How Magnetic Deflection Reveals the Charge Sign of Cathode Rays

Put a cathode-ray beam moving from left to right into a magnetic field directed into the page. The glowing beam bends downwards. What does that tell you about its charge?

It tells you the cathode rays are negatively charged. A positive charge would experience an upward magnetic force in this arrangement, so a negative charge must experience the opposite force.

The important evidence is the direction of the deflection, not simply the fact that the beam curves. Once you know the direction of the beam and the magnetic field, you can use the right-hand rule to determine the sign of the charge.

01The basic experiment: which way does the beam bend?

A cathode-ray tube contains two electrodes inside an evacuated glass tube. The cathode is negative, and the anode is positive. When a sufficiently large potential difference is applied, a beam of cathode rays travels away from the cathode towards the anode.

In modern terms, these cathode rays are electrons.

The beam can be made visible because it produces fluorescence when it strikes a suitable surface. This gives us a way to observe its path.

Now place a magnet or magnetic coils around the tube so that a magnetic field acts across the moving beam.

Before applying any equations, make a prediction.

If the beam contains positively charged particles, which way should it bend?

For a beam travelling rightwards with a magnetic field directed into the page, a positive charge experiences an upward force. A negative charge experiences a downward force.

Therefore, an observed downward deflection is evidence of negative charge.

Why does the direction matter?

A magnetic field exerts a force on a moving charged particle. This force acts perpendicular to both the particle’s velocity and the magnetic field.

Think of the field as steering the moving particle sideways. The particle’s path changes direction rather than simply speeding up or slowing down.

The direction of this force depends on three things:

• The direction in which the particle is travelling.
• The direction of the magnetic field.
• The sign of the particle’s charge.

The mathematical relationship is:

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

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 field, measured in tesla (T).
• The cross product determines the direction of the force.

The magnitude of the force is:

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

where \(\theta\) is the angle between the velocity and magnetic field.

Notice that the magnitude equation uses \(|q|\), the absolute value of the charge. It tells us how large the force is, but not whether the particle is positive or negative.

For the sign, we need the vector equation.

Applying the right-hand rule

The standard right-hand rule gives the magnetic force direction for a positive charge.

Use your right hand:

1. Point your index finger in the direction of the particle’s velocity.
2. Point your middle finger in the direction of the magnetic field.
3. Your thumb indicates the force on a positive charge.

For a negative charge, reverse the direction of the force.

Let’s establish a consistent sign convention:

• Right is the positive \(x\)-direction.
• Up is the positive \(y\)-direction.
• Out of the page is the positive \(z\)-direction.

A magnetic field into the page is therefore in the negative \(z\)-direction.

For a particle moving rightwards:

\[ \vec v\times\vec B = \hat x\times(-\hat z) = +\hat y \]

This means a positive charge experiences an upward force.

An electron has negative charge, so its force is downward.

The direction changes when you reverse either the beam’s velocity or the magnetic field.

This table is particularly useful when interpreting experimental diagrams. Always establish the beam direction and field direction before deciding on the charge sign.

A reliable method for determining charge sign

When given a cathode-ray deflection experiment, follow this sequence:

1. Identify the direction of the moving beam. Use the particle velocity, not conventional current.
2. Identify the magnetic field direction. Remember that crosses indicate a field into the page, while dots indicate a field out of the page.
3. Apply the right-hand rule for a positive charge.
4. Compare the predicted force direction with the observed deflection.
5. If the observed deflection is opposite to the positive-charge prediction, the particles are negatively charged.

One important limitation: a straight beam does not automatically mean the particles are neutral.

If the velocity is parallel or antiparallel to the magnetic field, then \(\sin\theta=0\), so the magnetic force is zero even for a charged particle.

The field might also be too weak to produce a detectable deflection.

02How Thomson used magnetic deflection to investigate cathode rays

The direction rule explains how we identify charge sign in a modern experiment. But how did scientists establish that cathode rays actually carried negative charge?

This was an important problem in the late nineteenth century.

Scientists disagreed about the nature of cathode rays. One possibility was that they were some kind of wave-like disturbance. Another was that they consisted of moving, negatively charged particles.

J. J. Thomson investigated this problem through a series of experiments in 1897.

The original experimental problem

Thomson needed evidence that the negative charge observed in an experiment was actually travelling with the cathode rays.

It was not enough to show that a discharge tube produced both a visible beam and negative electricity. Those two observations could potentially have unrelated causes.

The experiment needed to connect the visible path of the rays directly to the movement of negative charge.

Understanding the apparatus

Thomson developed an arrangement involving a discharge tube, a magnetic field, two metal cylinders, and an electrometer.

The apparatus can be understood by following the path of the rays.

The sequence was:

1. The cathode produced rays inside the low-pressure discharge tube.
2. The rays travelled through a narrow opening in a grounded metal plug.
3. They entered a glass bulb containing two coaxial metal cylinders.
4. The cylinders had narrow openings that allowed the rays to enter the inner cylinder when correctly aligned.
5. A magnetic field could bend the rays so that they passed through these openings.
6. The inner cylinder was connected to an electrometer, which detected the electrical charge arriving inside it.

The outer cylinder was grounded. It provided a grounded guard around the measuring cylinder, helping isolate the charge measurement.

The electrometer was particularly important because it could identify whether the collected charge was positive or negative.

What did Thomson predict?

If cathode rays were negatively charged particles, then bending the rays into the inner cylinder should deliver negative charge to it.

If the rays were diverted away from the opening, very little charge associated with the rays should reach the collector.

This gave Thomson a testable prediction.

What did he observe?

The observations supported the negatively charged particle model.

The key result was that the negative charge followed the same path as the magnetically deflected cathode rays.

This was much stronger evidence than simply observing that a beam curved in a magnetic field.

Why was this an improvement on earlier experiments?

Jean Perrin had previously used metal cylinders and an electroscope to show that cathode rays delivered negative charge.

However, an objection remained: perhaps the negative charge was produced by a separate stream of particles rather than being carried by the cathode rays themselves.

Thomson refined the arrangement so that the visible rays could only reach the inner collector when the magnetic field directed them through the aligned openings.

The close relationship between the magnetic deflection and the measured negative charge made the alternative explanation much less convincing.

What did the experiment establish, and what did it not establish?

The experiment established that negative electrical charge travelled with the cathode rays.

It supported the model that cathode rays were streams of negatively charged particles, rather than an unrelated wave-like phenomenon.

However, the experiment did not independently determine the mass or charge of each particle.

Thomson investigated those properties through further measurements involving magnetic and electric deflection. His wider work led to the determination of the charge-to-mass ratio and the identification of a particle much lighter than an atom.

There is an important historical distinction here:

• The magnetic field steered the rays and allowed Thomson to connect their path with the collected charge.
• The electrometer directly established that the collected charge was negative.
• The direction of magnetic deflection, when interpreted using the magnetic force law, also provides a way to determine charge sign.

Thomson’s separate electric-field experiments provided further confirmation. Earlier attempts to produce electric deflection had been unsuccessful because the remaining gas inside the tube became conductive under the action of the rays, reducing the electric field’s effect. More complete evacuation allowed the electric deflection to become observable.

His work is documented in his 1897 paper, Cathode Rays. The American Physical Society’s historical account also describes the experiments that established the electron.

03Worked examples: interpreting magnetic deflection

04Common mistakes when interpreting cathode-ray experiments

The most important distinction is between the direction of the force and its magnitude. Keep those two ideas separate when interpreting experimental evidence.

05Questions and solutions

Question 1

A beam of cathode rays travels horizontally from left to right through a uniform magnetic field directed out of the page.

The beam curves upwards.

Determine the sign of the charge carried by the particles. Explain what would happen if the magnetic field were reversed.

Solution 1

The particles are negatively charged, and reversing the magnetic field would make the beam curve downwards.

For a positive charge moving rightwards in a magnetic field directed out of the page, the right-hand rule predicts a downward force.

The observed beam curves upwards, so its force is opposite to the positive-charge prediction.

Therefore:

\[ q<0 \]

When the magnetic field is reversed, the magnetic force also reverses. The negatively charged particles will now experience a downward force.

The charge remains negative throughout both experiments.

Question 2

An unknown beam of charged particles travels horizontally from left to right through a uniform magnetic field.

Two experiments produce the following illustrative data.

Answer the following:

(a) Determine the sign of the particle charge.

(b) Calculate the magnitude of the charge-to-mass ratio using Experiment 1.

(c) Determine whether the second experiment is consistent with the relationship between magnetic field strength and radius of curvature.

(d) Predict the direction of deflection if the magnetic field in Experiment 1 is reversed.

Solution 2

The particles are negatively charged, their charge-to-mass ratio is approximately \(1.0\times10^{11}\ \mathrm{C\,kg^{-1}}\), and the second experiment is consistent with the magnetic deflection model.

(a) Determine the charge sign.

The beam travels rightwards, and the magnetic field points into the page.

The right-hand rule predicts an upward force for a positive charge.

The observed deflection is downwards, so the particles must be negatively charged.

(b) Calculate the charge-to-mass ratio.

For perpendicular motion through a uniform magnetic field:

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

Rearranging:

\[ \frac{|q|}{m}=\frac{v}{Br} \]

Substitute the Experiment 1 values:

\[ \frac{|q|}{m} = \frac{2.4\times10^7} {(0.60\times10^{-3})(0.40)} \]

\[ \frac{|q|}{m} = \frac{2.4\times10^7}{2.4\times10^{-4}} \]

\[ \boxed{\frac{|q|}{m}=1.0\times10^{11}\ \mathrm{C\,kg^{-1}}} \]

(c) Check the second experiment.

For the same particle speed and charge-to-mass ratio:

\[ r\propto\frac{1}{B} \]

Therefore:

\[ \frac{r_2}{r_1}=\frac{B_1}{B_2} \]

Substitute:

\[ r_2=0.40\times\frac{0.60}{0.90} \]

\[ r_2=0.267\ \mathrm{m} \]

The predicted radius is approximately \(0.27\ \mathrm{m}\), matching the measured value.

This supports the model that increasing magnetic field strength produces a tighter curved path when the other particle properties remain unchanged.

(d) Reverse the magnetic field.

Reversing the field reverses the magnetic force.

The beam will now curve upwards rather than downwards.

The radius remains unchanged because the speed and magnitude of the magnetic field are unchanged.

Question 3

A negatively charged particle enters a uniform magnetic field while travelling horizontally to the right.

The magnetic field has a magnitude of \(0.80\ \mathrm{mT}\) and makes an angle of \(30^\circ\) with the particle’s velocity. Its component perpendicular to the beam points into the page.

The particle has:

• Charge \(q=-1.6\times10^{-19}\ \mathrm{C}\).
• Speed \(v=2.5\times10^7\ \mathrm{m\,s^{-1}}\).

(a) Determine the direction of the magnetic force.

(b) Calculate the magnitude of the magnetic force.

(c) If the magnetic field were rotated to become perpendicular to the particle’s velocity, without changing its magnitude, what would happen to the force magnitude?

Explain why using the entire magnetic field strength without considering its angle would produce an incorrect result in the original arrangement.

Solution 3

The magnetic force acts downwards and has a magnitude of \(1.6\times10^{-15}\ \mathrm{N}\). Making the magnetic field perpendicular to the beam would double the force magnitude.

(a) Determine the force direction.

Only the component of the magnetic field perpendicular to the particle’s velocity contributes to the magnetic force.

This perpendicular component points into the page.

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

The particle is negatively charged, so its force acts in the opposite direction.

Therefore, the magnetic force acts downwards.

(b) Calculate the force magnitude.

Use:

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

Substitute:

\[ F_B= (1.6\times10^{-19}) (2.5\times10^7) (0.80\times10^{-3}) \sin30^\circ \]

Since \(\sin30^\circ=0.5\):

\[ F_B= (1.6\times10^{-19}) (2.5\times10^7) (8.0\times10^{-4}) (0.5) \]

\[ \boxed{F_B=1.6\times10^{-15}\ \mathrm{N}} \]

(c) Make the field perpendicular to the beam.

If the magnetic field is perpendicular to the velocity:

\[ \theta=90^\circ \]

Therefore:

\[ F_B=|q|vB \]

Substitute:

\[ F_B= (1.6\times10^{-19}) (2.5\times10^7) (8.0\times10^{-4}) \]

\[ \boxed{F_B=3.2\times10^{-15}\ \mathrm{N}} \]

The force doubles.

The original magnetic field was not completely perpendicular to the beam. Only its perpendicular component contributed to the force.

The component parallel to the velocity produces no magnetic force.

This is why the angle must be included when calculating the force magnitude.

Question 4

J. J. Thomson used a magnetic field to direct cathode rays into a metal cylinder connected to an electrometer.

The apparatus contained two coaxial cylinders. The outer cylinder was grounded, and the insulated inner cylinder was connected to the electrometer.

The following observations were recorded:

• With the magnetic field switched off, the cathode rays missed the cylinder openings and the electrometer showed only a small, irregular response.
• When the magnetic field was adjusted to direct the rays through the openings, the electrometer registered a substantial negative charge.
• When the magnetic field was increased further, the rays overshot the openings and the measured charge decreased.

Explain:

(a) Why the electrometer registered negative charge when the rays entered the inner cylinder.

(b) Why the measured charge decreased when the rays overshot the openings.

(c) How the experiment supported the negatively charged particle model.

(d) Why the experiment was stronger evidence than simply observing magnetic deflection of a glowing beam.

Solution 4

The experiment showed that negative electrical charge travelled along the same path as the magnetically deflected cathode rays.

(a) Explain the negative electrometer reading.

When the magnetic field directed the cathode rays through the aligned openings, the particles entered the inner cylinder.

If the rays consisted of negatively charged particles, they would transfer negative charge to the cylinder.

The electrometer detected this negative charge.

The observation therefore supported the idea that negative electricity was carried by the cathode rays.

(b) Explain the decreased reading when the rays overshot.

The stronger magnetic field changed the curvature of the rays.

Instead of passing through the openings, they travelled past the inner cylinder.

Consequently, fewer negatively charged particles reached the collector, so the electrometer registered a smaller charge.

This is important because it connects the charge measurement to the actual path of the visible rays.

(c) Explain how this supported the particle model.

The results were consistent with cathode rays being streams of negatively charged particles.

The magnetic field changed their trajectories, while the electrometer detected negative charge arriving along those trajectories.

The negative charge was therefore associated with the rays rather than being an unrelated electrical effect elsewhere in the tube.

(d) Explain why this was stronger evidence.

Simply observing a glowing beam bend in a magnetic field does not directly establish the sign of the charge.

Thomson’s arrangement added an independent measurement: the electrometer identified the sign of the charge reaching the collector.

Furthermore, the magnetic field controlled whether that charge could reach the collector.

The agreement between the observed ray path and the measured negative charge made the conclusion considerably stronger.

One limitation remains: the experiment established the presence of negative charge travelling with the rays, but it did not independently determine the mass and charge of each particle.

Question 5

An experimentalist investigates an unknown beam produced by a discharge tube. The beam contains three distinguishable components, labelled A, N, and C.

All three components enter the same uniform magnetic field with the same speed:

\[ v=2.0\times10^7\ \mathrm{m\,s^{-1}} \]

The magnetic field is:

\[ B=0.50\ \mathrm{mT} \]

It points into the page and is perpendicular to the initial direction of travel.

The magnetic-field region has a length of \(0.10\ \mathrm{m}\).

The following observations are made:

• Component A curves upwards with a radius of \(0.25\ \mathrm{m}\).
• Component C curves downwards with a radius of \(0.50\ \mathrm{m}\).
• Component N remains within \(0.50\ \mathrm{mm}\) of the original straight-line path.
• All three components produce equally bright traces on the fluorescent screen.

Further experiments reveal:

1. Reversing the magnetic field makes A and C exchange sides, while N remains straight.
2. Reversing the beam direction while keeping the original magnetic field makes A and C exchange sides again.
3. Making the magnetic field parallel to the beam causes all three components to travel straight.

Assume:

• All three components have the same initial speed and enter the field perpendicular to it in the first experiment.
• There are no significant electric fields or collisions.
• The magnetic field is uniform.
• The small-deflection approximation is valid for component N.
• The detector can resolve a vertical displacement of \(0.50\ \mathrm{mm}\) or greater.

Answer the following:

(a) Determine the charge signs of components A and C.

(b) Calculate the ratio of their charge-to-mass ratio magnitudes.

(c) Calculate an upper bound for the magnitude of the charge-to-mass ratio of component N.

(d) A student claims that N must be neutral because it does not deflect. Is this conclusion justified?

(e) Another student claims that the equal brightness of all three traces proves that the three components carry equal amounts of charge. Explain why this conclusion is unreliable.

(f) Explain why all three components travelling straight when the magnetic field is parallel to the beam does not establish that they are neutral. What would happen if both the beam direction and magnetic field direction were reversed simultaneously?

Solution 5

Component A is positive, component C is negative, and component N has a charge-to-mass ratio below the experiment’s detection limit. The observations do not establish that N is completely neutral, and equal brightness does not establish equal charge.

This question requires us to separate charge sign, charge-to-mass ratio, detector sensitivity, and the direction of the magnetic force.

(a) Determine the charge signs.

All three components initially travel rightwards, and the magnetic field points into the page.

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

For a negative charge, the force is downwards.

Component A curves upwards, so:

\[ \boxed{q_A>0} \]

Component C curves downwards, so:

\[ \boxed{q_C<0} \]

The opposite directions of curvature establish that A and C have opposite charge signs.

(b) Calculate the ratio of their charge-to-mass ratios.

For particles travelling perpendicular to a uniform magnetic field:

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

Rearranging:

\[ \frac{|q|}{m}=\frac{v}{Br} \]

Since both components have the same speed and experience the same magnetic field:

\[ \frac{(|q|/m)_A}{(|q|/m)_C} = \frac{r_C}{r_A} \]

Substitute:

\[ \frac{(|q|/m)_A}{(|q|/m)_C} = \frac{0.50}{0.25} \]

Therefore:

\[ \boxed{ (|q|/m)_A:(|q|/m)_C=2:1 } \]

Component A has twice the magnitude of the charge-to-mass ratio of component C.

Notice what we cannot conclude.

We cannot say that A has twice the charge of C because their masses are unknown.

A smaller radius indicates a larger charge-to-mass ratio under these conditions, not necessarily a larger charge.

(c) Calculate the upper bound for component N.

Component N has no measurable vertical displacement greater than:

\[ y_{\min}=0.50\ \mathrm{mm} \]

Convert to metres:

\[ y_{\min}=5.0\times10^{-4}\ \mathrm{m} \]

The magnetic force produces a transverse acceleration:

\[ a=\frac{|q|vB}{m} \]

Therefore:

\[ a=\frac{|q|}{m}vB \]

The time spent in the field is approximately:

\[ t=\frac{L}{v} \]

Using the constant-acceleration displacement equation:

\[ y=\frac12at^2 \]

Substituting the acceleration and time:

\[ y\approx \frac12 \left(\frac{|q|}{m}vB\right) \left(\frac{L}{v}\right)^2 \]

Simplifying:

\[ y\approx\frac{|q|}{m}\frac{BL^2}{2v} \]

Rearrange to find the maximum detectable charge-to-mass ratio:

\[ \frac{|q|}{m} < \frac{2vy_{\min}}{BL^2} \]

Substitute the values:

\[ \frac{|q_N|}{m_N} < \frac{ 2(2.0\times10^7)(5.0\times10^{-4}) }{ (5.0\times10^{-4})(0.10)^2 } \]

\[ \boxed{ \frac{|q_N|}{m_N} < 4.0\times10^9\ \mathrm{C\,kg^{-1}} } \]

This is an upper bound based on the detector’s resolution.

It does not mean the charge is exactly zero.

(d) Is N definitely neutral?

No. The experiment only establishes that N has no detectable magnetic deflection under the stated conditions.

It could be neutral, but it could also be a charged particle with a sufficiently small charge-to-mass ratio to produce a displacement below the detector’s resolution.

The calculated upper bound makes this limitation explicit.

A measurement can establish that an effect is smaller than the instrument can detect. It cannot automatically establish that the effect is exactly zero.

(e) Does equal brightness prove equal charge?

No. Fluorescent brightness is not a direct measurement of electrical charge.

The brightness depends on factors including:

• The number of particles reaching the screen.
• The kinetic energy transferred by each particle.
• The way the screen material responds to each type of particle.

Different particle populations can produce similar brightness even when their charge magnitudes and particle numbers differ.

To compare the electrical charge carried by the different components, a calibrated charge measurement would be required.

(f) Explain the parallel-field experiment and simultaneous reversal.

When the magnetic field is parallel to the beam:

\[ \theta=0^\circ \]

Therefore:

\[ F_B=|q|vB\sin0^\circ \]

\[ F_B=0 \]

This is true for every charged particle, regardless of whether its charge is positive or negative.

Consequently, all three components travelling straight in this arrangement tells us nothing new about whether they are neutral.

Now consider reversing both the beam direction and magnetic field direction simultaneously.

The magnetic force depends on:

\[ \vec v\times\vec B \]

Reversing both vectors gives:

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

The cross product is unchanged.

Therefore, the magnetic force remains in the same direction for each particle, and the curvature remains towards the same side in the laboratory reference frame.

This is a useful experimental check because it distinguishes the effect of reversing one direction from reversing both.

The deeper lesson is that magnetic deflection provides several different types of information, but they must not be confused:

• The direction of curvature reveals charge sign.
• The radius of curvature provides information about charge-to-mass ratio.
• A straight path only constrains the magnetic response under the particular experimental conditions.
• Fluorescent brightness is not a direct measurement of charge.

05Where to go next: from charge sign to particle motion

Once you can identify the sign of a moving charge, the next question is how strongly the magnetic field bends its path.

That depends on the particle’s speed, mass, charge magnitude, and magnetic field strength.

For motion perpendicular to a uniform magnetic field, the path is circular, with radius:

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

This relationship connects the direction-based experiment in this article to the quantitative study of charged-particle motion.

For further practice, continue with:

• Force on a Moving Charge in a Magnetic Field: HSC Physics – understanding the magnetic force, its direction, and its magnitude.
• Circular Motion of a Charge in a Magnetic Field: HSC Physics – using the magnetic force to calculate the radius and motion of charged particles.

Being able to distinguish charge sign from charge-to-mass ratio is the key to interpreting cathode-ray experiments correctly.