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.

The magnetic force: across both the velocity and the field	$\vec F=q\,\vec v\times\vec B,\qquad F=\|q\|vB\sin\theta$
Across a uniform field: a circle, and a lap time with no speed in it	$\begin{gathered}r=\frac{mv}{\|q\|B}\\ T=\frac{2\pi m}{\|q\|B},\qquad f=\frac{\|q\|B}{2\pi m}\end{gathered}$
Near the speed of light: the momentum is γmv	$\begin{gathered}p=\gamma mv,\qquad r=\frac{p}{\|q\|B}\\ T=\gamma\,\frac{2\pi m}{\|q\|B}\end{gathered}$

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 |q|vB 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, |q|B/(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

Where the formula comes from

The force. It is at right angles to the velocity, so it does no work: the speed never changes, only the direction.
$\vec F=q\,\vec v\times\vec B,\qquad\vec F\perp\vec v\ \Rightarrow\ W=0$
A circle. A force of constant size always across the motion is a centripetal force, and setting it equal to mv²/r gives the radius.
$|q|vB=\frac{mv^{2}}{r}\ \Rightarrow\ r=\frac{mv}{|q|B}$
The lap. Distance 2πr over speed v. The v in r cancels the v in the denominator.
$T=\frac{2\pi r}{v}=\frac{2\pi m}{|q|B}$
From the voltage. A charge that falls through ΔV from rest has kinetic energy |q|ΔV, which gives its speed and so its radius.
$|q|\,\Delta V=\tfrac12mv^{2}\ \Rightarrow\ r=\frac{\sqrt{2m|q|\,\Delta V}}{|q|B}$
Fast. The radius is set by the momentum, γmv, and the lap by γ times the slow lap. These are what the table works with at every speed.
$\begin{gathered}p=\gamma mv,\qquad r=\frac{p}{|q|B}\\ T=\gamma\,T_0\end{gathered}$

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.

Problems solved in full

From the accelerating voltage to the circle 3 steps
A proton is accelerated from rest through 5.0 kV and enters a 0.40 T field at right angles. Find its speed, the radius of its circle and the time for one lap.

Open this problem in the tool
1. The energy it gains falling through the voltage is all kinetic:
  
  $\tfrac12mv^{2}=|q|\,\Delta V\ \Rightarrow\ v=\sqrt{\frac{2|q|\,\Delta V}{m}}=\sqrt{\frac{2\,(1{.}602\times10^{-19})(5{.}0\times10^{3})}{1{.}673\times10^{-27}}}\approx9{.}787\times10^{5}\ \mathrm{m/s}$
2. The magnetic force supplies the centripetal force:
  
  $|q|vB=\frac{mv^{2}}{r}\ \Rightarrow\ r=\frac{mv}{|q|B}=\frac{(1{.}673\times10^{-27})(9{.}787\times10^{5})}{(1{.}602\times10^{-19})(0{.}40)}\approx25{.}54\ \mathrm{mm}$
3. One lap is the circumference over the speed, and the speed cancels:
  
  $T=\frac{2\pi r}{v}=\frac{2\pi m}{|q|B}=\frac{2\pi\,(1{.}673\times10^{-27})}{(1{.}602\times10^{-19})(0{.}40)}\approx164{.}0\ \mathrm{ns}$
Answer

v = 9.787 × 10⁵ m/s, r = 25.54 mm, T = 164.0 ns. At 5.0 kV the proton moves at a third of a per cent of the speed of light, and relativity shifts these figures by a few parts in a million.

$v\approx9{.}787\times10^{5}\ \mathrm{m/s},\quad r\approx25{.}54\ \mathrm{mm},\quad T\approx164{.}0\ \mathrm{ns}$
The force on an electron, and which way it turns 3 steps
An electron accelerated through 50 V moves along +x into a field of 0.50 mT along +z. Find the force on it as it enters, the radius of its circle, and which way it goes round, seen from +z.

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Show solution
1. Its speed, from the 50 eV it gained:
  
  $v=\sqrt{\frac{2e\,\Delta V}{m_e}}=\sqrt{\frac{2\,(1{.}602\times10^{-19})(50)}{9{.}109\times10^{-31}}}\approx4{.}194\times10^{6}\ \mathrm{m/s}$
2. The direction comes from the cross product. x̂ × ẑ = −ŷ, and the electron's charge is negative, which turns it round:
  
  $\vec F=q\,\vec v\times\vec B=(-e)\,vB\,(\hat x\times\hat z)=(-e)\,vB\,(-\hat y)=+evB\,\hat y$
3. The size of the force, and the radius it makes:
  
  $F=evB=(1{.}602\times10^{-19})(4{.}194\times10^{6})(5{.}0\times10^{-4})\approx3{.}359\times10^{-16}\ \mathrm{N},\qquad r=\frac{m_ev}{eB}\approx47{.}69\ \mathrm{mm}$
Answer

F = 3.359 × 10⁻¹⁶ N along +y; r = 47.69 mm; anticlockwise seen from +z. A proton in the same field would be pushed along −y and go round clockwise. Every right-hand rule works out the force on a positive charge, so for an electron reverse the answer.

$F\approx3{.}359\times10^{-16}\ \mathrm{N}\ (+y),\qquad r\approx47{.}69\ \mathrm{mm}$
Three particles through the same voltage 3 steps
A proton, a deuteron and an alpha particle are each accelerated through 10 kV into the same 1.5 T field. Find the radius of each circle.

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Show solution
1. Through the same voltage each gains |q|ΔV, so the momentum is √(2m|q|ΔV) and the radius goes as the square root of m/q:
  
  $r=\frac{\sqrt{2m|q|\,\Delta V}}{|q|B}=\frac{1}{B}\sqrt{\frac{2m\,\Delta V}{|q|}}$
2. The proton and the deuteron each carry one e; the deuteron has twice the mass:
  
  $r_p=\frac{1}{1{.}5}\sqrt{\frac{2\,(1{.}673\times10^{-27})(1{.}0\times10^{4})}{1{.}602\times10^{-19}}}\approx9{.}633\ \mathrm{mm},\qquad r_d=\sqrt{\frac{m_d}{m_p}}\;r_p\approx13{.}62\ \mathrm{mm}$
3. The alpha particle has four times the proton's mass and twice its charge, so the same m/q as the deuteron, nearly:
  
  $r_\alpha=\sqrt{\frac{m_\alpha/2e}{m_d/e}}\;r_d=\sqrt{\frac{6{.}645\times10^{-27}}{2\,(3{.}344\times10^{-27})}}\;r_d\approx0{.}9968\,r_d\approx13{.}58\ \mathrm{mm}$
Answer

9.633 mm, 13.62 mm and 13.58 mm. The alpha particle's circle is 0.32% smaller than the deuteron's. Its m/q is 0.64% below the deuteron's, because a helium nucleus is bound 23.8 MeV more tightly than two deuterons are, and that energy is missing from its mass.

$r_p\approx9{.}633,\quad r_d\approx13{.}62,\quad r_\alpha\approx13{.}58\ \mathrm{mm}$
A velocity selector and a mass spectrometer 3 steps
Singly charged carbon ions pass undeflected through crossed fields, E = 1.0 × 10⁵ V/m and B = 0.20 T, and then bend through a half circle in a field of 0.40 T. How far apart do carbon-12 and carbon-13 land?
Show solution
1. An ion goes straight through only if the electric and magnetic forces cancel, which fixes its speed whatever its mass:
  
  $qE=qvB_1\ \Rightarrow\ v=\frac{E}{B_1}=\frac{1{.}0\times10^{5}}{0{.}20}=5{.}0\times10^{5}\ \mathrm{m/s}$
2. In the second field each isotope moves on its own circle:
  
  $r=\frac{mv}{eB_2}:\qquad r_{12}=\frac{(12{.}000\,\mathrm u)(5{.}0\times10^{5})}{(1{.}602\times10^{-19})(0{.}40)}\approx155{.}5\ \mathrm{mm},\qquad r_{13}\approx168{.}5\ \mathrm{mm}$
3. After half a circle each lands a diameter from the entrance:
  
  $\Delta x=2\,(r_{13}-r_{12})\approx26{.}0\ \mathrm{mm}$
Answer

26.0 mm apart. With the speed fixed first, the radius is proportional to the mass itself rather than to its square root (problem 3), so the 8.4% difference in mass is an 8.4% difference in radius.

$\Delta x\approx26{.}0\ \mathrm{mm}$
A velocity at an angle to the field: the helix 4 steps
A proton moves at 5.0 × 10⁵ m/s at 60° to a 0.10 T field. Find the force on it, the radius of its path, and how far it moves along the field in one turn.
Show solution
1. Only the part of the velocity across the field feels a force:
  
  $F=|q|vB\sin\theta=(1{.}602\times10^{-19})(5{.}0\times10^{5})(0{.}10)\sin60^{\circ}\approx6{.}938\times10^{-15}\ \mathrm{N}$
2. Split the velocity into the part across the field and the part along it:
  
  $v_\perp=v\sin60^{\circ}\approx4{.}330\times10^{5},\qquad v_\parallel=v\cos60^{\circ}=2{.}5\times10^{5}\ \mathrm{m/s}$
3. The part across the field goes round a circle, in the slow lap time:
  
  $r=\frac{mv_\perp}{|q|B}\approx45{.}21\ \mathrm{mm},\qquad T=\frac{2\pi m}{|q|B}\approx655{.}9\ \mathrm{ns}$
4. The part along the field is untouched, and carries the proton forward one pitch every lap:
  
  $p=v_\parallel T=(2{.}5\times10^{5})(6{.}559\times10^{-7})\approx164{.}0\ \mathrm{mm}$
Answer

F = 6.938 × 10⁻¹⁵ N, r = 45.21 mm, pitch 164.0 mm. The proton winds round the field lines on a helix, one turn every 655.9 ns whatever the angle.

$F\approx6{.}938\times10^{-15}\ \mathrm{N},\quad r\approx45{.}21\ \mathrm{mm},\quad p\approx164{.}0\ \mathrm{mm}$
When relativity matters 3 steps
An electron with 1.0 MeV of kinetic energy moves across a 0.010 T field. Find the radius of its circle and the time for one lap. Press Electron, 1 MV.

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Show solution
1. The kinetic energy is nearly twice the rest energy, so γ is far from 1:
  
  $\gamma=1+\frac{K}{mc^{2}}=1+\frac{1{.}000}{0{.}5110}\approx2{.}957$
2. The momentum from the energy, then the radius:
  
  $pc=\sqrt{K^{2}+2Kmc^{2}}=\sqrt{1{.}000^{2}+2\,(1{.}000)(0{.}5110)}\approx1{.}422\ \mathrm{MeV},\qquad r=\frac{p}{eB}\approx474{.}3\ \mathrm{mm}$
3. The lap is γ times the slow lap:
  
  $T=\gamma\,\frac{2\pi m_e}{eB}\approx2{.}957\times3{.}572\ \mathrm{ns}\approx10{.}56\ \mathrm{ns}$
Answer

r = 474.3 mm, T = 10.56 ns. ½mv² = 1.0 MeV would give a speed of 5.931 × 10⁸ m/s, nearly twice the speed of light, and a radius of 337.2 mm.

$r\approx474{.}3\ \mathrm{mm},\qquad T\approx10{.}56\ \mathrm{ns}$
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.

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Show solution
1. The frequency is the cyclotron frequency of a slow proton, which the page prints at 10 kV:
  
  $f_0=\dfrac{qB}{2\pi m}=\dfrac{1{.}602177\times10^{-19}\times1{.}5}{2\pi\times1{.}672622\times10^{-27}}=22{.}87\ \mathrm{MHz}\qquad T_0=43{.}73\ \mathrm{ns}$
2. At the rim the radius is 0.5 m, so the momentum is qBR. Written as pc it is an energy:
  
  $pc=qBRc=1{.}602177\times10^{-19}\times1{.}5\times0{.}5\times2{.}997925\times10^{8}\ \mathrm{J}=224{.}84\ \mathrm{MeV}$
3. The kinetic energy is the total energy less the rest energy. Leaving relativity out would give 26.94 MeV, 1.4% too much:
  
  $K=\sqrt{(pc)^2+(mc^2)^2}-mc^2=\sqrt{224{.}84^2+938{.}27^2}-938{.}27=26{.}56\ \mathrm{MeV}\qquad\dfrac{(pc)^2}{2mc^2}=26{.}94\ \mathrm{MeV}$
4. At that energy γ is 1.028312, so a lap at the rim takes 2.83% longer than a lap at the centre:
  
  $\gamma=1+\dfrac{26{.}56}{938{.}27}=1{.}028312\quad\Rightarrow\quad T_{\mathrm{rim}}=1{.}028312\,T_0$
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:
  
  $n=\dfrac{26{.}56}{0{.}1}\approx266\qquad\sum_{k=1}^{n}\dfrac{k\,\Delta E}{mc^2}=\dfrac{\Delta E}{mc^2}\,\dfrac{n(n+1)}{2}=\dfrac{0{.}1}{938{.}27}\times\dfrac{266\times267}{2}=3{.}78$
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:
  
  $\dfrac{\Delta E}{mc^2}\,\dfrac{n(n+1)}{2}=\dfrac12\ \Rightarrow\ n(n+1)=\dfrac{938{.}27}{0{.}1}=9383\ \Rightarrow\ n=97,\ K=9{.}7\ \mathrm{MeV}\approx\sqrt{mc^2\,\Delta E}=9{.}69\ \mathrm{MeV}$
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 put back the γ a fixed frequency leaves out in one of two ways: the synchrocyclotron lowers the frequency as the proton gains energy, and the isochronous cyclotron makes the field grow with the radius so that the lap stays in step, a design L. H. Thomas showed in 1938 could still focus the beam.

$K_{\max}\approx\sqrt{mc^2\,\Delta E}:\qquad\Delta E=100\ \mathrm{keV}\Rightarrow9{.}7\ \mathrm{MeV},\qquad\Delta E=400\ \mathrm{keV}\Rightarrow19{.}2\ \mathrm{MeV}$

Supplementary problems, answers only

How long does an electron take to go once round a circle in the Earth's field, 50 μT, at a speed well below light's?

Show answer

714.5 ns, at any such speed.

$T=\frac{2\pi m_e}{eB}\approx714{.}5\ \mathrm{ns}$
A proton and a deuteron cross the same field at the same speed. What is the ratio of their radii?

Show answer

1.999: at the same speed the radius goes as m/q, not as its square root.

$\frac{r_d}{r_p}=\frac{m_d}{m_p}\approx1{.}999$
A proton moves at 3.0 × 10⁶ m/s at 30° to a 0.20 T field. Find the force on it.

Show answer

4.807 × 10⁻¹⁴ N.

$F=|q|vB\sin30^{\circ}\approx4{.}807\times10^{-14}\ \mathrm{N}$
A velocity selector has E = 2.4 × 10⁴ V/m and B = 0.060 T. What speed passes straight through?

Show answer

4.0 × 10⁵ m/s, for any charge and any mass.

$v=\frac{E}{B}=4{.}0\times10^{5}\ \mathrm{m/s}$
A cyclotron with a 1.5 T field and a 0.50 m radius accelerates deuterons. At what frequency must it run, and what energy do they leave with, leaving relativity out?

Show answer

11.44 MHz; 13.48 MeV.

$f=\frac{eB}{2\pi m_d}\approx11{.}44\ \mathrm{MHz},\qquad K=\frac{e^{2}B^{2}R^{2}}{2m_d}\approx13{.}48\ \mathrm{MeV}$
What field bends 10 keV electrons into a circle of radius 5.0 cm?

Show answer

6.777 mT; 6.744 mT if relativity is left out.

$B=\frac{p}{eR}\approx6{.}777\ \mathrm{mT}$

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/(|q|B), 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 last solved 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.

Learning path

Where fields come from

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 7 Raven Room team · Reviewed against published sources. Corrections welcome.