How to Find Mass-to-Charge Ratio from a Mass-Spectrometer Track

An ion enters a magnetic field and leaves a curved track. You measure the radius and know the magnetic field strength. Can you calculate its mass-to-charge ratio?

Not quite yet. You also need to know how fast the ion is moving perpendicular to the magnetic field. If the ion enters at an angle, part of its velocity is parallel to the field and does not contribute to the circular motion.

This gives us the central method:

\[ \boxed{\frac{m}{|q|}=\frac{BR}{v_\perp}} \]

If the ion’s total speed \(v\) and its entry angle \(\theta\) are known, then \(v_\perp=v\sin\theta\), giving

\[ \boxed{\frac{m}{|q|}=\frac{BR}{v\sin\theta}} \]

You can also identify whether the ion is positive or negative from the direction of its curvature, provided you know the magnetic field direction and the direction of travel.

Let’s build the method from the track itself, including what to do when the ion follows a helical path rather than a flat circle.

01What information can you get from a particle track?

A simplified magnetic-sector mass spectrometer sends ions through a sequence of stages:

1. An ion source produces charged particles.
2. An accelerating potential difference gives the ions kinetic energy.
3. The ions enter a region containing a uniform magnetic field.
4. A detector records their paths or arrival positions.

The magnetic field bends the moving ions. The resulting track provides information about their mass-to-charge ratios.

For the broader instrument arrangement and how different ions are separated, see How a Mass Spectrometer Separates Ions by Mass-to-Charge Ratio.

For this explanation, focus on one particular detail: the ion does not necessarily enter the magnetic field at right angles.

The important distinction: circular motion versus helical motion

Imagine an ion entering a uniform magnetic field at an angle of \(30^\circ\) to the field direction.

Would you expect it to travel in a flat circle?

Not necessarily.

Its velocity has two components:

• A component perpendicular to the magnetic field, which produces circular motion.
• A component parallel to the magnetic field, which remains unchanged.

The combination produces a helical path, similar to a corkscrew.

Think of the motion as a circle being carried steadily along a straight line. The circular motion determines the helix radius, while the forward motion determines how far the particle travels along the field during each revolution.

The analogy helps separate the two motions, although the actual magnetic force is what produces the circular component.

Resolving the velocity into components

Let:

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

Using basic trigonometry:

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

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

The perpendicular component is the one we need to calculate the mass-to-charge ratio.

Why?

The magnetic force on a moving charged particle is

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

Since \(v\sin\theta=v_\perp\), this becomes

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

The parallel component produces no magnetic force because it travels along the magnetic field lines.

Deriving the mass-to-charge equation

The perpendicular component produces circular motion around the magnetic field direction.

The magnetic force supplies the centripetal force required for this circular motion:

\[ F_B=F_c \]

The centripetal force is calculated using the perpendicular speed:

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

Equating the two forces:

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

Cancel one factor of \(v_\perp\):

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

Rearranging gives:

\[ \boxed{\frac{m}{|q|}=\frac{BR}{v_\perp}} \]

This is the equation you need when interpreting the radius of the circular component of a particle’s track.

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

\[ \boxed{\frac{m}{|q|}=\frac{BR}{v\sin\theta}} \]

Here, \(R\) is the radius of the circular component of the helix, measured perpendicular to the magnetic field axis. It is not the local radius of curvature of the complete three-dimensional helix.

Three important cases

The angle of entry changes the shape of the track.

Notice the boundary cases.

At \(90^\circ\), the entire velocity contributes to circular motion. The familiar equation \(R=mv/(|q|B)\) applies.

At \(0^\circ\), there is no magnetic deflection at all. You cannot determine the mass-to-charge ratio from a circular radius because no circular path exists.

02How to identify the charge sign from curvature

The radius tells you about the magnitude of the mass-to-charge ratio. It does not tell you whether the particle is positive or negative.

For that, you need the direction of the magnetic force.

The magnetic force direction is described by

\[ \boxed{\mathbf F_B=q(\mathbf v\times\mathbf B)} \]

This is the magnetic part of the Lorentz force equation.

The sign of \(q\) determines which way the force acts:

• Positive charge: use the right-hand rule.
• Negative charge: the force acts in the opposite direction.

The magnetic force always points towards the centre of the circular component of the trajectory.

A useful reference arrangement

Suppose:

• The magnetic field points into the page.
• The particle initially moves to the right.
• The velocity is perpendicular to the page in the field direction only through its parallel component; its perpendicular component points right.

What happens?

For a positive charge, the magnetic force points upwards.

For a negative charge, the magnetic force points downwards.

This gives you a useful reference table.

The directions reverse if you reverse either the magnetic field or the particle’s direction of travel.

Why clockwise and anticlockwise are not enough

A common mistake is to look at a track and say:

“The particle curves clockwise, so it must be negative.”

That conclusion is incomplete.

Clockwise and anticlockwise describe the direction of rotation, but the charge sign also depends on the direction of travel and the magnetic field direction.

You must know all three:

1. The direction of the magnetic field.
2. The direction in which the particle travels.
3. The direction of the track’s curvature.

If a photograph shows a curved track but gives no indication of which way the particle travelled, you cannot reliably identify its charge sign from the curvature alone.

03Choosing the correct equation for mass-to-charge ratio

There are two common ways an HSC Physics question supplies the information needed to calculate the ratio.

Method A: The particle’s speed is known

If the question gives the particle’s total speed, magnetic field strength, entry angle, and track radius, use

\[ \boxed{\frac{m}{|q|}=\frac{BR}{v\sin\theta}} \]

Remember to calculate the perpendicular velocity first.

Method B: The accelerating voltage is known

Sometimes the question does not give the particle’s speed. Instead, it tells you that the ion starts nearly from rest and passes through an accelerating potential difference.

In this case, use conservation of energy.

The electrical energy transferred to the ion becomes kinetic energy:

\[ |q|V_a=\frac12 mv^2 \]

where \(V_a\) is the accelerating potential difference, measured in volts.

Rearranging:

\[ v^2=\frac{2|q|V_a}{m} \]

We already know that

\[ R=\frac{mv\sin\theta}{|q|B} \]

Squaring this expression and substituting the energy equation gives

\[ R^2=\frac{2mV_a\sin^2\theta}{|q|B^2} \]

Therefore,

\[ \boxed{\frac{m}{|q|} =\frac{B^2R^2}{2V_a\sin^2\theta}} \]

This is particularly useful when the mass spectrometer provides an accelerating voltage rather than a measured particle speed.

Which equation should you use?

The final row is important. Finding the magnitude of the ratio and identifying the charge sign are two separate jobs.

We normally calculate the positive magnitude \(m/|q|\), then report the charge sign separately. If a question explicitly requests the signed ratio \(m/q\), attach the appropriate positive or negative sign.

The SI unit is

\[ \mathrm{kg\,C^{-1}} \]

04What if the track is helical and the entry angle is unknown?

There is another useful piece of information that can be extracted from a helical track: its pitch.

The pitch, \(p\), is the distance the particle travels parallel to the magnetic field during one complete revolution.

This provides a way to determine the entry angle even when it is not supplied directly.

The time taken for one complete revolution is the period:

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

Using the magnetic circular-motion equation, this can also be written as

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

The particle travels at a constant parallel speed during this time, so

\[ p=v_\parallel T \]

Combining these expressions gives

\[ \frac{p}{2\pi R} = \frac{v_\parallel}{v_\perp} \]

Since \(v_\parallel=v\cos\theta\) and \(v_\perp=v\sin\theta\),

\[ \boxed{\frac{p}{2\pi R}=\cot\theta} \]

This means you can determine the angle from the geometry of the track.

For example, if the pitch equals \(2\pi R\), then

\[ \cot\theta=1 \]

and therefore

\[ \theta=45^\circ \]

The pitch and radius together tell you the direction of the particle’s motion relative to the magnetic field.

This becomes particularly useful when a question gives you a track image or measurements but does not directly provide the entry angle.

05Important limitations when interpreting a track

Before using these equations in an exam, check the assumptions behind them.

The magnetic field is uniform.

The circular-motion model assumes a uniform magnetic field in the region where the particle travels.

The particle’s initial kinetic energy is negligible when using the voltage equation.

The equation

\[ \frac12 mv^2=|q|V_a \]

assumes the ion starts nearly from rest.

If it already has significant kinetic energy \(K_0\), the correct energy equation becomes

\[ \frac12 mv^2=K_0+|q|V_a \]

You cannot ignore the initial energy without justification.

The angle is measured relative to the magnetic field.

If a question gives the angle between the velocity and the plane perpendicular to the field, call that angle \(\alpha\). In that case,

\[ v_\perp=v\cos\alpha \]

Do not automatically use sine without checking the angle’s reference direction.

The track radius must be interpreted correctly.

For an oblique entry, the particle follows a helix. The radius \(R\) in our equations is the radius of the circular motion around the magnetic field axis, not the local radius of curvature of the full helical line.

The track alone does not always determine the charge sign.

You need the direction of travel and the magnetic field direction. A curved line without those details is insufficient.

These limitations explain why a track is a measurement that needs a physical model, rather than a direct reading of mass or charge.

06Questions and solutions

The following questions progress from a standard radius calculation to more demanding problems involving velocity components, charge direction, and helical geometry.

Question 1

A positively or negatively charged ion enters a uniform magnetic field of strength \(0.25\,\mathrm{T}\).

Its speed is \(4.0\times10^5\,\mathrm{m\,s^{-1}}\), and its velocity makes an angle of \(30^\circ\) with the magnetic field.

The radius of its helical track is \(2.0\,\mathrm{cm}\).

In an end-on view, the magnetic field points into the page, the initial perpendicular velocity points to the right, and the track curves upwards.

Calculate:

(a) The perpendicular velocity component.

(b) The magnitude of the mass-to-charge ratio.

(c) The sign of the ion’s charge.

Solution 1

The ion has a mass-to-charge ratio magnitude of \(2.5\times10^{-8}\,\mathrm{kg\,C^{-1}}\) and is positively charged.

(a) Calculate the perpendicular velocity.

Use

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

Substitute:

\[ v_\perp=(4.0\times10^5)\sin30^\circ \]

\[ \boxed{v_\perp=2.0\times10^5\,\mathrm{m\,s^{-1}}} \]

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

Use

\[ \frac{m}{|q|}=\frac{BR}{v_\perp} \]

Convert the radius:

\[ R=0.020\,\mathrm{m} \]

Substitute:

\[ \frac{m}{|q|} = \frac{(0.25)(0.020)} {2.0\times10^5} \]

\[ \boxed{\frac{m}{|q|} =2.5\times10^{-8}\,\mathrm{kg\,C^{-1}}} \]

(c) Determine the charge sign.

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

A positive charge experiences an upward magnetic force under these conditions.

Since the track curves upwards, the ion must be positively charged.

The signed ratio is therefore

\[ \boxed{\frac{m}{q} =+2.5\times10^{-8}\,\mathrm{kg\,C^{-1}}} \]

Question 2

Two different ions, A and B, enter the same magnetic analyser.

Both ions:

• Start nearly from rest.
• Are accelerated through a potential difference of \(900\,\mathrm{V}\).
• Enter the same uniform magnetic field of \(0.30\,\mathrm{T}\).

Their measured track data are:

Calculate:

(a) The mass-to-charge ratio magnitude of each ion.

(b) The ratio \((m/|q|)_B/(m/|q|)_A\).

(c) Whether you can conclude that Ion B has a smaller mass than Ion A.

Solution 2

Ion A has a mass-to-charge ratio of \(8.0\times10^{-8}\,\mathrm{kg\,C^{-1}}\), while Ion B has a ratio of \(6.0\times10^{-8}\,\mathrm{kg\,C^{-1}}\). Ion B’s ratio is 75% of Ion A’s, but this does not establish that Ion B has a smaller mass.

(a) Calculate each ratio.

Since both ions start from rest and experience the same accelerating voltage, use

\[ \frac{m}{|q|} = \frac{B^2R^2}{2V_a\sin^2\theta} \]

For Ion A:

\[ \frac{m_A}{|q_A|} = \frac{(0.30)^2(0.020)^2} {2(900)\sin^2 30^\circ} \]

\[ = \frac{3.6\times10^{-5}}{450} \]

\[ \boxed{\frac{m_A}{|q_A|} =8.0\times10^{-8}\,\mathrm{kg\,C^{-1}}} \]

For Ion B:

\[ \frac{m_B}{|q_B|} = \frac{(0.30)^2(0.030)^2} {2(900)\sin^2 60^\circ} \]

\[ = \frac{8.1\times10^{-5}}{1350} \]

\[ \boxed{\frac{m_B}{|q_B|} =6.0\times10^{-8}\,\mathrm{kg\,C^{-1}}} \]

(b) Compare the ratios.

Divide the two results:

\[ \frac{(m/|q|)_B}{(m/|q|)_A} = \frac{6.0\times10^{-8}} {8.0\times10^{-8}} \]

\[ \boxed{\frac{(m/|q|)_B}{(m/|q|)_A}=0.75} \]

Ion B has a mass-to-charge ratio magnitude 25% smaller than Ion A’s.

(c) Can we compare their masses?

No. We know that

\[ \frac{m_B}{|q_B|} < \frac{m_A}{|q_A|} \]

But we do not know whether the ions have equal charge magnitudes.

For example, Ion B could have a smaller mass but also a smaller charge magnitude. Alternatively, it could have a larger mass and a sufficiently larger charge magnitude.

The track measurements establish the mass-to-charge ratios, not the individual masses.

The tempting mistake is to compare the radii alone. Ion B has the larger radius, but it also enters at a larger angle, giving it a greater perpendicular velocity for a given total speed. Both effects must be included.

Question 3

A detector records a curved particle track in a uniform magnetic field directed into the page.

The track bends upwards.

However, the detector image does not show an arrow indicating the direction in which the particle travelled.

A student concludes that the particle must be positively charged.

(a) Is this conclusion justified? Explain.

(b) A second measurement establishes that the particle was initially travelling to the right. Determine its charge sign.

(c) The same particle is then sent through a magnetic field of equal strength but opposite direction. Its initial velocity and entry angle remain unchanged. Predict what happens to:

• The direction of curvature.
• The radius of the circular component of the track.

Solution 3

The charge sign cannot be determined from the original image alone. With the additional information that the particle initially travels to the right, it must be positively charged. Reversing the magnetic field reverses the curvature but leaves the radius unchanged.

(a) The original image is insufficient.

The magnetic force depends on the direction of travel as well as the magnetic field:

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

If the particle travels to the right while the field points into the page, an upward force indicates positive charge.

However, if it travels to the left, a negative charge can also produce an upward force.

The curved line alone does not distinguish these possibilities.

This is why a direction-of-travel arrow is essential when identifying charge sign from curvature.

(b) Determine the charge sign.

Now we know:

• Magnetic field: into the page.
• Initial velocity: to the right.
• Curvature: upwards.

Using the right-hand rule, a positive charge experiences an upward force.

Therefore,

\[ \boxed{q>0} \]

The particle is positively charged.

(c) Reverse the magnetic field.

The magnetic force is

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

Reversing the magnetic field reverses the force direction because the cross product changes direction.

The particle will therefore curve downwards instead of upwards.

However, the radius is

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

The magnitude of the magnetic field, particle mass, charge magnitude, and perpendicular speed are unchanged.

Therefore,

\[ \boxed{R_{\text{new}}=R_{\text{original}}} \]

This is an important distinction: reversing the field changes the direction of bending, not the amount of bending.

Question 4

A charged particle travels through a uniform magnetic field of strength \(0.30\,\mathrm{T}\).

Its total speed is \(2.0\times10^5\,\mathrm{m\,s^{-1}}\).

The particle follows a helical path with:

• Circular radius \(R=2.0\,\mathrm{cm}\).
• Pitch \(p=0.126\,\mathrm{m}\).

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

Calculate:

(a) The angle between the particle’s velocity and the magnetic field.

(b) The perpendicular velocity component.

(c) The magnitude of the mass-to-charge ratio.

Solution 4

The entry angle is approximately \(45^\circ\), and the mass-to-charge ratio magnitude is approximately \(4.2\times10^{-8}\,\mathrm{kg\,C^{-1}}\).

(a) Calculate the entry angle.

For a helical path,

\[ \frac{p}{2\pi R}=\cot\theta \]

Rearrange:

\[ \theta=\tan^{-1}\left(\frac{2\pi R}{p}\right) \]

Convert the radius:

\[ R=0.020\,\mathrm{m} \]

Substitute:

\[ \theta= \tan^{-1} \left( \frac{2\pi(0.020)}{0.126} \right) \]

\[ \boxed{\theta\approx45.0^\circ} \]

The pitch is approximately one circumference of the circular component. This tells us that the parallel and perpendicular velocity components are approximately equal.

(b) Calculate the perpendicular velocity.

Use

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

Substitute:

\[ v_\perp=(2.0\times10^5)\sin45^\circ \]

\[ \boxed{v_\perp=1.41\times10^5\,\mathrm{m\,s^{-1}}} \]

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

Use

\[ \frac{m}{|q|}=\frac{BR}{v_\perp} \]

Substitute:

\[ \frac{m}{|q|} = \frac{(0.30)(0.020)} {1.41\times10^5} \]

Therefore,

\[ \boxed{\frac{m}{|q|} \approx4.2\times10^{-8}\,\mathrm{kg\,C^{-1}}} \]

The key was extracting the entry angle from the helix geometry before calculating the perpendicular velocity.

Using the total speed directly would have overestimated the perpendicular component and underestimated the mass-to-charge ratio.

Question 5

An ion starts nearly from rest and is accelerated through a potential difference of \(1.0\,\mathrm{kV}\).

It then enters a uniform magnetic field of strength \(0.50\,\mathrm{T}\) at an unknown angle.

The ion produces a helical track with:

• Circular radius \(R=2.0\,\mathrm{cm}\).
• Pitch \(p=0.126\,\mathrm{m}\).

In the end-on view, the magnetic field points into the page, the initial perpendicular velocity points to the right, and the track initially curves upwards.

Determine:

(a) The entry angle between the velocity and the magnetic field.

(b) The magnitude of the mass-to-charge ratio.

(c) The charge sign.

(d) A student ignores the pitch and assumes the particle’s entire speed is perpendicular to the field. Calculate the ratio they would obtain and explain why it is incorrect.

Solution 5

The ion enters at approximately \(45^\circ\), has a mass-to-charge ratio magnitude of \(1.0\times10^{-7}\,\mathrm{kg\,C^{-1}}\), and is positively charged. Ignoring the pitch produces a result of \(5.0\times10^{-8}\,\mathrm{kg\,C^{-1}}\), which is half the correct value.

(a) Determine the entry angle.

We know that

\[ \frac{p}{2\pi R}=\cot\theta \]

Substitute the measured track dimensions:

\[ \cot\theta= \frac{0.126}{2\pi(0.020)} \]

\[ \cot\theta\approx1.00 \]

Therefore,

\[ \boxed{\theta\approx45^\circ} \]

This means that approximately half the ion’s kinetic energy is associated with its perpendicular motion, while the other half is associated with its parallel motion.

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

Since the ion starts nearly from rest and is accelerated through a known voltage, use

\[ \frac{m}{|q|} = \frac{B^2R^2}{2V_a\sin^2\theta} \]

Substitute:

\[ \frac{m}{|q|} = \frac{(0.50)^2(0.020)^2} {2(1.0\times10^3)\sin^245^\circ} \]

Since

\[ \sin^245^\circ=0.50 \]

we obtain

\[ \frac{m}{|q|} = \frac{1.0\times10^{-4}}{1.0\times10^3} \]

Therefore,

\[ \boxed{\frac{m}{|q|} =1.0\times10^{-7}\,\mathrm{kg\,C^{-1}}} \]

(c) Determine the charge sign.

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

A positive charge experiences an upward magnetic force.

The observed curvature is upwards, so

\[ \boxed{q>0} \]

The ion is positively charged.

(d) Investigate the student’s assumption.

The student assumes that

\[ v_\perp=v \]

This would only be true if the velocity were perpendicular to the magnetic field.

Using the perpendicular-entry formula instead gives

\[ \frac{m}{|q|} = \frac{B^2R^2}{2V_a} \]

Substitute:

\[ \frac{m}{|q|} = \frac{(0.50)^2(0.020)^2} {2(1.0\times10^3)} \]

\[ \boxed{\frac{m}{|q|} =5.0\times10^{-8}\,\mathrm{kg\,C^{-1}}} \]

This is half the correct result.

Why does this happen?

The accelerating voltage supplies kinetic energy to the particle’s entire velocity, not just its perpendicular component.

At an entry angle of \(45^\circ\),

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

The perpendicular component is smaller than the total speed. The remaining component is parallel to the field and contributes to the pitch rather than the circular radius.

The student has effectively assumed that all the kinetic energy contributes to circular motion. That assumption is inconsistent with the measured helical track.

There is another way to solve this problem that makes the role of the pitch particularly clear.

From

\[ \frac{p}{2\pi R}=\cot\theta \]

we obtain

\[ \frac{1}{\sin^2\theta} = 1+\left(\frac{p}{2\pi R}\right)^2 \]

Substitute this into the accelerating-voltage equation:

\[ \boxed{ \frac{m}{|q|} = \frac{B^2}{2V_a} \left[ R^2+\left(\frac{p}{2\pi}\right)^2 \right] } \]

This expression allows you to calculate the mass-to-charge ratio directly from the helix radius, pitch, magnetic field strength, and accelerating voltage.

It also reveals something important: both the circular motion and the motion along the field contribute to the particle’s total kinetic energy.

07Where to go next

You now have three ways to extract information from a particle track:

• The circular radius tells you about the perpendicular velocity and mass-to-charge ratio.
• The direction of curvature identifies the charge sign when the field and travel directions are known.
• The pitch reveals the parallel velocity component and can help determine an otherwise unknown entry angle.

A useful next step is understanding the cyclotron period of a charged particle. The period determines how long the particle takes to complete one revolution, which is exactly the relationship needed to connect helix radius, pitch, and parallel motion.

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