Motion of charges in magnetic and electric fields

A path in the shape of a circle

The magnetic force does no work, so the speed of a charge in a magnetic field stays the same. Only the direction turns.

A force of constant size that stays at right angles to the velocity is exactly what a circle needs. Think of a stone on a string. In a magnetic field, the magnetic force plays the part of the string.

Radius, period and frequency

Let a charge $q$ of mass $m$ enter a field $B$ at right angles with speed $v$. The magnetic force $qvB$ is the centripetal force, so

We want $r$, so

where $p = mv$ is the momentum. A fast or a heavy particle needs a big circle. A strong field makes the circle small.

The time for one turn is the circumference divided by the speed:

The speed has gone out of the formula. The number of turns each second is $f = \dfrac{1}{T} = \dfrac{qB}{2\pi m}$. The angular speed is $\omega = \dfrac{qB}{m}$.

This is a surprise. A fast ion goes round a big circle and a slow ion goes round a small one, but they take the same time for a turn. The cyclotron uses this fact.

For a positive charge in a field that points into the page, the circle is anticlockwise. A negative charge goes round clockwise.

Double the speed. What happens to the radius? What happens to the period? Then change the mass and the charge. Which of them changes the period?

The speed from a voltage

Often a charge is sped up by a voltage $V$ before it enters the field. If it starts almost at rest, the work $qV$ becomes kinetic energy:

We want $v$, so $v = \sqrt{\dfrac{2qV}{m}}$. Put this in $r = \dfrac{mv}{qB}$:

For particles sped up through the same voltage, $r$ goes as $\sqrt{m/q}$.

A worked example. Electrons are sped up through $250$ V and then enter a field of $1.5$ mT at right angles. Find the speed, the radius and the time for a turn.

1. The speed. $v = \sqrt{\dfrac{2eV}{m}} = \sqrt{\dfrac{2 \times 1.6 \times 10^{-19}\ \text{C} \times 250\ \text{V}}{9.1 \times 10^{-31}\ \text{kg}}} = 9.4 \times 10^{6}\ \text{m/s}$.
2. The radius. The top of $r = \dfrac{mv}{eB}$ is $mv = 9.1 \times 10^{-31}\ \text{kg} \times 9.4 \times 10^{6}\ \text{m/s} = 8.55 \times 10^{-24}\ \text{kg m/s}$. The bottom is $eB = 1.6 \times 10^{-19}\ \text{C} \times 1.5 \times 10^{-3}\ \text{T} = 2.4 \times 10^{-22}\ \text{C T}$. So $r = \dfrac{8.55 \times 10^{-24}\ \text{kg m/s}}{2.4 \times 10^{-22}\ \text{C T}} = 3.6 \times 10^{-2}\ \text{m}$, that is $3.6$ cm.
3. The period. $T = \dfrac{2\pi m}{eB} = \dfrac{2\pi \times 9.1 \times 10^{-31}\ \text{kg}}{2.4 \times 10^{-22}\ \text{C T}} = 2.4 \times 10^{-8}\ \text{s}$, that is $24$ ns.

When the velocity is at an angle: a helix

Now let the velocity make an angle $\theta$ with the field. Split it in two parts.

• Along the field: $v_\parallel = v\cos\theta$. The field has no force on this part, so it stays the same.
• Across the field: $v_\perp = v\sin\theta$. This part turns round a circle, so the radius is $r = \dfrac{mv_\perp}{qB} = \dfrac{mv\sin\theta}{qB}$.

The charge goes round and along at the same time. Its path is a helix, like a spring. The distance it moves along the field in one turn is the pitch:

The period is still $T = 2\pi m/(qB)$, since only the part across the field turns.

A worked example. An electron moves at $6.0 \times 10^6$ m/s at $40^\circ$ to a field of $2.0$ mT. Find the radius and the pitch of its helix.

1. The radius. The bottom of $r = \dfrac{mv\sin\theta}{eB}$ is $eB = 1.6 \times 10^{-19}\ \text{C} \times 2.0 \times 10^{-3}\ \text{T} = 3.2 \times 10^{-22}\ \text{C T}$. The top is $mv\sin\theta = 9.1 \times 10^{-31}\ \text{kg} \times 6.0 \times 10^{6}\ \text{m/s} \times 0.643 = 3.51 \times 10^{-24}\ \text{kg m/s}$. So $r = \dfrac{3.51 \times 10^{-24}\ \text{kg m/s}}{3.2 \times 10^{-22}\ \text{C T}} = 1.1 \times 10^{-2}\ \text{m}$, that is $1.1$ cm.
2. The pitch. The top of $p = \dfrac{2\pi m\,v\cos\theta}{eB}$ is $2\pi \times 9.1 \times 10^{-31}\ \text{kg} \times 6.0 \times 10^{6}\ \text{m/s} \times 0.766 = 2.63 \times 10^{-23}\ \text{kg m/s}$. So $p = \dfrac{2.63 \times 10^{-23}\ \text{kg m/s}}{3.2 \times 10^{-22}\ \text{C T}} = 8.2 \times 10^{-2}\ \text{m}$, that is $8.2$ cm.

The Earth's field does this to charged particles from the Sun. They spiral along the field lines towards the poles and make the aurora.

Electric and magnetic fields together: the velocity selector

Now put an electric field $E$ and a magnetic field $B$ in the same region, at right angles to each other and to the velocity of a beam. The electric force is $qE$ and the magnetic force is $qvB$. We arrange them to point in opposite directions.

For one speed they are equal. Then

Particles of this speed go straight. Slower ones bend one way and faster ones bend the other way. The charge and the mass have cancelled, so the device passes any particle of the right speed. We call it a velocity selector.

Find the speed that goes straight. Then make the proton slower and then faster. The path is a loop that moves on at the speed $E/B$. What does the readout of the ratio of the two forces tell you at the straight path?

Measuring $e/m$. In 1897 J. J. Thomson used this idea to measure the ratio of charge to mass of the electron. First he found the straight path with both fields on, so $v = E/B$. Then he switched $E$ off. The electrons went round a circle of radius $r = \dfrac{mv}{eB}$. Put $v = E/B$ in it:

The accepted value is $1.76 \times 10^{11}$ C/kg.

The cyclotron

How can we give a particle a very high energy without a very high voltage? A cyclotron does it in small steps.

Two hollow D-shaped metal boxes, the dees, stand in a vacuum with a gap between them. A strong magnet makes a field $B$ at right angles to the dees. An alternating voltage acts across the gap. Inside a dee there is no electric field, so the ion simply goes round a half circle.

Each time the ion reaches the gap, the voltage has the right direction to give it a push of $qV$. Then it goes round a larger half circle in the other dee. The time for a half circle is $T/2 = \pi m/(qB)$. It does not depend on the radius. So the voltage can alternate at one fixed frequency,

and the ion arrives at the gap at the right moment every time.

The ion leaves at the edge of the dees, at the radius $R$. There $v_{\max} = \dfrac{qBR}{m}$, so the final energy is

The energy depends on $B$ and $R$, but not on the voltage. A larger voltage only means fewer turns. The ion crosses the gap twice in each turn, and gains $qV$ each time, so the number of turns is $N = \dfrac{K_{\max}}{2qV}$.

There is a limit. At speeds near that of light the mass of the particle grows, so its time for a turn changes and it falls out of step with the voltage. A simple cyclotron is not used for electrons for this reason.

A worked example. A cyclotron has a field of $1.4$ T and the dees have a radius of $0.35$ m. It accelerates deuterons. A deuteron has a charge $e$ and a mass of $3.34 \times 10^{-27}$ kg. Find the frequency and the final energy.

1. The frequency. $f = \dfrac{qB}{2\pi m} = \dfrac{1.6 \times 10^{-19}\ \text{C} \times 1.4\ \text{T}}{2\pi \times 3.34 \times 10^{-27}\ \text{kg}} = 1.07 \times 10^{7}\ \text{Hz}$, that is $10.7$ MHz.
2. The final energy. First $qBR = 1.6 \times 10^{-19}\ \text{C} \times 1.4\ \text{T} \times 0.35\ \text{m} = 7.84 \times 10^{-20}\ \text{kg m/s}$. Then $K_{\max} = \dfrac{(qBR)^2}{2m} = \dfrac{(7.84 \times 10^{-20}\ \text{kg m/s})^2}{2 \times 3.34 \times 10^{-27}\ \text{kg}} = 9.2 \times 10^{-13}\ \text{J}$. With $1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$, that is $5.75$ MeV.

Practise with new numbers

Try it in Python

The program turns the velocity of a proton a little at each step, as the field does. It finds the radius from the steps, and compares it with $r = mv/(qB)$. Then it draws the path.

Common mistakes

• Thinking a faster particle takes less time for a turn. The period $2\pi m/(qB)$ has no speed in it. A faster particle goes round a bigger circle in the same time.
• Forgetting the charge in the radius. An ion with a charge of $2e$ goes round a circle half as big as an ion with $e$, for the same mass and speed.
• Using the whole speed in the helix. The radius uses $v\sin\theta$. The pitch uses $v\cos\theta$.
• Saying the cyclotron energy grows with the voltage. It depends on $B$ and $R$. The voltage only sets the number of turns.
• Mixing up the direction of the two forces. A velocity selector works only when the electric force and the magnetic force are opposite.

Remember these

• At right angles to the field: $r = \dfrac{mv}{qB} = \dfrac{p}{qB}$, $T = \dfrac{2\pi m}{qB}$, $f = \dfrac{qB}{2\pi m}$, $\omega = \dfrac{qB}{m}$.
• The period does not depend on the speed or the radius.
• After a voltage $V$: $r = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$.
• At an angle $\theta$: a helix with $r = \dfrac{mv\sin\theta}{qB}$ and pitch $p = \dfrac{2\pi m\,v\cos\theta}{qB}$.
• Velocity selector: $v = \dfrac{E}{B}$. Thomson's ratio: $\dfrac{e}{m} = \dfrac{E}{B^2 r}$.
• Cyclotron: $f = \dfrac{qB}{2\pi m}$ and $K_{\max} = \dfrac{q^2B^2R^2}{2m}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A proton goes round a circle in a field. Its speed is doubled. By what factor do the radius and the period change?

2. A proton and a deuteron enter the same field at the same speed, at right angles to it. A deuteron has the same charge and twice the mass. Which goes round the bigger circle, and how much bigger?

3. A beam of electrons and a beam of protons, both of the same speed, enter a velocity selector. Both go straight. Why?

4. An electron moves at $4.0 \times 10^6$ m/s at $60^\circ$ to a field of $5.0$ mT. Find the radius and the pitch of its helix.