Lorentz Force and Cyclotron Motion

Derivation

Lorentz Force and Cyclotron Motion

D-055 Home PU-102 Threads force · fields Depends on Electric Field from Coulomb's Law, Biot–Savart Law from the Current Force Law

Statement

A point charge q moving with velocity v in the presence of an electric field E and a magnetic field B experiences the force F = q(E + v × B). For uniform, static E and B the resulting equation of motion is linear, and its solution is a helix: uniform circular motion at the cyclotron angular frequency ω_c = |q|B/m in the plane perpendicular to B, superposed with uniform drift along B plus the E×B drift.

Why it matters

The Lorentz force is the operational definition of the electromagnetic field: it is how E and B are measured, and it is the single equation that couples Maxwell's fields back to matter. Everything from a cathode-ray tube to a mass spectrometer to a tokamak is an exercise in choosing E and B so that charges follow useful trajectories.

Cyclotron motion is the prototype: it fixes the frequency of particle accelerators, sets the gyroradius that determines magnetic confinement of fusion plasmas, and — through ω_c being independent of speed at low energy — is what makes the classical cyclotron work at all. The same ω_c reappears quantum-mechanically as the spacing of Landau levels.

Assumptions

Fields are uniform and static.If E or B vary in space or time the equation of motion is no longer linear with constant coefficients; the clean helix is replaced by drifts (grad-B, curvature) and the guiding-centre approximation is needed.

The charge is a structureless point (no spin, no radius).A real current loop or spinning particle also feels a torque and a force −∇(m·B); dropping this omits magnetic-moment dynamics and spin precession.

Speeds are non-relativistic, v ≪ c.At relativistic speeds m must be replaced by the relativistic momentum p = γmv; the gyrofrequency becomes ω_c = |q|B/(γm) and falls with energy — the reason the plain cyclotron fails and the synchrotron is needed.

Radiation reaction is neglected.An accelerating charge radiates (Larmor); the resulting energy loss slowly shrinks the orbit. Ignoring it is excellent for heavy particles at modest B but fails for electrons in strong fields (synchrotron radiation).

Derivation

F = q(E + v × B)

Empirical Lorentz force law; taken as the definition of E and B acting on charge q. A

m dv/dt = qE + q v × B

Newton's second law with this force; non-relativistic so m is constant and p = mv. A

Choose B = B ẑ, E = E_x x̂ + E_z ẑ

A uniform B defines a natural axis; align ẑ with it, and drop any E_y by rotating axes about ẑ. No loss of generality. B

v × B = (v_yB) x̂ − (v_xB) ŷ

Cross product of v = (v_x,v_y,v_z) with Bẑ; the ẑ component vanishes because ẑ×ẑ = 0. A

v̇_x = (q/m)E_x + (qB/m)v_y
v̇_y = −(qB/m)v_x
v̇_z = (q/m)E_z

Take components of step 2 with step 4 substituted. The three axes decouple into a 2×2 rotational block (x,y) and a free 1-D block (z). B

Define ω_c ≡ qB/m

The combination qB/m has units of inverse time and is the only frequency in the problem; naming it is a definition, not a step. (Signed here; the observed rate is |ω_c|.) A

v̇_z = (q/m)E_z ⇒ v_z(t) = v_z0 + (qE_z/m) t

The parallel equation is a constant acceleration; integrate once. Motion along B is unaffected by the magnetic force. A

Let u_x = v_x − v_d, with v_d = −E_x/B

Shift to the frame drifting at v_d so the constant E_x term is absorbed: with this choice v̇_x = ω_cv_y becomes u̇_x = ω_cv_y and the perpendicular block is homogeneous. C

u̇_x = ω_cv_y, v̇_y = −ω_cu_x

Substitute step 8 into step 5. Check: −ω_cu_x = −ω_c(v_x+E_x/B) = −ω_cv_x since ω_cE_x/B = (qE_x/m) is cancelled — this is exactly the E_x term of step 5. C

d/dt(u_x + iv_y) = −iω_c(u_x + iv_y)

Form the complex combination w = u_x + iv_y: adding step 9's first equation to i times the second gives ẇ = ω_cv_y − iω_cu_x = −iω_cw. A standard trick to diagonalise the 2×2 rotation. C

w(t) = w₀ e^−iω_ct

First-order linear ODE with constant coefficient; the solution is a pure exponential of constant modulus |w| = v_⊥. Constant modulus already shows the perpendicular speed is conserved. C

u_x = v_⊥ cos(ω_ct − φ), v_y = −v_⊥ sin(ω_ct − φ)

Take real and imaginary parts of step 11 with w₀ = v_⊥e^iφ. This is uniform rotation of the velocity vector at rate ω_c. B

x(t) = v_dt + r_c sin(ω_ct−φ) + x_g
y(t) = r_c cos(ω_ct−φ) + y_g

Integrate v_x = u_x+v_d and v_y once, with gyroradius r_c ≡ v_⊥/ω_c = mv_⊥/(qB). The (x_g,y_g) is the fixed guiding centre. B

v_d = −E_x/B ⇒ v_d = (E × B)/B²

Rewrite the scalar drift of step 8 in vector form: with E = E_xx̂ and B = Bẑ, (E×B)/B² = E_x(x̂×ẑ)/B = −(E_x/B)ŷ. Note this drift is independent of q, m and sign. C

Result

m dv/dt = q(E + v×B) ⟹ helix: ω_c = |q|B/m, r_c = mv_⊥/(|q|B), v_d = (E×B)/B²

Reading. Perpendicular to B the charge circles at the cyclotron frequency ω_c, tracing a circle of radius r_c whose centre (the guiding centre) drifts at v_d = (E×B)/B². Parallel to B it moves freely, accelerating only if E has a component along B. The full path is a helix (a drifting cycloid when a transverse E is present). Crucially ω_c is independent of speed and of r_c: faster particles trace bigger circles in the same time. The magnetic force does no work, so |v_⊥| is constant.

Units check. [ω_c] = C·T/kg = C·(kg·s⁻¹·C⁻¹)/kg = s⁻¹ ✓ (using 1 T = 1 kg·s⁻¹·C⁻¹). [r_c] = kg·(m·s⁻¹)/(C·T) = kg·m·s⁻¹/(C·kg·s⁻¹·C⁻¹) = m ✓. [v_d] = (V·m⁻¹)/T = (kg·m·s⁻³·A⁻¹)/(kg·s⁻²·A⁻¹) = m·s⁻¹ ✓.

Limiting cases

E = 0: pure helix — circular gyration at ω_c plus uniform glide along B; guiding centre stationary in the perpendicular plane.
B = 0: parabolic motion under the constant force qE — ordinary projectile-like kinematics; ω_c and r_c undefined.
v_⊥ = 0: charge released along B stays on the field line (no gyration, r_c = 0); a magnetic field alone cannot turn it.
E ⊥ B with v(0)=v_d: the gyration amplitude vanishes and the charge moves in a straight line at v_d — the principle of the velocity selector (v_d = E/B).
Strong-field / small r_c: gyration is fast and tight; only the guiding-centre drift survives on macroscopic scales — the drift approximation of plasma physics.

Breaks when

Fields are non-uniform. A gradient in B makes r_c vary around one orbit, producing a grad-B drift ∝ (B×∇B); field-line curvature adds a curvature drift. The single helix is replaced by guiding-centre theory, valid only when r_c ≪ scale of variation.
Speeds approach c. p = γmv makes the effective mass energy-dependent, so ω_c = |q|B/(γm) drops as the particle speeds up. A fixed-frequency cyclotron falls out of phase — the historical reason for the synchrocyclotron and synchrotron.
Radiation is significant. Circular motion is accelerated motion, so the charge radiates (synchrotron/cyclotron radiation). For light particles in strong B the orbit slowly decays, violating the conservation of v_⊥ assumed here.
Time-varying fields. A changing B induces an E (Faraday), and resonant time-varying E at ω_c pumps energy in (cyclotron resonance / betatron acceleration) — the orbit is no longer a closed circle of fixed radius.
Collisions / many particles. In a dense plasma inter-particle forces and collisions randomise the phase before an orbit completes; single-particle helices are meaningless and a kinetic or fluid description is required.

Failure modes

Sign confusion in v×B: forgetting that v×B is antisymmetric and getting the sense of rotation backwards. Electrons and positive ions gyrate in opposite senses about the same B.
Thinking the magnetic force does work: since F ⊥ v always, F·v = 0 and magnetic forces never change kinetic energy. Only E can. Students routinely try to "accelerate" a particle with B alone.
Believing ω_c depends on speed or radius: the whole point is that it does not (non-relativistically). Confusing period with radius leads to wrong cyclotron-timing arguments.
Putting v_d = E/B in the wrong direction: the drift is along E×B, perpendicular to both fields — not along E.
Assuming v_d depends on charge: the E×B drift is identical for electrons and ions (no q, m), so it drives no current — unlike grad-B and curvature drifts.
Using non-relativistic r_c for fast particles: forgetting the γ in p = γmv underestimates the gyroradius of MeV electrons badly.

Discussion

The Lorentz force is where the fields become physics: Maxwell's equations tell you how E and B are produced and propagate, but only F = q(E+v×B) tells you what they do to matter. It is manifestly not derivable from the earlier electrostatic result alone — the v×B piece is a genuinely new, velocity-dependent, geometry-mixing term. That the two combine into a single Lorentz-covariant object F^μ = qF^μνu_ν is one of the strongest hints, historically, that space and time must mix: what one observer calls a pure B another calls a mixture of E and B.

The cyclotron frequency is remarkable for what it omits. Two particles of the same q/m circle in lockstep regardless of how fast they move or how big their orbits are, because a faster particle has proportionately more inertia to bend. This speed-independence is the design principle of Lawrence's cyclotron: a fixed-frequency accelerating gap kicks the particle every half-turn no matter its energy — until relativity spoils the constancy of ω_c. The gyroradius, by contrast, grows with momentum, which is why r_c = p_⊥/(|q|B) is used to measure momentum in bubble chambers and tracking detectors: photograph the curvature, read off the momentum.

The E×B drift carries a deep lesson about magnetised matter. Because it is independent of charge, mass and sign, an entire plasma — electrons and ions alike — drifts together across the fields, carrying no net current. This is why cross-field transport in fusion devices and in planetary magnetospheres is dominated by E×B convection, and why the Hall thruster and the magnetron both exploit it. The velocity selector is the same physics run in reverse: tune E/B so that only particles with v = E/B pass undeflected.

Quantum-mechanically the classical circular orbits become the Landau levels: a charged particle in a uniform B has quantised perpendicular energy E_n = (n+½)ℏω_c, exactly the harmonic-oscillator ladder with the classical cyclotron frequency as its rung spacing. The macroscopic degeneracy of these levels underlies the de Haas–van Alphen effect and the integer quantum Hall effect, where the Hall conductance is quantised in units of e²/h. Even the semiclassical guiding-centre picture survives: the magnetic moment μ = ½mv_⊥²/B associated with the gyration is an adiabatic invariant, conserved when the fields change slowly compared with ω_c — the principle behind magnetic mirrors and the trapping of particles in the Van Allen belts.

Common misconceptions. A magnetic field cannot speed a particle up (it does no work), it can only redirect it; "energy from a cyclotron" comes entirely from the electric gap voltage, not from B. The E×B drift is not "the particle being pushed by E" — it is perpendicular to E, and its speed E/B can be tiny even for a strong E. And the cyclotron period is not the orbital period one would naively get from "circumference over speed" unless you use the correct r_c — the two are consistent precisely because r_c = v_⊥/ω_c.

Worked examples

Example 1 — Gyrofrequency and gyroradius of a proton. A proton (q = 1.60×10⁻¹⁹ C, m = 1.67×10⁻²⁷ kg) moves with perpendicular speed v_⊥ = 1.0×10⁵ m/s in a uniform field B = 0.50 T. Find ω_c, the period, and r_c.

ω_c = |q|B/m

Result of the derivation; symbols first. A

ω_c = (1.60×10⁻¹⁹)(0.50)/(1.67×10⁻²⁷) = 4.79×10⁷ rad/s

Insert numbers; units C·T/kg = s⁻¹. A

T = 2π/ω_c = 2π/(4.79×10⁷) = 1.31×10⁻⁷ s

Period of one gyration; independent of v_⊥. A

r_c = v_⊥/ω_c = (1.0×10⁵)/(4.79×10⁷) = 2.09×10⁻³ m

Gyroradius from r_c = v_⊥/ω_c. A

ω_c ≈ 4.8×10⁷ rad/s, T ≈ 0.13 μs, r_c ≈ 2.1 mm

Reading. The proton circles about 7.6 million times per second on a millimetre-scale orbit; doubling its speed would double r_c but leave ω_c and T unchanged.

Example 2 — Velocity selector. Crossed fields E = 3.0×10⁴ V/m (along x̂) and B = 0.20 T (along ẑ) are arranged so a positive ion travelling along ŷ passes undeflected. What ion speed is selected, and what is the E×B drift speed?

Undeflected ⇒ net force zero ⇒ qE = qvB

Electric and magnetic forces must cancel; for v along ŷ, v×B is along −x̂, opposing E. B

v = E/B

Cancel q — the selected speed is independent of charge and mass. A

v = (3.0×10⁴)/(0.20) = 1.5×10⁵ m/s

Insert numbers; (V·m⁻¹)/T = m·s⁻¹. A

v_d = |E×B|/B² = E/B = 1.5×10⁵ m/s

The E×B drift equals the selected speed — a particle launched at exactly v_d has zero gyration amplitude and moves straight. B

v = E/B = 1.5×10⁵ m/s

Reading. Only ions at 150 km/s survive regardless of their charge or mass; slower ions are bent toward −x̂, faster ions toward +x̂. The selector is exactly the special case of cyclotron motion where the initial velocity equals the E×B drift.

Problems

An electron (m = 9.11×10⁻³¹ kg, |q| = 1.60×10⁻¹⁹ C) moves in B = 0.010 T. Find its cyclotron frequency f_c = ω_c/2π.
Solution
ω_c = |q|B/m = (1.60×10⁻¹⁹)(0.010)/(9.11×10⁻³¹) = 1.756×10⁹ rad/s. f_c = ω_c/2π = 2.80×10⁸ Hz ≈ 280 MHz. (This is the ~28 GHz/T electron cyclotron frequency scaled to 0.01 T.)
A proton and an alpha particle (charge 2e, mass ≈ 4m_p) enter the same B with the same speed v_⊥. Find the ratio of their gyroradii and of their cyclotron periods.
Solution
r_c = mv_⊥/(|q|B). Ratio r_α/r_p = (m_α/m_p)(q_p/q_α) = (4)(1/2) = 2. Period T = 2πm/(|q|B), so T_α/T_p = (4)(1/2) = 2. The alpha traces a circle twice as large and takes twice as long.
A proton with kinetic energy K = 1.0 MeV moves perpendicular to B = 1.2 T. Find its gyroradius. (Treat non-relativistically; m_p = 1.67×10⁻²⁷ kg, 1 MeV = 1.60×10⁻¹³ J.)
Solution
v_⊥ = √(2K/m) = √(2·1.60×10⁻¹³/1.67×10⁻²⁷) = √(1.916×10¹⁴) = 1.384×10⁷ m/s (≈0.046c, so non-relativistic is fine). r_c = mv_⊥/(|q|B) = (1.67×10⁻²⁷)(1.384×10⁷)/[(1.60×10⁻¹⁹)(1.2)] = 2.311×10⁻²⁰/1.92×10⁻¹⁹ = 0.120 m ≈ 12 cm.
In crossed fields E = 5.0×10³ V/m and B = 0.25 T (mutually perpendicular), an electron starts from rest. Describe its trajectory and find the maximum speed it reaches and the drift speed.
Solution
Starting from rest is v(0) = 0, which differs from the drift v_d = E/B = (5.0×10³)/(0.25) = 2.0×10⁴ m/s. The motion is a cycloid: guiding centre drifts at v_d = 2.0×10⁴ m/s along E×B, with gyration of amplitude equal to v_d superposed. In the drift frame the speed is constant at v_d, so in the lab frame the speed oscillates between 0 (at cusps) and 2v_d = 4.0×10⁴ m/s (at the tops of the arches). Max speed = 4.0×10⁴ m/s; average drift = 2.0×10⁴ m/s.
A charged particle has velocity components v_∥ = 3.0×10⁵ m/s along B and v_⊥ = 4.0×10⁵ m/s across it, with B = 0.30 T and q/m = 9.6×10⁷ C/kg. Find the pitch of the helix (axial advance per gyration).
Solution
ω_c = (q/m)B = (9.6×10⁷)(0.30) = 2.88×10⁷ rad/s. Period T = 2π/ω_c = 2.182×10⁻⁷ s. Pitch = v_∥·T = (3.0×10⁵)(2.182×10⁻⁷) = 6.5×10⁻² m ≈ 6.5 cm. (The gyroradius, for comparison, is r_c = v_⊥/ω_c = 1.39 cm; v_⊥ plays no role in the pitch.)