Why an Electron and a Proton Curve in Opposite Directions

An electron and a proton enter the same uniform magnetic field, both travelling to the right. The magnetic field points into the page. The proton initially curves upwards, while the electron curves downwards.

Why do they bend in opposite directions?

The answer is their opposite electric charge signs. A proton has positive charge, while an electron has negative charge. The magnetic force on a negative particle acts in the opposite direction to the force on a positive particle moving with the same velocity through the same magnetic field.

But there’s another interesting detail. The electron also follows a much tighter curve than the proton if both particles have the same speed. That difference comes from their masses, not their charge signs.

Keep these two effects separate throughout this topic:

• Charge sign determines the direction of curvature.
• Mass, perpendicular speed, and magnetic field strength determine the radius of curvature.

Let’s see why.

01The experiment: what happens when the particle changes?

Imagine an experiment designed to compare an electron and a proton travelling through the same magnetic field.

The question is simple: does the direction of curvature depend on the particle’s mass, its charge, or both?

Setting up the apparatus

An idealised beam experiment could investigate this using the following arrangement.

1. A particle source produces an electron beam or a proton beam. The two particles are tested in separate runs.
2. A narrow slit collimates each beam, giving the particles a well-defined direction of travel.
3. The beam enters a chamber between the pole pieces of an electromagnet. The magnetic field is uniform and directed into the page.
4. A tracking detector records the particle’s trajectory as it passes through the magnetic field.

The sources are adjusted so that both particles enter the field travelling to the right, with the same speed. Their initial velocities are perpendicular to the magnetic field.

What should the experiment show?

Before looking at the predicted tracks, consider what stays the same and what changes.

The particles have equal charge magnitudes and identical initial velocities. Their charge signs and masses are different.

For a positive particle travelling to the right with the magnetic field directed into the page, the right-hand rule predicts an upward magnetic force.

Therefore:

• The proton should initially curve upwards.
• The electron should initially curve downwards.

The detector should record two tracks curving in opposite directions.

There is also a prediction about their curvature. The electron should turn much more sharply because its mass is considerably smaller.

This is the result expected from the idealised model, not a claim about measurements from a particular historical experiment.

Testing the explanation

A useful control is to reverse the current through the electromagnet, which reverses the magnetic field.

The predicted result is that both particles reverse their initial directions of curvature.

The proton now bends downwards, and the electron bends upwards.

This supports the explanation that magnetic deflection depends on the relationship between the particle’s velocity, the magnetic field direction, and its charge sign.

There is one experimental detail worth controlling. If the two particles are accelerated through the same potential difference, they will not have the same speed. This matters when comparing their radii, although it does not change the basic rule for determining their charge signs.

We’ll return to this distinction when calculating particle trajectories.

02Why opposite charge signs produce opposite curvature

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

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

Here:

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

The important feature is the sign of \(q\).

For a positive particle, the right-hand rule gives the direction of \(\vec v\times\vec B\).

For a negative particle, the magnetic force acts in the opposite direction.

Apply the right-hand rule

Return to our original arrangement:

• Particle velocity points right.
• Magnetic field points into the page.
• The particle is moving perpendicular to the magnetic field.

For a positive charge, point your right-hand fingers in the direction of velocity and orient them towards the magnetic field. Your thumb indicates the direction of the magnetic force.

The resulting force points upwards.

A proton has positive charge, so it experiences an upward magnetic force.

An electron has negative charge, so its magnetic force points downwards.

A direction table for common arrangements

The following table assumes the particle initially travels to the right and its velocity is perpendicular to the magnetic field.

The symbols have a simple meaning:

• A cross (\(\times\)) represents a magnetic field directed into the page.
• A dot (\(\bullet\)) represents a magnetic field directed out of the page.

Don’t memorise the table as a universal rule. If the magnetic field or the direction of particle travel changes, the force direction changes too.

The reliable method is to apply the right-hand rule to the actual directions given in the question.

For a more general method of interpreting particle tracks, see How to Find a Particle’s Charge Sign from Its Curved Path.

03Why only the perpendicular velocity component produces curvature

So far, we’ve assumed that the particle travels at right angles to the magnetic field.

But what happens if an electron or proton enters the field at an angle?

The answer requires one extra step: resolving the velocity into two components.

Imagine a particle travelling diagonally relative to the magnetic field. Its velocity can be separated into:

• A perpendicular component, \(v_\perp\), directed at right angles to the magnetic field.
• A parallel component, \(v_\parallel\), directed along the magnetic field.

The perpendicular component produces the magnetic curvature.

The parallel component experiences no magnetic force because it travels along the magnetic field direction.

This is why we cannot always use the particle’s total speed when calculating its magnetic force.

Resolving the velocity

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

Using basic trigonometry:

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

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

where:

• \(v\) is the total speed, in \(\mathrm{m\,s^{-1}}\).
• \(v_\perp\) is the perpendicular velocity component.
• \(v_\parallel\) is the parallel velocity component.
• \(\theta\) is the angle between the velocity and magnetic field.

The magnitude of the magnetic force is:

\[ F_B=|q|vB\sin\theta \]

Substituting \(v_\perp=v\sin\theta\):

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

Notice that the parallel component has disappeared.

It contributes to the particle’s motion, but it does not produce magnetic force.

Three possible particle trajectories

The relationship between the two velocity components determines the shape of the trajectory.

A helix is similar to a spring wrapped around a straight axis. The circular motion comes from the perpendicular component, while the parallel component carries the particle along the axis.

The analogy helps picture the shape, although the particle is not physically following a spring.

One important consequence follows: even when a particle follows a helix, its charge sign still determines the direction of its circular motion around the magnetic field.

If the particle travels exactly parallel to the field, however, there is no curvature at all. You cannot determine its charge sign from a straight trajectory alone.

04Why an electron has a tighter curve than a proton

We’ve established why an electron and proton bend in opposite directions.

Now consider their different masses.

If both particles have the same perpendicular velocity and enter the same magnetic field, their magnetic force magnitudes are equal because their charge magnitudes are equal.

However, equal forces do not produce equal accelerations.

Newton’s second law tells us:

\[ F=ma \]

The electron has a much smaller mass, so the same magnetic force produces a much greater acceleration.

This makes the electron change its direction of travel more rapidly.

Deriving the radius of curvature

For circular motion, the magnetic force provides the centripetal force.

The magnetic force magnitude is:

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

The required centripetal force is:

\[ F_c=\frac{mv_\perp^2}{r} \]

Equating the two:

\[ |q|v_\perp B=\frac{mv_\perp^2}{r} \]

Rearranging:

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

This equation tells us how the radius depends on the particle’s properties.

• Increasing mass increases the radius.
• Increasing perpendicular velocity increases the radius.
• Increasing magnetic field strength decreases the radius.
• Increasing charge magnitude decreases the radius.

The radius depends on the magnitude of charge, not its sign.

Comparing electron and proton radii

An electron and a proton have approximately equal charge magnitudes:

\[ |q_e|=|q_p|=1.60\times10^{-19}\,\mathrm C \]

Their masses are very different:

\[ m_e=9.11\times10^{-31}\,\mathrm{kg} \]

\[ m_p=1.67\times10^{-27}\,\mathrm{kg} \]

If both particles have the same perpendicular velocity and experience the same magnetic field:

\[ \frac{r_p}{r_e}=\frac{m_p}{m_e} \]

Therefore:

\[ \frac{r_p}{r_e}\approx1.83\times10^3 \]

The proton’s radius is approximately 1,836 times greater than the electron’s radius.

This is a useful comparison, but notice the condition: both particles must have the same perpendicular velocity.

If they have instead been accelerated through the same potential difference, their speeds will be different. Their radius ratio will change.

This distinction becomes important in mass spectrometry and questions involving accelerating voltages. See How to Compare Ion Path Radii After the Same Accelerating Voltage.

05Worked examples

06Questions and solutions

The following questions move from basic charge identification to more demanding situations involving velocity components, accelerating voltages, and changes in magnetic field direction.

Assume uniform magnetic fields, negligible forces other than the magnetic force, and non-relativistic particle speeds unless stated otherwise. All numerical values are illustrative.

Question 1

An electron enters a uniform magnetic field travelling horizontally to the right. The magnetic field points into the page.

Determine the initial direction of the electron’s magnetic force and explain why its trajectory curves in that direction.

Solution 1

The electron initially curves downwards.

First, apply the right-hand rule as though the particle were positively charged.

With velocity directed right and magnetic field directed into the page, the force on a positive particle would point upwards.

However, an electron has negative charge.

Its magnetic force therefore acts in the opposite direction:

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

The force points downwards, so the electron initially curves downwards.

The important step is to reverse the right-hand-rule result because the particle is negatively charged.

Question 2

An unknown charged particle enters a uniform magnetic field of \(0.25\,\mathrm T\), directed out of the page.

Its total speed is \(4.0\times10^6\,\mathrm{m\,s^{-1}}\), and its velocity makes an angle of \(60^\circ\) with the magnetic field.

The perpendicular velocity component points east, while the parallel component points out of the page.

The particle’s projected trajectory initially curves upwards.

Given that the magnitude of its charge is \(1.60\times10^{-19}\,\mathrm C\):

a) Determine the charge sign.

b) Calculate the perpendicular velocity component.

c) Calculate the magnetic force magnitude.

d) Explain what happens to the magnetic force and projected radius if a second run increases only the parallel velocity component, leaving the perpendicular component unchanged.

Solution 2

The particle is negatively charged, and its magnetic force magnitude is approximately \(1.39\times10^{-13}\,\mathrm N\). Increasing only its parallel velocity component does not change its magnetic force or projected radius.

Part a: Determine the charge sign.

The perpendicular velocity points east, and the magnetic field points out of the page.

Using the right-hand rule, a positive charge would experience a force downwards.

The actual particle curves upwards.

Its force therefore acts opposite to the positive-charge prediction.

The particle is negatively charged.

Part b: Calculate the perpendicular velocity component.

Use:

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

Substitute:

\[ \begin{aligned} v_\perp&=(4.0\times10^6)\sin60^\circ\\ &=(4.0\times10^6)(0.866)\\ &=3.46\times10^6\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part c: Calculate the magnetic force.

Use:

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

Substitute:

\[ \begin{aligned} F_B&=(1.60\times10^{-19})(3.46\times10^6)(0.25)\\ &=1.38\times10^{-13}\,\mathrm N \end{aligned} \]

Therefore:

\[ F_B\approx1.39\times10^{-13}\,\mathrm N \]

Part d: Increase only the parallel velocity component.

The magnetic force depends on:

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

The perpendicular velocity component remains unchanged, as does the magnetic field strength and charge magnitude.

Therefore, the magnetic force remains unchanged.

The projected radius also remains unchanged because:

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

However, the particle now travels faster along the magnetic field direction.

It therefore advances a greater distance along the field during each revolution, stretching out its helical trajectory.

The tempting assumption is that increasing total speed must increase magnetic force. That is only true if the perpendicular component increases.

Question 3

An electron and a proton are each accelerated from rest through a potential difference of \(500\,\mathrm V\) in separate beamlines.

Both particles then enter the same uniform magnetic field of \(0.20\,\mathrm T\), directed into the page. Their initial velocities point to the right and are perpendicular to the field.

Given:

\[ \begin{aligned} m_e&=9.11\times10^{-31}\,\mathrm{kg}\\ m_p&=1.67\times10^{-27}\,\mathrm{kg}\\ |q|&=1.60\times10^{-19}\,\mathrm C \end{aligned} \]

Calculate:

a) The speed of each particle after acceleration.

b) The radius of each particle’s trajectory.

c) The ratio of the proton’s radius to the electron’s radius.

d) The initial direction of curvature of each particle.

Assume non-relativistic motion.

Solution 3

The proton curves upwards, the electron curves downwards, and the proton’s radius is approximately 42.8 times larger.

Part a: Calculate the speeds.

The electrical potential energy gained by a particle becomes kinetic energy.

For a particle accelerated from rest:

\[ |q|\Delta V=\frac12 mv^2 \]

Rearrange:

\[ v=\sqrt{\frac{2|q|\Delta V}{m}} \]

For the electron:

\[ \begin{aligned} v_e&=\sqrt{\frac{2(1.60\times10^{-19})(500)} {9.11\times10^{-31}}}\\ &=1.33\times10^7\,\mathrm{m\,s^{-1}} \end{aligned} \]

For the proton:

\[ \begin{aligned} v_p&=\sqrt{\frac{2(1.60\times10^{-19})(500)} {1.67\times10^{-27}}}\\ &=3.10\times10^5\,\mathrm{m\,s^{-1}} \end{aligned} \]

The electron travels considerably faster.

Although both particles gain the same kinetic energy, their masses are different.

Part b: Calculate the radii.

Use:

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

For the electron:

\[ \begin{aligned} r_e&=\frac{(9.11\times10^{-31})(1.33\times10^7)} {(1.60\times10^{-19})(0.20)}\\ &=3.77\times10^{-4}\,\mathrm m \end{aligned} \]

For the proton:

\[ \begin{aligned} r_p&=\frac{(1.67\times10^{-27})(3.10\times10^5)} {(1.60\times10^{-19})(0.20)}\\ &=1.62\times10^{-2}\,\mathrm m \end{aligned} \]

Part c: Calculate the radius ratio.

\[ \begin{aligned} \frac{r_p}{r_e} &=\frac{1.62\times10^{-2}}{3.77\times10^{-4}}\\ &\approx42.8 \end{aligned} \]

The proton’s radius is approximately 42.8 times greater.

Notice that this is not the mass ratio of approximately 1,836.

Why?

The particles have the same kinetic energy, not the same speed. The proton moves more slowly, which reduces its radius compared with the equal-speed comparison.

In fact, when both particles have the same kinetic energy and charge magnitude:

\[ r\propto\sqrt m \]

This is why the radius ratio is the square root of the mass ratio.

Part d: Determine the directions of curvature.

Both particles travel right, and the magnetic field points into the page.

A positive particle experiences an upward force.

Therefore:

• The proton curves upwards.
• The electron curves downwards.

The accelerating voltage affects their speeds and radii, but their opposite charge signs still determine their opposite initial directions of curvature.

Question 4

An unknown charged particle enters a uniform magnetic field directed into the page. Its velocity points to the right, and its initial trajectory curves upwards.

A technician repeats the experiment under three different conditions:

• Run A: The magnetic field direction is reversed, but the particle’s velocity is unchanged.
• Run B: The particle’s velocity is reversed, but the magnetic field direction is unchanged.
• Run C: Both the magnetic field and the particle’s velocity are reversed.

For each run:

a) Determine the initial direction of magnetic curvature.

b) Explain whether the particle’s charge sign changes.

Assume the same particle is used in every run and the magnetic field has the same magnitude.

Solution 4

The particle is positively charged. Its initial curvature is downwards in runs A and B, and upwards in run C. Its charge sign does not change.

First, identify the particle’s charge from the original experiment.

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

A positive charge would experience an upward magnetic force.

The particle actually curves upwards, so it must be positively charged.

Now consider the three changes.

Run A: Reverse the magnetic field only.

The velocity remains directed right, but the field now points out of the page.

Reversing the magnetic field reverses the magnetic force.

The particle therefore curves downwards.

Run B: Reverse the velocity only.

The particle now travels left, while the magnetic field still points into the page.

Reversing the velocity also reverses the direction of the cross product.

The particle again curves downwards.

Run C: Reverse both velocity and magnetic field.

This time, both vectors are reversed.

The cross product becomes:

\[ (-\vec v)\times(-\vec B) =\vec v\times\vec B \]

The two reversals cancel.

The magnetic force therefore retains its original direction, and the particle curves upwards.

The results are:

The particle remains positively charged in every experiment.

The important point is that changing the magnetic field or velocity changes the force direction, but it does not change the particle’s electric charge.

Question 5

An electron and a proton enter the same uniform magnetic field in separate runs.

The magnetic field points into the page and has a magnitude of \(0.25\,\mathrm T\).

Both particles have:

• A perpendicular velocity component of magnitude \(2.0\times10^6\,\mathrm{m\,s^{-1}}\).
• A parallel velocity component of magnitude \(1.0\times10^6\,\mathrm{m\,s^{-1}}\), initially directed into the page.
• A charge magnitude of \(1.60\times10^{-19}\,\mathrm C\).

However, their perpendicular velocity components point in opposite directions:

• The electron’s perpendicular velocity points east.
• The proton’s perpendicular velocity points west.

Answer the following:

a) Determine the initial direction of magnetic curvature for each particle. Do they curve in opposite directions?

b) Calculate the magnetic force magnitude on each particle.

c) Compare their initial acceleration magnitudes.

d) Determine the ratio of their helical radii.

e) The proton’s parallel velocity component is then reversed so that it points out of the page. Its perpendicular component remains unchanged. Explain what happens to its magnetic force, initial curvature, radius, and direction of travel along the magnetic field.

Assume both particles remain in the same uniform magnetic field, and no other forces act.

Solution 5

Both particles initially curve downwards, despite having opposite charges. Their magnetic force magnitudes are equal, but the electron has a much greater acceleration and a much smaller helical radius.

This question changes an assumption that is easy to overlook: the two particles have opposite perpendicular velocity directions as well as opposite charge signs.

Part a: Determine the initial curvature directions.

Start with the electron.

Its perpendicular velocity points east, and the magnetic field points into the page.

A positive particle would experience an upward magnetic force.

The electron is negative, so its force points downwards.

Now consider the proton.

Its perpendicular velocity points west, while the magnetic field still points into the page.

Applying the right-hand rule to the westward velocity gives a downward force for a positive particle.

The proton is positive, so its force also points downwards.

Therefore, both particles initially curve downwards.

This is the important conceptual result: opposite charges do not automatically produce opposite curvature directions.

They do so when their perpendicular velocity directions are the same. Here, both the charge sign and perpendicular velocity direction are reversed, so the two reversals cancel.

Part b: Calculate the magnetic force magnitude.

For both particles:

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

Substitute:

\[ \begin{aligned} F_B&=(1.60\times10^{-19})(2.0\times10^6)(0.25)\\ &=8.0\times10^{-14}\,\mathrm N \end{aligned} \]

Both particles experience a magnetic force magnitude of:

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

Their force magnitudes are equal because their charge magnitudes, perpendicular speeds, and magnetic field strengths are equal.

Part c: Compare their accelerations.

Use Newton’s second law:

\[ a=\frac{F}{m} \]

Since both particles experience equal force magnitudes:

\[ \frac{a_e}{a_p}=\frac{m_p}{m_e} \]

Substituting:

\[ \frac{a_e}{a_p} = \frac{1.67\times10^{-27}} {9.11\times10^{-31}} \]

\[ \frac{a_e}{a_p}\approx1833 \]

The electron’s initial acceleration magnitude is approximately 1,836 times greater than the proton’s.

The electron therefore changes its perpendicular velocity direction much more rapidly.

Part d: Compare their helical radii.

The radius of the circular component is:

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

Both particles have equal perpendicular velocity magnitudes, equal charge magnitudes, and experience the same magnetic field.

Therefore:

\[ \frac{r_p}{r_e}=\frac{m_p}{m_e} \]

\[ \frac{r_p}{r_e}\approx1836 \]

The proton’s helical radius is approximately 1,836 times larger than the electron’s.

The electron follows a much tighter circular path around the magnetic field direction.

Part e: Reverse the proton’s parallel velocity component.

The proton’s parallel velocity originally points into the page.

It is now reversed so that it points out of the page.

Does this reverse the magnetic force?

No.

The parallel velocity component does not contribute to magnetic force. Only the perpendicular component matters:

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

The proton’s perpendicular velocity is still directed west, and the magnetic field still points into the page.

Therefore:

• The magnetic force remains downwards.
• The initial transverse curvature remains downwards.
• The helical radius remains unchanged.
• The particle now advances out of the page rather than into it.

The direction of travel along the magnetic field has changed, but the circular motion has not.

07What to remember when interpreting electron and proton trajectories

When you see an electron or proton entering a magnetic field in an HSC Physics question, separate the problem into three decisions.

1. Which way does the particle curve?

Use the right-hand rule for a positive charge, then reverse the result if the particle is negative.

2. Which part of its velocity produces the curvature?

Resolve the velocity into perpendicular and parallel components.

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

Only \(v_\perp\) contributes to the magnetic force and circular motion.

3. How tightly does the particle curve?

Use:

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

The charge magnitude, mass, perpendicular velocity, and magnetic field determine the radius.

The central distinction is simple: charge sign determines which way the particle turns, while mass and perpendicular velocity determine how tightly it turns.

Once you can separate these effects, the next useful step is interpreting measured particle radii to calculate mass-to-charge ratios, or using the circular motion to calculate a particle’s period in a magnetic field.