Why Heavier Ions Have Larger Magnetic Orbits at Equal Speeds

Two neon ions enter the same magnetic field at exactly the same speed. Both carry one positive elementary charge, but one is neon-20 and the other is the heavier neon-22. Which ion follows the wider curve?

Neon-22 does. Although the magnetic field exerts the same sideways force on both ions, the heavier ion has more inertia. Its direction changes more slowly, so it follows a circular path with a larger radius.

That result depends on three conditions: the ions must have the same charge magnitude, experience the same magnetic field, and have the same velocity component perpendicular to that field.

01Why a heavier ion follows a larger magnetic orbit

Imagine steering two shopping trolleys around equally tight corners at the same speed. A heavily loaded trolley needs more inward force to make the turn.

An ion in a magnetic field faces a similar problem. Circular motion requires an inward force that continually changes the particle’s direction.

The analogy has a limit: a trolley experiences friction and forces from the person steering it. A magnetic force acts continuously perpendicular to an ion’s velocity, so it changes the direction without increasing or decreasing the ion’s speed.

For an ion moving perpendicular to a uniform magnetic field, the magnetic force has magnitude

\[ F_B=|q|vB \]

where:

• \(F_B\) is the magnetic force in newtons (N).
• \(|q|\) is the magnitude of the ion’s charge in coulombs (C).
• \(v\) is its speed in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(B\) is the magnetic flux density in tesla (T).

If two ions have the same charge magnitude and speed, the magnetic force on each is equal.

But equal force does not mean equal acceleration.

Newton’s second law tells us that acceleration is force divided by mass. The heavier ion therefore has a smaller acceleration.

For a circular path, the required centripetal force is

\[ F_c=\frac{mv^2}{r} \]

Here \(m\) is mass in kilograms (kg), and \(r\) is the radius of the circular trajectory in metres (m).

The magnetic force supplies this centripetal force:

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

Cancelling one factor of \(v\) and rearranging gives

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

This equation explains the entire comparison.

At the same speed, charge magnitude, and magnetic field strength, the orbital radius is directly proportional to mass:

\[ r\propto m \]

If one ion has twice the mass, it follows a circle with twice the radius.

The radius here describes the ion’s trajectory, not the physical size of the ion itself. An ion can be incredibly small while following a magnetic orbit several centimetres across.

The direction of the curve reveals the charge sign

The radius equation uses \(|q|\), the magnitude of charge. It does not tell us whether the ion is positive or negative.

Charge sign determines which way the path bends.

Suppose an ion initially travels to the right, and the magnetic field points into the page.

Predict the initial deflection.

A positive ion bends upwards. A negative ion bends downwards.

For a positive charge, the right-hand rule gives the direction of the magnetic force from the velocity and magnetic field. For a negative charge, reverse that direction.

In vector form,

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

The arrow above each symbol indicates a vector, which has both magnitude and direction.

If two ions have equal masses, equal speeds, and equal charge magnitudes but opposite signs, their orbital radii are equal. Their paths simply curve in opposite directions.

You must know both the magnetic field direction and the particle’s direction of travel before inferring charge sign from a track.

What if the ion enters at an angle?

So far, we’ve assumed the entire velocity is perpendicular to the magnetic field.

That is not always true.

If an ion enters at an angle \(\theta\) to the magnetic field, split its velocity into:

• \(v_\perp\): the component perpendicular to the magnetic field.
• \(v_\parallel\): the component parallel to the magnetic field.

The perpendicular component is

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

Only this component produces magnetic curvature.

The more general radius equation is therefore

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

If both components are present, the particle travels in a helix: circular motion around the magnetic field combined with steady movement along it.

If \(v_\perp=0\), the ion travels straight along the field, with no magnetic orbit at all.

This creates an important qualification to our original answer. A heavier ion does not necessarily have a larger radius just because its total speed is the same. The perpendicular velocity components must also be equal.

For the more general explanation of why the magnetic force changes direction without changing speed, see Why Magnetic Fields Change a Charged Particle’s Direction, Not Speed.

02The experimental problem: how do you guarantee equal speeds?

Our calculation assumes the ions entered at the same speed.

But how would an experimenter make that happen?

Imagine building a simple apparatus to compare ion masses.

An ion source produces positive ions. An electric field between accelerating plates gives the ions kinetic energy. They then pass through a narrow slit, enter a region containing a uniform magnetic field, and curve towards a detector.

In spatial order, the beam follows this route:

1. Ion source.
2. Accelerating plates separated by a known potential difference.
3. Collimating slit, which narrows the beam.
4. Magnetic analysing region, where the trajectories curve.
5. Detector, which records the arrival positions.

The apparatus is operated in a vacuum to reduce collisions with gas particles.

Here is the tempting assumption: if the same accelerating voltage acts on both ions, surely they leave with the same speed.

They don’t.

Equal voltage means equal energy gain, not equal speed

If an ion starts from rest and is accelerated through a potential difference of magnitude \(U\), the electrical energy transferred to its kinetic energy is

\[ |q|U=\frac12mv^2 \]

Rearranging,

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

For ions with the same charge magnitude, the heavier ion leaves the accelerating region more slowly.

The reason is that the same kinetic energy is distributed over more mass.

Substitute this speed into the magnetic radius equation:

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

The result is

\[ \boxed{ r=\frac{1}{B}\sqrt{\frac{2mU}{|q|}} } \]

This is the important difference between the two experimental conditions.

In both cases, the heavier singly charged ion follows a wider orbit.

But the size of the difference is not the same.

A detector showing a larger radius for the heavier ion would support the general magnetic-motion model. It would not, by itself, establish a linear relationship between mass and radius unless the speed had been controlled.

03The revised experiment: select the speed before comparing masses

To measure mass differences using the direct relationship \(r\propto m\), we need a more controlled beam.

A crossed-field velocity selector provides a way to do that.

This is the principle used in a Bainbridge-type mass spectrometer. The simple accelerating-only apparatus described above is a teaching comparison, not a claim about the historical sequence in which Bainbridge developed his instrument.

How the velocity selector works

Picture a positive ion moving horizontally to the right between two parallel plates.

The upper plate is positive and the lower plate is negative, so the electric field points downwards.

A magnetic field also acts in the selector, pointing into the page.

For the positive ion:

• The electric force acts downwards.
• The magnetic force acts upwards.
• The two forces cancel at one selected speed.

The ion then travels straight through the exit slit.

The force magnitudes balance when

\[ |q|E=|q|vB_s \]

Here \(E\) is the electric field strength in newtons per coulomb (\(\mathrm{N\,C^{-1}}\)), and \(B_s\) is the magnetic field strength inside the selector.

Cancelling \(|q|\),

\[ \boxed{v=\frac{E}{B_s}} \]

This means a narrow, correctly aligned beam leaving the selector has a speed determined by the electric and magnetic fields, rather than by the ion’s mass.

An ion moving too quickly or too slowly experiences an unbalanced sideways force and is deflected away from the exit slit.

There is one subtle limitation.

If two different-mass ions both start from rest and accelerate through one fixed voltage, they generally leave with different speeds. Simply inserting a very narrow speed selector afterwards will not make both pass.

To compare both species at the selected speed, the source must provide ions of each species at that speed. This can be achieved through an appropriate spread of initial energies or by adjusting the accelerating voltage between measurements.

The selector transmits matching ions; it does not accelerate all ions to the same speed.

Once the selected ions enter the analysing field, the electric field is removed. The magnetic force alone curves their paths.

For equal charge magnitudes and perpendicular speeds,

\[ \frac{r_2}{r_1}=\frac{m_2}{m_1} \]

This gives a much cleaner comparison of masses.

What would the detector show?

Using the approximate neon masses from our calculations, the expected radius ratios are:

The difference matters because a mass spectrometer does not measure mass by watching an ion struggle to turn. It measures a track position or curvature and infers the mass-to-charge relationship from the known conditions.

For this ideal model, we assume uniform magnetic fields, narrow beams, negligible collisions, non-relativistic speeds, and no significant energy losses.

Real instruments have finite slit widths, velocity spreads, and imperfect field boundaries. Those effects can broaden the detected signals and limit how precisely two nearby masses can be distinguished.

04Three checks before comparing magnetic radii

When you see two curved ion tracks in an HSC Physics question, don’t immediately conclude that the wider one belongs to the heavier ion.

First establish what is actually being held constant.

A second check is whether you’re comparing the radius or the full diameter. If an ion completes a semicircle and returns to a detector along the line through its entry point, its displacement along that detector is \(2r\).

Finally, remember that a radius measurement alone usually determines \(m/|q|\), not mass by itself. Charge sign can be inferred from curvature if the field direction and direction of travel are known, but that does not establish charge magnitude.

05Questions and solutions

Question 1

A singly charged positive ion has a mass of \(3.20\times10^{-26}\ \mathrm{kg}\).

It enters a uniform magnetic field of strength \(0.400\ \mathrm{T}\), perpendicular to the field, at a speed of \(1.00\times10^5\ \mathrm{m\,s^{-1}}\).

The ion initially moves to the right. The magnetic field points into the page.

Calculate its orbital radius and determine its initial direction of deflection.

Use \(e=1.60\times10^{-19}\ \mathrm{C}\).

Solution 1

The orbital radius is \(0.0500\ \mathrm{m}\), or \(5.00\ \mathrm{cm}\), and the ion initially bends upwards.

For perpendicular entry,

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

Substitute the given values:

\[ \begin{aligned} mv&=3.20\times10^{-21}\\ |q|B&=6.40\times10^{-20}\\ r&=\frac{3.20\times10^{-21}} {6.40\times10^{-20}}\ \mathrm{m}\\ &=0.0500\ \mathrm{m} \end{aligned} \]

For the direction, apply the right-hand rule to a positive charge moving right through a magnetic field directed into the page.

The magnetic force acts upwards, so the centre of the circular path lies above the entry point.

A common mistake is to use the right-hand rule to find the curvature correctly, but then confuse the radius with the diameter. The full circular diameter would be \(0.100\ \mathrm{m}\), not the requested radius.

Question 2

Two positive ions, A and B, start from rest and accelerate through the same potential difference \(U\).

Both carry charge \(+e\), but ion B has four times the mass of ion A.

They enter the same uniform magnetic field perpendicularly. Ion A follows a circular path with radius \(3.0\ \mathrm{cm}\).

A student proposes adding a narrow velocity selector tuned to A’s speed. They claim this will automatically allow both ions through and give ion B a radius of \(12\ \mathrm{cm}\).

Determine:

(a) Ion B’s radius before the velocity selector is added.

(b) Whether B can pass a narrow selector tuned to A’s speed while both still accelerate from rest through the same voltage \(U\).

(c) The accelerating voltage required for B, in a separate run, to reach A’s original speed. Calculate B’s new magnetic radius.

Assume the magnetic field remains unchanged.

Solution 2

Ion B initially has a radius of \(6.0\ \mathrm{cm}\), not \(12\ \mathrm{cm}\). It cannot pass the same narrow speed selector as A under the original fixed-voltage conditions. B requires an accelerating voltage of \(4U\) to reach A’s speed, after which its radius becomes \(12\ \mathrm{cm}\).

Part (a): Initial radius

For ions starting from rest and accelerated through the same potential difference,

\[ |q|U=\frac12mv^2 \]

The speed ratio is

\[ \begin{aligned} \frac{v_B}{v_A} &=\sqrt{\frac{m_A}{m_B}}\\ &=\sqrt{\frac14}\\ &=\frac12 \end{aligned} \]

B is four times as massive but travels at half A’s speed.

Using \(r=mv/(|q|B)\),

\[ \begin{aligned} \frac{r_B}{r_A} &=\frac{m_Bv_B}{m_Av_A}\\ &=4\left(\frac12\right)\\ &=2 \end{aligned} \]

Therefore,

\[ r_B=2(3.0)=6.0\ \mathrm{cm} \]

The tempting answer of \(12\ \mathrm{cm}\) incorrectly assumes equal speeds.

Part (b): Passing the selector

B cannot pass a narrow selector tuned to A’s speed, assuming negligible velocity spread.

Its speed is only \(v_A/2\).

The electric and magnetic forces in the selector therefore do not balance for B. It is deflected away from the exit slit.

A selector rejects particles with the wrong speed. It does not bring them to the required speed.

Part (c): Adjusting the accelerating voltage

To make B travel at \(v_A\), its kinetic energy must be

\[ \frac12(4m_A)v_A^2 \]

This is four times A’s original kinetic energy.

Since both have charge \(+e\),

\[ eU_B=4eU \]

Thus,

\[ U_B=4U \]

With equal speeds and charges in the same magnetic field,

\[ \begin{aligned} r_B&=\frac{m_B}{m_A}r_A\\ &=4(3.0)\\ &=12\ \mathrm{cm} \end{aligned} \]

The student’s predicted radius becomes correct only after the experimental conditions are changed to produce equal speeds.

Question 3

Two ions enter a uniform magnetic field pointing into the page.

Both have the same total speed \(v\) and the same charge magnitude \(e\).

Ion C has mass \(m\) and charge \(+e\). It enters perpendicular to the field, with its velocity directed right.

Ion D has mass \(2m\) and charge \(-e\). Its velocity makes an angle of \(30^\circ\) with the magnetic field, and its perpendicular velocity component initially points right.

(a) Compare their magnetic orbital radii.

(b) Determine their initial curvature directions when viewed along the magnetic field.

(c) Describe their actual three-dimensional trajectories.

(d) Explain why a measurement of orbital radius alone could mislead someone comparing their masses.

Solution 3

The ions have equal projected orbital radii, despite D having twice C’s mass. C bends upwards in a circular path, while D bends downwards as it follows a helix.

Part (a): Compare the radii

For ion C,

\[ v_{\perp,C}=v \]

Therefore,

\[ r_C=\frac{mv}{eB} \]

For ion D,

\[ \begin{aligned} v_{\perp,D}&=v\sin30^\circ\\ &=\frac12v \end{aligned} \]

Its radius is

\[ \begin{aligned} r_D&=\frac{(2m)(v/2)}{eB}\\ &=\frac{mv}{eB} \end{aligned} \]

Thus,

\[ \boxed{r_D=r_C} \]

The factor of two in mass is exactly cancelled by the factor of one-half in perpendicular speed.

Part (b): Curvature direction

Both perpendicular velocity components initially point right, and the magnetic field points into the page.

A positive charge experiences an upward magnetic force, so C bends upwards.

D has negative charge, so its magnetic force acts downwards.

The signs change the bending directions, not the magnitudes of the radii.

Part (c): Three-dimensional trajectories

C has no parallel velocity component, so it follows a circle.

D has a non-zero component parallel to the field. That component carries it into the page while its perpendicular component undergoes circular motion.

Its actual path is therefore a helix, not a flat circle.

Part (d): Why radius alone is insufficient

An observer might see equal radii and conclude that the ions have equal masses.

That would ignore the different entry angles.

The governing equation is

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

Mass is only one factor. Different perpendicular velocities can compensate for different masses.

Even with equal total speeds and equal charge magnitudes, equal orbital radii do not establish equal masses.

Question 4

A mass-spectrometer experiment uses a velocity selector to give ions the same perpendicular speed of \(2.00\times10^5\ \mathrm{m\,s^{-1}}\).

They enter a uniform analysing magnetic field moving right, with no velocity component parallel to the field.

The analysing field points into the page.

A reference ion R has:

• Mass \(m_R=4.00\times10^{-26}\ \mathrm{kg}\)
• Charge \(q_R=+e\)

An unknown ion X has mass \(3m_R\) and charge \(-ne\), where \(n\) is either 1, 2, or 3.

Each ion completes a semicircular path and reaches a detector on the vertical line through its entry point.

The measured positions are:

• Ion R: \(10.0\ \mathrm{cm}\) above the entry point.
• Ion X: \(15.0\ \mathrm{cm}\) below the entry point.

(a) Calculate the analysing magnetic field strength.

(b) Determine \(n\) and the charge of ion X.

(c) Predict both detector positions if the analysing magnetic field is reversed but its magnitude, the selected speed, and the entry directions remain unchanged.

Use \(e=1.60\times10^{-19}\ \mathrm{C}\).

Solution 4

The analysing field is \(1.00\ \mathrm{T}\), ion X has charge \(-2e=-3.20\times10^{-19}\ \mathrm{C}\), and reversing the analysing field moves R to \(10.0\ \mathrm{cm}\) below entry and X to \(15.0\ \mathrm{cm}\) above entry.

Part (a): Determine the magnetic field

The detector displacement after a semicircle is the diameter, not the radius.

For R,

\[ \begin{aligned} r_R&=\frac{0.100}{2}\\ &=0.0500\ \mathrm{m} \end{aligned} \]

Use the radius formula, rearranged for \(B\):

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

For reference ion R,

\[ \begin{aligned} m_Rv&=8.00\times10^{-21}\\ er_R&=8.00\times10^{-21} \end{aligned} \]

The products use the given values in SI units.

Therefore,

\[ \begin{aligned} B&=\frac{8.00\times10^{-21}} {8.00\times10^{-21}}\ \mathrm{T}\\ &=1.00\ \mathrm{T} \end{aligned} \]

Part (b): Identify X’s charge

Its semicircular detector displacement is \(15.0\ \mathrm{cm}\), so

\[ r_X=0.0750\ \mathrm{m} \]

The velocity and magnetic field are the same for both ions.

Therefore,

\[ \frac{r_X}{r_R} =\frac{m_X/|q_X|}{m_R/|q_R|} \]

Substituting \(m_X=3m_R\) and \(|q_X|=ne\),

\[ \begin{aligned} \frac{0.0750}{0.0500}&=\frac{3}{n}\\ 1.50&=\frac{3}{n}\\ n&=2 \end{aligned} \]

So the charge magnitude is \(2e\).

The path bends downwards while a positive reference ion bends upwards. Because the ions initially move right and the magnetic field points into the page, X must be negative.

Thus,

\[ q_X=-2e \]

or

\[ q_X=-3.20\times10^{-19}\ \mathrm{C} \]

Part (c): Reverse the analysing magnetic field

Reversing the field reverses the magnetic force direction on each ion.

The field magnitude, ion speeds, masses, and charge magnitudes remain unchanged. Therefore the orbital radii remain unchanged.

R now bends downwards and returns \(10.0\ \mathrm{cm}\) below entry.

X now bends upwards and returns \(15.0\ \mathrm{cm}\) above entry.

The key distinction is that the sign of the charge determines the side of the detector, while the charge magnitude contributes to the radius.

Had X’s mass been unknown as well, the radius comparison would only determine its mass-to-charge ratio. The independent mass information is what makes the charge magnitude uniquely identifiable here.

Question 5

A redesigned mass spectrometer has a wide-aperture crossed-field velocity selector followed by a magnetic analyser.

Use these coordinate directions:

• \(+x\): right.
• \(+y\): up.
• \(+z\): out of the page.

The selector has an electric field pointing downwards and a magnetic field \(B_s\) pointing into the page.

Its fields satisfy

\[ \frac{E}{B_s}=v_0 \]

where \(v_0\) is a positive speed.

Two positive ions enter the selector at \(y=0\), with these masses, charges, and velocity components:

Assume the selector openings and analysing chamber are wide enough in the \(z\)-direction that neither ion is physically blocked by its depth.

Both then enter the analyser at the same point \((0,0,0)\). The analyser contains only a uniform magnetic field \(B_a\) pointing into the page.

A tiny detector is centred at \((0,2r_A,0)\), where A would arrive after a semicircular turn. It has negligible thickness in the \(z\)-direction.

Determine:

(a) Whether both ions pass undeflected through the velocity selector.

(b) Their total speeds, the radii of their paths projected onto the \(xy\)-plane, and their times to complete half a turn in that projection.

(c) Which ion strikes the detector, and why.

(d) Whether both ions could have reached their stated velocities by starting from rest and accelerating through one common potential difference.

(e) Whether changing only \(B_a\) could separate their projected circular tracks.

Assume ideal uniform fields, negligible interactions between the ions, and non-relativistic motion.

Solution 5

Both ions pass the selector. B is twice as fast overall, yet the ions have identical projected magnetic radii and half-turn times. Only A hits the thin detector. Their specified velocities cannot both result from starting at rest under one common accelerating voltage, and changing the analysing magnetic field strength alone cannot separate their projected tracks.

This problem is difficult because three comparisons that usually seem interchangeable are not: equal total speed, equal perpendicular speed, and equal mass-to-charge ratio.

Part (a): Which ions pass the selector?

Both pass undeflected.

The electric field points downwards. For a positive ion with rightward velocity component \(v_x\), the magnetic force points upwards.

The part of velocity parallel or antiparallel to the magnetic field produces no magnetic force.

Since both ions have \(v_x=v_0\) and \(v_y=0\), the upward magnetic force magnitude is

\[ F_B=|q|v_0B_s \]

The downward electric force magnitude is

\[ F_E=|q|E \]

The selected condition is

\[ E=v_0B_s \]

So the forces cancel for both ions.

The factor \(|q|\) cancels, even though B carries twice A’s charge.

The surprising result is that this wide-aperture selector does not require equal total speeds. Under this geometry, it selects the velocity component perpendicular to its magnetic field.

Part (b): Compare total speeds, projected radii, and half-turn times

For A,

\[ v_A=v_0 \]

For B, use the perpendicular velocity components to calculate its total speed:

\[ \begin{aligned} v_B&=\sqrt{v_0^2+(\sqrt3v_0)^2}\\ &=\sqrt{4v_0^2}\\ &=2v_0 \end{aligned} \]

B is moving twice as fast overall.

But both ions have the same perpendicular speed:

\[ v_{\perp,A}=v_{\perp,B}=v_0 \]

Their projected orbital radii are

\[ r_A=\frac{mv_0}{eB_a} \]

and

\[ \begin{aligned} r_B&=\frac{(2m)v_0}{(2e)B_a}\\ &=\frac{mv_0}{eB_a} \end{aligned} \]

Therefore,

\[ \boxed{r_A=r_B} \]

To compare turning times, divide the circumference of a projected circle by the perpendicular speed:

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

Substituting the magnetic radius,

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

The full-turn times are

\[ T_A=\frac{2\pi m}{eB_a} \]

and

\[ \begin{aligned} T_B&=\frac{2\pi(2m)}{(2e)B_a}\\ &=\frac{2\pi m}{eB_a} \end{aligned} \]

They are equal.

The half-turn time for each ion is therefore

\[ t_{\text{half}}=\frac{\pi m}{eB_a} \]

The larger charge on B gives it a larger magnetic force, but B also has twice the mass. Its charge-to-mass ratio is identical to A’s, so its perpendicular motion is identical.

Part (c): Which ion hits the detector?

Only A hits the detector at \((0,2r_A,0)\).

A has no \(z\)-velocity, so it remains in the \(xy\)-plane while following its semicircular path.

B has a non-zero \(z\)-velocity and moves out of the page while curving.

Its \(z\)-displacement during the half-turn is

\[ \Delta z_B=v_{z,B}t_{\text{half}} \]

Substitute the given parallel velocity and half-turn time:

\[ \begin{aligned} \Delta z_B &=(\sqrt3v_0)\frac{\pi m}{eB_a}\\ &=\frac{\sqrt3\pi mv_0}{eB_a} \end{aligned} \]

This displacement is positive and non-zero.

When B’s projected path reaches the same \(x\)- and \(y\)-coordinates as A’s detector position, B is no longer in the plane \(z=0\).

It passes the detector’s projected position at a different depth.

The two projected circular tracks coincide, but the three-dimensional trajectories do not.

Part (d): Could both start from rest under one common accelerating voltage?

No.

For an ion starting from rest,

\[ |q|U=\frac12mv^2 \]

Rearranging,

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

The ions have equal charge-to-mass ratios:

\[ \frac{|q_A|}{m_A} =\frac{e}{m} \]

and

\[ \frac{|q_B|}{m_B} =\frac{2e}{2m} =\frac{e}{m} \]

Under one common accelerating voltage, they would therefore reach the same total speed.

But the stated velocities give

\[ v_B=2v_A \]

So the proposed common-voltage origin is inconsistent with the given final velocities.

For example, B could have had additional initial kinetic energy in its \(z\)-motion, or the ions could have been prepared under different acceleration conditions.

The velocity selector alone does not tell us which preparation occurred.

Part (e): Can changing \(B_a\) separate the projected tracks?

No.

The radius is

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

For these ions, the value of \(mv_\perp/|q|\) is identical.

Increasing \(B_a\) reduces both radii by the same factor. Decreasing \(B_a\) increases both by the same factor.

The projected circles remain coincident for any common, uniform non-zero analysing field strength.

The key experimental limitation is therefore not the precision of the radius measurement. It is that the ions have the same mass-to-charge ratio.

A magnetic analyser cannot determine their separate masses from these trajectories without additional information about their charge states.

06What the magnetic orbit actually tells you

A heavier ion has a larger magnetic orbit at the same perpendicular speed only when its charge magnitude and magnetic field strength are also the same.

The useful physical reason is simple: equal sideways force produces less turning acceleration in a heavier particle.

But a real detector does not directly read out mass. It records a trajectory whose radius depends on mass, charge magnitude, perpendicular speed, and magnetic field strength.

Once the velocity and magnetic field are controlled, the next useful step is to calculate the mass-to-charge ratio from the measured track. That is the central method developed in How to Find Mass-to-Charge Ratio from a Mass-Spectrometer Track.