Physics • Year 12 • Module 6: Electromagnetism • Lesson 5
Circular Motion in Magnetic Fields
Apply your understanding of r = mv/qB and T = 2πm/qB to real data, multi-step calculations, and a comparison between particle orbits.
1. Multi-particle orbit comparison
An electron, a proton, and an alpha particle each enter the same uniform magnetic field of B = 0.050 T with the same speed v = 2.0 × 106 m/s, perpendicular to the field. Use the data table and formulas r = mv/qB and T = 2πm/qB to complete the table below. 10 marks (1 per cell)
Particle data: me = 9.11 × 10−31 kg, mp = 1.67 × 10−27 kg, mα = 6.64 × 10−27 kg, e = 1.60 × 10−19 C, qα = 3.20 × 10−19 C.
| Particle | Charge (C) | Mass (kg) | Radius r (m) | Period T (s) |
|---|---|---|---|---|
| Electron | 1.60 × 10−19 | 9.11 × 10−31 | ||
| Proton | 1.60 × 10−19 | 1.67 × 10−27 | ||
| Alpha | 3.20 × 10−19 | 6.64 × 10−27 |
1.1 Show your working for the electron’s radius. 2 marks
1.2 Without further calculation, state the ratio of the proton’s radius to the electron’s radius. Explain your reasoning in terms of the formula r = mv/qB. 2 marks
1.3 Compare the period of the proton and the alpha particle. Which is longer, and by what factor? 2 marks
2. Interpret a radius-vs-speed graph
A student plots the orbital radius of a proton in a fixed magnetic field of B = 0.10 T against the proton’s speed. The graph below shows the results. 7 marks
Figure 2.1. Orbital radius of a proton in B = 0.10 T plotted against speed. Illustrative data based on r = mpv/eB.
2.1 Describe the relationship shown in the graph between orbital radius and speed. 2 marks
2.2 Use the graph to determine the orbital radius of the proton at a speed of 5.0 × 106 m/s. Verify this using r = mv/qB. 2 marks
2.3 A student claims: “If the speed doubles, the period also doubles.” Use the graph and the formula T = 2πm/qB to explain whether this claim is correct. 3 marks
3. Compare circular motion: charged particle in B-field vs satellite in orbit
Complete the comparison table. For each feature, write a concise description that contrasts or connects the two systems. 10 marks (1 per cell)
| Feature | Charged particle in B-field | Satellite in orbit |
|---|---|---|
| Source of centripetal force | ||
| Expression equated to mv²/r | ||
| Radius formula | ||
| How radius depends on speed | ||
| Does the centripetal force do work? |
4. Predict and justify — the synchrotron upgrade
The Australian Synchrotron in Clayton, Victoria, accelerates electrons in a circular storage ring. The electrons travel at very high (but constant) speed in a horizontal ring of fixed radius. A team of engineers proposes to increase the ring radius by 20% to allow higher-energy electrons. 6 marks
4.1 For a fixed particle speed v, what change to the magnetic field strength B would be needed to increase the orbital radius by 20%? Justify using r = mv/qB, and state whether B should increase or decrease. 3 marks
4.2 If the magnetic field strength is reduced as in 4.1, predict what happens to the period T and the cyclotron frequency f of the electrons. Justify quantitatively using the relevant formula. 3 marks
Q1 — Multi-particle orbit comparison table
Calculations (B = 0.050 T, v = 2.0 × 106 m/s):
Electron: r = (9.11 × 10−31)(2.0 × 106) / (1.60 × 10−19 × 0.050) = 2.28 × 10−4 m (0.228 mm). T = 2π(9.11 × 10−31) / (1.60 × 10−19 × 0.050) = 7.15 × 10−10 s.
Proton: r = (1.67 × 10−27)(2.0 × 106) / (1.60 × 10−19 × 0.050) = 0.418 m. T = 2π(1.67 × 10−27) / (1.60 × 10−19 × 0.050) = 1.31 × 10−6 s.
Alpha: r = (6.64 × 10−27)(2.0 × 106) / (3.20 × 10−19 × 0.050) = 0.830 m. T = 2π(6.64 × 10−27) / (3.20 × 10−19 × 0.050) = 2.60 × 10−6 s.
Marking criteria: 1 mark per correct r value; 1 mark per correct T value (5 cells × 1 = 10 marks maximum, with the given cells already filled).
Q1.1 — Electron radius working
re = mev / (eB) = (9.11 × 10−31 × 2.0 × 106) / (1.60 × 10−19 × 0.050) = 1.822 × 10−24 / 8.00 × 10−21 = 2.28 × 10−4 m. 1 mark for correct substitution; 1 mark for correct answer with unit.
Q1.2 — Ratio of proton to electron radius
rp/re = mp/me because v, q, and B are identical for both particles. mp/me = 1.67 × 10−27 / 9.11 × 10−31 ≈ 1834. The proton’s radius is approximately 1834 times larger. 1 mark for explaining the proportionality r ∝ m (with q, v, B fixed); 1 mark for the correct numerical factor.
Q1.3 — Proton vs alpha period
Tp = 1.31 × 10−6 s; Tα = 2.60 × 10−6 s. The alpha’s period is longer. Ratio: Tα/Tp = (mα/qα) / (mp/qp) = (6.64 × 10−27 / 3.20 × 10−19) / (1.67 × 10−27 / 1.60 × 10−19) ≈ 2.0. The alpha’s period is twice the proton’s because its m/q ratio is double. 1 mark for identifying the alpha has a longer period; 1 mark for the factor of 2 with correct reasoning via T ∝ m/q.
Q2.1 — Graph relationship
The graph shows a linear (directly proportional) relationship between orbital radius and speed: as speed doubles, radius doubles. The line passes through the origin. This is consistent with r = mv/qB — for fixed m, q, B, radius is directly proportional to speed (r ∝ v). 1 mark for “linear / directly proportional”; 1 mark for referencing r ∝ v from the formula.
Q2.2 — Radius at v = 5.0 × 106 m/s
From graph: at v = 5 units the radius reads approximately 0.52 m. Verification: r = (1.67 × 10−27 × 5.0 × 106) / (1.60 × 10−19 × 0.10) = 8.35 × 10−21 / 1.60 × 10−20 = 0.522 m. 1 mark for correctly reading the graph; 1 mark for a correct calculation that agrees.
Q2.3 — Period when speed doubles
The student’s claim is incorrect. While doubling the speed does double the orbital radius (from the graph, r ∝ v), the period is T = 2πm/qB — it contains no v and is therefore independent of speed. A faster particle travels a larger circle in exactly the same time. Doubling the speed does not change the period at all. 1 mark for identifying the claim as incorrect; 1 mark for citing T = 2πm/qB and noting no v dependence; 1 mark for the physical explanation (larger circle at the same period).
Q3 — Comparison table
Source of centripetal force: Charged particle: magnetic force F = qvB. Satellite: gravitational force F = GMm/r².
Expression equated to mv²/r: Charged particle: qvB = mv²/r. Satellite: GMm/r² = mv²/r.
Radius formula: Charged particle: r = mv/qB. Satellite: r = GM/v².
How radius depends on speed: Charged particle: r ∝ v (radius increases with speed). Satellite: r ∝ 1/v² (radius decreases as speed increases — lower orbit = higher speed).
Does centripetal force do work? Both: No. In both cases the force is perpendicular to the velocity at every instant, so no work is done and the speed remains constant.
Marking criteria: 1 mark per correctly completed cell (5 rows × 2 cells = 10).
Q4.1 — Increasing radius by 20%
From r = mv/qB, rearranging gives B = mv/qr. With m, v, q fixed, B ∝ 1/r. If r increases by a factor of 1.2 (i.e. r’ = 1.2r), then B’ = B/1.2. The magnetic field strength must decrease by a factor of 1/1.2, i.e. decrease to approximately 83% of its original value. 1 mark for correct rearrangement B ∝ 1/r; 1 mark for stating B must decrease; 1 mark for the correct factor of 1/1.2 or equivalent.
Q4.2 — Effect on period and frequency
T = 2πm/qB. Since B decreases by factor 1/1.2, T increases by factor 1.2: T’ = 1.2T. The period increases by 20%. Correspondingly, f = qB/2πm: since B decreases, f decreases by factor 1/1.2, i.e. f’ = f/1.2. The cyclotron frequency decreases by approximately 17%. 1 mark for T increasing (T ∝ 1/B); 1 mark for f decreasing; 1 mark for the correct factor (1.2 for T, 1/1.2 for f) with formula citation.