Charged-particle orbits in uniform fields

Only one force acts on a free charge in a magnetic field, F = qv×B, and a cross product is perpendicular to both of its factors. Take the rate of change of kinetic energy: dK/dt = F⋅v = q(v×B)⋅v, and that dot product is zero because F is at right angles to v at every instant. So |v| is constant for all time. Two things follow at once. The size of the force, |q|v⊥B, is constant too, since neither v⊥ nor B changes — and a force of fixed size always pulling at right angles to the motion is exactly the condition for circular motion. And no magnetic field, however strong, can raise a particle's energy: a track in a detector curves without slowing, and a beam line steers with magnets but accelerates with electric fields.

Write v = v∥ + v⊥, the parts along B and across it. Then v×B = (v∥ + v⊥)×B = v⊥×B, because the cross product of parallel vectors vanishes. Two facts drop out. The force has no component along B, so v∥ is untouched and the particle coasts along the field line at constant speed forever. And the force on the perpendicular part always lies in the plane across B, so v⊥ can only rotate, never grow or shrink. The motion is therefore a uniform circle in the plane perpendicular to B superposed on uniform straight-line travel along B — a helix wound on a cylinder whose axis is a field line. Because the total speed is fixed, the split is fixed too: with v = 2.4 × 10⁶ m s⁻¹ at 35° to B, v⊥ = v sin 35° = 1.38 × 10⁶ m s⁻¹ and v∥ = v cos 35° = 1.97 × 10⁶ m s⁻¹, and those two numbers hold for the whole trajectory.

For the circular part the magnetic force is the centripetal force: |q|v⊥B = mv⊥²/r, so r = mv⊥/(|q|B). Read that as a statement about momentum, mv⊥ = |q|Br, which is why a curved track in a known field measures momentum per unit charge. Now take the turning rate: ω_c = v⊥/r = |q|B/m, and one power of v⊥ has cancelled. The angular frequency, and with it the period T = 2πm/(|q|B) and the frequency f_c = |q|B/(2πm), contain no speed at all. A fast particle runs a proportionally longer circumference at a proportionally higher speed and arrives back at the same instant. An electron in 3.0 mT circles with T = 11.9 ns, f_c = 84.0 MHz, whether its transverse speed is 2.0 × 10⁶ m s⁻¹, giving r = 3.79 mm, or ten times that, giving r = 37.9 mm. Swap in a proton and only the mass changes: T = 21.9 μs, longer by the mass ratio 1836.

The two motions share one clock. In the time T that the circle takes, the coasting part advances p = v∥T = 2πmv∥/(|q|B), and that advance is the pitch of the helix. Notice what the expression leaves out: v⊥. Raise the transverse speed and the helix grows fatter while its turns stay exactly as far apart as before. The pitch angle α between v and B collects both effects — at α = 90° there is no v∥ and the path closes into a circle, while as α → 0 the helix stretches towards a straight line. For the electron above in 1.8 mT, T = 19.8 ns and v∥ = 1.97 × 10⁶ m s⁻¹ give p = 39.0 mm, while the orbit is only 8.7 mm across: about 4.5 diameters of advance per turn. It is also why a magnetic field guides particles without confining them along itself — they stream freely down a field line.

Real fields fill finite regions, and a beam usually crosses a slab of one. Take a slab of thickness d with a uniform field inside and none outside, entered at right angles to the face. Inside, the path is a circular arc of radius r whose centre lies on the entry face, a distance r from the entry point. After turning through θ the particle has advanced r sin θ into the slab, so it reaches the far face when sin θ = d/r, leaving at that same angle θ to its original direction and displaced sideways by r(1 − cos θ). With r = 12 mm and d = 6.0 mm, θ = 30° and the sideways shift is 1.6 mm. The relation also announces its own limit: if r < d there is no solution, the arc turns back before reaching the far face, and the particle re-emerges from the entry face travelling backwards, displaced by 2r. Outside the field it resumes a straight line, at exactly its entry speed.

Three assumptions carried everything above. First, non-relativistic motion: at high speed the mass in ω_c = |q|B/m becomes γm, so the circulation rate falls as the energy rises while the speed itself stays fixed — a 5.0 MeV proton has γ = 1.0053 and takes 0.53% longer per turn, a small effect but a fatal one for any scheme that must stay in step with a fixed-frequency drive. Second, uniformity. Where B varies in space the orbit is still a fast tight circle, but its guiding centre drifts sideways, and along a field that strengthens the quantity mv⊥²/(2B) stays nearly constant, so v⊥ grows at the expense of v∥ until the particle reflects — the magnetic mirror that traps particles in the Van Allen belts. Third, an undisturbed particle: collisions in a gas interrupt the orbit, and an accelerating charge radiates, so a real high-energy orbit spirals slowly inwards instead of closing on itself.