Charged Particles in a Magnetic Field

Choose a particle, the voltage that accelerated it and the strength of the field it crosses. The page works out the circle it moves on, how long one lap takes with relativity included, and what the same voltage and field do to five other particles.

Worked example — Charged Particles in a Magnetic Field

Rest energy	938.272 MeV
Kinetic energy from the voltage	10.00 keV
Lorentz factor γ	1.000011
Speed	1.384×10⁶ m/s
Momentum	2.315×10⁻²¹ kg·m/s
Radius of the circle	9.633 mm
Lap time with relativity left out	43.73 ns
Lap time	43.73 ns

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Charged Particles in a Magnetic Field

A charge q moving with velocity v through a magnetic field B feels the magnetic part of the Lorentz force, F = q v × B: at right angles to both the velocity and the field, of size qvB when the two are perpendicular. A force that is always at right angles to the motion does no work, so the speed never changes and only the direction does, and the path is a circle. Its radius is r = p/(|q|B), the momentum over charge times field, and one lap takes T = 2πγm/(|q|B). At speeds well below light's, γ is 1 to many decimal places and the lap time has no speed in it at all; its inverse, qB/(2πm), is the cyclotron frequency.

What each symbol means

q: the particle's charge, in coulombs: a whole number of elementary charges e = 1.602176634×10⁻¹⁹ C. Its sign sets the direction of turn
B: the magnetic field, in tesla, the same everywhere the particle goes
r: radius of the circle. It is set by the momentum, not by the speed or the energy on their own
T_0: the lap time with relativity left out, 2πm/(|q|B)
γ: the Lorentz factor, 1 + K/(mc²). The true lap time is γT₀

Assumes: A field that is uniform and steady, and a particle moving exactly across it: any part of the velocity along the field is untouched by it, and the circle becomes a helix. A vacuum, so nothing slows the particle. The particle is taken not to radiate, although a circling charge does; for the electron preset the energy it radiates in a lap is below a millionth of an electronvolt, but at a gigaelectronvolt in a strong field it becomes a real loss, which is why electron synchrotrons have to keep feeding energy back in. The uranium ions are singly charged atoms, their mass the atomic mass less one electron.
Breaks when: Relativity is built in, so a fast particle does not break the page. What breaks it is the magnet: a real field weakens towards its edge, and in a cyclotron it has to be shaped both to keep the beam focused and to stay close to resonance, which is the subject of Rose's 1938 paper. In a working cyclotron the electric field in the gap adds energy every half lap, so the real path is a spiral of half circles of growing radius rather than one circle. And many ions in one beam repel each other and spread the beam, a problem large enough to be in the title of William Parkins's account of the wartime calutron.

Twice as fast, a circle twice as big, the same 43.73 ns 🖖

Press Proton, 10 kV and then Proton, 40 kV. Four times the energy doubles the speed, and the circle doubles with it, from 9.633 to 19.27 mm. The lap takes 43.73 ns both times. The speed cancels: the radius is mv/(qB), the lap is 2πr/v, and what is left is 2πm/(qB), with no v in it. Ernest Lawrence and Stanley Livingston's 1932 paper on the cyclotron rests on that cancellation. Because a faster proton needs no less time for its half circle, one fixed radio frequency across the gap between two hollow electrodes keeps kicking it at the right moment as it spirals outwards, and they produced 1.22 MeV protons from a tube to which no more than 4,000 V was ever applied. In the panel of three orbits the dots mark each orbit every quarter of the classical lap, and at these energies the dots on all three circles sit on the same four spokes.

A 20 MeV proton laps 2.132% late, and a fixed frequency cannot follow 🖖

The exact lap time is 2πγm/(qB), and γ grows with the energy. Proton, 20 MV has γ = 1.021316, so each lap takes 44.66 ns instead of 43.73, while the radio frequency driving it is still tuned to the slow proton. Every lap it arrives a little later at the gap, and the lateness adds up until it arrives while the field is pointing the wrong way. For an electron the useful range is tiny. Set it to 20 kV and its lap is already 3.914% long; Electron, 1 MV has γ = 2.956951, a lap nearly three times its slow value. The second panel draws T ÷ T₀ for four particles; each curve turns up when the kinetic energy reaches a few per cent of the rest energy, which for an electron is tens of keV and for a proton tens of MeV. M. E. Rose worked out in 1938 that with 50 kV on the electrodes and a carefully shaped field a fixed-frequency cyclotron could bring protons to 13.9 MeV, or 14.9 with a more complicated field. The worked problem below makes a rougher estimate of that ceiling, and finds why it rises only as the square root of the voltage.

Uranium-238 is 1.28% heavier and its circle only 0.638% wider 🖖

Every particle through the same voltage gets the same energy per unit of charge, and a slow particle's momentum is √(2mK), so for the same charge the radius goes as the square root of the mass. Uranium-235, 35 kV circles at 1.2145 m. In the table beside it, uranium-238, 238.0502 u against 235.0434, circles at 1.2223 m: 1.28% more mass, a radius 1.00638 times as large. In an instrument that bends the beam through half a circle to a collector, the two isotopes land 15.5 mm apart, twice the difference of the radii. That is the whole of the separation a magnet gives them at these settings, and it is the method the wartime calutrons used to separate uranium-235.

Problem solved in full

How far can a fixed-frequency cyclotron push a proton? 6 steps
A cyclotron has a 1.5 T magnet, hollow electrodes 0.5 m in radius, and gives a proton 100 keV every lap. Find the radio frequency it has to run at, the energy of a proton that reaches the rim, and whether a fixed frequency can get it there. Press Proton, 10 kV to check the first step.

Open this problem in the tool
1. The frequency is the cyclotron frequency of a slow proton, which the page prints at 10 kV:
  
  $f_{0} = \frac{q B}{2 π m} = \frac{1 . 602177 \times 1 0 ^{- 19} \times 1 . 5}{2 π \times 1 . 672622 \times 1 0 ^{- 27}} = 22.87 MHz T_{0} = 43.73 ns$
2. At the rim the radius is 0.5 m, so the momentum is qBR. Written as pc it is an energy:
  
  $p c = q B R c = 1.602177 \times 1 0^{- 19} \times 1.5 \times 0.5 \times 2.997925 \times 1 0^{8} J = 224.84 MeV$
3. The kinetic energy is the total energy less the rest energy. Leaving relativity out would give 26.94 MeV, 1.4% too much:
  
  $K = (p c)^{2} + (m c^{2})^{2} - m c^{2} = 224.8 4^{2} + 938.2 7^{2} - 938.27 = 26.56 MeV \frac{( p c ) ^{2}}{2 m c ^{2}} = 26.94 MeV$
4. At that energy γ is 1.028312, so a lap at the rim takes 2.83% longer than a lap at the centre:
  
  $γ = 1 + \frac{26 . 56}{938 . 27} = 1.028312 \Rightarrow T_{rim} = 1.028312 T_{0}$
5. At 100 keV a lap the proton needs 266 laps to get there. On lap k it is γ − 1 = kΔE/mc² late, as a fraction of a lap, and the lateness adds up, to 3.78 whole cycles of the radio frequency by the time it would reach the rim:
  
  $n = \frac{26 . 56}{0 . 1} \approx 266 k = 1 \sum n \frac{k Δ E}{m c ^{2}} = \frac{Δ E}{m c ^{2}} \frac{n ( n + 1 )}{2} = \frac{0 . 1}{938 . 27} \times \frac{266 \times 267}{2} = 3.78$
6. It is pushed forward only while the field in the gap points forwards, half of each cycle, so it stops gaining once the accumulated lateness reaches half a cycle, π of phase:
  
  $\frac{Δ E}{m c ^{2}} \frac{n ( n + 1 )}{2} = \frac{1}{2} \Rightarrow n (n + 1) = \frac{938 . 27}{0 . 1} = 9383 \Rightarrow n = 97, K = 9.7 MeV \approx m c^{2} Δ E = 9.69 MeV$
Answer

The magnet can hold a 26.56 MeV proton at its rim; a fixed frequency can accelerate one to about 9.7 MeV at 100 keV a lap, and then the proton falls out of step. Quadruple the energy per lap to 400 keV and the ceiling only doubles, to 19.2 MeV, because it goes as √(mc²·ΔE). This estimate gives every lap the full 100 keV, which is generous; Rose's 1938 calculation, with a shaped field, reached 13.9 MeV with 50 kV on the electrodes and found the same square-root law. The machines that went further lower the frequency as the proton gains energy, which puts back the γ that a fixed frequency leaves out.

$K_{m a x} \approx m c^{2} Δ E : Δ E = 100 keV \Rightarrow 9.7 MeV, Δ E = 400 keV \Rightarrow 19.2 MeV$

References (3)

The cyclotron paper, and claim 1 in its authors' words: because the radii are proportional to the velocities, the time for a half circle is independent of them. It reports 1.22 MeV protons from a tube with at most 4,000 V applied: E. O. Lawrence and M. S. Livingston, "The Production of High Speed Light Ions Without the Use of High Voltages." Physical Review 40(1), 1932, pp. 19-35. DOI 10.1103/PhysRev.40.19
Why a fixed-frequency cyclotron tops out: the field has to be shaped for focusing and for resonance at once, and with 50 kV on the dees the best fields give protons of 13.9 and 14.9 MeV, with maximum energies rising as the square root of the dee voltage: M. E. Rose, "Focusing and Maximum Energy of Ions in the Cyclotron." Physical Review 53(5), 1938, pp. 392-408. DOI 10.1103/PhysRev.53.392
Separating uranium-235 from uranium-238 with magnets during the Second World War, told by one of the physicists who did it, and the beam-spreading problem that dominated the work: W. E. Parkins, "The Uranium Bomb, the Calutron, and the Space-Charge Problem." Physics Today 58(5), 2005, pp. 45-51. DOI 10.1063/1.1995747

By the 42Math team · Reviewed against published sources. Corrections welcome.