Year 12 Physics Module 6 ⏱ ~45 min 5 MC · 2 Short Answer Lesson 5 of 21 IQ1: Charged Particles in Fields

Circular Motion in Magnetic Fields

Derive the radius, period and frequency of a charge moving perpendicular to a uniform magnetic field, then extend the model to oblique entry and its non-relativistic limits.

Today's hook: A proton and an electron enter the same uniform magnetic field at the same perpendicular speed. Which curves more tightly, by what factor, and which way does each path bend?
0/5TASKS
Learn 0 of 6 steps complete
1
Retrieve the magnetic-force direction rule and predict the orbit model.
Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

An electron enters a uniform magnetic field pointing out of the page, moving perpendicular to the field.

  1. What shape do you think the electron's path will be: straight line, parabola, circle, or spiral?
  2. If you doubled the magnetic field strength, would the circle become larger or smaller?
  3. If you doubled the electron's speed, would it complete one full circle faster, slower, or in the same time?

Write your predictions before reading on, you will revisit them at the end.

Warm-up, when a charged particle moves perpendicular to a magnetic field, the magnetic force acts in which direction relative to the velocity?

Learning Intentions
goals

Know, Circular Motion Equations

  • $r=mv_\perp/(|q|B)$ follows by equating magnetic-force magnitude to centripetal force
  • $T=2\pi m/(|q|B)$ is speed-independent in the non-relativistic model
  • $f=|q|B/(2\pi m)$ is the classical cyclotron frequency

Understand, Why a Circle?

  • Magnetic force is always perpendicular to velocity, so it acts as centripetal force
  • Speed is constant (no work done), so radius is constant
  • The period is independent of speed: faster particles travel larger circles in the same time

Can Do, Calculate & Predict

  • Calculate radius and period given $m$, $v$, $q$, $B$
  • Predict how changing $v$, $B$, or $m$ affects the path
  • Explain why the period is independent of speed
Scan these before reading
vocab
Radius of curvature$r=mv_\perp/(|q|B)$ for the circular component in a uniform B-field.
Period (T)Classical time for one revolution: $T=2\pi m/(|q|B)$.
Cyclotron frequency$f=|q|B/(2\pi m)$ in the non-relativistic model.
Centripetal force$F_c = mv^2/r$, the inward force required for circular motion.
Charge-to-mass ratio$q/m$ determines how tightly a particle curves in a given B-field.
Cross-lesson links: L04 showed a single magnetic deflection. L05 shows how repeated deflections produce circular motion: the magnetic-force magnitude $|q|vB$ supplies the centripetal force, while electric-field crossings can increase kinetic energy in a cyclotron.

Core Content

2
Derive radius, period and frequency for perpendicular non-relativistic entry.
1
Derivation: Why the Path is Circular
+5 XP

Equating magnetic force to centripetal force

Fire a proton sideways into a region of uniform magnetic field, and instead of continuing in a straight line or following a parabola, it sweeps in a perfect circle and returns exactly to where it started. The magnetic force at every instant points inward toward the centre of that circle, always perpendicular to the proton's motion, always the same magnitude. This is uniform circular motion, with the magnetic force playing the role of the centripetal force.

We equate the magnetic force to the required centripetal force:

Deriving the Radius

$F_{\text{magnetic}} = F_{\text{centripetal}}$

$|q|v_\perp B = \dfrac{mv_\perp^2}{r}$

$r = \dfrac{mv_\perp}{|q|B}$, radius increases with perpendicular momentum and decreases with field strength

The period of revolution is the circumference divided by speed:

Period and Frequency

$T = \dfrac{2\pi r}{v_\perp} = \dfrac{2\pi m}{|q|B}$, period is independent of speed in the non-relativistic model

$f = \dfrac{1}{T} = \dfrac{|q|B}{2\pi m}$, cyclotron frequency depends only on $|q|/m$ and $B$ classically

This classical independence of period from speed is the principle behind the cyclotron: particles are accelerated by an electric field and steered by a uniform magnetic field. At relativistic speeds the effective inertia increases and the fixed-frequency cyclotron loses synchronism.

centre - v F - v F - v F - v F r B out of page

Figure 1, Electron with $\vec B$ out of the page: the velocity is tangent to the counterclockwise path and the magnetic force points to the centre at every instant.

Stop & Check

A proton and an electron have the same speed and enter the same uniform magnetic field perpendicular to their velocities. Which one travels in the larger circle? By what factor is the radius larger?

For perpendicular entry, setting $|q|vB=mv^2/r$ gives $r=mv/(|q|B)$ and $T=2\pi m/(|q|B)$. These magnitudes use $|q|$; the sign determines curvature direction. The period is speed-independent only in the uniform-field, non-relativistic model.

Pause, copy the highlighted formulas into your book before moving on.

The classical period of a charged particle orbiting in a magnetic field is $T=2\pi m/(|q|B)$. Which statement follows?

3
Explain classical speed-independent period and test it in the circular-motion mode.
Beyond the syllabus. Deriving and using $r = \frac{mv}{qB}$ for a charge in a magnetic field is core. Cyclotron frequency, period derivations and helical/cyclotron engineering are extension — exam questions ask for the radius reasoning, not accelerator design.
2
Cyclotrons and the Independence of Period
+5 XP

Why faster particles take the same time to complete a loop

We just saw that $r=mv/(|q|B)$ and $T=2\pi m/(|q|B)$ for perpendicular, non-relativistic motion. That raises a question: how can two particles with different speeds complete a loop in the same time? This card answers it → the faster particle has a proportionally larger circumference in this model.

The fact that the classical $T=2\pi m/(|q|B)$ contains no $v$ is remarkable. A fast non-relativistic particle travels a larger circle at higher speed; a slower one travels a smaller circle, with the same period for equal $m/|q|$ and $B$.

This is the operating principle of the cyclotron. Particles are injected near the centre and accelerated by an electric field every half-revolution. A fixed frequency $f=|q|B/(2\pi m)$ remains synchronous only while the non-relativistic approximation is adequate.

HSC Tip

When comparing two particles in the same B-field, remember that $r \propto mv/qB$ while $T \propto m/qB$. The period depends only on mass-to-charge ratio, not on speed.

Interactive Tool, circular motion in a magnetic field Open fullscreen ↗

Use in this lesson: select the circular-motion controls and compare radius and period as $v$, $B$, particle mass and charge magnitude change. Selector and analyser controls belong to Lessons 6 and 18.

Stop & Check

In a cyclotron, protons are accelerated to higher and higher speeds. A student worries that as they speed up, they will fall out of sync with the alternating voltage. Explain why this does not happen in an ideal cyclotron.

In an ideal non-relativistic cyclotron, faster particles spiral into larger circles while $T=2\pi m/(|q|B)$ stays constant, so a fixed alternating voltage remains in sync. At relativistic speeds this approximation fails and synchronism is lost.

Add the highlighted principle to your notes before the check below.

In a cyclotron, the period of a proton's circular orbit is independent of its speed.

Doubling the speed of a particle in a uniform magnetic field doubles its orbital period.

The magnetic force on a moving charged particle does no work on the particle.

4
Compare electron and proton orbit magnitudes while separating charge sign from radius.
3
Worked Example, Comparing Electron and Proton Orbits
+5 XP

Radius, period, and the power of charge-to-mass ratio

We just saw the cyclotron principle, period is independent of speed. That raises a question: how different are the actual orbits of an electron versus a proton in the same field? This card answers it → the proton's radius is 1836× larger and its period 1836× longer due to its greater mass.

Problem

An electron and a proton each enter a uniform magnetic field of 0.020 T perpendicular to their velocity. Both have speed 4.0 × 106 m/s.

  1. (a) Calculate the radius of the electron's path.
  2. (b) Calculate the radius of the proton's path.
  3. (c) Calculate the period of each particle's motion.
  4. (d) Explain why the proton's period is much longer even though both travel at the same speed.
Step 1, Part (a): Electron radius

$r_e = \dfrac{m_e v}{eB} = \dfrac{(9.11 \times 10^{-31})(4.0 \times 10^6)}{(1.60 \times 10^{-19})(0.020)}$

$r_e = \mathbf{1.14 \times 10^{-3}}$ m = 1.14 mm

Step 2, Part (b): Proton radius

$r_p = \dfrac{m_p v}{eB} = \dfrac{(1.67 \times 10^{-27})(4.0 \times 10^6)}{(1.60 \times 10^{-19})(0.020)}$

$r_p = \mathbf{2.09}$ m

The proton's radius is about 1836 times larger than the electron's because the proton is 1836 times more massive.

Step 3, Part (c): Periods

Electron: $T_e = \dfrac{2\pi m_e}{eB} = \dfrac{2\pi(9.11 \times 10^{-31})}{(1.60 \times 10^{-19})(0.020)} = \mathbf{1.79 \times 10^{-9}}$ s

Proton: $T_p = \dfrac{2\pi m_p}{eB} = \dfrac{2\pi(1.67 \times 10^{-27})}{(1.60 \times 10^{-19})(0.020)} = \mathbf{3.28 \times 10^{-6}}$ s

Step 4, Part (d): Why is the proton's period longer?

The period depends only on $m/qB$, not on speed. The proton has the same charge magnitude as the electron but is 1836 times more massive. Therefore its period is 1836 times longer. Even though both particles travel at the same speed, the proton's much greater inertia means it resists the centripetal change in direction, taking far longer to complete each loop.

Using $r=mv/(|q|B)$ and $T=2\pi m/(|q|B)$ in the non-relativistic model: proton radius and period are about $1836\times$ the electron values at equal speed, charge magnitude and field.

Pause, write the highlighted ratio into your book before moving on.

An electron travels at 2.0 × 106 m/s perpendicular to a 0.010 T field. The radius of its path is approximately:

5
Compare circular-force models and extend perpendicular entry to helical motion.
4
Comparing Magnetic and Gravitational Circular Motion
+5 XP

A charged particle in a B-field and a satellite in orbit follow the same mathematical logic

We just saw how to calculate radius and period for charged particles in B-fields. That raises a question: how is this the same as, or different from, a satellite orbiting Earth? This card answers it → both use real force = $mv^2/r$, but magnetic radius $\propto v$ while gravitational radius $\propto 1/v^2$.

The syllabus asks you to compare charged particle motion in magnetic fields to other examples of uniform circular motion. The most important comparison is with a satellite in orbit.

Feature Charged Particle in B-Field Satellite in Orbit
Centripetal force Magnetic-force magnitude: $F_B=|q|vB$ Gravitational force: $F = GMm/r^2$
Setting equal to $mv^2/r$ $|q|vB = mv^2/r$ $GMm/r^2 = mv^2/r$
Radius result $r=mv/(|q|B)$ $r = GM/v^2$
Key difference Radius increases with speed ($r \propto v$) Radius decreases with speed ($r \propto 1/v^2$)
Key similarity Both derive radius by equating the real force to the required centripetal force $mv^2/r$
HSC Tip

When asked to compare, always state the similarity (both use $F = mv^2/r$) and the difference (how radius depends on speed). Do not just describe each situation separately.

Similarity: both set real force = $mv^2/r$. Difference: magnetic $r \propto v$ (faster → bigger circle); gravitational $r \propto 1/v^2$ (faster → lower orbit). Always start any circular-motion problem with: real force = $mv^2/r$.

Add the highlighted comparison to your notes before the check below.

Three of these are correct comparisons between a charged particle in a B-field and a satellite in orbit. Pick the odd one out.

Oblique entry and model boundary

A circle requires $\vec v\perp\vec B$. For oblique entry, resolve $\vec v$ into $v_\perp$ and $v_\parallel$: the perpendicular component produces circular motion with $r=mv_\perp/(|q|B)$, while the parallel component is unchanged. Their combination is a helix whose pitch is $v_\parallel T$. These results assume a uniform field and non-relativistic speed.

B and v∥ v v∥ v⊥ pitch = v∥T

Figure 2, Oblique entry: $v_\perp$ sets the helix radius, while unchanged $v_\parallel$ carries the particle along the uniform field.

6
Apply proportional reasoning, calculate frequency and consolidate model limits.
Activity 1, Cyclotron Simulator Exploration
ApplyBand 3

Predict, then verify using the interactive above

  1. Set B = 50 mT and v = 3.0 × 106 m/s with an electron. Predict the radius using $r=mv/(|q|B)$. Check the simulator readout.
  2. Double the speed to 6.0 × 106 m/s. What happens to the radius? What happens to the period? Explain why the period stays the same even though the particle is moving faster.
  3. Switch to a proton at the same speed and field. Predict the radius and period by what factor they change compared to the electron. Verify.
  4. Switch to an alpha particle (2× charge, ~4× proton mass). Predict the radius and period. Verify your prediction with the simulator.

Fill the gap. For a proton (mass $m_p$, charge $e$) in a field $B$, the cyclotron frequency is $f = eB / (\_\_\_ \cdot m_p)$. The missing factor is _____.

Activity 2, Comparing Circular Motion Systems
UnderstandBand 4

Link charged particle motion to satellite orbital mechanics

Two electrons enter the same magnetic field at different speeds. Electron A has twice the speed of Electron B. Compare: (i) their orbital radii, (ii) their periods, (iii) their cyclotron frequencies, and (iv) the centripetal force required for each.

Which set correctly describes how the radius of a charged particle's circular path changes when each variable is doubled (all others constant)?

7
Retrieve the orbit equations and explain their conditions independently.
Quick recall, circular motion in magnetic fields
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 7 marks
+5 XP

ApplyBand 4(3 marks) 1. A proton enters a uniform magnetic field of 0.030 T perpendicular to its velocity of 5.0 × 105 m/s. (a) Calculate the radius of the proton's circular path. (b) Calculate the period of the proton's motion. (c) An electron enters the same field at the same speed. Without calculating, explain how its radius and period compare to the proton's.

1 mark: correct radius with working · 1 mark: correct period with working · 1 mark: correct comparison with explanation

EvaluateBand 6(4 marks) 2. A student claims: "In a cyclotron, faster protons must spin at a higher frequency to complete each smaller circle in less time." Evaluate this claim. In your answer, derive the expression for cyclotron frequency and explain what it depends on.

1 mark: identifies the claim as incorrect · 1 mark: derives $f=|q|B/(2\pi m)$ from $|q|vB=mv^2/r$ · 1 mark: states classical frequency is speed-independent and radius increases with speed · 1 mark: states the non-relativistic limitation

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (a): $r=mv/(|q|B)=(1.67 \times 10^{-27} \times 5.0 \times 10^5)/(1.60 \times 10^{-19} \times 0.030)=0.174$ m (1 mark).

Q1 (b): $T=2\pi m/(|q|B)=2\pi \times 1.67 \times 10^{-27}/(1.60 \times 10^{-19} \times 0.030)=2.19 \times 10^{-6}$ s (1 mark).

Q1 (c): The electron's radius is much smaller than the proton's (by a factor of $m_p/m_e \approx 1836$) because $r \propto m$ and the electron is 1836 times less massive. The electron's period is also much smaller (by the same factor) because $T \propto m$, the electron zips around its tiny circle in far less time (1 mark).

Q2 (4 marks): The claim is incorrect in the ideal non-relativistic model: faster protons follow larger, not smaller, circles (1 mark). From $|q|vB=mv^2/r$, $r=mv/(|q|B)$ and $T=2\pi r/v=2\pi m/(|q|B)$ (1 mark), so $f=|q|B/(2\pi m)$ has no speed term classically (1 mark). At relativistic speed the classical expression fails; the orbital frequency decreases as relativistic momentum effects grow, so a fixed-frequency cyclotron loses synchronism (1 mark).

8
Check retention and revisit radius, direction and model limitations.
Check what actually stuck
Take the full module quiz
quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →
How did your thinking change?

At the start, a proton and an electron entered the same uniform magnetic field at the same perpendicular speed. You were asked which curves more tightly, by what factor, and how their curvature directions compare.

The answer: the electron curves far more tightly. Since $r=mv/(|q|B)$ and both particles have the same charge magnitude but the proton is 1836 times more massive, the proton's radius is 1836 times larger at equal speed and field. Charge sign changes curvature direction, not radius magnitude.

← Previous: Magnetic Fields and the Lorentz Force Next: Electric vs Magnetic Fields →

Module overview · Physics · Checkpoint 1 · Module quiz