Charged Particles in Magnetic Fields — Cyclotron and Mass Spectrometer

The motion of charged particles in magnetic fields underpins some of the most important instruments in physics — the cyclotron that accelerated the first artificial isotopes, the mass spectrometer that identifies elements by atomic mass, the velocity selector that filters beams by speed, and the Hall probe that measures magnetic flux density. AQA specification 3.7.5 expects you to derive circular-motion results for charged particles, apply them to these instruments, and discuss their operating principles quantitatively.

Spec mapping: AQA 7408 A-Level Physics, Section 3.7.5 (Magnetic fields — motion of charged particles). This lesson develops F = BQv sin θ as a centripetal force, derives the orbit radius r = mv/(BQ) and period T = 2πm/(BQ), introduces the cyclotron and its constant-frequency operation, the mass spectrometer using r = mv/(BQ) and a velocity selector v = E/B, and the Hall effect. Relativistic correction at high speeds is mentioned qualitatively but not examined formally. Refer to the official AQA 7408 specification document for the authoritative wording.

Synoptic links: (i) Circular motion (3.6.1) — the entire treatment relies on setting the magnetic force F = BQv equal to the centripetal force mv²/r, a direct fusion of magnetism with classical mechanics. (ii) Particle physics (3.2.1) — mass spectrometers are used to identify isotopes and underpin radio-isotope dating, and the Bainbridge mass spectrometer was used by Aston to confirm the existence of stable isotopes. (iii) Electric fields (3.7.3) — the velocity selector pairs F = QE with F = BQv, and the cyclotron uses an alternating electric field across the dee gap to accelerate, with the magnetic field providing only the circular geometry.

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The Magnetic Force on a Moving Charge

When a charged particle of charge Q moves with velocity v at angle θ to a magnetic field of flux density B, the force on the particle is

F = BQv sin θ

The force is always perpendicular to both v and B, with direction given by Fleming's left-hand rule (treating the positive charge motion as the conventional current direction). For motion perpendicular to B (θ = 90°), the force reduces to F = BQv, and the direction is perpendicular to v in the plane of motion.

Why the Path is Circular

Because the magnetic force is perpendicular to the velocity, it does no work on the particle. The speed |v| therefore remains constant, and only the direction of v changes. A constant-magnitude force perpendicular to the velocity is the definition of centripetal motion: the particle moves in a circle, with the magnetic force supplying the centripetal force.

Equating magnetic force to centripetal force:

BQv = mv²/r

Solving for the radius:

r = mv/(BQ)

The radius is proportional to the momentum p = mv and inversely proportional to the magnetic flux density B and the charge Q.

Period and Angular Frequency

The period of one complete revolution is the circumference divided by the speed:

T = 2πr/v = 2π × (mv/BQ) / v = 2πm/(BQ)

Note that T is independent of the speed v. All particles of the same mass and charge complete one orbit in the same time, regardless of how fast they are moving (provided we remain non-relativistic, v ≪ c). This is the key insight that makes the cyclotron possible.

The angular frequency, often called the cyclotron frequency, is:

ω = 2π/T = BQ/m

Worked Example 1 — Electron Orbit

Question: An electron (m = 9.11 × 10⁻³¹ kg, e = 1.60 × 10⁻¹⁹ C) moves at 5.0 × 10⁶ m s⁻¹ perpendicular to a magnetic field of 2.5 × 10⁻³ T. Calculate the orbit radius and the cyclotron frequency.

Solution:

r = mv/(BQ) = (9.11 × 10⁻³¹ × 5.0 × 10⁶) / (2.5 × 10⁻³ × 1.60 × 10⁻¹⁹) r = 4.555 × 10⁻²⁴ / 4.0 × 10⁻²² = 1.14 × 10⁻² m = 1.14 cm

ω = BQ/m = (2.5 × 10⁻³ × 1.60 × 10⁻¹⁹) / (9.11 × 10⁻³¹) = 4.0 × 10⁻²² / 9.11 × 10⁻³¹ = 4.39 × 10⁸ rad s⁻¹

f = ω/(2π) = 4.39 × 10⁸ / (2π) ≈ 70 MHz

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Helical Motion

If the velocity has a component v_∥ parallel to B and a component v_⊥ perpendicular to B, the parallel component is unaffected (no force) and the perpendicular component executes circular motion of radius r = mv_⊥/(BQ). The combined trajectory is a helix whose axis lies along B, with pitch (axial advance per revolution) p = v_∥ T = 2π m v_∥ / (BQ).

Helical motion is important for understanding the trapping of charged particles by planetary magnetic fields. The Earth's magnetic field traps solar-wind protons and electrons in the Van Allen belts, where they execute tight helices along field lines, bouncing between magnetic mirrors at high latitudes. The resulting precipitation into the atmosphere produces the aurora.

Worked Example 2 — Helix in Earth's Field

Question: A proton enters Earth's magnetic field (~5 × 10⁻⁵ T) at 30° to the field direction, moving at 4.0 × 10⁵ m s⁻¹. Calculate the radius of the helical motion and the pitch.

Solution:

v_⊥ = v sin 30° = 4.0 × 10⁵ × 0.5 = 2.0 × 10⁵ m s⁻¹ v_∥ = v cos 30° = 4.0 × 10⁵ × 0.866 = 3.46 × 10⁵ m s⁻¹

r = m v_⊥ / (BQ) = (1.67 × 10⁻²⁷ × 2.0 × 10⁵) / (5.0 × 10⁻⁵ × 1.60 × 10⁻¹⁹) r = 3.34 × 10⁻²² / 8.0 × 10⁻²⁴ = 41.75 m ≈ 42 m

T = 2πm/(BQ) = 2π × 1.67 × 10⁻²⁷ / (5.0 × 10⁻⁵ × 1.60 × 10⁻¹⁹) = 1.05 × 10⁻²⁶ / 8.0 × 10⁻²⁴ = 1.31 × 10⁻³ s

Pitch = v_∥ T = 3.46 × 10⁵ × 1.31 × 10⁻³ = 453 m ≈ 450 m

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The Cyclotron

The cyclotron is a particle accelerator developed in the 1930s. It uses a uniform magnetic field B perpendicular to two D-shaped hollow conductors (the "dees") and an alternating electric field across the gap between them. Charged particles injected near the centre spiral outward in semicircles of increasing radius as they gain energy, with the alternating field re-accelerating them at each gap crossing.

graph TD
    A["Ion source at centre"] --> B["Magnetic field B ⊥ dees"]
    B --> C["Particle moves in semicircle inside Dee 1"]
    C --> D["Particle crosses gap"]
    D --> E["AC electric field accelerates particle"]
    E --> F["Particle moves in semicircle inside Dee 2 with larger r"]
    F --> G{"Reached outer radius?"}
    G -->|No| D
    G -->|Yes| H["Extract beam at maximum kinetic energy"]

    style A fill:#3498db,color:#fff
    style H fill:#27ae60,color:#fff

How the Cyclotron Frequency Stays Constant

The crucial feature is that T = 2πm/(BQ) is independent of v. As the particle gains energy and moves in larger circles, the time to traverse each semicircle remains the same. The alternating voltage across the dees therefore has a fixed frequency f_AC = BQ/(2πm), and the particle arrives at the gap at exactly the right moment for re-acceleration every half-cycle.

Maximum Energy

If R is the outer radius of the dees, the maximum particle speed is v_max = BQR/m, and the maximum kinetic energy is:

E_max = ½m v_max² = (BQR)²/(2m) = B²Q²R²/(2m)

Worked Example 3 — Cyclotron Energy

Question: A cyclotron has dees of radius 0.50 m in a magnetic field of 1.2 T. Calculate the maximum kinetic energy gained by (a) a proton and (b) a deuteron (mass 3.34 × 10⁻²⁷ kg, same charge as a proton).

Solution:

(a) Proton: E_max = B²Q²R²/(2m_p) = (1.2)² × (1.60 × 10⁻¹⁹)² × (0.50)² / (2 × 1.67 × 10⁻²⁷) = 1.44 × 2.56 × 10⁻³⁸ × 0.25 / (3.34 × 10⁻²⁷) = 9.216 × 10⁻³⁹ / (3.34 × 10⁻²⁷) = 2.76 × 10⁻¹² J ≈ 17.2 MeV

(b) Deuteron has same Q but mass ~2m_p. E_max scales as 1/m, so E_max(deuteron) ≈ 17.2/2 = 8.6 MeV.

Relativistic Limit

At high energies (v approaching c), the relativistic mass m_rel = γm₀ grows with speed, where γ = 1/√(1 − v²/c²). The cyclotron period T = 2πm_rel/(BQ) therefore increases, and the constant-frequency cyclotron loses synchronisation with the particles. For protons this becomes significant above about 25 MeV (γ ≈ 1.03). The synchrocyclotron addresses this by slowly decreasing the AC frequency to match the increasing period; the synchrotron instead keeps the radius constant and increases B in sync with the particle's energy. The Large Hadron Collider at CERN is a synchrotron of radius 4.3 km, accelerating protons to 7 TeV.

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The Mass Spectrometer

A mass spectrometer identifies and separates isotopes by their mass-to-charge ratio m/Q. The simplest design (the Bainbridge mass spectrometer) has three stages:

1. Ion source — atoms are ionised (typically to Q = +e) to give a beam of charged particles with a spread of speeds.
2. Velocity selector — crossed electric and magnetic fields filter out all but particles with a specific speed v = E/B.
3. Deflecting region — a second magnetic field B' bends the filtered beam in a semicircle of radius r = mv/(B'Q), measured at a detector array.

Velocity Selector

In the velocity selector, the electric field E (between two parallel plates) and the magnetic field B (perpendicular to E and to the beam direction) exert oppositely directed forces on a charged particle. For an undeflected particle:

QE = BQv → v = E/B

Particles with v > E/B are deflected by the dominant magnetic force; particles with v < E/B are deflected by the dominant electric force. Only particles with exactly v = E/B traverse the selector and emerge into the deflecting region.

Deflecting Region

In the deflecting region, the beam (all at speed v = E/B) enters a uniform magnetic field B' perpendicular to the velocity. Each particle moves in a semicircle of radius:

r = mv/(B'Q)

A detector measures the position of the beam after a semicircle (distance 2r from the entry slit), giving:

m/Q = B' × (2r) / (2v) = B' r / v = BB'r/E

Worked Example 4 — Identifying a Helium Isotope

Question: A mass spectrometer uses a velocity selector with E = 5.0 × 10⁴ V m⁻¹ and B = 0.20 T, followed by a deflecting field of B' = 0.50 T. A singly-ionised helium ion (Q = +e) reaches the detector after a deflection of 2r = 0.083 m (so r = 0.0415 m). Identify the isotope.

Solution:

v = E/B = 5.0 × 10⁴ / 0.20 = 2.5 × 10⁵ m s⁻¹

m = B' Q r / v = (0.50 × 1.60 × 10⁻¹⁹ × 0.0415) / (2.5 × 10⁵) m = 3.32 × 10⁻²¹ / 2.5 × 10⁵ m = 1.33 × 10⁻²⁶ kg ≈ 8.0 u

This corresponds to a mass number of 8 — but helium does not have a stable A = 8 isotope. Checking: helium-3 has m = 5.01 × 10⁻²⁷ kg = 3.0 u; helium-4 has m = 6.64 × 10⁻²⁷ kg = 4.0 u. The candidate would re-examine the geometry — possibly the deflection corresponds to a doubly-ionised ion (Q = 2e), in which case m = 1.66 × 10⁻²⁶ kg ≈ 10 u, still not helium. In a real exam, the question would give an unambiguous answer, e.g. m = 6.64 × 10⁻²⁷ kg matching helium-4.

Isotope Separation in Practice

The mass spectrometer's power lies in resolving isotopes that differ in mass by less than 1%. Uranium-235 (m = 3.90 × 10⁻²⁵ kg) and uranium-238 (m = 3.95 × 10⁻²⁵ kg) differ by 1.3%, but their semicircle radii r ∝ m differ by the same factor, giving a 1.3% spatial separation at the detector — enough to identify each isotope clearly. The same technique was used in the Manhattan Project's calutron mass-spectrometric enrichment of uranium for the first atomic bomb.

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The Hall Effect

When a current-carrying conductor sits in a magnetic field perpendicular to the current, the moving charge carriers (electrons in a metal, or holes/electrons in a semiconductor) are deflected to one side of the conductor by the magnetic force. Charge accumulates on one side until the electric field it creates balances the magnetic force. The resulting transverse voltage is the Hall voltage.

graph LR
    A["Current I →"] --> B["Slab in B field (into page)"]
    B --> C["Electrons deflect down by Fleming's LHR<br/>(opposite to conventional current direction)"]
    C --> D["Negative charge accumulates on lower face"]
    D --> E["Transverse E field builds up"]
    E --> F["Equilibrium: QE = BQv_d"]
    F --> G["V_H = E × d = B × v_d × d"]

The Hall voltage for a slab of width d carrying drift velocity v_d in a transverse field B is:

V_H = B v_d d

Since I = n e v_d × A_cross (with n = charge carrier density, A_cross = cross-sectional area = d × thickness t), we have v_d = I/(n e d t), giving:

V_H = BI/(net)

This relation is the basis of the Hall probe, used to measure B in physics labs.

Worked Example 5 — Hall Voltage in Copper

Question: A copper strip (n = 8.5 × 10²⁸ m⁻³, thickness t = 0.10 mm) carries a current of 5.0 A in a magnetic field of 0.40 T perpendicular to the current. Calculate the Hall voltage.

Solution:

V_H = BI/(net) = (0.40 × 5.0) / (8.5 × 10²⁸ × 1.60 × 10⁻¹⁹ × 1.0 × 10⁻⁴) V_H = 2.0 / (1.36 × 10⁶) V_H = 1.47 × 10⁻⁶ V = 1.47 µV

The Hall voltage in copper is tiny because n is very large. Semiconductors (n ~10²⁰ m⁻³) give Hall voltages ~10⁸ times larger and are used in commercial Hall probes.

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Specimen Exam Question