How to Find a Particle’s Charge Sign from Its Curved Path

A detector records a charged particle as a curved track. The magnetic field is known, but the photograph has no arrow showing which way the particle travelled. Is the particle positively or negatively charged?

You can work it out, but there is an important catch: you need to know both the magnetic field direction and the particle’s direction of travel.

The method is to use the magnetic force direction. For a positive particle, the force acts in the direction of \(\vec v \times \vec B\). If the observed path curves in that direction, the particle is positive. If it curves in the opposite direction, the particle is negative.

There is one more detail. Only the component of velocity perpendicular to the magnetic field produces this curvature. If the particle also moves parallel to the field, its trajectory can become helical rather than circular.

Let’s build a reliable method for identifying the charge sign, including the situations where a curved track can be misleading.

01Start with the direction of the magnetic force

Imagine a particle travelling horizontally towards the right of a detector. A uniform magnetic field points into the page.

Predict which way a positively charged particle will bend: upwards or downwards?

The answer is upwards.

For a positive charge, the right-hand rule gives a magnetic force upwards. A negative particle travelling through exactly the same field would experience a force downwards.

This is the basic principle behind identifying charge from a curved particle track.

The right-hand rule

The magnetic force on a moving charged particle is described by:

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

where:

• \(\vec F_B\) is the magnetic force, measured in newtons (N).
• \(q\) is the particle’s electric charge, measured in coulombs (C).
• \(\vec v\) is the particle’s velocity, measured in metres per second (m s\(^{-1}\)).
• \(\vec B\) is the magnetic flux density, measured in tesla (T).

For a positive charge, use your right hand to find the direction of \(\vec v\times\vec B\).

For a negative charge, the force acts in the opposite direction.

Two symbols are particularly useful when reading particle-track diagrams:

• \(\times\) means the magnetic field points into the page. Imagine looking at the tail feathers of an arrow moving away from you.
• \(\bullet\) means the magnetic field points out of the page. Imagine looking directly at the tip of an arrow travelling towards you.

A useful direction table

Suppose the particle travels towards the right (east).

Notice something important: reversing the magnetic field reverses the direction of the magnetic force.

This means that an upward-curving track does not automatically indicate a positive particle. You must know the field direction.

02Why only the perpendicular velocity component matters

So far, we’ve considered particles travelling perpendicular to the magnetic field.

Real particle trajectories aren’t always that simple.

Imagine a particle entering a magnetic field at an angle. Its velocity can be separated into two components:

• A perpendicular component, \(v_\perp\), directed across the magnetic field.
• A parallel component, \(v_\parallel\), directed along the magnetic field.

The magnetic force acts on the perpendicular component only.

Let \(\theta\) represent the angle between the particle’s velocity and the magnetic field.

The perpendicular component is:

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

The parallel component is:

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

The magnetic force magnitude is therefore:

\[
\boxed{F_B=|q|vB\sin\theta}
\]

Or, using the perpendicular velocity component:

\[
\boxed{F_B=|q|v_\perp B}
\]

This gives us three useful situations.

The parallel component remains unchanged in a uniform magnetic field. The perpendicular component changes direction continuously.

The combination produces a helix: circular motion around the field direction while the particle continues travelling along it.

Why this matters when finding the charge sign

The perpendicular component determines the direction of the magnetic force:

\[
\vec F_B=q(\vec v_\perp\times\vec B)
\]

You should therefore use \(v_\perp\) when applying the right-hand rule.

You do not need to treat the parallel component as another source of magnetic force.

A common mistake is to assume that a particle travelling faster must experience a larger magnetic force. Not necessarily. If its extra speed comes entirely from motion parallel to the field, the magnetic force does not increase.

03The experimental problem: how do you know which way the particle travelled?

There’s a problem we haven’t solved yet.

Suppose you receive a photograph showing a curved particle track. You know the magnetic field direction, but there is no arrow indicating the direction of travel.

Can you determine the charge sign?

Not from the curvature alone.

A curved path has a tangent at every point, but a photograph of that path does not necessarily tell you which way along the tangent the particle was moving.

Reverse the direction of travel and the magnetic force direction required for the same charge sign also reverses.

This was more than a theoretical issue. It played an important role in the experimental discovery of the positron.

Carl Anderson’s cloud chamber experiment

In 1932, physicist Carl Anderson investigated cosmic rays using a cloud chamber surrounded by a powerful electromagnet.

The cloud chamber contained a vapour mixture under conditions that allowed tiny droplets to form along the paths of ionising particles.

Here’s how the apparatus worked.

1. A charged particle entered the chamber and passed through the vapour.
2. As it travelled, it ionised the gas along its path.
3. Tiny droplets formed around the ions, creating a visible track.
4. A camera photographed the track.
5. The surrounding electromagnet produced a magnetic field perpendicular to the plane of the photograph.

The photograph made the particle’s otherwise invisible motion observable.

What should the experiment show?

If the direction of travel and magnetic field were known, the curvature would reveal the charge sign.

Anderson observed tracks curving in both positive-charge and negative-charge directions.

However, there was an important ambiguity.

A positive particle travelling downwards could produce a track with the same geometric shape as a negative particle travelling upwards.

The photograph showed the path, but not the direction of travel.

This meant that the interpretation depended on establishing which way the particle was actually moving.

The lead plate solved the problem

Anderson introduced a thin lead plate across the cloud chamber.

Why lead?

A particle passing through the plate would lose kinetic energy. It would emerge travelling more slowly, assuming it continued through the magnetic field without a major change in direction.

For a particle moving perpendicular to the field, the radius of its circular trajectory is:

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

Here:

• \(r\) is the radius of the circular path, in metres.
• \(m\) is the particle’s mass, in kilograms.
• \(v\) is its speed, in metres per second.
• \(|q|\) is its charge magnitude, in coulombs.
• \(B\) is the magnetic flux density, in tesla.

For the same particle in the same magnetic field, a lower speed produces a smaller radius.

In other words, the track becomes more tightly curved after the particle loses energy.

Anderson could use this change to establish the direction of travel.

In his famous 1932 photograph, the upper part of the track was more tightly curved than the lower part. This indicated that the particle had passed upwards through the lead plate, losing energy along the way.

With the direction of travel established and the magnetic field direction known, the curvature showed that the particle was positively charged.

The result provided evidence for the positron, the positively charged antiparticle of the electron.

The historical account is described by the American Physical Society’s article on the discovery of the positron.

The important experimental lesson is that the lead plate did not directly measure the charge sign. It established the direction of travel. The magnetic field and the direction of curvature then revealed the sign.

A quick scale estimate: how much does the radius change?

Consider a hypothetical non-relativistic particle with:

• Mass \(m=2.0\times10^{-27}\,\mathrm{kg}\).
• Charge magnitude \(|q|=1.60\times10^{-19}\,\mathrm C\).
• Perpendicular velocity \(v_\perp=4.0\times10^6\,\mathrm{m\,s^{-1}}\).
• Magnetic field \(B=0.25\,\mathrm T\).

Its radius is:

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

Substituting:

\[
r=
\frac{(2.0\times10^{-27})(4.0\times10^6)}
{(1.60\times10^{-19})(0.25)}
\]

\[
r=0.20\,\mathrm m
\]

Now suppose the particle loses energy and its perpendicular speed falls to \(3.0\times10^6\,\mathrm{m\,s^{-1}}\).

Its new radius is:

\[
r=
\frac{(2.0\times10^{-27})(3.0\times10^6)}
{(1.60\times10^{-19})(0.25)}
\]

\[
r=0.15\,\mathrm m
\]

The radius decreases from \(0.20\,\mathrm m\) to \(0.15\,\mathrm m\).

The track becomes more tightly curved, providing evidence that the particle has slowed down.

These are illustrative values, not measurements from Anderson’s experiment.

For the non-relativistic HSC model, the momentum is \(p=mv\). At relativistic speeds, the more general relationship is \(r=p_\perp/(|q|B)\), where \(p_\perp\) is the perpendicular component of momentum.

04A reliable method for HSC questions

When you are given a curved particle track, work through the following steps.

Step 1: Identify the magnetic field direction.

Look for crosses and dots, or an explicitly stated field direction.

Step 2: Establish the particle’s direction of travel.

Look for a velocity arrow or information about how the particle entered the detector.

If the direction is unknown, check whether a lead plate, energy loss, or another experimental observation can establish it.

Step 3: Identify the perpendicular velocity component.

If the particle travels at an angle to the field, use:

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

Use this component when determining the magnetic force.

Step 4: Predict the force direction for a positive charge.

Apply the right-hand rule to \(\vec v_\perp\times\vec B\).

Step 5: Compare the prediction with the observed curvature.

• Same direction: positive charge.
• Opposite direction: negative charge.

One final distinction is worth remembering: identifying a particle’s charge sign does not identify the particle itself. A negative track could belong to an electron, a negative muon, or another negatively charged particle.

05Common mistakes to avoid

06Questions and solutions

The following questions build from a straightforward charge-sign identification towards more complicated particle trajectories and experimental evidence.

Question 1

A particle travels westwards at \(2.5\times10^6\,\mathrm{m\,s^{-1}}\) through a uniform magnetic field of \(0.20\,\mathrm T\), directed out of the page.

The particle’s path initially curves northwards.

Determine the sign of the particle’s charge and calculate the magnitude of the magnetic force if its charge magnitude is \(1.60\times10^{-19}\,\mathrm C\).

Solution 1

The particle is positively charged, and the magnetic force has magnitude \(8.0\times10^{-14}\,\mathrm N\).

First, apply the right-hand rule.

The particle travels westwards, and the magnetic field points out of the page.

For a positive charge, the magnetic force points northwards.

This matches the observed curvature, so the particle must be positive.

Now calculate the force magnitude.

Since the velocity is perpendicular to the magnetic field:

\[
F_B=|q|vB
\]

Substituting:

\[
\begin{aligned}
F_B&=(1.60\times10^{-19})(2.5\times10^6)(0.20)\\
&=8.0\times10^{-14}\,\mathrm N
\end{aligned}
\]

Therefore:

\[
\boxed{F_B=8.0\times10^{-14}\,\mathrm N}
\]

The important part is getting the direction right before calculating. The magnetic force magnitude alone cannot tell you whether the particle is positive or negative.

Question 2

A particle enters a uniform magnetic field of \(0.30\,\mathrm T\), directed northwards.

Its speed is \(5.0\times10^6\,\mathrm{m\,s^{-1}}\), and its velocity makes an angle of \(35^\circ\) with the magnetic field.

The perpendicular component of its velocity points eastwards. Its trajectory initially curves into the page.

The particle has a charge magnitude of \(3.2\times10^{-19}\,\mathrm C\).

Determine:

a) The perpendicular component of its velocity.

b) The magnitude of the magnetic force.

c) The sign of its charge.

Solution 2

The perpendicular velocity is \(2.87\times10^6\,\mathrm{m\,s^{-1}}\), the magnetic force is approximately \(2.75\times10^{-13}\,\mathrm N\), and the particle is negatively charged.

Part a: Calculate the perpendicular velocity.

The perpendicular component is:

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

Substitute:

\[
\begin{aligned}
v_\perp&=(5.0\times10^6)\sin35^\circ\\
&=(5.0\times10^6)(0.574)\\
&=2.87\times10^6\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

Therefore:

\[
\boxed{v_\perp=2.87\times10^6\,\mathrm{m\,s^{-1}}}
\]

Part b: Calculate the magnetic force.

Use:

\[
F_B=|q|v_\perp B
\]

Substituting:

\[
\begin{aligned}
F_B&=(3.2\times10^{-19})(2.87\times10^6)(0.30)\\
&=2.75\times10^{-13}\,\mathrm N
\end{aligned}
\]

Therefore:

\[
\boxed{F_B=2.75\times10^{-13}\,\mathrm N}
\]

Part c: Determine the charge sign.

The perpendicular velocity points east, and the magnetic field points north.

For a positive particle, the right-hand rule predicts a force out of the page.

However, the observed trajectory curves into the page.

The force is therefore opposite to the direction predicted for a positive charge.

Hence:

\[
\boxed{q<0}
\]

The particle is negatively charged.

Notice that we used the perpendicular velocity for both the force calculation and the direction analysis. The parallel component does not contribute to the magnetic force.

Question 3

A particle travels through a uniform magnetic field directed into the page.

Its trajectory is circular, and it crosses a thin lead plate positioned vertically through the centre of the detector.

The measured radii are:

The particle loses kinetic energy when passing through the plate. Assume that:

• The particle’s motion remains perpendicular to the magnetic field.
• The plate causes negligible scattering.
• The particle’s mass and charge remain unchanged.
• The magnetic field is uniform on both sides.

At the crossing point, the trajectory is horizontal and curves northwards.

Determine:

a) The direction in which the particle travelled through the plate.

b) The ratio of its speed after the plate to its speed before the plate.

c) The sign of its charge.

Solution 3

The particle travelled from left to right, its speed decreased to two-thirds of its original value, and it is positively charged.

Part a: Determine the direction of travel.

The particle loses kinetic energy when passing through the lead plate.

Therefore, its speed must be lower after crossing the plate.

For circular motion perpendicular to a uniform magnetic field:

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

Since the mass, charge magnitude, and magnetic field remain unchanged:

\[
r\propto v
\]

A lower speed produces a smaller radius.

The radius is smaller on the right-hand side of the plate:

\[
12\,\mathrm{cm}<18\,\mathrm{cm}
\]

Therefore, the right-hand side represents the particle’s motion after crossing the plate.

The particle must have travelled from left to right.

Part b: Calculate the speed ratio.

Because the radius is directly proportional to speed:

\[
\frac{v_{\text{after}}}{v_{\text{before}}}
=
\frac{r_{\text{after}}}{r_{\text{before}}}
\]

Substitute:

\[
\frac{v_{\text{after}}}{v_{\text{before}}}
=
\frac{12}{18}
\]

\[
\boxed{\frac{v_{\text{after}}}{v_{\text{before}}}=0.667}
\]

The particle retains approximately \(66.7\%\) of its original speed.

Part c: Determine the charge sign.

The particle travels left to right, so its velocity at the crossing points eastwards.

The magnetic field points into the page.

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

The observed trajectory curves northwards, matching this prediction.

Therefore, the particle is positively charged.

The interesting part of this question is that the photograph initially provides no direction arrow. The change in radius supplies the missing information, allowing the charge sign to be determined.

Question 4

Two particles, A and B, have identical masses and charge magnitudes.

They enter the same uniform magnetic field, which has a flux density of \(0.30\,\mathrm T\) and points out of the page.

Both particles have a perpendicular velocity component of \(2.0\times10^6\,\mathrm{m\,s^{-1}}\), directed eastwards.

Their parallel velocity components are different:

Both projected trajectories curve southwards.

Each particle has a mass of \(3.0\times10^{-27}\,\mathrm{kg}\) and a charge magnitude of \(1.60\times10^{-19}\,\mathrm C\).

Determine:

a) The charge sign of each particle.

b) The radius of each trajectory.

c) Whether each particle follows a circular or helical path.

A student argues that particle B must have a larger radius because its total speed is greater. Explain why this prediction is incorrect.

Solution 4

Both particles are positively charged and have the same trajectory radius of \(0.125\,\mathrm m\). Particle A follows a circular path, while particle B follows a helical path.

Part a: Determine the charge signs.

Both particles have perpendicular velocity components directed eastwards.

The magnetic field points out of the page.

For a positive charge, the magnetic force points southwards.

Both trajectories curve southwards, so both particles must be positively charged.

Part b: Calculate the trajectory radii.

The radius of the circular component is:

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

Substitute the given values:

\[
r=
\frac{(3.0\times10^{-27})(2.0\times10^6)}
{(1.60\times10^{-19})(0.30)}
\]

\[
r=\frac{6.0\times10^{-21}}{4.8\times10^{-20}}
\]

Therefore:

\[
\boxed{r=0.125\,\mathrm m}
\]

Both particles have the same mass, charge magnitude, magnetic field strength, and perpendicular velocity.

Their radii must therefore be identical.

Part c: Identify the trajectories.

Particle A has no parallel velocity component.

Its entire velocity is perpendicular to the magnetic field, so it follows a circular path.

Particle B has a nonzero parallel velocity component.

It continues moving along the magnetic field while simultaneously undergoing circular motion around the field direction.

Its trajectory is therefore helical.

Why is the student’s prediction incorrect?

The student’s mistake is using total velocity instead of perpendicular velocity.

Particle B does have a greater total speed:

\[
v_B=\sqrt{v_\perp^2+v_\parallel^2}
\]

\[
\begin{aligned}
v_B&=\sqrt{(2.0\times10^6)^2+(4.0\times10^6)^2}\\
&=4.47\times10^6\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

However, its perpendicular velocity remains exactly the same as particle A’s.

The radius depends on \(v_\perp\), not the total speed.

Adding velocity parallel to the magnetic field changes the helix’s pitch, not its radius.

This is a useful distinction when interpreting three-dimensional particle tracks.

Question 5

A particle travels through a uniform magnetic field directed out of the page. Its trajectory is helical, and it passes through a thin foil separating two regions, labelled A and B.

A detector reconstructs the following measurements:

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

The foil removes kinetic energy and also scatters the particle, potentially changing the direction of its velocity.

Assume:

• The same particle is observed in both regions.
• Its mass and charge magnitude remain unchanged.
• The magnetic field is uniform and unchanged.
• The particle is non-relativistic.
• There is no electric field between the foil and the measurement regions.
• The helical period is therefore the same in both regions.

At a reference point in region A, the perpendicular component of the particle’s velocity points eastwards. Its projected trajectory curves southwards.

A student makes the following argument:

“The radius is larger in region B, so the particle must be travelling faster there. Therefore, it travelled from B to A, losing energy as it moved.”

Is the student correct?

Determine:

a) The actual direction of travel through the foil.

b) The sign of the particle’s charge.

Your explanation must account for both the helix radius and pitch. A comparison of radii alone is insufficient.

Solution 5

The particle travels from region A to region B, and it is positively charged. The student’s conclusion is incorrect because the helix radius depends on perpendicular velocity, not total speed.

This question requires us to consider both components of the particle’s motion.

A particle can lose kinetic energy while its perpendicular velocity increases, provided its parallel velocity decreases sufficiently.

The scattering at the foil makes this possible.

Part a: Determine the direction of travel.

We need to compare the particle’s total speed in the two regions.

For a non-relativistic particle moving in a uniform magnetic field, the period of its circular motion is:

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

The particle’s mass, charge magnitude, and magnetic field strength are unchanged.

Therefore, its period \(T\) is the same in both regions.

During one complete revolution, the particle travels a circular distance of \(2\pi r\) due to its perpendicular velocity.

Therefore:

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

During the same period, it travels a distance equal to the helix pitch \(h\) along the magnetic field.

Therefore:

\[
v_\parallel=\frac{h}{T}
\]

The total speed is found by combining the perpendicular and parallel components:

\[
v=\sqrt{v_\perp^2+v_\parallel^2}
\]

Substituting the expressions above:

\[
v=\frac{1}{T}\sqrt{(2\pi r)^2+h^2}
\]

Since \(T\) is identical in both regions, we only need to compare:

\[
S=\sqrt{(2\pi r)^2+h^2}
\]

Region A:

Using \(r_A=5.0\,\mathrm{cm}\) and \(h_A=30\,\mathrm{cm}\):

\[
\begin{aligned}
S_A&=\sqrt{(2\pi\times5.0)^2+30^2}\\
&=\sqrt{986.96+900}\\
&=43.4\,\mathrm{cm}
\end{aligned}
\]

Region B:

Using \(r_B=6.0\,\mathrm{cm}\) and \(h_B=10\,\mathrm{cm}\):

\[
\begin{aligned}
S_B&=\sqrt{(2\pi\times6.0)^2+10^2}\\
&=\sqrt{1421.22+100}\\
&=39.0\,\mathrm{cm}
\end{aligned}
\]

Since:

\[
S_B<S_A
\]

the particle’s total speed is lower in region B.

We can even compare the speeds directly:

\[
\frac{v_B}{v_A}=\frac{S_B}{S_A}
\]

\[
\frac{v_B}{v_A}=\frac{39.0}{43.4}
\]

\[
\boxed{\frac{v_B}{v_A}\approx0.90}
\]

The particle travels approximately 10% more slowly in region B.

The foil removes kinetic energy, so the lower-speed region must be the region after the foil.

Therefore, the particle travels:

\[
\boxed{\mathrm{A\longrightarrow B}}
\]

Why does the radius increase if the particle slows down?

This is the central trap in the question.

The perpendicular velocity actually increases because the radius increases:

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

Meanwhile, the parallel velocity decreases substantially because the pitch decreases:

\[
v_\parallel=\frac{h}{T}
\]

The scattering at the foil has redirected the particle’s motion towards the plane perpendicular to the magnetic field.

Its perpendicular component has increased, but its parallel component has decreased enough that its total speed is lower.

The student assumed that a larger radius must mean a greater total speed.

That assumption only works if the direction of the velocity relative to the magnetic field remains unchanged, or if the motion is entirely perpendicular to the field.

Here, the pitch angle changes, so radius alone cannot establish the direction of travel.

Part b: Determine the charge sign.

At the reference point in region A:

• The perpendicular velocity points east.
• The magnetic field points out of the page.
• The observed trajectory curves southwards.

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

This matches the observed curvature.

Therefore:

\[
\boxed{q>0}
\]

The particle is positively charged.

This question combines three ideas that are easy to treat separately: charge-sign identification, perpendicular velocity, and helical motion. The key is recognising that the radius tells you about the perpendicular component, while the total kinetic energy depends on both velocity components.

Once you can determine the charge sign reliably, the next useful step is extracting more information from the trajectory. The radius and period can be used to investigate a particle’s momentum and mass-to-charge relationship.

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