 keV proton long before it completed a turn, which is why cyclotrons and storage rings are pumped empty.

Where this leads

The force $q\vec{v}\times \vec{B}$ runs through the rest of electromagnetism. Push a metal rod of length $L$ across a field at speed $v$, and the force drives its free charges along the rod: that is the EMF $\epsilon =BLv$ of Electromagnetic Induction. Summed over all the moving charges in a wire carrying a current $I$, the same force gives $F=BIL\sin \theta$ on a length $L$ of wire at the angle $\theta$ to the field, and the torque on a coil, the subjects of Motors, Generators and Transformers.

In a conductor carrying a current across a field, the sideways push piles charge up on one edge until the electric field of that charge balances it. The sign of the resulting Hall voltage tells whether the carriers are electrons or holes, as in Semiconductors, Diodes and LEDs, and Hall sensors measure magnetic fields in phones and cars.

The frame dependence of $\vec{E}$ and $\vec{B}$ belongs to Special Relativity, and fields that change in time, each creating the other, make up Electromagnetic Waves. Finally, $p=qBr$ is how particle momenta are measured in Particles and Forces: the Standard Model, from the spirals in bubble-chamber photographs to the detectors of the Large Hadron Collider.