Motion of Charged Particles in Magnetic Fields

In short

A charge q with mass m moving with speed v perpendicular to a uniform magnetic field \vec B traces a circle of radius

\boxed{\;r \;=\; \frac{mv}{qB}\;}.

The circulation is at a fixed frequency, independent of the speed and of the radius:

\boxed{\;f_c \;=\; \frac{qB}{2\pi m},\qquad \omega_c \;=\; \frac{qB}{m}\;}\qquad\text{(the cyclotron frequency).}

If the velocity has a component along \vec B, that component is untouched by the field (magnetic force is perpendicular to velocity, so it cannot change the parallel component), and the perpendicular component still drives circular motion. The two together make a helix: circles drawn while drifting sideways.

The pitch of the helix — how far the charge advances along \vec B per revolution — is

\boxed{\;p \;=\; v_\parallel \cdot T \;=\; \frac{2\pi m\,v_\parallel}{qB}\;}.

These three results — circle, cyclotron frequency, helix pitch — underlie the mass spectrometer at BARC in Trombay that separates uranium isotopes, the aurora that paints the Ladakh sky when solar-wind protons corkscrew down along Earth's field lines, the classroom cathode-ray tube with its electron gun and deflection coils, and the cyclotron accelerator in a medical-physics department. Every one of them is an application of two facts — magnetic force is perpendicular to velocity, and perpendicular force of constant magnitude gives circular motion.

Here is something that should surprise you. Fire a proton into a region of uniform magnetic field, at right angles to the field. The proton does not go in a straight line. It does not slow down. It does not speed up. It runs in a perfect circle — forever, or at least until it hits something.

No force you met earlier does this. Gravity pulls a cricket ball into a parabola, not a circle. The Coulomb force pushes two charges apart on a ray. Friction drags a sliding puck to a stop along a straight deceleration. But the magnetic force is genuinely unusual: it is always perpendicular to the velocity. It is the purest centripetal force in all of physics. A perpendicular force of constant magnitude produces one unique trajectory, and that trajectory is the circle.

This article takes that single observation — \vec F_\text{mag} = q\vec v \times \vec B is perpendicular to \vec v — and extracts everything: the radius of the circle, the period of one revolution (which turns out to be independent of how fast the particle is going, a fact called isochronism that runs the cyclotron), what happens when the velocity has a component along the field, and why a shower of protons from the sun ends up painting the sky over Leh in green and red.

The circle — derivation from first principles

Start with the one law you need, the Lorentz force on a moving charge in a pure magnetic field:

\vec F \;=\; q\vec v \times \vec B.

Take \vec B = B\hat z (pointing in the +z direction, say, out of the page) and a charge moving in the xy-plane with speed v perpendicular to \vec B. Then:

Step 1. Write the two features of the magnetic force that matter.

(a) The magnitude: |\vec F| = qvB\sin 90° = qvB — constant, because v and B are constant.

(b) The direction: perpendicular to \vec v, and perpendicular to \vec B. Since \vec B is out of the page and \vec v lies in the page, \vec F lies in the page too, at right angles to \vec v.

Why: this is the defining feature of the magnetic force. Constant magnitude, perpendicular to the instantaneous velocity — that is the exact recipe for uniform circular motion.

Step 2. Recognise that "constant-magnitude force perpendicular to velocity" is the signature of circular motion.

Recall centripetal acceleration: a particle moving at speed v on a circle of radius r has acceleration v^2/r directed toward the centre of the circle. The force required is

F_\text{centripetal} \;=\; \frac{mv^2}{r},

and it must point toward the centre at every instant. The magnetic force does both: constant magnitude qvB, always perpendicular to \vec v (and therefore always pointing somewhere in the plane perpendicular to the motion — in fact, always toward a specific point, as we are about to see).

Step 3. Set magnetic force equal to centripetal force.

qvB \;=\; \frac{mv^2}{r}.

Why: Newton's second law. If the circle has constant v and constant r, the net force is the centripetal force. The only force here is magnetic, so the magnetic force must equal the centripetal force. One equation in one unknown (r).

Step 4. Solve for r.

\boxed{\;r \;=\; \frac{mv}{qB}\;}

Why: cancel one v from both sides and rearrange. Notice that the mass appears in the numerator — heavier particles sweep larger circles at the same speed. This is the core of the mass spectrometer (worked example 2 below).

Step 5. Derive the period of one revolution.

The circumference of the circle is 2\pi r; the particle traces it at speed v; so the time for one revolution is

T \;=\; \frac{2\pi r}{v} \;=\; \frac{2\pi}{v}\cdot \frac{mv}{qB} \;=\; \frac{2\pi m}{qB}.

Why: the speed drops out completely. A fast particle makes a big circle and a slow particle makes a small one, but both complete one loop in the same time. This is isochronism — the most important single fact about motion in a uniform magnetic field.

The frequency and angular frequency (the cyclotron frequency and cyclotron angular frequency) are:

\boxed{\;f_c \;=\; \frac{1}{T} \;=\; \frac{qB}{2\pi m},\qquad \omega_c \;=\; 2\pi f_c \;=\; \frac{qB}{m}\;}.

Which way round the circle?

The sign of the charge and the direction of \vec B together decide the sense of rotation (clockwise or anticlockwise). For a positive charge with \vec B pointing out of the page, \vec v\times\vec B is perpendicular to \vec v and rotated 90° clockwise from \vec v. So the force points to the clockwise-perpendicular direction of motion — which means the charge curves clockwise as seen from above (i.e., looking along -\hat z). For a negative charge, the force reverses, and the circulation is anticlockwise.

With $\vec B$ out of the page, a positive charge (red) circulates clockwise and a negative charge (dark) circulates anticlockwise. Both have the same radius $r = mv/(|q|B)$ and the same period $T = 2\pi m/(|q|B)$ — the direction is the only thing the sign of the charge affects.

A Hindi-film mnemonic that actually works: the positive charge is conservative and curls to the right (clockwise in the usual orientation); the negative charge is contrarian and curls left. Or just draw \vec v, draw \vec B, compute \vec v \times \vec B, multiply by the sign of q, and you will get the direction right every time.

Why the frequency does not depend on speed — isochronism

This is worth pausing over. A particle moving twice as fast sweeps a circle with twice the radius (because r \propto v). The circumference is doubled. But the speed is also doubled. So the time per lap — circumference divided by speed — is the same. The particle that is moving faster does not complete its loop any faster, because the loop it is on is proportionally bigger.

Isochronism is what makes the cyclotron possible. In a cyclotron, an oscillating electric field kicks the particle once per half-orbit — always in the same direction relative to its motion — and the particle gains speed continuously. Because the revolution time does not depend on speed, the same fixed-frequency oscillator can kick the particle efficiently at every stage of its acceleration, right from rest up to relativistic speeds. No frequency-tracking needed. That is the engineering miracle at the heart of the cyclotron.

Explore the radius and period — interactive

The formula r = mv/(qB) and T = 2\pi m/(qB) both have B in the denominator: double the field and the radius halves, and the period halves. The interactive below fixes v = 10^6 m/s (a hot-thermal-electron speed) and m/q = m_p/e for a proton (about 1.04\times 10^{-8} kg/C), and lets you drag B from 0.001 T to 1 T to see the radius collapse with increasing field.

Drag to choose the field strength $B$. At 0.001 T (Earth-field class), the proton's circle has a radius over 10 m — macroscopic, the size of a room. At 1 T (an MRI-class field), the radius shrinks to about 1 cm. The period $T = 2\pi m/(qB)$ also scales as $1/B$ — at 1 T a proton makes a revolution in about 66 nanoseconds, the kind of timing a modern oscilloscope can follow with ease.

When velocity is not perpendicular to the field — the helix

Suppose the velocity has both a perpendicular part v_\perp (across the field) and a parallel part v_\parallel (along the field). Write

\vec v \;=\; \vec v_\perp \;+\; \vec v_\parallel,

where \vec v_\parallel \parallel \vec B and \vec v_\perp \perp \vec B. The magnetic force is

\vec F \;=\; q(\vec v_\perp + \vec v_\parallel) \times \vec B \;=\; q\vec v_\perp \times \vec B \;+\; q\vec v_\parallel \times \vec B.

The second term is zero because \vec v_\parallel \parallel \vec B and a vector crossed with itself is zero. Only the \vec v_\perp part feels a force. Consequently:

The parallel component is unaffected. v_\parallel remains constant — the charge slides along the field at whatever speed it had in that direction.
The perpendicular component undergoes uniform circular motion at radius r = mv_\perp/(qB) and frequency \omega_c = qB/m, exactly as before but with v_\perp instead of v.

Combining a uniform circle in the plane perpendicular to \vec B with a uniform drift along \vec B gives you a helix — a corkscrew pattern, like the thread on a bolt.

A positive charge with both perpendicular and parallel velocity components traces a helix. The radius is set by the perpendicular speed; the pitch — how far it advances along $\vec B$ per revolution — is set by the parallel speed. The charge completes the same number of revolutions per second (the cyclotron frequency) regardless of its parallel speed.

The pitch of the helix

The pitch p is the axial distance the helix advances per revolution. Since the period is T = 2\pi m/(qB) and the axial speed is v_\parallel:

\boxed{\;p \;=\; v_\parallel \cdot T \;=\; \frac{2\pi m\,v_\parallel}{qB}\;}.

Why: in one period, the charge completes one loop around the axis and drifts a distance v_\parallel T along the axis. That drift is the pitch. The formula is a two-line consequence of the period formula combined with the definition of v_\parallel.

If v_\parallel = 0, the pitch is zero and the helix collapses to a circle (all perpendicular motion, no advance along the field). If v_\perp = 0, the radius is zero and the helix collapses to a straight line along the field (no circulation, no perpendicular motion). The helix interpolates smoothly between these two limits as you vary the angle \alpha between \vec v and \vec B.

In terms of the total speed v and this angle:

v_\perp \;=\; v\sin\alpha,\qquad v_\parallel \;=\; v\cos\alpha,

r \;=\; \frac{mv\sin\alpha}{qB},\qquad p \;=\; \frac{2\pi mv\cos\alpha}{qB}.

For \alpha = 90°: pure circle, zero pitch. For \alpha = 0°: straight line along \vec B, pitch infinite (the helix is stretched out flat). In between: a tight tall helix for small \alpha, a shallow wide helix for \alpha near 90°.

Applications

The mass spectrometer — separating isotopes by their radius

A mass spectrometer is the single most important practical application of r = mv/(qB). The idea: if you can give every ion the same speed, then the radius of its circle in a magnetic field is directly proportional to its mass. Heavier ions curve less; lighter ions curve more. By catching ions at different positions on a detector plate downstream, you separate them by mass.

Step 1: the velocity selector. The first stage of the spectrometer sends an ion beam through a region of crossed electric and magnetic fields. Only ions with one specific speed, v = E/B_1, pass straight through undeflected; all the others curve away (this is the velocity selector, discussed in the parallel currents article).

Step 2: the mass analyser. Ions that survive the selector enter a region of pure magnetic field B_2, perpendicular to their velocity. They circle. Radius:

r \;=\; \frac{mv}{qB_2} \;=\; \frac{m\,(E/B_1)}{qB_2} \;=\; \frac{mE}{qB_1 B_2}.

For ions of fixed charge q and known E, B_1, B_2, the radius depends linearly on the mass m. A detector plate placed half a semicircle downstream catches different isotopes at different positions, separated by 2\Delta r per mass step.

At BARC Trombay, mass spectrometers of exactly this kind are used to analyse isotope ratios in yellowcake (uranium oxide) fuel for India's thermal reactors, and in the thorium programme that will define the second half of India's nuclear future. The mass difference between ^{238}\text{U} and ^{235}\text{U} is about 1.3%; the corresponding radius difference is 1.3%; at a 1 m radius, that is a 1.3 cm separation at the detector — easily resolved. This is how India (and the world) makes enriched uranium: electromagnetic separation using r = mv/(qB), though modern facilities prefer centrifuge methods which are more efficient per kilogram.

Aurora over Ladakh — charges spiralling down field lines

When the sun is violent — during a coronal mass ejection — a gust of solar-wind plasma hits Earth's magnetic field. The field lines funnel the charged particles (mostly protons and electrons) down toward the magnetic poles. The particles spiral along the field lines on tight helices, with pitch determined by their parallel velocity and radius determined by their perpendicular velocity.

As they plunge into Earth's upper atmosphere at 100–300 km altitude, they collide with oxygen and nitrogen molecules, exciting them. The molecules then relax back to their ground state by emitting characteristic colours — green (oxygen at 557.7 nm), red (oxygen at 630 nm), blue and violet (nitrogen). The result is the aurora borealis in the northern hemisphere, or the aurora australis in the south.

India sees aurorae rarely — we are at low magnetic latitude — but when a strong solar storm hits, Ladakh and parts of Himachal do occasionally light up with faint red glows. The 2024 May G5 storm produced auroras visible as far south as Leh and Pangong Tso. Every one of those photons was launched by a helix: a solar-wind proton corkscrewing down Earth's field line, smashing into an oxygen atom at 200 km altitude.

Cathode-ray tubes and the fine-tuning of the older school television

Before LCDs took over, every Indian home had a cathode-ray tube (CRT) television. The picture was drawn by a narrow beam of electrons fired from a heated cathode, steered across a phosphor screen by magnetic deflection coils. The coils produce a magnetic field perpendicular to the beam; the beam follows a tiny arc of a circle inside the field (the radius is large, so the arc is nearly straight, but it is still a circle in disguise); and by varying the field strength 50 times a second (or faster), the beam scans the entire screen line by line.

Every picture you ever saw on a CRT TV was built out of electrons on circular arcs, one pixel at a time, at about 10^7 m/s. The deflection field was weak — a few milliteslas — but precise; the radius of curvature varied smoothly with the coil current; and the beam hit the right place on the screen for the right colour. All an application of r = mv/(qB).

Worked examples

Example 1: Proton in Earth's field

A proton is fired at v = 2\times 10^5 m/s perpendicular to a magnetic field of strength B = 3\times 10^{-5} T (roughly Earth's field at the magnetic equator). Find the radius of the proton's circular path and the period of revolution. Mass of proton m_p = 1.673\times 10^{-27} kg; charge q = e = 1.602\times 10^{-19} C.

At a proton's speed of $2\times 10^5$ m/s in Earth's field, the circular radius is about 70 metres — a large circle, because Earth's field is weak. The diagram is not to scale.

Step 1. Apply r = mv/(qB).

r \;=\; \frac{(1.673\times 10^{-27})(2\times 10^5)}{(1.602\times 10^{-19})(3\times 10^{-5})}.

Why: direct substitution. All numbers in SI units, so the answer will come out in metres.

Step 2. Work out the numerator and denominator separately.

Numerator: 1.673 \times 2 \times 10^{-27+5} = 3.346\times 10^{-22} kg·m/s.

Denominator: 1.602 \times 3 \times 10^{-19-5} = 4.806\times 10^{-24} C·T.

Step 3. Divide.

r \;=\; \frac{3.346\times 10^{-22}}{4.806\times 10^{-24}} \;=\; 69.6\;\text{m}.

Why: 10^{-22}/10^{-24} = 10^{2}; and 3.346/4.806 \approx 0.696. So the answer is about 70 m.

Step 4. Compute the period.

T \;=\; \frac{2\pi m}{qB} \;=\; \frac{2\pi (1.673\times 10^{-27})}{(1.602\times 10^{-19})(3\times 10^{-5})}.

Numerator: 2\pi \times 1.673\times 10^{-27} = 1.051\times 10^{-26}.

Denominator: 4.806\times 10^{-24}.

T \;=\; \frac{1.051\times 10^{-26}}{4.806\times 10^{-24}} \;=\; 2.19\times 10^{-3}\;\text{s} \;=\; 2.2\;\text{ms}.

Result. Radius r \approx 70 m; period T \approx 2.2 ms (frequency \approx 457 Hz — in the audio range, roughly the pitch of the A just above middle C).

What this shows. Even a fast proton in Earth's weak field makes a big loop — large enough that a cosmic-ray proton arriving in our atmosphere spirals in arcs tens of metres wide. But it cycles every 2 milliseconds, so over the roughly 10 km a cosmic ray might travel in the ionosphere, it completes thousands of revolutions. This is why cosmic-ray tracks bend noticeably in Earth's field, and why cosmic-ray observatories at high altitudes (LeHAASO in Tibet, GRAPES-3 at Ooty) have to correct for geomagnetic bending when reconstructing particle trajectories.

Example 2: Mass spectrometer separating sodium isotopes

A beam of singly-ionised sodium atoms is passed through a velocity selector that sets v = 1.5\times 10^5 m/s. The ions then enter a mass analyser with B = 0.50 T perpendicular to the beam. Find the radius of the circular path for (a) ^{23}\text{Na}^+ and (b) ^{22}\text{Na}^+, and the separation between where they hit a detector plate placed half a circle downstream. Atomic mass unit u = 1.66\times 10^{-27} kg; e = 1.6\times 10^{-19} C.

The two sodium isotopes follow slightly different radii in the mass analyser. The lighter isotope ($^{22}$Na) curves tighter; the heavier ($^{23}$Na) curves wider. The detector plate sees the two beams separated by $2\Delta r$.

Step 1. Apply r = mv/(qB) to ^{23}\text{Na}^+. Mass is 23u = 23 \times 1.66\times 10^{-27} = 3.818\times 10^{-26} kg.

r_{23} \;=\; \frac{(3.818\times 10^{-26})(1.5\times 10^5)}{(1.6\times 10^{-19})(0.50)} \;=\; \frac{5.727\times 10^{-21}}{8.0\times 10^{-20}}.

r_{23} \;=\; 0.0716\;\text{m} \;=\; 71.6\;\text{mm}.

Step 2. Apply r = mv/(qB) to ^{22}\text{Na}^+. Mass is 22u = 3.652\times 10^{-26} kg.

r_{22} \;=\; \frac{(3.652\times 10^{-26})(1.5\times 10^5)}{8.0\times 10^{-20}} \;=\; \frac{5.478\times 10^{-21}}{8.0\times 10^{-20}} \;=\; 0.0685\;\text{m} \;=\; 68.5\;\text{mm}.

Why: same formula, smaller mass, smaller radius. The arithmetic scales linearly in m because v, q, and B are identical for both.

Step 3. Compute the separation at the detector. The detector is placed half a circle downstream (at the bottom of each path), and the distance from the source to where each isotope lands is 2r. So the separation is

\Delta x \;=\; 2(r_{23} - r_{22}) \;=\; 2(71.6 - 68.5)\;\text{mm} \;=\; 6.2\;\text{mm}.

Why: both isotopes start at the same point and complete a half-circle before hitting the detector. The geometric separation of the landing points is 2\Delta r, twice the radius difference.

Result. r_{23} = 71.6 mm, r_{22} = 68.5 mm, separation at detector = 6.2 mm.

What this shows. A 4.5% mass difference between the two isotopes gives a 4.5% radius difference — cleanly resolvable by a detector with 1 mm resolution. This is how isotope ratios are measured in geology (dating volcanic rocks by ^{40}Ar/^{39}Ar), in medicine (tracer studies using ^{11}C and ^{18}F), and in nuclear-materials accounting (uranium enrichment verification by the IAEA). The same physics, dressed differently for each application.

Example 3: Helix pitch of a beta particle

A \beta-particle (electron) with kinetic energy 10 keV is emitted at 30° to a uniform magnetic field of 0.02 T. Find the radius and pitch of its helical trajectory. Electron mass m_e = 9.11\times 10^{-31} kg; e = 1.6\times 10^{-19} C.

A 10 keV electron at 30° to a 0.02 T field traces a helix of radius 9.4 mm and pitch 102 mm — very loose. The pitch is more than ten times the radius because $\cos 30° / \sin 30° = \sqrt{3} \approx 1.73$, and the charge moves mostly along the field.

Step 1. Find the total speed from the kinetic energy.

\tfrac{1}{2} m_e v^2 \;=\; 10\;\text{keV} \;=\; 10\times 10^3 \times 1.6\times 10^{-19}\;\text{J} \;=\; 1.6\times 10^{-15}\;\text{J}.

v \;=\; \sqrt{\frac{2\times 1.6\times 10^{-15}}{9.11\times 10^{-31}}} \;=\; \sqrt{3.513\times 10^{15}} \;=\; 5.93\times 10^7\;\text{m/s}.

Why: KE = ½mv², solved for v. Note: v is about 20% of the speed of light, so a small relativistic correction creeps in at this energy. For an introductory calculation we ignore it — the classical answer is good to a few percent.

Step 2. Decompose into perpendicular and parallel components.

v_\perp \;=\; v\sin 30° \;=\; 5.93\times 10^7 \times 0.5 \;=\; 2.97\times 10^7\;\text{m/s},

v_\parallel \;=\; v\cos 30° \;=\; 5.93\times 10^7 \times 0.866 \;=\; 5.14\times 10^7\;\text{m/s}.

Step 3. Compute the radius.

r \;=\; \frac{m_e v_\perp}{eB} \;=\; \frac{(9.11\times 10^{-31})(2.97\times 10^7)}{(1.6\times 10^{-19})(0.02)}.

Numerator: 9.11 \times 2.97 \times 10^{-24} = 2.705\times 10^{-23}.

Denominator: 1.6 \times 0.02 \times 10^{-19} = 3.2\times 10^{-21}.

r \;=\; \frac{2.705\times 10^{-23}}{3.2\times 10^{-21}} \;=\; 8.45\times 10^{-3}\;\text{m} \;=\; 8.45\;\text{mm}.

Why: plug into r = mv_\perp/(qB) with v_\perp, not the total v. A common slip is to use v and forget the \sin\alpha; that would give a radius 2× too big.

Step 4. Compute the period and pitch.

T \;=\; \frac{2\pi m_e}{eB} \;=\; \frac{2\pi(9.11\times 10^{-31})}{(1.6\times 10^{-19})(0.02)} \;=\; \frac{5.72\times 10^{-30}}{3.2\times 10^{-21}} \;=\; 1.79\times 10^{-9}\;\text{s}.

p \;=\; v_\parallel T \;=\; (5.14\times 10^7)(1.79\times 10^{-9}) \;=\; 9.2\times 10^{-2}\;\text{m} \;=\; 92\;\text{mm}.

Why: the period depends only on m, q, B (not on any velocity). The pitch is the axial distance in that period, which is v_\parallel T.

Result. Radius r \approx 8.5 mm, pitch p \approx 92 mm. The electron traces a fast, loose helix — 23 nanoseconds per revolution, 9 cm advance per loop.

What this shows. The helix geometry is fully determined by three numbers: the field, the charge-to-mass ratio, and the two components of velocity. Once you know the angle \alpha between \vec v and \vec B, the radius and pitch are set. This is exactly the physics of a solar-wind proton heading toward Earth's magnetic pole — except the solar proton has a much smaller v/B and so a much larger helix.

Common confusions

"The magnetic force speeds up the particle." No. The magnetic force is always perpendicular to the velocity, so it does no work. The particle's speed is constant throughout the circle (or helix). What changes is the direction, not the magnitude. If the speed of a charged particle in a magnetic field appears to change, there is also an electric field somewhere.
"The cyclotron frequency depends on the speed." No — that is precisely what makes it useful. f_c = qB/(2\pi m) depends only on the charge, mass, and field — not on the kinetic energy. This is why a single fixed-frequency oscillator can accelerate a particle in a cyclotron from very slow speeds to near-relativistic speeds.
"The radius depends on the speed but the period does not — that's a paradox." It is not a paradox; it is a consequence. Think of two particles on the same cyclotron, one fast and one slow. The fast one sweeps a big circle; the slow one sweeps a small circle. Both arrive back at the starting point at the same time, because the fast one has further to go but is moving faster, and the ratio works out to the same time exactly. Isochronism.
"For a helix, the radius is mv/(qB)." This is the classic slip. The radius is mv_\perp/(qB), where v_\perp is the perpendicular component of velocity. The parallel component does not contribute to the radius — only to the pitch.
"If \vec v is exactly parallel to \vec B, the charge moves in a straight line without feeling anything." Correct, for the magnetic force alone — zero by the cross product. But the charge still has inertia, so if there is an electric field or any other force, that still acts. Only the magnetic force vanishes.
"The cyclotron frequency of an electron is the same as that of a proton." Wrong: the mass is 1836 times smaller for the electron, so the electron's cyclotron frequency is 1836 times higher. In a 1 T field, proton f_c \approx 15.3 MHz, but electron f_c \approx 28 GHz (in the microwave range). This mass-dependence is how electron-cyclotron-resonance heating works in tokamaks at the Institute for Plasma Research in Gandhinagar: a microwave at 28 GHz resonantly pumps energy into electrons without touching ions.

If you came to understand r = mv/(qB), the cyclotron frequency, the helix, and how a mass spectrometer works — you are done. What follows is the deeper structure: the full vector equation of motion, the special-relativistic correction that limits the cyclotron, the motion in non-uniform fields (magnetic mirrors and the Van Allen belts), and the connection to quantum mechanics (Landau levels).

Solving the equations of motion directly

For full rigor, write Newton's second law component by component. Take \vec B = B\hat z; let the charge have position (x, y, z) and velocity (\dot x, \dot y, \dot z).

\vec v \times \vec B \;=\; (\dot x, \dot y, \dot z) \times (0, 0, B) \;=\; (\dot y B, -\dot x B, 0).

Equation of motion m\ddot{\vec r} = q\vec v \times \vec B becomes:

m\ddot x \;=\; q B \dot y,\qquad m\ddot y \;=\; -qB\dot x,\qquad m\ddot z \;=\; 0.

The z-equation says \ddot z = 0, so \dot z = constant — the parallel velocity is unchanged, as we argued.

The two coupled equations in x and y can be decoupled by introducing \omega_c = qB/m. Differentiating the first and substituting:

\dddot x \;=\; \omega_c \ddot y \;=\; -\omega_c^2 \dot x,

so \dot x satisfies the harmonic-oscillator equation \dddot x + \omega_c^2 \dot x = 0, with solution

\dot x(t) \;=\; v_\perp\cos(\omega_c t + \phi).

Integrating once more gives

x(t) \;=\; x_0 + \frac{v_\perp}{\omega_c}\sin(\omega_c t + \phi),\quad y(t) \;=\; y_0 - \frac{v_\perp}{\omega_c}\cos(\omega_c t + \phi).

This is a circle of radius v_\perp/\omega_c = mv_\perp/(qB) — exactly our earlier r — traced at angular speed \omega_c. Adding the straight-line motion in z gives a helix. The algebra confirms what we argued from centripetal reasoning.

Relativistic cyclotron — why the frequency drifts

The non-relativistic period T = 2\pi m/(qB) is, strictly, only the low-speed limit. At relativistic speeds the mass is replaced by \gamma m where \gamma = 1/\sqrt{1 - v^2/c^2}, so

T_\text{rel} \;=\; \frac{2\pi \gamma m}{qB}.

The period grows with \gamma. For the non-relativistic cyclotron with a fixed-frequency oscillator, this is catastrophic: as the particle speeds up, its period stretches, and it falls out of phase with the oscillator. After about 50 MeV kinetic energy for a proton, the period has grown enough that the particle no longer arrives at the gap in time for the next kick.

Two fixes:

Synchrocyclotron: decrease the oscillator frequency as the particle speeds up. Only one bunch at a time.
Synchrotron: keep the particle on a fixed-radius ring by ramping the magnetic field. This is how the Large Hadron Collider and India's INDUS-2 synchrotron at RRCAT Indore work.

Both are sophisticated applications of the same basic cyclotron principle, with engineering around the relativistic mass.

Non-uniform fields — magnetic mirrors and the Van Allen belts

If the magnetic field strengthens along the field-line direction (like Earth's field as you head toward the poles, where field lines converge), a helical charge experiences an additional effect: the perpendicular component of the field at the bulging part of the helix produces a component of force along the field direction, opposing the motion. The charge decelerates in the parallel direction, stops, and eventually turns around — it reflects.

This is the magnetic mirror effect. Earth's field is strong at the poles and weaker at the equator; charged particles in the magnetosphere bounce between northern and southern hemispheres, trapped on field lines. The region of space filled with these trapped particles is the Van Allen belt — a doughnut of high-energy protons and electrons encircling Earth. Satellites in low-earth orbit (like India's GSAT series) are mostly below the inner belt; satellites in geostationary orbit (like INSAT-3D) sit at the outer edge. Crossing the belts is a radiation hazard that ISRO engineers shield against carefully.

The condition for reflection involves the magnetic moment \mu = mv_\perp^2/(2B), which is an adiabatic invariant — conserved when the field varies slowly compared to the gyration period. At the turning point, all kinetic energy is parallel-converted, and the perpendicular speed drops to zero. That is the fundamental principle behind plasma confinement in tokamaks, stellarators, and magnetic bottles.

Quantisation — Landau levels

In quantum mechanics, the circular motion becomes quantised. The energy of an electron in a uniform magnetic field takes discrete values

E_n \;=\; \hbar\omega_c \left(n + \tfrac{1}{2}\right),\qquad n = 0, 1, 2, \ldots

These are the Landau levels, and they are the magnetic analogue of the quantum-harmonic-oscillator spectrum. For strong enough fields and cold enough temperatures, the discreteness becomes observable — it is the foundation of the quantum Hall effect, one of the most precisely measured phenomena in physics, and the basis of the modern resistance standard in the SI system.

The spacing \hbar\omega_c = \hbar eB/m_e is exactly the cyclotron frequency times \hbar — quantum mechanics laid on top of the classical cyclotron. In a 1 T field, the spacing is about 1.16\times 10^{-4} eV, which is room-temperature thermal energy (k_B T \approx 0.025 eV) divided by a factor of 200. You need milli-kelvin temperatures to resolve Landau levels cleanly.

Why the magnetic force does no work — the deeper view

A persistent worry: the magnetic force makes the particle move in a circle, so doesn't that require work? No. Work is force dotted with displacement, and \vec F \cdot d\vec r = \vec F \cdot \vec v\, dt = q(\vec v\times\vec B)\cdot\vec v\,dt = 0, identically, because the cross product is perpendicular to \vec v. The force changes the direction of momentum, which is vector-valued, but does not change the magnitude of momentum or the kinetic energy, which are scalars.

The deep reason: the magnetic force is a gauge-invariant piece of the Lorentz force, and at a still deeper level it is a consequence of the U(1) gauge symmetry of electromagnetism. The fact that it does no work on a free charge is a consequence of that symmetry. It is not an accident; it is structural.

Where this leads next

The Cyclotron — the particle accelerator that exploits the speed-independence of the cyclotron period to accelerate ions to MeV energies.
Force Between Parallel Currents and Moving Charges — the velocity selector, the Hall effect, and the fine print of moving charges in combined \vec E and \vec B fields.
Force on a Moving Charge — the Lorentz Force — the one-line equation \vec F = q\vec E + q\vec v\times\vec B from which everything in this article descends.
Force on a Current-Carrying Conductor — what this article's physics looks like when the charges are confined to a wire.
Magnetic Dipole Moment — the adiabatic invariant \mu = mv_\perp^2/(2B) and the connection to spin angular momentum.
Centripetal Force — the kinematics of circular motion, which the magnetic force realises with no need for tension, friction, or gravity.