How to Calculate a Charged Particle’s Period in a Magnetic Field

A proton and an electron enter the same uniform magnetic field at the same speed. Both follow curved paths, but the electron completes each revolution much sooner. Now tilt the particles’ initial velocities so that neither enters perpendicular to the field. Does the period change?

The key result is surprisingly simple: the period of a charged particle’s circular motion in a uniform magnetic field does not depend on its speed or entry angle, provided the particle is moving non-relativistically.

The equation you need is

\[
\boxed{T=\frac{2\pi m}{|q|B}}
\]

where:

• \(T\) is the period, in seconds (s),
• \(m\) is the particle’s mass, in kilograms (kg),
• \(q\) is its electric charge, in coulombs (C), and
• \(B\) is the magnetic field strength, in tesla (T).

Notice the absolute value around \(q\). The charge’s magnitude determines the period, while its sign determines the direction of curvature.

Let’s work out why this equation works, how to use it when the velocity is at an angle to the field, and how to identify whether a particle is positive or negative from its path.

01What does the period actually measure?

Imagine watching a charged particle move through a uniform magnetic field. If its velocity is perpendicular to the field, it follows a circular path.

The period is the time taken to complete one full revolution: 360° of circular motion.

A useful way to picture the period is to imagine two runners going around circular tracks. One runs on a small track, the other on a much larger track. The runner on the larger track takes longer if both run at the same speed.

A charged particle in a magnetic field behaves differently when its speed changes. A faster particle follows a larger circular path, but it also travels faster. These two changes cancel when we calculate the time for one revolution.

That’s why speed disappears from the final period equation.

02Deriving the period equation

Start with a particle moving perpendicular to a uniform magnetic field.

The magnetic force continually points towards the centre of the circular path. It therefore supplies the centripetal force needed to keep the particle moving in a circle.

The magnitude of the magnetic force is

\[
F_{B}=|q|vB
\]

The centripetal force required for circular motion is

\[
F_{c}=\frac{mv^{2}}{r}
\]

Here, \(v\) is the particle’s speed and \(r\) is the radius of its circular path, measured in metres.

Since the magnetic force supplies the centripetal force, equate the two magnitudes:

\[
|q|vB=\frac{mv^{2}}{r}
\]

Rearranging gives

\[
r=\frac{mv}{|q|B}
\]

This tells us that a faster particle has a larger orbit, provided its mass, charge magnitude, and magnetic field strength remain constant.

Now consider the distance travelled in one complete revolution. It is the circumference of the circle, \(2\pi r\).

Since period equals distance divided by speed,

\[
T=\frac{2\pi r}{v}
\]

Substitute the expression for the radius:

\[
T=\frac{2\pi}{v}
\left(\frac{mv}{|q|B}\right)
\]

The speed cancels:

\[
\boxed{T=\frac{2\pi m}{|q|B}}
\]

There is the result.

A faster particle travels around a larger circle, but it also covers that circle at a greater speed. The time for one revolution stays the same.

This is also the principle behind the cyclotron, a particle accelerator developed by Ernest Lawrence around 1930. Its design used a magnetic field to bend charged particles around repeated paths, while an alternating electric field accelerated them each time they crossed a gap. The fact that the period is independent of speed, within the non-relativistic model, allowed the accelerating voltage to operate at a fixed frequency.

03What if the particle enters at an angle?

So far, we’ve assumed the particle enters perpendicular to the magnetic field. Real questions can be less convenient.

Suppose a particle enters at \(30^\circ\) to the magnetic field. Only part of its velocity is perpendicular to the field. What happens to the rest?

Resolve the initial velocity into two components:

\[
v_{\perp}=v\sin\theta
\]

\[
v_{\parallel}=v\cos\theta
\]

Here, \(\theta\) is the angle between the particle’s velocity and the magnetic field.

The two components behave differently:

• \(v_{\perp}\) is perpendicular to the field. It experiences a magnetic force and produces circular motion.
• \(v_{\parallel}\) is parallel to the field. It experiences no magnetic force and remains constant.

The combination produces a helical path: the particle circles around the magnetic field lines while simultaneously travelling along them.

Does the helical path change the period?

No. The period remains

\[
T=\frac{2\pi m}{|q|B}
\]

The important detail is that we must use the perpendicular velocity component when calculating the radius:

\[
r=\frac{mv_{\perp}}{|q|B}
\]

The time for one complete revolution is the circumference divided by the speed around that circle:

\[
T=\frac{2\pi r}{v_{\perp}}
\]

Substituting the radius:

\[
T=\frac{2\pi}{v_{\perp}}
\left(\frac{mv_{\perp}}{|q|B}\right)
\]

Again, the velocity cancels.

The parallel component does not affect the time taken to complete a revolution. It simply determines how far the particle advances along the magnetic field during that time.

This distance is called the pitch of the helix:

\[
p=v_{\parallel}T
\]

There are two useful boundary cases:

At exactly \(0^\circ\), the particle experiences no magnetic force and does not visibly revolve. The calculated cyclotron period is then a characteristic frequency, not an observable time between successive revolutions.

04How to identify the charge sign from curvature

The period equation uses the magnitude of the charge. It cannot tell you whether a particle is positive or negative.

To identify the charge sign, you need to examine the direction in which the particle curves.

The magnetic force on a moving charge is

\[
\vec F=q(\vec v\times\vec B)
\]

Use the right-hand rule to determine the force direction for a positive charge. For a negative charge, reverse that direction.

Consider a particle moving to the right into a magnetic field directed into the page.

We represent a magnetic field into the page using crosses (\(\times\)), as though we are looking at the tail of an arrow.

For this particular arrangement:

If the magnetic field instead points out of the page, the directions reverse.

The important thing is to establish the field direction before interpreting the curve. Don’t memorise that electrons always curve downwards or protons always curve upwards. The result depends on the arrangement.

Why charge sign does not change the period

Suppose two particles have identical masses and equal charge magnitudes, but opposite charge signs.

They curve in opposite directions, but their radii have the same magnitude if their perpendicular speeds are equal.

Their periods are also identical:

\[
T=\frac{2\pi m}{|q|B}
\]

This gives you a useful exam distinction:

• Curvature direction tells you the charge sign.
• Orbit radius depends on mass, charge magnitude, perpendicular speed, and field strength.
• Period depends on mass, charge magnitude, and field strength.

For a fuller explanation of the magnetic force and why it changes direction rather than speed, see Why Magnetic Fields Change a Charged Particle’s Direction, Not Its Speed.

05The cyclotron: Why the period matters

The period equation is more than a calculation. It explains how an important particle accelerator works.

In the late 1920s, physicist Ernest Lawrence was considering how to accelerate charged particles to high energies without constructing an extremely long linear accelerator.

A linear accelerator uses electric fields to accelerate particles through a sequence of stages. Lawrence considered whether particles could be made to pass through the same accelerating gap repeatedly instead.

The magnetic field provided a way to bend their paths back towards that gap.

Understanding the apparatus

A cyclotron uses two hollow, semicircular metal electrodes called dees, because each resembles the letter D.

Picture the apparatus from above:

• The two dees sit inside an evacuated chamber, with a narrow gap between them.
• The chamber sits between the poles of a large electromagnet.
• The magnetic field points perpendicular to the plane of the dees.
• An alternating voltage is connected across the two dees, creating an electric field in the gap.
• A source near the centre introduces charged particles.

Now follow one positively charged particle through the apparatus.

Step 1: The particle enters the gap.

The electric field in the gap accelerates the positive particle towards the negatively charged dee.

An electric field can transfer energy to a charged particle, increasing its speed.

Step 2: The particle travels inside the first dee.

Once inside the hollow metal electrode, the particle is shielded from the accelerating electric field.

The magnetic field continues to act, bending the particle into a semicircular path.

Step 3: The particle returns to the gap.

After half a revolution, the particle reaches the gap again.

By this time, the alternating voltage has reversed its polarity. The electric field now points in the opposite direction, allowing it to accelerate the positive particle across the gap again.

Step 4: The process repeats.

The particle gains more kinetic energy each time it crosses the gap.

Its speed increases, so its orbit radius increases:

\[
r=\frac{mv}{|q|B}
\]

However, under the non-relativistic model, its period remains constant:

\[
T=\frac{2\pi m}{|q|B}
\]

The particle therefore spirals outwards while remaining synchronised with the alternating voltage.

The prediction and the experimental result

Lawrence’s central prediction was that a particle’s revolution frequency would remain constant as its speed increased, provided relativistic effects were negligible.

That meant the accelerating voltage could operate at a fixed frequency rather than needing to be continually adjusted.

The early cyclotron development provided a practical test of this idea. Berkeley Lab’s historical account records that the first working model accelerated hydrogen ions to an energy of approximately \(80{,}000\ \mathrm{eV}\) in January 1931.

The significance was not simply that the ions gained energy. They gained energy through repeated acceleration across the same gap, supporting the timing principle behind the design.

You can read more about this development in Berkeley Lab’s history of the 1930s.

Why the voltage frequency is not twice the particle frequency

Here’s a detail that can catch students out.

A particle crosses the accelerating gap twice during one complete revolution. It is tempting to conclude that the alternating voltage must therefore have twice the particle’s frequency.

It doesn’t.

The voltage must reverse polarity between successive crossings because the particle approaches the gap from the opposite side.

The voltage completes one full cycle during one complete particle revolution:

\[
f=\frac{1}{T}
\]

where \(f\) is the cyclotron frequency, measured in hertz (Hz).

The two gap crossings occur half a period apart, matching the two opposite halves of the alternating voltage cycle.

Where the original model reaches its limit

The fixed-frequency cyclotron works well when the particles are non-relativistic, meaning their speeds are much smaller than the speed of light.

At very high speeds, relativistic effects become important. The particle’s relativistic momentum is \(\gamma mv\), where

\[
\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}
\]

Here, \(c\) is the speed of light.

The relativistic period becomes

\[
T=\frac{2\pi\gamma m}{|q|B}
\]

As the particle gains energy, \(\gamma\) increases. Its revolution period therefore increases, and a fixed-frequency accelerating voltage eventually falls out of synchronisation.

Later designs, including the synchrocyclotron, addressed this problem by adjusting the accelerating frequency as the particles gained energy.

This is an extension beyond the basic non-relativistic calculation. For HSC-style questions using the classical period equation, use the supplied mass and assume relativistic effects are negligible unless the question states otherwise. The underlying accelerator physics is discussed in the CERN accelerator-physics course notes.

06A reliable method for exam questions

When you see a question asking for the period of a charged particle in a magnetic field, work through this decision process.

A useful comparison is:

One final warning: if the question supplies a speed, don’t automatically substitute it into the period equation. Speed is needed for the radius and pitch calculations, not for the non-relativistic period.

All numerical values in the following questions are illustrative.

07Questions and solutions

Question 1

A proton enters a uniform magnetic field of strength \(0.30\ \mathrm T\), with its velocity perpendicular to the field.

Calculate the time taken for one complete revolution.

Use:

• Proton mass: \(1.67\times10^{-27}\ \mathrm{kg}\)
• Proton charge: \(+1.60\times10^{-19}\ \mathrm C\)

Assume the proton is non-relativistic.

Solution 1

The proton’s period is \(2.2\times10^{-7}\ \mathrm s\).

Since the velocity is perpendicular to the field, the particle follows circular motion. We can apply the period equation directly.

Step 1: Write the equation.

\[
T=\frac{2\pi m}{|q|B}
\]

Step 2: Substitute the given values.

\[
T=
\frac{2\pi(1.67\times10^{-27})}
{(1.60\times10^{-19})(0.30)}
\]

Step 3: Calculate.

\[
T=\frac{1.05\times10^{-26}}
{4.80\times10^{-20}}
\]

\[
T=2.19\times10^{-7}\ \mathrm s
\]

Step 4: Round to two significant figures.

\[
\boxed{T=2.2\times10^{-7}\ \mathrm s}
\]

The proton takes approximately \(0.22\) microseconds to complete one revolution.

Notice that its speed was never needed. If the proton travelled faster, its orbit would become larger, but its period would remain unchanged under the same conditions.

Question 2

An ion has a mass of \(6.64\times10^{-27}\ \mathrm{kg}\) and a positive charge of \(3.20\times10^{-19}\ \mathrm C\).

It enters a uniform magnetic field of \(0.30\ \mathrm T\) at an angle of \(45^\circ\) to the field, travelling at \(3.0\times10^6\ \mathrm{m\,s^{-1}}\).

Calculate:

a) The perpendicular and parallel velocity components.

b) The radius of the helical path.

c) The period of its circular motion.

d) The pitch of the helix.

A student uses the total speed in the radius equation and obtains \(0.21\ \mathrm m\). Explain why this result is incorrect.

Solution 2

The correct radius is \(0.15\ \mathrm m\), the period is \(4.3\times10^{-7}\ \mathrm s\), and the pitch is \(0.92\ \mathrm m\). The student’s radius is incorrect because only the perpendicular velocity contributes to circular motion.

Step 1: Resolve the velocity.

\[
v_{\perp}=v\sin45^\circ
\]

\[
v_{\perp}=(3.0\times10^6)(0.707)
\]

\[
v_{\perp}=2.12\times10^6\ \mathrm{m\,s^{-1}}
\]

Similarly,

\[
v_{\parallel}=v\cos45^\circ
\]

\[
v_{\parallel}=2.12\times10^6\ \mathrm{m\,s^{-1}}
\]

Step 2: Calculate the radius.

Use the perpendicular component:

\[
r=\frac{mv_{\perp}}{|q|B}
\]

Substitute:

\[
r=
\frac{(6.64\times10^{-27})(2.12\times10^6)}
{(3.20\times10^{-19})(0.30)}
\]

\[
r=0.147\ \mathrm m
\]

Therefore,

\[
\boxed{r=0.15\ \mathrm m}
\]

Step 3: Calculate the period.

\[
T=\frac{2\pi m}{|q|B}
\]

\[
T=
\frac{2\pi(6.64\times10^{-27})}
{(3.20\times10^{-19})(0.30)}
\]

\[
\boxed{T=4.3\times10^{-7}\ \mathrm s}
\]

Step 4: Calculate the pitch.

The pitch is the distance travelled parallel to the magnetic field during one period.

\[
p=v_{\parallel}T
\]

\[
p=(2.12\times10^6)(4.35\times10^{-7})
\]

\[
\boxed{p=0.92\ \mathrm m}
\]

Step 5: Identify the student’s mistake.

The student used

\[
r=\frac{mv}{|q|B}
\]

with the total speed of \(3.0\times10^6\ \mathrm{m\,s^{-1}}\).

That equation would overestimate the radius because the parallel component does not contribute to the magnetic force.

The correct equation uses \(v_{\perp}\), giving a radius of \(0.15\ \mathrm m\), rather than \(0.21\ \mathrm m\).

The key distinction is that the particle’s total speed determines its overall motion, but only its perpendicular velocity determines the circular component.

Question 3

An unknown charged particle enters a uniform magnetic field directed into the page.

Its initial velocity is horizontally to the right, and its observed trajectory curves downwards.

The particle has:

• Mass: \(3.34\times10^{-27}\ \mathrm{kg}\)
• Charge magnitude: \(1.60\times10^{-19}\ \mathrm C\)
• Magnetic field strength: \(0.50\ \mathrm T\)

Determine:

a) Whether the particle is positively or negatively charged.

b) Its period of revolution.

Solution 3

The particle is negatively charged, and its period is \(2.6\times10^{-7}\ \mathrm s\).

Step 1: Determine the charge sign.

Apply the right-hand rule for a positive charge.

With velocity to the right and magnetic field into the page, a positive particle would experience an upward magnetic force.

It would therefore curve upwards.

However, the observed particle curves downwards. Its magnetic force must point in the opposite direction.

Therefore, the particle is negatively charged.

Step 2: Calculate the period.

Use the magnitude of the charge:

\[
T=\frac{2\pi m}{|q|B}
\]

Substitute:

\[
T=
\frac{2\pi(3.34\times10^{-27})}
{(1.60\times10^{-19})(0.50)}
\]

\[
T=
\frac{2.10\times10^{-26}}
{8.00\times10^{-20}}
\]

\[
\boxed{T=2.6\times10^{-7}\ \mathrm s}
\]

The negative sign determines the direction of curvature, not the period.

A common mistake is to substitute a negative charge into the period equation and obtain a negative answer. A period is a positive time, so always use \(|q|\).

Question 4

A proton is accelerated inside an ideal cyclotron containing a uniform magnetic field of \(0.80\ \mathrm T\).

Use:

• Proton mass: \(1.67\times10^{-27}\ \mathrm{kg}\)
• Proton charge: \(1.60\times10^{-19}\ \mathrm C\)

Calculate:

a) The proton’s period.

b) The frequency of the alternating voltage required to keep the proton synchronised.

c) Explain why the proton crosses the accelerating gap twice per revolution without requiring the alternating voltage to operate at twice the revolution frequency.

d) The proton gains energy and its orbit radius increases by 25%. Assuming the non-relativistic model remains valid, determine the percentage change in its period.

Solution 4

The proton’s period is \(8.2\times10^{-8}\ \mathrm s\), and the required alternating voltage frequency is approximately \(12.2\ \mathrm{MHz}\). The period does not change when the orbit radius increases under the non-relativistic model.

Step 1: Calculate the period.

\[
T=\frac{2\pi m}{|q|B}
\]

\[
T=
\frac{2\pi(1.67\times10^{-27})}
{(1.60\times10^{-19})(0.80)}
\]

\[
T=8.20\times10^{-8}\ \mathrm s
\]

Step 2: Calculate the frequency.

Frequency is the number of revolutions per second.

\[
f=\frac{1}{T}
\]

\[
f=\frac{1}{8.20\times10^{-8}}
\]

\[
\boxed{f=1.22\times10^7\ \mathrm{Hz}}
\]

Or:

\[
\boxed{f=12.2\ \mathrm{MHz}}
\]

Step 3: Explain the two gap crossings.

The proton crosses the gap twice per revolution, with half a revolution between successive crossings.

During that half-revolution, the alternating voltage reverses polarity.

This reversal is essential because the proton approaches the gap from the opposite side. The electric field must reverse so that the proton continues gaining kinetic energy rather than being slowed down.

One complete voltage cycle therefore corresponds to one complete particle revolution.

The tempting mistake is to double the frequency because there are two crossings. However, the two crossings occur at opposite voltage polarities. They are two half-cycles of one complete alternating voltage cycle.

Step 4: Determine the percentage change in period.

The period equation is

\[
T=\frac{2\pi m}{|q|B}
\]

There is no radius term in this equation.

Increasing the radius means the proton’s speed has increased, but the period remains constant because the larger circumference is matched by the greater speed.

Therefore,

\[
\boxed{\text{Percentage change in period}=0\%}
\]

This assumes the particle remains non-relativistic and the magnetic field remains constant.

Question 5

An unknown ion moves through a uniform magnetic field directed into the page.

Its initial velocity has two components:

• A perpendicular component directed horizontally to the right.
• A parallel component directed into the page, along the magnetic field.

The ion’s total speed is \(3.0\times10^6\ \mathrm{m\,s^{-1}}\), and its velocity makes an angle of \(60^\circ\) with the magnetic field.

Its projected circular trajectory initially curves upwards.

The ion has:

• Mass: \(6.64\times10^{-27}\ \mathrm{kg}\)
• Charge magnitude: \(3.20\times10^{-19}\ \mathrm C\)
• Magnetic field strength: \(0.25\ \mathrm T\)

Answer the following:

a) Determine the sign of the ion’s charge.

b) Calculate its period of circular motion.

c) Calculate the radius of its helical path.

d) Calculate the pitch of its helix.

e) A student claims that the ion does not have a period because it never returns to its original position. Evaluate this claim. Explain what has returned to its original state after one period and what has not.

f) Suppose the magnetic field is switched off immediately after one complete revolution. Describe the ion’s subsequent motion, assuming no other forces act.

Solution 5

The ion is positively charged. Its period is \(5.2\times10^{-7}\ \mathrm s\), its radius is \(0.22\ \mathrm m\), and its pitch is \(0.78\ \mathrm m\).

The student is incorrect: the ion completes one full revolution after one period, even though it does not return to its original three-dimensional position.

Step 1: Determine the charge sign.

The magnetic field points into the page, and the perpendicular velocity points right.

For a positive charge, the right-hand rule predicts an upward magnetic force.

The observed circular component initially curves upwards.

Therefore, the ion is positively charged.

Step 2: Resolve the velocity.

The perpendicular component is

\[
v_{\perp}=v\sin60^\circ
\]

\[
v_{\perp}=(3.0\times10^6)(0.866)
\]

\[
v_{\perp}=2.60\times10^6\ \mathrm{m\,s^{-1}}
\]

The parallel component is

\[
v_{\parallel}=v\cos60^\circ
\]

\[
v_{\parallel}=(3.0\times10^6)(0.50)
\]

\[
v_{\parallel}=1.50\times10^6\ \mathrm{m\,s^{-1}}
\]

Step 3: Calculate the period.

\[
T=\frac{2\pi m}{|q|B}
\]

\[
T=
\frac{2\pi(6.64\times10^{-27})}
{(3.20\times10^{-19})(0.25)}
\]

\[
T=\frac{4.17\times10^{-26}}
{8.00\times10^{-20}}
\]

\[
\boxed{T=5.2\times10^{-7}\ \mathrm s}
\]

Step 4: Calculate the radius.

Use the perpendicular velocity:

\[
r=\frac{mv_{\perp}}{|q|B}
\]

\[
r=
\frac{(6.64\times10^{-27})(2.60\times10^6)}
{(3.20\times10^{-19})(0.25)}
\]

\[
r=0.216\ \mathrm m
\]

Therefore,

\[
\boxed{r=0.22\ \mathrm m}
\]

Step 5: Calculate the pitch.

The pitch is the distance travelled parallel to the magnetic field during one complete revolution.

\[
p=v_{\parallel}T
\]

\[
p=(1.50\times10^6)(5.22\times10^{-7})
\]

\[
\boxed{p=0.78\ \mathrm m}
\]

Step 6: Evaluate the student’s claim.

The student is confusing one complete revolution with returning to the original three-dimensional position.

After one period:

• The perpendicular component of the particle’s position has returned to its starting position.
• Its perpendicular velocity has returned to its original direction.
• Its parallel velocity is unchanged.
• Its position along the magnetic field has advanced by one pitch, approximately \(0.78\ \mathrm m\).

The particle has therefore completed a full revolution, but it has not returned to its starting point.

This is the important distinction between circular motion and helical motion. The circular component is periodic, even though the complete three-dimensional trajectory does not close.

Step 7: Describe the motion after the magnetic field is switched off.

Once the magnetic field is removed, the magnetic force becomes zero.

With no other forces acting, Newton’s first law tells us that the particle continues in a straight line at constant velocity.

Its velocity at the instant the field is removed is the same as its initial velocity because one complete circular revolution has restored the perpendicular velocity direction, while the parallel component has remained unchanged.

Therefore, it continues along a straight line in its original velocity direction.

The key lesson is that a particle can complete a full period of circular motion without returning to its original position in three-dimensional space.

07What to learn next

The period equation connects magnetic force, circular motion, and the charge-to-mass ratio of a particle.

The next useful comparison is with electric fields.

A magnetic field can change the direction of a moving charged particle without changing its speed, while an electric field can do work and change its kinetic energy.

This difference becomes particularly important when analysing a cyclotron, where the magnetic field bends the particle’s path and the electric field accelerates it across the gap.

For the electric-field side of charged-particle motion, continue with Charged-Particle Trajectories in Electric Fields: HSC Physics.