ARCHETYPE: Cyclotron & Mass-Spectrometer Numericals

In 1929, a young Berkeley professor named Ernest Lawrence was flipping through a German journal when a sketch stopped him: a long line of electrodes, each giving an ion a small push. Wideröe's linear accelerator needed metres of hardware for modest energies. Lawrence's insight fit on the back of an envelope — fold the line into a spiral with a magnetic field, and one pair of electrodes can deliver every push. The trick only works because of a coincidence you have already met: the period of a charge circling in a magnetic field, $T = 2\pi m/qB$ , does not depend on its speed. Slow inner laps and fast outer laps finish on the same clock, so one fixed-frequency voltage stays forever in step. The first working cyclotron was 11 cm of brass and sealing wax, cost about $25, and accelerated protons to 80,000 volts' worth of energy. A Nobel Prize followed within a decade. This archetype drills the two machines that turn$ r = mv/qB$ into numbers: the cyclotron, which reads the equation as "each kick makes the circle grow", and the mass spectrometer, which reads it backwards as "the circle's size weighs the particle". Between them they own a compact family of exam numericals — oscillator frequency, maximum energy, number of turns, isotope separation on a photographic plate. One honesty note before the machinery: CBSE deleted the cyclotron in its 2023–24 rationalisation, and today's JEE and NEET syllabi followed. It survives on the ISC syllabus — and in decades of archived exam papers — which is exactly where this module aims.

Key takeaways

A cyclotron is two hollow D-shaped electrodes in a uniform magnetic field: the field bends the ion on half-circles inside the dees, the oscillating gap voltage kicks it between them.
Synchronism is the whole trick: the oscillator must run at the cyclotron frequency $f = \dfrac{qB}{2\pi m}$ — fixed once, valid for every lap, because the period ignores speed.
Each full turn crosses the gap twice: energy gained per turn $= 2qV$ ; the number of turns to reach kinetic energy $K$ is $N = K/2qV$ .
The exit energy is set by the magnet, not the voltage: $K_{\max} = \dfrac{q^2B^2R^2}{2m}$ at the dee radius $R$ — a bigger $V$ just gets there in fewer turns.
A mass spectrometer runs the logic in reverse: send ions of known speed (or known accelerating voltage) into a field and the landing radius weighs them — $r = mv/qB$ , or $r = \dfrac{1}{B}\sqrt{2mV_{\text{acc}}/q}$ , so $r \propto \sqrt{m}$ .
Key idea: both machines are the same equation $r = mv/qB$ — the cyclotron grows $r$ by feeding $v$ , the mass spectrometer reads $m$ from $r$ .

The core insight

Because a charge's orbital period in a magnetic field is speed-independent, a fixed-frequency voltage at $f = qB/2\pi m$ can kick an ion every half-turn until the spiral fills the magnet ( $K_{\max} = q^2B^2R^2/2m$ ); the same radius equation, read in reverse, makes the mass spectrometer weigh ions by where they land.

Where this lives today

The cyclotron never retired — it moved to the hospital basement. Around 1,500 compact medical cyclotrons run worldwide, most of them making fluorine-18 before dawn for the day's PET scans: protons at 10–20 MeV (exactly the Board-example energy band) strike oxygen-18-enriched water, and the resulting tracer is in a patient by mid-morning, its 110-minute half-life ticking. Proton-therapy centres push the same physics to 230 MeV with superconducting magnets. The mass spectrometer had an even bigger afterlife: every pharmaceutical lab, doping-control facility and planetary probe carries one — the Curiosity rover weighs Martian isotopes with a magnetic sector, $r = mv/qB$ at work 200 million kilometres from the nearest examiner.

The setup

Geometry first — the machine anatomy is the physics. Two hollow, D-shaped copper electrodes (the dees) face each other across a narrow gap, the whole assembly sandwiched between the poles of a large electromagnet so a uniform field $\vec B$ threads it vertically. An ion source sits at the centre; a high-frequency oscillator drives the gap voltage $V$ , flipping its polarity on a fixed schedule.

Figure

Cyclotron anatomy — dees, gap and the growing spiral of half-circles

Two design facts carry all the marks, and both trace to modules you have already banked:

Inside a dee, the ion coasts on a half-circle. The hollow conductor screens the electric field (the shielding logic of Electrostatics of Conductors & Shielding), so only $\vec B$ acts. Circular & Helical Paths in Magnetic Fields then guarantees the half-circle takes a time $\pi m/qB$ — independent of how fast the ion is going.

In the gap, the ion gets kicked. If the polarity has flipped since the last crossing, the gap field pushes the ion forward, adding $qV$ of kinetic energy. Faster now, the ion rides a bigger half-circle in the next dee ( $r = mv/qB$ grows with $v$ ), but finishes it in exactly the same time. The path is a spiral of half-circles, growing lap by lap — the field is the steering, never the pedal. The synchronism condition is therefore a single number, fixed before the machine is ever switched on:

f_{\text{osc}} \;=\; f_c \;=\; \frac{qB}{2\pi m} \qquad (1)

where: $f_c$ — the cyclotron frequency of the ion in field $B$ (quoted from Circular & Helical Paths in Magnetic Fields, where the speed-independence is earned); $f_{\text{osc}}$ — the frequency of the gap voltage. Read it physically: the oscillator is a metronome set to the ion's lap time. Because the lap time never drifts as the ion speeds up, one setting serves the whole acceleration — this single cancellation is the entire reason the machine exists.

The energy ledger

Each full turn crosses the gap twice (once in each direction, each time with freshly flipped polarity), so the ion banks

\Delta K_{\text{per turn}} = 2qV \qquad (2)

where: $V$ — the peak gap voltage. Read it as a savings account with fixed deposits: to reach a target kinetic energy $K$ , the ion needs $N = K/2qV$ turns. A 50 kV gap feeds a proton 100 keV per turn; twenty million electron-volts is two hundred turns — all completed in a few tens of microseconds, because each turn takes the same $T = 2\pi m/qB$ . But here is the design punchline, and the most-tested idea in the whole module: the voltage never appears in the final energy. The ion leaves when its spiral fills the magnet — when $r$ reaches the dee radius $R$ . At that moment $v_{\max} = qBR/m$ , and the exit energy is fixed by the hardware:

K_{\max} = \tfrac12 m v_{\max}^2 = \frac{q^2 B^2 R^2}{2m} \qquad (3)

where: $R$ — the outermost usable radius; note the charge squared and the mass downstairs. Read it physically: $B$ and $R$ set the biggest circle that fits, and the biggest circle sets the speed. Doubling the gap voltage changes none of that — it halves the number of turns and the time in the machine, nothing more. The magnet writes the cheque; the oscillator only decides how many instalments it is paid in.

Derivation AThe maximum energy from the largest circle that fits

We derive equation (3), the cyclotron's exit energy, from the radius equation and the machine's one geometric limit — the ion must fit inside the dees. The strategy is two lines of algebra: solve $r = mv/qB$ for the speed at the moment the orbit radius equals the dee radius $R$ , then substitute into the kinetic energy. Watch what does NOT appear in the result: the gap voltage $V$ and the number of turns both cancel out of the story entirely — they set the schedule, not the destination.

Step 1. The radius equation (quoted from Circular & Helical Paths in Magnetic Fields — we apply it, we do not re-derive it) at the exit moment:

R = \frac{m v_{\max}}{qB} \quad\Longrightarrow\quad v_{\max} = \frac{qBR}{m}

Why this is the exit condition: each kick fattens the spiral; the last usable half-circle is the one that just grazes the dee rim. Beyond r = R there is no copper to coast in — a deflector electrode peels the beam out.

Step 2. Kinetic energy at that speed:

K_{\max} = \frac12 m v_{\max}^2 = \frac12 m \left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}

Step 3. Read the scalings like an engineer. Double the field: four times the energy ( $B^2$ ). Double the magnet radius: four times the energy ( $R^2$ ) — which is why cyclotron upgrades are magnet projects, not electronics projects. Heavier ion at the same charge: less energy per exit ( $1/m$ ), though a deuteron ( $q = e$ , $m \approx 2m_p$ ) still beats a proton in momentum. Verified: for protons in $B = 1.2$ T with $R = 0.60$ m: $v_{\max} = (1.6\times10^{-19})(1.2)(0.60)/(1.67\times10^{-27}) = 6.9\times10^{7}$ m/s, and $K_{\max} = q^2B^2R^2/2m = 4.0\times10^{-12}$ J $= 24.8$ MeV — the same number both ways, recomputed in Board example 1. Also checked: $v_{\max}/c \approx 0.23$ , so the non-relativistic clock is already straining — see the warning below.

The mass spectrometer: the same equation, read backwards

Where the cyclotron feeds $v$ to grow $r$ , the spectrometer measures $r$ to find $m$ . The classic layout has two stages, and the first belongs to a module you already own: a velocity selector (Crossed Fields & the Velocity Selector) passes only ions with $v = E/B_1$ , whatever their mass or charge. The purified beam then enters a second, pure magnetic field $B_2$ , bends through a semicircle, and lands on a detector plate at a distance of one diameter, $2r$ , from the slit:

r = \frac{mv}{qB_2} \qquad\text{so, at fixed } v, \qquad r \propto \frac{m}{q} \qquad (4)

where: everything on the right is either fixed by the machine ( $v$ , $B_2$ ) or the property being weighed ( $m/q$ ). Read it physically: two isotopes — same charge, different mass — enter together and land apart. The plate is a set of scales marked in kilograms.

Figure

Bainbridge mass spectrometer — velocity selector in, semicircles out, the plate weighs the ion

A common exam variant skips the selector and instead accelerates the ions from rest through a potential difference $V_{\text{acc}}$ , so that $qV_{\text{acc}} = \tfrac12 mv^2$ sets the speed. Then the landing radius obeys

r = \frac{1}{B}\sqrt{\frac{2mV_{\text{acc}}}{q}} \qquad\Longrightarrow\qquad r \propto \sqrt{m} \qquad (5)

where: the square root is the fingerprint of this variant — accelerated-through-a-voltage means $r \propto \sqrt{m}$ , selector-fed means $r \propto m$ . Spotting which regime the question set up is half the marks.

Go deeper · Derivation B · 1 minIsotope separation on the detector plate

Derivation BIsotope separation on the detector plate

We derive equation (5) and the plate separation of two isotopes in the accelerated variant — the geometry Aston used to split neon in 1919. The strategy: energy bookkeeping fixes the entry speed from the accelerating voltage, the radius equation converts speed to landing radius, and the semicircle doubles radii into plate positions. The watch-point is that the two isotopes do NOT enter at the same speed here (unlike the selector version): the same energy $qV_{acc}$ makes the lighter one faster, and the square root in the result remembers that.

Step 1. Speed from the accelerating gap, for an ion of mass $m$ and charge $q$ starting from rest:

qV_{\text{acc}} = \tfrac12 mv^2 \quad\Longrightarrow\quad v = \sqrt{\frac{2qV_{\text{acc}}}{m}}

Why the lighter isotope exits faster: same energy purse, smaller mass — speed scales as $1/\sqrt{m}$ .

Step 2. Landing radius from $r = mv/qB$ , substituting the speed:

r = \frac{m}{qB}\sqrt{\frac{2qV_{\text{acc}}}{m}} = \frac{1}{B}\sqrt{\frac{2mV_{\text{acc}}}{q}}

The mass survives only under the square root — the faster speed of the light isotope partially offsets its smaller mass.

Step 3. Separation on the plate. Each ion performs a half-circle and lands one diameter from the entry slit, so two isotopes of masses $m_1 < m_2$ land apart by

\Delta x = 2r_2 - 2r_1 = \frac{2\sqrt{2V_{\text{acc}}/q}}{B}\left(\sqrt{m_2} - \sqrt{m_1}\right)

Why the factor 2: the semicircle geometry converts each radius into a diameter on the plate — drawing the two concentric semicircles from the same slit makes it obvious. *Verified: singly charged Ne-20 and Ne-22 ( $m = 20u, 22u$ , $u = 1.66\times10^{-27}$ kg) accelerated through $2.0$ kV into $B = 0.50$ T: $r_{20} = 5.76$ cm, $r_{22} = 6.04$ cm, plate separation $2\Delta r = 5.6$ mm — a comfortable gap for a 1919-vintage photographic plate, recomputed for this edition. The selector-fed variant with $v = 3.0\times10^{5}$ m/s and $B_2 = 0.40$ T gives $r_{20} = 15.6$ cm, $r_{22} = 17.1$ cm: separation $3.1$ cm, linear in the mass difference.*

The variants examiners actually pose

variant	what is asked	the one-line key
oscillator matching	find $f_{\text{osc}}$ for a given ion and $B$	$f = qB/2\pi m$ — no speed anywhere
reverse matching	find $B$ for a given oscillator	$B = 2\pi m f/q$
exit energy	$K_{\max}$ from $B$ , $R$	$q^2B^2R^2/2m$ ; convert J → MeV by $1.6\times10^{-13}$
turn counting	turns and time to a target energy	$N = K/2qV$ ; time $= N\,T$
species swap	same machine, new ion (deuteron, α)	rescale by $q/m$ : frequency; by $q^2/m$ : energy
selector-fed spectrometer	landing positions of isotopes	$r \propto m$ at fixed $v$ ; plate gap $= 2\Delta r$
accelerated spectrometer	landing positions after $V_{\text{acc}}$	$r \propto \sqrt{m}$ — the square-root fingerprint

Limiting cases

case	result	meaning
$V$ doubled	same $K_{\max}$ , half the turns	voltage sets the schedule, magnet sets the ceiling
$B$ doubled	$f$ doubles, $K_{\max}$ quadruples	retune the oscillator or lose the beam
proton → deuteron ( $q$ same, $m\times 2$ )	$f$ halves, $K_{\max}$ halves	every species needs its own frequency
proton → α ( $q\times2$ , $m\times4$ )	$f$ halves, $K_{\max}$ same	deuteron and α share a frequency — a classic MCQ
electron injected	$f_c$ larger by $m_p/m_e \approx 1836$	and relativity wrecks the clock almost immediately
$v \to$ near $c$	period grows, kicks fall out of step	the cyclotron's ceiling; synchrotrons take over
spectrometer, same $m/q$	identical landing spot	a spectrometer weighs $m/q$ , not $m$ alone

Worked examples

Board level 1 — the machine's data sheet (ISC pattern)

A cyclotron has a magnetic field of $B = 1.2$ T and dees of radius $R = 0.60$ m. For protons ( $m = 1.67\times10^{-27}$ kg, $q = 1.6\times10^{-19}$ C), find (a) the oscillator frequency, (b) the maximum speed, and (c) the maximum kinetic energy in MeV.

Step 1. Oscillator frequency = cyclotron frequency, equation (1): $f = \dfrac{qB}{2\pi m} = \dfrac{(1.6\times10^{-19})(1.2)}{2\pi(1.67\times10^{-27})} = \boxed{18.3\ \mathrm{MHz}}$ — set once, correct for every lap.

Step 2. Maximum speed at the rim: $v_{\max} = \dfrac{qBR}{m} = \dfrac{(1.6\times10^{-19})(1.2)(0.60)}{1.67\times10^{-27}} = \boxed{6.9\times10^{7}\ \mathrm{m/s}}$ .

Step 3. Maximum kinetic energy, equation (3): $K_{\max} = \dfrac{q^2B^2R^2}{2m} = \dfrac{\left[(1.6\times10^{-19})(1.2)(0.60)\right]^2}{2(1.67\times10^{-27})} = 4.0\times10^{-12}$ J.

Step 4. Convert: $K_{\max} = \dfrac{4.0\times10^{-12}}{1.6\times10^{-13}} = \boxed{24.8\ \mathrm{MeV}}$ — hospital-cyclotron territory. Sanity: $\tfrac12 m v_{\max}^2$ with Step 2's speed gives the same $4.0\times10^{-12}$ J ✓.

Board level 2 — reverse-engineering the magnet

A cyclotron's oscillator runs at $f = 10$ MHz. (a) What magnetic field keeps protons in step? (b) With dees of radius $R = 0.50$ m, what is the exit energy?

Step 1. Invert the synchronism condition: $B = \dfrac{2\pi m f}{q} = \dfrac{2\pi(1.67\times10^{-27})(1.0\times10^{7})}{1.6\times10^{-19}} = \boxed{0.66\ \mathrm{T}}$ .

Step 2. Exit energy from equation (3): $K_{\max} = \dfrac{q^2B^2R^2}{2m} = \dfrac{\left[(1.6\times10^{-19})(0.656)(0.50)\right]^2}{2(1.67\times10^{-27})} = 8.2\times10^{-13}$ J $= \boxed{5.2\ \mathrm{MeV}}$ .

Step 3. Cross-check via the speed: $v_{\max} = qBR/m = 3.1\times10^{7}$ m/s, and $\tfrac12mv^2 = 8.2\times10^{-13}$ J ✓. (This is the NCERT-vintage exercise pattern — the numbers land within a whisker of $\pi\times10^{7}$ m/s, a coincidence students have noticed for forty years.)

Competitive level 1 — the kick ledger (legacy-JEE pattern, ISC-hard)

The machine of Board example 1 ( $B = 1.2$ T, $R = 0.60$ m) runs its gap at a peak voltage $V = 50$ kV. A proton is to be delivered at $K = 20$ MeV. Find (a) the energy gained per turn, (b) the number of complete turns required, (c) the total time spent spiralling, and (d) verify the proton still fits inside the dees at 20 MeV.

Step 1. Two gap crossings per turn, equation (2): $\Delta K = 2qV = 2(1.6\times10^{-19})(5.0\times10^{4}) = 1.6\times10^{-14}$ J $= \boxed{100\ \mathrm{keV\ per\ turn}}$ .

Step 2. Turns: $N = \dfrac{K}{\Delta K} = \dfrac{20\ \mathrm{MeV}}{0.100\ \mathrm{MeV}} = \boxed{200\ \mathrm{turns}}$ .

Step 3. Every turn costs the same $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(1.2)} = 5.5\times10^{-8}$ s, so $t = NT = 200 \times 5.5\times10^{-8} = \boxed{1.1\times10^{-5}\ \mathrm{s}}$ — twenty MeV in eleven microseconds.

Step 4. Radius at 20 MeV: $v = \sqrt{2K/m} = \sqrt{2(20\times10^{6})(1.6\times10^{-19})/(1.67\times10^{-27})} = 6.2\times10^{7}$ m/s, giving $r = \dfrac{mv}{qB} = 0.54$ m $< R = 0.60$ m ✓ — the spiral has room to spare (its ceiling is the 24.8 MeV of Board example 1). Timing note: ~3 minutes. Parts (a)–(c) are pure ledger arithmetic; the standard slip is using $qV$ per turn instead of $2qV$ — the gap is crossed twice per lap, and the polarity flip is what makes both crossings push forward.

Competitive level 2 — weighing neon (mass-spectrometer numerical)

In a Bainbridge-type spectrometer, singly ionised neon ions pass a velocity selector set to $v = 3.0\times10^{5}$ m/s and enter a uniform field $B = 0.40$ T, landing on a photographic plate after a semicircle. For the isotopes Ne-20 and Ne-22 (masses $20u$ and $22u$ , $u = 1.66\times10^{-27}$ kg), find the landing radii and the separation of the two lines on the plate.

Step 1. Both isotopes arrive at the same speed — that is the selector's job (Crossed Fields & the Velocity Selector owns the why). At fixed $v$ , equation (4) says $r \propto m$ .

Step 2. Ne-20: $r_{20} = \dfrac{mv}{qB} = \dfrac{(20)(1.66\times10^{-27})(3.0\times10^{5})}{(1.6\times10^{-19})(0.40)} = \boxed{15.6\ \mathrm{cm}}$ .

Step 3. Ne-22 by proportion (save the calculator): $r_{22} = r_{20}\times\dfrac{22}{20} = \boxed{17.1\ \mathrm{cm}}$ .

Step 4. Each ion lands one diameter from the slit, so the plate separation is $2(r_{22} - r_{20}) = 2(1.56\ \mathrm{cm}) = \boxed{3.1\ \mathrm{cm}}$ — two clean lines, the heavier isotope farther out. Timing note: ~2.5 minutes. The factor of 2 from radius to plate position is the classic dropped mark — sketch the two semicircles from the same slit and it can't be missed. If the question had accelerated the ions through a voltage instead of a selector, the ratio would be $\sqrt{22/20}$ , not $22/20$ : check which machine you're in before Step 3.

Exam traps

Which boards can actually ask this. CBSE deleted the cyclotron in the 2023–24 rationalisation, and today's JEE-Main, JEE-Advanced and NEET syllabi do not list it. ISC retains it — and question banks recycle decades of archived items — so treat this module as ISC-examinable and archive-practice everywhere else.

$2qV$ per turn, not $qV$ . One turn = two gap crossings, and the oscillator's polarity flip makes both accelerate. Halving this doubles your turn count and wrecks parts (b) and (c) of ledger questions.

$K_{\max}$ has no $V$ in it. The single most-tested conceptual point: gap voltage changes the number of instalments, never the final amount. If your answer to "what happens to the exit energy when the voltage is doubled?" is anything but "nothing", re-read equation (3).

Frequency vs angular frequency. $f = qB/2\pi m$ in hertz; $\omega = qB/m$ in rad/s. Quoting $\omega$ where $f$ was asked is a factor-of- $6.28$ error the options will be waiting for.

Radius grows with $\sqrt{K}$ , not $K$ . Since $K = q^2B^2r^2/2m$ , the orbit radius scales as $\sqrt{K}$ — doubling the energy multiplies the radius by $1.41$ , not 2. Turn-ledger questions that also ask for the radius test exactly this.

Electrons are not cyclotron material. At even ~1 MeV an electron's speed is a large fraction of $c$ ; its lap time stretches (the honest statement of the period picks up the relativistic factor), the fixed-frequency kicks fall out of step, and the machine fails. State it qualitatively — "relativistic mass increase breaks synchronism" earns the mark; no formula expected at this level.

Constants and conversions. $e = 1.6\times10^{-19}$ C, $m_p = 1.67\times10^{-27}$ kg, $1u = 1.66\times10^{-27}$ kg; energy conversions $1\ \mathrm{eV} = 1.6\times10^{-19}$ J and $1\ \mathrm{MeV} = 1.6\times10^{-13}$ J. ISC papers state values used; keep the MeV conversion at your fingertips — it appears in nearly every cyclotron numerical.

Origins · 1 minErnest Orlando Lawrence and how this was discovered

Ernest Orlando Lawrence

1901–1958 · Berkeley

Ernest Lawrence found the cyclotron principle in 1929 in the Berkeley library, reading Wideröe's paper on a two-stage linear accelerator. His leap: a magnetic field could fold the acceleration line into a spiral, and the speed-independent period meant one radio-frequency voltage would stay in step forever. The first working model, built in 1931 with his student M. Stanley Livingston, was 4.5 inches of brass, wax and window glass — about $25 of hardware — and reached 80 keV. Within a decade his lab's machines were making radioisotopes for medicine and mapping the nucleus; the 1939 Nobel Prize in Physics followed, the first ever awarded for a machine. Lawrence's deeper invention was the style of science the machine demanded — teams of physicists, engineers and physicians around one big instrument — which is how physics has been done ever since. Element 103, lawrencium, carries his name.

History

1919

— Aston's mass spectrograph splits neon into Ne-20 and Ne-22: the first isotopes of a stable element, weighed by landing position

1929

— Lawrence, reading Wideröe's linac paper, sketches the magnetic spiral that reuses one gap over and over

1931

— the first cyclotron: 4.5 inches, ~$25, protons to 80 keV; the 27-inch follows within three years

1939

— Nobel Prize to Lawrence; cyclotrons worldwide are making phosphorus-32 and iodine-131 for medicine

1945

— phase stability (McMillan, Veksler) breaks the relativistic ceiling: synchrocyclotrons, then synchrotrons

2020

s — roughly 1,500 hospital cyclotrons produce PET isotopes every morning; mass spectrometry weighs molecules in every biochemistry lab

Going further

Texts. ISC Physics Class 12 (Nootan/S. Chand), Moving Charges & Magnetism · NCERT Class 12 Ch. 4 (pre-2023 editions carry the cyclotron section this module preserves) · HC Verma, Concepts of Physics Vol. 2, Ch. 34.

Provenance. All machine physics standard; every number recomputed for this edition. A verifier should recheck: Board 1 — $f = 18.3$ MHz, $v_{\max} = 6.9\times10^{7}$ m/s, $K_{\max} = 4.0\times10^{-12}$ J $= 24.8$ MeV (both routes); Board 2 — $B = 0.656$ T, $K_{\max} = 5.2$ MeV, $v_{\max} = 3.1\times10^{7}$ m/s; Competitive 1 — $100$ keV/turn, $N = 200$ , $T = 5.5\times10^{-8}$ s, $t = 1.1\times10^{-5}$ s, $r(20\ \mathrm{MeV}) = 0.54$ m; Competitive 2 — $r_{20} = 15.6$ cm, $r_{22} = 17.1$ cm, plate gap $3.1$ cm; accelerated variant (derivation B) — $r_{20} = 5.76$ cm, $r_{22} = 6.04$ cm, gap $5.6$ mm.

Prerequisites. Circular & Helical Paths in Magnetic Fields · Crossed Fields & the Velocity Selector. Leads to. Charged-Particle Trajectories in E & B (the geometry gym these machines industrialise) · Cyclotron Motion (UG deep dive).

If the magnetic field does no work, what exactly is it for?

Geometry. The magnet folds a kilometres-long acceleration path into a half-metre spiral, so a single 50 kV gap can be reused hundreds of times instead of building hundreds of gaps in a row (which is what a linear accelerator does). Every joule still comes from the electric field; the magnet's contribution is that the ion keeps coming back for more. That is why cyclotron upgrades chase bigger magnets, not bigger voltages — the magnet also sets the exit energy through $K_{\max} = q^2B^2R^2/2m$ .

Why does the ion feel no electric force inside the dees?

The dees are hollow conductors, and the field inside a hollow conductor's cavity is screened — the same shielding physics that Electrostatics of Conductors & Shielding banks for Faraday cages. The gap between the dees is the only place the oscillator's field lives. So the motion cleanly alternates: pure magnetic half-circle inside a dee, pure electric kick in the gap. That separation is what makes the machine analysable with two formulas.

What actually goes wrong if the oscillator frequency is slightly off?

The kicks drift out of phase. On each gap crossing the voltage has flipped slightly early or late, so the push is a little weaker; the error accumulates lap after lap until the ion arrives when the gap field opposes it and starts losing energy. The beam simply fails to build. This is also exactly how relativity kills the machine at high speed: the ion's lap time stretches as it approaches $c$ , which is indistinguishable from the oscillator running fast — synchronism is the cyclotron's whole spine, and it is brittle.

Why exactly can't a cyclotron accelerate electrons?

Because electrons get relativistic almost immediately. An electron carrying just 1 MeV of kinetic energy moves at about 94% of light speed; its effective inertia has nearly tripled, so its lap time is no longer $2\pi m/qB$ with the rest mass, and the fixed-frequency kicks fall hopelessly out of step within a few turns. Protons, 1836 times heavier, stay comfortably non-relativistic up to tens of MeV — which is precisely the working range of hospital cyclotrons. Machines that do accelerate electrons (betatrons, synchrotrons) change the field or the frequency on the fly.

Can a cyclotron or a mass spectrometer handle neutral particles?

No — both machines grip the particle through its charge. A neutron feels neither the gap's electric kick nor the magnet's turning force: $q = 0$ zeroes every formula in this module. This is why neutron beams are made by collisions (firing accelerated charged particles at targets) rather than by direct acceleration, and why a mass spectrometer must ionise its sample before weighing it — the "singly ionised" in every question is a load-bearing phrase.

Two isotopes with the same m/q would land on the same spot — does that actually happen?

Yes, and it is a genuine limitation: a spectrometer weighs the ratio $m/q$ , not the mass alone. A doubly ionised mass-44 ion and a singly ionised mass-22 ion land together. Real laboratories break the tie by looking at multiple charge states or fragmenting the molecule. In exams the phrase "singly ionised" is doing exactly this job — pinning $q = e$ so the landing position reads pure mass.

Is the path inside the cyclotron a circle or a spiral?

Both, depending on how you zoom. Inside each dee the ion travels a genuine half-circle — constant speed, constant radius, no electric field. Each gap crossing bumps the speed, so the next half-circle is slightly larger: stitched together, a spiral of growing semicircles. Exam figures drawing a smooth spiral are cartooning; the physics is "circle, kick, bigger circle".