How to Calculate Magnetic Field Strength from a Particle Orbit

A proton enters a magnetic field at a known speed and follows a circular path with a radius of 12 cm. The magnetic field strength isn’t given. Can you work it out from the particle’s orbit?

Yes. If you know the particle’s mass, charge magnitude, perpendicular velocity, and orbit radius, you can calculate the magnetic field strength directly.

The equation is:

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

If the particle enters at an angle \(\theta\) to the magnetic field, use:

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

The important detail is that you must use the velocity component perpendicular to the magnetic field, not automatically the particle’s total speed.

This method also lets you work backwards from the direction of a particle’s curvature to identify whether its charge is positive or negative.

01What information can you get from a particle’s orbit?

Imagine an experiment in which a beam of charged particles enters a region containing a uniform magnetic field.

The particles do not continue in straight lines. Their paths curve because the magnetic field exerts a force on moving charges.

The interesting part is that the amount of curvature depends on the magnetic field strength.

For the same particle travelling at the same perpendicular speed:

• A stronger magnetic field produces a tighter curve and a smaller orbit radius.
• A weaker magnetic field produces a gentler curve and a larger orbit radius.

This means the orbit contains information about the magnetic field, even if you cannot measure the field directly.

How the measurement works

Picture an idealised particle-tracking experiment.

1. A particle source produces a beam of charged particles.
2. The particle passes through two timing detectors, which independently establish its speed.
3. The beam enters a region between the poles of an electromagnet.
4. A uniform magnetic field acts on the moving particles.
5. A detector records their curved trajectories.
6. The radius of each trajectory is measured.

The particle’s identity provides its mass and charge. The timing detectors provide its speed, and the recorded trajectory provides its radius.

Together, these measurements allow you to calculate the magnetic field strength.

There is an important experimental limitation here.

A photograph of a curved track alone is not enough to calculate \(B\). Different combinations of particle speed and magnetic field strength can produce the same radius.

The original measurement therefore needs an independent speed measurement and an identified particle species. If the particle enters at an angle, its entry angle must also be known.

For example, suppose two timing gates are separated by \(0.30\ \mathrm m\), and a particle takes \(1.0\times10^{-7}\ \mathrm s\) to travel between them.

Its speed is:

\[ v=\frac{d}{\Delta t} \]

\[ v=\frac{0.30}{1.0\times10^{-7}} \]

\[ v=3.0\times10^6\ \mathrm{m\,s^{-1}} \]

This independently measured speed can now be combined with the orbit radius to calculate the field.

02Why only the perpendicular velocity component matters

Before using the equation, consider a particle entering a magnetic field at \(30^\circ\).

Should you use its entire speed when calculating the orbit radius?

It seems reasonable. After all, the particle is moving at that total speed.

However, only the component of velocity perpendicular to the magnetic field contributes to the circular motion.

To understand why, split the particle’s velocity into two components.

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

If the angle between the velocity and magnetic field is \(\theta\), then:

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

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

What happens to each component?

The perpendicular component experiences a magnetic force. This force continually changes its direction, producing circular motion.

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

It therefore continues unchanged in the ideal uniform-field model.

The combination produces three possible trajectories:

For a helical trajectory, the radius in our calculation is the radius of the circular motion perpendicular to the magnetic field. It is not the length of the three-dimensional helical path.

Deriving the magnetic field equation

Now connect the particle’s motion to the forces acting on it.

The magnitude of the magnetic force is:

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

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

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

The perpendicular component undergoes circular motion, so the required centripetal force is:

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

The magnetic force supplies this centripetal force. Equating their magnitudes:

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

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

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

Rearrange to make \(B\) the subject:

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

Substituting \(v_\perp=v\sin\theta\) gives the more general expression:

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

This is the main equation for finding magnetic field strength from a particle orbit.

Understanding the symbols and units

Notice that the equation uses \(|q|\), not \(q\).

This is because we are calculating the magnitude of the magnetic field. The charge sign determines the direction of curvature, which is a separate calculation.

The result must be expressed in tesla.

03Worked examples: calculating magnetic field strength

04The common mistake: using total speed instead of perpendicular speed

Return to the alpha particle example.

Its total speed was \(4.0\times10^6\ \mathrm{m\,s^{-1}}\), but its perpendicular speed was only \(2.0\times10^6\ \mathrm{m\,s^{-1}}\).

What happens if you use the total speed directly?

You would calculate:

\[ B_{\text{wrong}}= \frac{ (6.64\times10^{-27})(4.0\times10^6) }{ (3.20\times10^{-19})(0.20) } \]

\[ B_{\text{wrong}}=0.415\ \mathrm T \]

This is exactly twice the correct result.

Why?

Because only half of the total speed contributes to the circular motion in this example.

The magnetic force does not act on the parallel velocity component. Including it in the radius equation incorrectly assumes that the entire velocity is being bent around the magnetic field.

A useful angle check

Make sure you know which angle the question gives you.

If \(\theta\) is measured between the velocity and the magnetic field:

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

If \(\alpha\) is measured between the velocity and the plane perpendicular to the magnetic field:

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

These expressions describe the same geometry from different reference directions.

A quick sketch of the velocity vector will help you avoid choosing the wrong trigonometric function.

05Identifying the charge sign from the particle’s curvature

Calculating the magnetic field strength tells you how strong the field is. It does not tell you whether the particle is positively or negatively charged.

For that, you need the direction of the particle’s curvature.

The magnetic force on a moving charge is described by:

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

The cross product gives the force direction for a positive charge. A negative charge experiences a force in the opposite direction.

You can determine this using the right-hand rule.

A specific example: Magnetic field into the page

Imagine a particle travelling horizontally to the right.

The magnetic field points into the page, represented by crosses (\(\times\)).

What happens to a positive particle?

Point your right-hand fingers in the direction of the velocity, then curl them towards the magnetic field. Your thumb indicates the magnetic force direction for a positive charge.

In this arrangement, the force points upwards.

A positive particle therefore curves upwards.

A negative particle experiences the opposite force, so it curves downwards.

The result can be summarised:

Do not memorise that electrons always curve downwards or protons always curve upwards.

The direction depends on both the magnetic field orientation and the particle’s initial direction of travel.

One more important detail: You must know which way the particle is travelling

A photograph of a curved track does not always tell you the direction of motion.

Imagine seeing a circular arc without knowing which end is the entry point.

The same geometric path could be travelled in either direction. Reversing the velocity reverses the direction of \(\vec v\times\vec B\).

Consequently, the charge sign you infer would also reverse.

To identify the charge reliably, you need:

• The magnetic field direction.
• The particle’s direction of travel.
• The direction of its initial curvature.

Time-resolved detector images or a clearly identified entry point can provide the missing direction information.

06A reliable exam method

When a question asks you to calculate magnetic field strength from a particle’s orbit, follow this sequence.

Step 1: Identify the particle.

Find its mass and charge magnitude. Check whether the question provides the values or expects you to recognise the particle.

Step 2: Establish the angle.

Determine the angle between the particle’s velocity and the magnetic field.

Step 3: Calculate perpendicular velocity.

Use:

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

If the particle enters perpendicular to the field, simply use \(v_\perp=v\).

Step 4: Identify the orbit radius.

If the question gives a diameter, divide it by two.

If the question gives a helical trajectory, use the radius of its circular projection perpendicular to the field.

Step 5: Calculate the magnetic field strength.

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

Step 6: Identify the charge sign separately, if required.

Use the right-hand rule and reverse the force direction for a negative charge.

Assumptions and limitations

The calculation assumes:

• The magnetic field is uniform throughout the measured orbit.
• The particle’s mass and charge are known.
• Its perpendicular velocity is known or can be determined.
• Other forces, such as electric forces, are negligible.
• The particle is travelling at a non-relativistic speed.
• Collisions and energy losses are negligible, so the particle’s speed remains constant.

If the magnetic field changes along the trajectory, the radius may change. Similarly, if the particle loses energy through collisions, its speed and orbit radius may change.

The equation describes the ideal circular motion, or the circular component of helical motion.

07Questions and solutions

The following questions progress from a straightforward field-strength calculation to problems involving multiple magnetic regions and three-dimensional particle motion.

Question 1

An alpha particle enters a uniform magnetic field perpendicular to the field direction.

Its speed is \(2.0\times10^6\ \mathrm{m\,s^{-1}}\), and the radius of its circular trajectory is \(0.15\ \mathrm m\).

Calculate the magnetic field strength.

Use:

• Mass of alpha particle: \(6.64\times10^{-27}\ \mathrm{kg}\)
• Charge of alpha particle: \(+3.20\times10^{-19}\ \mathrm C\)

Solution 1

The magnetic field strength is approximately \(0.28\ \mathrm T\).

Because the particle enters perpendicular to the field, its entire speed contributes to the circular motion.

Step 1: Identify the perpendicular velocity.

\[ v_\perp=v=2.0\times10^6\ \mathrm{m\,s^{-1}} \]

Step 2: Apply the magnetic field equation.

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

Step 3: Substitute the values.

\[ B= \frac{ (6.64\times10^{-27})(2.0\times10^6) }{ (3.20\times10^{-19})(0.15) } \]

\[ B=\frac{1.328\times10^{-20}}{4.80\times10^{-20}} \]

\[ B=0.2767\ \mathrm T \]

Therefore:

\[ \boxed{B=0.28\ \mathrm T} \]

The result is physically reasonable: a larger magnetic field would produce a smaller radius if the particle’s speed and charge remained unchanged.

Question 2

A singly charged positive ion has a mass of \(3.34\times10^{-27}\ \mathrm{kg}\) and enters a uniform magnetic field at an angle of \(30^\circ\).

Its total speed is \(5.0\times10^6\ \mathrm{m\,s^{-1}}\), and the radius of its helical trajectory is \(0.16\ \mathrm m\).

Calculate:

(a) The perpendicular velocity component.

(b) The magnetic field strength.

(c) The parallel velocity component, and explain why the trajectory is helical.

The magnitude of the ion’s charge is \(1.60\times10^{-19}\ \mathrm C\).

Solution 2

The perpendicular velocity is \(2.5\times10^6\ \mathrm{m\,s^{-1}}\), and the magnetic field strength is approximately \(0.33\ \mathrm T\). The particle follows a helix because it also has a non-zero parallel velocity component.

(a) Perpendicular velocity

Step 1: Resolve the total velocity.

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

\[ v_\perp=(5.0\times10^6)\sin30^\circ \]

\[ \boxed{v_\perp=2.5\times10^6\ \mathrm{m\,s^{-1}}} \]

(b) Magnetic field strength

Step 2: Substitute the perpendicular velocity into the orbit equation.

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

\[ B= \frac{ (3.34\times10^{-27})(2.5\times10^6) }{ (1.60\times10^{-19})(0.16) } \]

\[ B=\frac{8.35\times10^{-21}}{2.56\times10^{-20}} \]

\[ \boxed{B=0.33\ \mathrm T} \]

(c) Parallel velocity

Step 3: Calculate the parallel component.

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

\[ v_\parallel=(5.0\times10^6)\cos30^\circ \]

\[ \boxed{v_\parallel=4.3\times10^6\ \mathrm{m\,s^{-1}}} \]

The parallel component is not affected by the magnetic force.

The particle therefore moves along the field while simultaneously undergoing circular motion perpendicular to it. These two motions combine to produce a helix.

Question 3

An unknown particle has a mass of \(9.11\times10^{-31}\ \mathrm{kg}\) and a charge magnitude of \(1.60\times10^{-19}\ \mathrm C\).

It enters a uniform magnetic field at \(1.8\times10^7\ \mathrm{m\,s^{-1}}\), travelling perpendicular to the field.

The magnetic field points into the page. The particle initially travels to the right and its trajectory curves downwards.

A detector records a circular track with a diameter of \(24\ \mathrm{cm}\).

Determine:

(a) The magnetic field strength.

(b) Whether the particle is positively or negatively charged.

Solution 3

The magnetic field strength is approximately \(8.5\times10^{-4}\ \mathrm T\), and the particle is negatively charged.

(a) Magnetic field strength

Step 1: Convert the diameter into a radius.

The measured diameter is \(24\ \mathrm{cm}\).

\[ r=\frac{24}{2}=12\ \mathrm{cm} \]

Convert to metres:

\[ r=0.12\ \mathrm m \]

Step 2: Identify the perpendicular velocity.

The particle enters perpendicular to the magnetic field, so:

\[ v_\perp=1.8\times10^7\ \mathrm{m\,s^{-1}} \]

Step 3: Calculate the field strength.

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

\[ B= \frac{ (9.11\times10^{-31})(1.8\times10^7) }{ (1.60\times10^{-19})(0.12) } \]

\[ B=\frac{1.6398\times10^{-23}}{1.92\times10^{-20}} \]

\[ \boxed{B=8.5\times10^{-4}\ \mathrm T} \]

(b) Charge sign

Step 4: Apply the right-hand rule.

The magnetic field points into the page, and the particle initially travels to the right.

For a positive charge, the magnetic force would point upwards.

However, the particle curves downwards.

Its force must therefore point opposite to the positive-charge prediction.

\[ \boxed{q<0} \]

The particle is negatively charged.

Notice that calculating the field strength and identifying the charge sign require two separate pieces of reasoning. The radius determines the field magnitude, while the curvature direction determines the charge sign.

Question 4

A proton travels perpendicular to the magnetic field through two adjacent regions, A and B.

The magnetic fields in both regions are uniform and point into the page in the same direction.

The proton’s speed remains constant at \(4.0\times10^6\ \mathrm{m\,s^{-1}}\).

Its measured orbit radii are:

The proton has a mass of \(1.67\times10^{-27}\ \mathrm{kg}\) and a charge of \(+1.60\times10^{-19}\ \mathrm C\).

(a) Calculate the magnetic field strength in region A.

(b) Calculate the magnetic field strength in region B.

(c) Explain why the proton’s radius changes even though its speed remains constant.

Solution 4

The magnetic field strength is approximately \(0.21\ \mathrm T\) in region A and \(0.35\ \mathrm T\) in region B. Region B has the stronger magnetic field, which produces the smaller orbit radius.

(a) Field strength in region A

Step 1: Apply the orbit equation.

\[ B_A=\frac{mv}{|q|r_A} \]

The proton travels perpendicular to the field, so \(v_\perp=v\).

\[ B_A= \frac{ (1.67\times10^{-27})(4.0\times10^6) }{ (1.60\times10^{-19})(0.20) } \]

\[ B_A=\frac{6.68\times10^{-21}}{3.20\times10^{-20}} \]

\[ \boxed{B_A=0.21\ \mathrm T} \]

(b) Field strength in region B

Step 2: Use the same equation with the new radius.

\[ B_B=\frac{mv}{|q|r_B} \]

\[ B_B= \frac{ (1.67\times10^{-27})(4.0\times10^6) }{ (1.60\times10^{-19})(0.12) } \]

\[ B_B=\frac{6.68\times10^{-21}}{1.92\times10^{-20}} \]

\[ \boxed{B_B=0.35\ \mathrm T} \]

(c) Explaining the change in radius

The magnetic field in region B is stronger.

Because the particle’s mass, charge magnitude, and speed remain constant:

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

The radius is inversely proportional to the magnetic field strength.

The magnetic force changes the particle’s direction, not its speed, because the force is perpendicular to its velocity.

Therefore, the smaller radius does not indicate that the proton has slowed down. It indicates that the magnetic field has become stronger.

This is an important distinction when interpreting a particle track that passes between different magnetic-field regions.

Question 5

A charged particle follows a helical trajectory through a uniform magnetic field.

Its mass is \(3.34\times10^{-27}\ \mathrm{kg}\), its charge magnitude is \(1.60\times10^{-19}\ \mathrm C\), and its total speed is \(4.0\times10^6\ \mathrm{m\,s^{-1}}\).

A detector records the following measurements:

• Radius of the circular projection: \(0.10\ \mathrm m\).
• Pitch of the helix: \(0.63\ \mathrm m\).

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

The magnetic field points out of the page. Time-stamped observations establish that the particle moves along the magnetic field direction. At the labelled starting point, its perpendicular velocity component points to the right, and its projected trajectory curves downwards.

Determine:

(a) The perpendicular velocity component.

(b) The magnetic field strength.

(c) Whether the particle is positively or negatively charged.

(d) Explain why an ordinary photograph without time information would not necessarily be enough to determine the charge sign.

Solution 5

The perpendicular velocity is approximately \(2.82\times10^6\ \mathrm{m\,s^{-1}}\), the magnetic field strength is approximately \(0.59\ \mathrm T\), and the particle is positively charged.

The key is to use the measured pitch and radius to determine the perpendicular velocity rather than incorrectly substituting the total speed.

(a) Perpendicular velocity component

Step 1: Understand what the pitch and radius tell us.

During one complete revolution, the particle travels:

• A distance of \(2\pi r\) due to its circular motion.
• A distance of \(p\) parallel to the magnetic field.

Both distances are covered in the same time.

Therefore, the ratio of the parallel and perpendicular velocity components is:

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

Step 2: Substitute the measured geometry.

\[ \frac{v_\parallel}{v_\perp} = \frac{0.63}{2\pi(0.10)} \]

\[ \frac{v_\parallel}{v_\perp} \approx1.00 \]

The two velocity components are approximately equal.

Step 3: Use the total velocity.

The perpendicular and parallel components are at right angles, so:

\[ v^2=v_\perp^2+v_\parallel^2 \]

Since the components are approximately equal:

\[ v^2\approx2v_\perp^2 \]

Rearranging:

\[ v_\perp\approx\frac{v}{\sqrt2} \]

\[ v_\perp\approx\frac{4.0\times10^6}{\sqrt2} \]

\[ \boxed{v_\perp=2.82\times10^6\ \mathrm{m\,s^{-1}}} \]

(b) Magnetic field strength

Step 4: Apply the particle orbit equation.

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

Substitute:

\[ B= \frac{ (3.34\times10^{-27})(2.82\times10^6) }{ (1.60\times10^{-19})(0.10) } \]

\[ B=\frac{9.42\times10^{-21}}{1.60\times10^{-20}} \]

\[ \boxed{B=0.59\ \mathrm T} \]

The magnetic field strength is approximately \(0.59\ \mathrm T\).

(c) Charge sign

Step 5: Establish the force direction for a positive charge.

The magnetic field points out of the page.

The perpendicular velocity component points to the right.

Using the right-hand rule:

\[ \vec v_\perp\times\vec B \]

points downwards.

Therefore, a positive particle experiences a downward magnetic force.

The observed projected trajectory curves downwards, matching the positive-charge prediction.

\[ \boxed{q>0} \]

The particle is positively charged.

(d) Why time information matters

A static photograph records the shape of the helix, but it does not necessarily establish which way the particle travelled along that path.

If the direction of travel is reversed, the cross product \(\vec v\times\vec B\) also reverses.

A particle with the opposite charge could therefore produce the same geometric trajectory when travelling in the opposite direction.

The time-stamped observations remove this ambiguity by establishing the direction of motion.

This is why identifying charge sign requires more than simply looking at whether a track appears clockwise or anticlockwise.

The geometry determines the orbit radius and pitch. The velocity direction and magnetic field direction determine the force direction, which finally allows you to identify the charge sign.

07What to learn next

Finding magnetic field strength from a particle orbit brings together three important ideas:

• Magnetic force supplies the centripetal force.
• Only the perpendicular velocity component determines the orbit radius.
• The direction of curvature, interpreted using the right-hand rule, reveals the charge sign when the direction of travel is known.

Once you can calculate \(B\) from an orbit, the next useful step is understanding how the same field determines the time taken for one complete revolution.

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

For a broader explanation of how magnetic field direction and strength are represented, see Magnetic Field Models: From Field Lines to Field Strength.