University Physics — Year 1 · Bachelor Year 1

17Charged Particles in E and B Fields

Inside the oscilloscope that drew the traces of Chapter 7, a beam of electrons is steered across a screen by the voltage between two plates; inside the magnet of a mass spectrometer, ions of two isotopes separate by millimeters because one is eight percent heavier than the other; inside a hospital’s cyclotron, protons spiral outward through two hundred turns before leaving at a third of the speed of light to burn a tumor. All three are the same dynamics: a charge in a uniform field, electric or magnetic, and Newton’s law. This chapter works it out — the parabola in an electric field, the circle and the helix in a magnetic field — and turns it into the instruments that sort, accelerate and measure charged particles.

Lawrence’s 60-inch cyclotron at Berkeley, 1939 (US Department of Energy): between the poles of the magnet, protons spiral outward as they are kicked twice per turn, exactly as in the weekend problem.

17.1 The Lorentz force

Definition 17.1 (Lorentz force)

A particle of charge $q$ moving at velocity $\vect v$ in an electric field $\vect E$ and a magnetic field $\vect B$ receives the force

\vect F = q\big(\vect E + \vect v\wedge\vect B\big) .

The electric part is along $\vect E$ (or opposite, for $q < 0$ ) and independent of the motion; the magnetic part is perpendicular both to $\vect v$ and to $\vect B$ , proportional to the speed, and vanishes for a particle at rest. Units: $\vect E$ in $\mathrm{V}/\mathrm{m}$ (or $\mathrm{N}/\mathrm{C}$ ), $\vect B$ in teslas ( $\mathrm{T}$ ): $1\,\mathrm{T} = 1\,\mathrm{N}/(\mathrm{A}\,\mathrm{m})$ .

Proposition 17.2 (Work of the Lorentz force)

The magnetic force does no work: its power $q(\vect v\wedge\vect B)\cdot\vect v$ is zero, so a magnetic field alone changes the direction of the velocity but never the speed or the kinetic energy. The electric force has power $q\vect E\cdot\vect v$ , and between two points its work is $q(V_A - V_B) = qU_{AB}$ (Chapter 27): a particle of charge $q$ crossing a voltage $U$ gains the kinetic energy $\abs{qU}$ .

Proof. $\vect v\wedge\vect B \perp \vect v$ . The electrostatic field derives from the potential, $\vect E = -\overrightarrow{\operatorname{grad}}V$ , so $q\vect E$ is conservative with $E_p = qV$ . ∎

Definition 17.3 (The electron-volt)

The electron-volt is the kinetic energy gained by an elementary charge crossing one volt: $1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$ . Chemistry lives at a few eV, X-ray tubes at tens of keV, nuclear physics at MeV, particle physics at GeV and TeV.

Remark 17.4 (Why gravity is ignored)

For an electron in a field of $1\,\mathrm{kV}/\mathrm{m}$ — a modest laboratory value — the electric force is $eE = 1.6 \times 10^{-16}\,\mathrm{N}$ , against a weight $mg = 9 \times 10^{-30}\,\mathrm{N}$ : $10^{13}$ times smaller. Weight is dropped from every calculation of this chapter without a second thought. (The relativistic limit is another matter: the formulas below hold for $\tfrac12 mv^2 \ll mc^2$ , i.e. below about $50\,\mathrm{keV}$ for electrons and $100\,\mathrm{MeV}$ for protons; see the Year 3 volume.)

17.2 Uniform electric field

Proposition 17.5 (Acceleration and deflection)

In a uniform field $\vect E$ the acceleration $\vect a = q\vect E/m$ is constant: the motion is that of a projectile, with $q\vect E/m$ in place of $\vect g$ .

Released from rest and accelerated through a voltage $U$ , the particle reaches $v = \sqrt{2\abs{qU}/m}$ .
Entering at speed $v_0$ along $x$ a region of length $L$ where $\vect E = E\vect e_y$ , it follows the parabola $y = (qE/2mv_0^2)x^2$ and leaves deflected by the angle $\alpha$ with $\tan\alpha = qEL/mv_0^2$ , as if it had come in a straight line from the middle of the plates.

Proof. Energy: $\tfrac12 mv^2 = \abs{qU}$ . Deflection: $x = v_0t$ , $y = \tfrac12(qE/m)t^2$ ; at the exit, $\dd y/\dd x = (qE/m)(x/v_0^2)$ evaluated at $L$ ; the tangent at the exit, $y = (qEL/mv_0^2)(x - L/2)$ , crosses the axis at $x = L/2$ . ∎

Electrostatic deflection: between the plates the electron (here q < 0, deflected toward the positive plate) follows a parabola; beyond them it flies straight, as if coming from the midpoint of the plates — the geometry of the cathode-ray oscilloscope. — Electrostatic deflection: between the plates the electron (here $q < 0$ , deflected toward the positive plate) follows a parabola; beyond them it flies straight, as if coming from the midpoint of the plates — the geometry of the cathode-ray oscilloscope.

Example 17.6 (The oscilloscope tube)

Electrons accelerated through $U_0 = 2.0\,\mathrm{kV}$ ( $v_0 = 2.65 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ , a tenth of $c$ — just non-relativistic) cross $4.0\,\mathrm{cm}$ plates $1.0\,\mathrm{cm}$ apart carrying $100\,\mathrm{V}$ : $E = 1.0 \times 10^{4}\,\mathrm{V}/\mathrm{m}$ , $\tan\alpha = eEL/mv_0^2 = EL/2U_0 = 0.10$ , and on a screen $30\,\mathrm{cm}$ past the plates’ center the spot moves by $3.0\,\mathrm{cm}$ — $0.3\,\mathrm{mm}$ per volt, the tube’s sensitivity, set purely by geometry and the accelerating voltage.

17.3 Uniform magnetic field

Theorem 17.7 (Motion in a uniform magnetic field)

A particle of charge $q$ , mass $m$ , velocity $\vect v_0$ perpendicular to a uniform $\vect B$ , describes a circle at constant speed $v_0$ , of radius and angular frequency

R = \frac{mv_0}{\abs q B}, \qquad \omega_c = \frac{\abs q B}{m},

the cyclotron frequency $\omega_c$ being independent of the speed; the sense of rotation depends on the sign of $q$ . If $\vect v_0$ also has a component $v_\parallel$ along $\vect B$ , that component is unchanged and the trajectory is a helix of radius $mv_\perp/\abs qB$ and pitch $2\pi mv_\parallel/\abs qB$ wound around the field lines.

Proof. The force $q\vect v\wedge\vect B$ is perpendicular to $\vect v$ and to $\vect B$ ; the speed is constant (no work) and the component along $\vect B$ sees no force. In the plane perpendicular to $\vect B$ the force has the constant norm $\abs qv_\perp B$ , always perpendicular to the velocity: uniform circular motion (Theorem 11.13) with $mv_\perp^2/R = \abs qv_\perp B$ . Then $\omega_c = v_\perp/R$ ; the period $2\pi m/\abs qB$ times $v_\parallel$ is the pitch. ∎

Left: a charge whose velocity is perpendicular to B circles at the cyclotron frequency qB/m, the magnetic force supplying the centripetal acceleration. Right: with a velocity component along B, a helix winds around the field line — the motion of electrons trapped in the Earth’s field. — Left: a charge whose velocity is perpendicular to $\vect B$ circles at the cyclotron frequency $\abs qB/m$ , the magnetic force supplying the centripetal acceleration. Right: with a velocity component along $\vect B$ , a helix winds around the field line — the motion of electrons trapped in the Earth’s field.

Example 17.8 (Orders of magnitude)

A $1\,\mathrm{keV}$ electron ( $v = 1.9 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ ) in $1\,\mathrm{mT}$ : $R = 11\,\mathrm{cm}$ , $\omega_c = 1.8 \times 10^{8}\,\mathrm{rad}/\mathrm{s}$ , one turn in $36\,\mathrm{ns}$ — the same period at any energy, which is what makes the cyclotron possible. A proton of the solar wind ( $1\,\mathrm{keV}$ , $4.4 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ ) in the Earth’s $50\,\text{µ}\mathrm{T}$ : $R = 92\,\mathrm{m}$ ; it cannot cross the field lines, only spiral along them toward the poles — the aurora.

17.4 Instruments

Proposition 17.9 (Velocity selector and mass spectrometer)

In crossed uniform fields $\vect E \perp \vect B$ , a particle moving perpendicular to both passes undeflected iff $v = E/B$ , whatever its charge and mass (velocity selector). In a magnetic field alone a particle of known speed describes a semicircle of diameter $2mv/\abs qB$ : measuring it gives $m/q$ (mass spectrometer). Ions accelerated through a voltage $U$ then bent in $B$ land at

R = \frac{1}{B}\sqrt{\frac{2mU}{\abs q}} :

two isotopes of masses $m$ and $m + \Delta m$ separate by $\Delta R \approx R\,\Delta m/2m$ .

Proof. $q\vect E + q\vect v\wedge\vect B = \vect 0$ requires $vB = E$ with the right orientation. $R = mv/qB$ with $v = \sqrt{2qU/m}$ ; differentiate $\ln R = \tfrac12\ln m + \text{const}$ . ∎

Example 17.10 (Carbon isotopes)

Singly charged $^{12}$ C and $^{13}$ C ions accelerated through $20\,\mathrm{kV}$ in $0.50\,\mathrm{T}$ : $R = 14.1\,\mathrm{cm}$ and $14.7\,\mathrm{cm}$ , $5.8\,\mathrm{mm}$ apart — two separate spots on the detector, and the ratio of their intensities is the isotopic abundance, the raw material of radiocarbon dating and of every isotope analysis.

Proposition 17.11 (Cyclotron)

Two hollow half-cylinders (“dees”) in a uniform $\vect B$ , with an alternating voltage $U$ between them at the cyclotron frequency: the particle circles inside each dee (no field there), is accelerated by $\abs qU$ at each of the two gap crossings per turn, and spirals outward with $R \propto v$ ; at the extraction radius $R_{\max}$ its kinetic energy is

E_k = \frac{q^2B^2R_{\max}^2}{2m},

independent of the voltage, which only sets the number of turns.

Proof. Because $\omega_c$ does not depend on the speed, a fixed-frequency voltage stays in step with the particle turn after turn; $v = \abs qBR/m$ at radius $R$ . ∎

The cyclotron: two dees in a uniform magnetic field; the particle circles inside the dees and is kicked by the gap voltage twice per turn, always in phase because its period does not depend on its speed. The energy at the rim depends only on B and the radius. — The cyclotron: two dees in a uniform magnetic field; the particle circles inside the dees and is kicked by the gap voltage twice per turn, always in phase because its period does not depend on its speed. The energy at the rim depends only on $B$ and the radius.

Example 17.12 (A hospital cyclotron)

Protons, $B = 1.5\,\mathrm{T}$ , $R_{\max} = 0.50\,\mathrm{m}$ : $E_k = (1.6\times10^{-19} \times 1.5 \times 0.5)^2/(2 \times 1.67\times10^{-27}) = 4.3 \times 10^{-12}\,\mathrm{J} = 27\,\mathrm{MeV}$ , at $f_c = \omega_c/2\pi = 23\,\mathrm{MHz}$ . With $50\,\mathrm{kV}$ on the dees, $100\,\mathrm{keV}$ per turn: $270$ turns in $12\,\text{µ}\mathrm{s}$ . Above a few tens of MeV the relativistic increase of the mass (the Year 3 volume) desynchronizes the protons; therapy machines at $230\,\mathrm{MeV}$ modulate the frequency or the field to compensate — the weekend problem’s subject.

Proposition 17.13 (Hall effect)

A ribbon of thickness $d$ carrying a current $I$ of carriers of charge $q$ and density $n$ , in a field $B$ perpendicular to the ribbon, develops across its width the Hall voltage

U_H = \frac{IB}{nqd} :

proportional to $B$ (a magnetometer) and inversely to $n$ (a probe of the carrier density and sign).

Proof. The magnetic force $qv_dB$ on the carriers (drift speed $v_d$ ) pushes them to one edge until the transverse electric field $E_H$ balances it: $qE_H = qv_dB$ , and $U_H = E_Hw$ across the width $w$ ; with $I = nqv_dwd$ , $E_Hw = v_dBw = IB/(nqd)$ . ∎

17.5 Exercises

Exercise 17.1 ★

Convert $1\,\mathrm{eV}$ , $13.6\,\mathrm{eV}$ , $1\,\mathrm{MeV}$ to joules. Speed of an electron and of a proton accelerated from rest through $1.0\,\mathrm{kV}$ .

Solution

Solution of Exercise 17.1.

$1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$ , $13.6\,\mathrm{eV} = 2.18 \times 10^{-18}\,\mathrm{J}$ , $1\,\mathrm{MeV} = 1.60 \times 10^{-13}\,\mathrm{J}$ . $v = \sqrt{2eU/m}$ : electron $1.9 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ ; proton $4.4 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ ( $\sqrt{1836}$ times slower).

Exercise 17.2 ★

A $1.0\,\mathrm{keV}$ electron enters a $1.0\,\mathrm{mT}$ field perpendicularly. Radius, cyclotron angular frequency, period.

Solution

Solution of Exercise 17.2.

$v = 1.88 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ ; $R = mv/eB = 9.11\times10^{-31} \times 1.88\times10^7/ (1.6\times10^{-19} \times 10^{-3}) = 0.11\,\mathrm{m}$ ; $\omega_c = eB/m = 1.8 \times 10^{8}\,\mathrm{rad}/\mathrm{s}$ ; $T = 2\pi/\omega_c = 36\,\mathrm{ns}$ .

Exercise 17.3 ★

A proton at $1.0 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ in the Earth’s field ( $50\,\text{µ}\mathrm{T}$ ): radius of its circle. What happens if its velocity makes $30^\circ$ with the field?

Solution

Solution of Exercise 17.3.

$R = mv/qB = 1.67\times10^{-27} \times 10^5/(1.6\times10^{-19} \times 5\times 10^{-5}) = 21\,\mathrm{m}$ . At $30^\circ$ : a helix of radius $R\sin 30^\circ = 10\,\mathrm{m}$ and pitch $2\pi(m/qB)v\cos 30^\circ = 114\,\mathrm{m}$ , drifting along the field line.

Exercise 17.4 ★

A velocity selector has $E = 10\,\mathrm{kV}/\mathrm{m}$ and $B = 20\,\mathrm{mT}$ . Which speed passes? What happens to faster ions, and to the same speed with the opposite charge?

Solution

Solution of Exercise 17.4.

$v = E/B = 10^4/0.020 = 5.0 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ . Faster ions: the magnetic force wins, they curve toward the magnetic side; opposite charge at the same speed: both forces reverse, still balanced — the selector is blind to the sign and to the mass.

Exercise 17.5 ★★

Oscilloscope: electrons at $2.0\,\mathrm{kV}$ , plates $4.0\,\mathrm{cm}$ long, $1.0\,\mathrm{cm}$ apart, $100\,\mathrm{V}$ between them, screen $30\,\mathrm{cm}$ past the plates’ center. Deflection on the screen and sensitivity in mm/V. Why does a higher accelerating voltage reduce the sensitivity?

Solution

Solution of Exercise 17.5.

$\tan\alpha = EL/2U_0 = 10^4 \times 0.04/4000 = 0.10$ ; deflection $0.30 \times 0.10 = 3.0\,\mathrm{cm}$ ; sensitivity $0.30\,\mathrm{mm}/\mathrm{V}$ . Faster electrons spend less time between the plates: $\tan\alpha \propto 1/U_0$ .

Exercise 17.6 ★★

Singly charged $^{12}$ C and $^{13}$ C ions, accelerated through $20\,\mathrm{kV}$ , enter a $0.50\,\mathrm{T}$ field ( $1\,\mathrm{u} = 1.66 \times 10^{-27}\,\mathrm{kg}$ ). Radii and separation of the two spots after a half turn.

Solution

Solution of Exercise 17.6.

$R = \sqrt{2mU/e}/B$ : $m_{12} = 1.99 \times 10^{-26}\,\mathrm{kg}$ , $R_{12} = \sqrt{2 \times 1.99\times10^{-26} \times 2\times10^4/1.6\times10^{-19}}/0.5 = 14.1\,\mathrm{cm}$ ; $R_{13} = R_{12}\sqrt{13/12} = 14.7\,\mathrm{cm}$ ; separation $2\Delta R = 1.2\,\mathrm{cm}$ between the landing points (radii differ by $5.8\,\mathrm{mm}$ , diameters by twice that).

Exercise 17.7 ★★

A proton cyclotron has $B = 1.5\,\mathrm{T}$ and $R_{\max} = 0.50\,\mathrm{m}$ . Frequency of the dee voltage, maximal kinetic energy, number of turns for $50\,\mathrm{kV}$ on the dees, time spent inside. By what fraction has the proton’s mass increased at the exit ( $\gamma - 1 \approx E_k/mc^2$ , $mc^2 = 938\,\mathrm{MeV}$ )?

Solution

Solution of Exercise 17.7.

$f_c = eB/2\pi m = 23\,\mathrm{MHz}$ ; $E_k = e^2B^2R^2/2m = 4.3 \times 10^{-12}\,\mathrm{J} = 27\,\mathrm{MeV}$ ; $100\,\mathrm{keV}$ per turn: $270$ turns; time $270/f_c = 12\,\text{µ}\mathrm{s}$ ; $\gamma - 1 = 27/938 = 2.9\%$ .

Exercise 17.8 ★★

An electron at $1.0 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ enters a $2.0\,\mathrm{mT}$ field at $30^\circ$ to the field lines. Radius and pitch of its helix; number of turns per meter along the field.

Solution

Solution of Exercise 17.8.

$v_\perp = 5.0 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ , $v_\parallel = 8.7 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ : $R = mv_\perp/eB = 9.11\times10^{-31} \times 5\times10^5/(3.2\times10^{-22}) = 1.4\,\mathrm{mm}$ ; period $2\pi m/eB = 18\,\mathrm{ns}$ , pitch $v_\parallel T = 1.5\,\mathrm{cm}$ : about $64$ turns per meter.

Exercise 17.9 ★★

A copper strip $1.0\,\mathrm{mm}$ thick carries $5.0\,\mathrm{A}$ in $1.0\,\mathrm{T}$ ( $n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$ ). Hall voltage. Why are Hall sensors made of semiconductors ( $n \sim 1 \times 10^{22}\,\mathrm{m}^{-3}$ )?

Solution

Solution of Exercise 17.9.

$U_H = IB/nqd = 5.0/(8.5\times10^{28} \times 1.6\times10^{-19} \times 10^{-3}) = 3.7 \times 10^{-7}\,\mathrm{V}$ : sub-microvolt. With $n$ ten million times smaller, a semiconductor gives millivolts — a usable sensor.

Exercise 17.10 ★★★

Prove from Newton’s law that a charge in any magnetic field (uniform or not), with no electric field, keeps a constant speed. A $10\,\mathrm{keV}$ electron enters a region of strong, non-uniform field: what can and cannot change?

Solution

Solution of Exercise 17.10.

$m\,\dd\vect v/\dd t = q\vect v\wedge\vect B$ ; dot with $\vect v$ : $\dd(\tfrac12 mv^2)/ \dd t = 0$ whatever $\vect B(\vect r, t)$ , provided there is no $\vect E$ . The direction of the velocity, the radius of curvature, the pitch can all change; the speed, the kinetic energy ( $10\,\mathrm{keV}$ ) cannot.

Exercise 17.11 ★★★

A charge $q$ starts from rest at the origin in $\vect E = E\vect e_y$ and $\vect B = B\vect e_z$ . Write the equations of motion, show that the particle drifts on average at the velocity $E/B$ along $\vect e_x$ (hint: look for a solution of the form $\vect v = \vect u + \vect w$ with $\vect u$ constant), and describe the trajectory (a cycloid). Numbers for an electron with $E = 1.0 \times 10^{4}\,\mathrm{V}/\mathrm{m}$ , $B = 0.10\,\mathrm{T}$ .

Solution

Solution of Exercise 17.11.

$m\dot v_x = qv_yB$ , $m\dot v_y = qE - qv_xB$ , $v_z = 0$ . With $\vect v = (E/B)\vect e_x + \vect w$ : $m\dot w_x = qw_yB$ , $m\dot w_y = -qw_xB$ — a pure cyclotron rotation of $\vect w$ at $\omega_c$ . Starting from rest, $\vect w(0) = -(E/B)\vect e_x$ : $\vect w$ turns at constant norm $E/B$ , so the velocity is the drift $E/B$ plus a rotating vector of the same norm — a cycloid, with cusps where $\vect v = \vect 0$ . Electron: drift $E/B = 1.0 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ , $\omega_c = 1.8 \times 10^{10}\,\mathrm{rad}/\mathrm{s}$ , arch height $2mE/qB^2 = 11\,\text{µ}\mathrm{m}$ .

Exercise 17.12 ★★★

In the oscilloscope of Exercise 17.5, replace the plates by a magnetic deflection coil producing $1.0\,\mathrm{mT}$ over the same $4.0\,\mathrm{cm}$ . Deflection angle (small-angle: the arc of radius $R$ over a length $L$ ), and deflection on the screen. Why do television tubes, with their large angles, use magnetic rather than electric deflection?

Solution

Solution of Exercise 17.12.

$R = mv_0/eB = 9.11\times10^{-31} \times 2.65\times10^7/(1.6\times10^{-22}) = 0.15\,\mathrm{m}$ ; $\alpha \approx L/R = 0.04/0.15 = 0.27$ (about $15^\circ$ ): deflection $\approx 0.30 \times 0.27 = 8\,\mathrm{cm}$ for a modest millitesla. Large angles with plates would need kilovolts and long plates; a coil bends the beam through tens of degrees with no voltage at all.

17.6 Problem: Protons against a tumor

Problem 17.1

Weekend problem — the cyclotron behind the wall of a hospital: how protons are spun up to a third of the speed of light, why the simple machine stops working before the energy a therapist needs, and how many protons it takes to treat a tumor

Protons: $m = 1.67 \times 10^{-27}\,\mathrm{kg}$ , $q = e = 1.60 \times 10^{-19}\,\mathrm{C}$ , $mc^2 = 938\,\mathrm{MeV}$ ; $c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ . Cyclotron: $B = 1.5\,\mathrm{T}$ , extraction radius $R_{\max} = 0.50\,\mathrm{m}$ , dee voltage amplitude $U = 50\,\mathrm{kV}$ .

Part I — One turn.

Write the Lorentz force on a proton moving at $\vect v$ perpendicular to $\vect B$ , and show that its speed cannot change inside a dee.
Derive the radius of its circle and its period; show that the period does not depend on the speed.
Compute the cyclotron frequency $f_c$ for these protons.
Energy gained at each gap crossing; per turn.
Why must the dee voltage alternate at exactly $f_c$ ? What happens if it is slightly off?
The protons are injected at the center with negligible speed: radius after the first turn? After $n$ turns?

Part II — To the rim.

Kinetic energy at extraction (in J and MeV), and the speed, as a fraction of $c$ .
Number of turns needed and time spent in the machine.
Total length of the spiral path (sum of the circumferences — approximate the sum by an integral).
Does the result of question 7 depend on $U$ ? On what does $U$ act?
The vacuum inside must be good: estimate the distance a proton travels between collisions with residual gas molecules if the mean free path must exceed the spiral length; the mean free path scales as $1/p$ and is about $70\,\mathrm{nm}$ at atmospheric pressure. What pressure is needed?
Compute $\gamma - 1 \approx E_k/mc^2$ at extraction and the resulting relative change of the period. Over the number of turns found above, by what fraction of a period has the proton slipped relative to the voltage? Comment.

Part III — 230 MeV. Proton therapy needs about $230\,\mathrm{MeV}$ to reach tumors $30\,\mathrm{cm}$ deep.

With the non-relativistic formula, what product $BR$ would give $230\,\mathrm{MeV}$ ? With $B = 2.0\,\mathrm{T}$ , what radius?
At that energy $\gamma - 1 = 0.245$ : what happens to the cyclotron period as the proton gains energy, and why does a fixed-frequency cyclotron fail?
Two cures exist: vary the voltage frequency during the acceleration (synchrocyclotron), or make $B$ grow with radius so that $B/\gamma$ stays constant (isochronous cyclotron). Explain each in one sentence and name a drawback of the first.
The relativistic momentum is $p = \gamma mv$ and the radius $R = p/qB$ still holds: with $\gamma = 1.245$ and $v = 0.60c$ , compute $R$ for $B = 2.0\,\mathrm{T}$ and compare with question 13.
The extracted beam passes between two plates $1.0\,\mathrm{m}$ long with $E = 1.0 \times 10^{5}\,\mathrm{V}/\mathrm{m}$ to be steered by a small angle: estimate the deflection angle (non-relativistically, with $p$ from the previous question as $mv$ replacement). Is electric steering practical here?
The same steering with a magnet of $1.0\,\mathrm{T}$ over $1.0\,\mathrm{m}$ : angle? Conclude on why beam lines use magnets.

Part IV — The dose.

A $230\,\mathrm{MeV}$ proton deposits most of its energy near the end of its range (the Bragg peak). If all of it is absorbed in a tumor of mass $0.50\,\mathrm{kg}$ , how many protons deliver a dose of $2.0\,\mathrm{Gy}$ ( $1\,\mathrm{Gy} = 1\,\mathrm{J}/\mathrm{kg}$ )?
The beam current is $1.0\,\mathrm{nA}$ : protons per second, and duration of the session.
Power carried by the beam; compare with a light bulb.
The protons are slowed to a stop inside the patient: what is the energy deposited per proton in the last centimeter, roughly, if a quarter of the energy is lost there? In electron-volts per micrometer?
Each ionization in tissue costs about $30\,\mathrm{eV}$ : how many ionizations does one proton cause along its whole path?
Why is a proton beam preferable to X-rays for a deep tumor, in one sentence about where the energy goes?
Summarize the chain from the Lorentz force to the dose: which equations fixed the energy, the radius, the number of turns and the treatment time.

Solution

Solution of Problem 17.1.

1. $\vect F = e\vect v\wedge\vect B \perp \vect v$ : no power, constant speed.

2. $mv^2/R = evB$ : $R = mv/eB$ ; $T = 2\pi R/v = 2\pi m/eB$ , free of $v$ .

3. $f_c = eB/2\pi m = 1.6\times10^{-19} \times 1.5/(2\pi \times 1.67 \times10^{-27}) = 23\,\mathrm{MHz}$ .

4. $eU = 50\,\mathrm{keV}$ per crossing, $100\,\mathrm{keV}$ per turn.

5. The gap must be at the accelerating polarity each time the proton arrives, half a period apart; off frequency, the phase slips and the kicks eventually decelerate.

6. After one turn $E_k = 100\,\mathrm{keV}$ , $v = \sqrt{2E_k/m} = 4.4 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ , $R = mv/eB = 3.0\,\mathrm{cm}$ ; after $n$ turns $E_k = n \times 100\,\mathrm{keV}$ and $R = 3.0\,\mathrm{cm}\sqrt n$ .

7. $E_k = e^2B^2R^2/2m = 4.3 \times 10^{-12}\,\mathrm{J} = 27\,\mathrm{MeV}$ ; $v = \sqrt{2E_k/m} = 7.2 \times 10^{7}\,\mathrm{m}/\mathrm{s} = 0.24c$ .

8. $270$ turns; $270 \times 43.5\,\mathrm{ns} = 12\,\text{µ}\mathrm{s}$ .

9. $\sum 2\pi R_n = 2\pi \times 0.030\sum\sqrt n \approx 2\pi \times 0.030 \times \tfrac23 \times 270^{3/2} = 560\,\mathrm{m}$ .

10. No: $E_k$ depends on $B$ and $R_{\max}$ only; $U$ sets the number of turns and the time.

11. Path $560\,\mathrm{m}$ : need $\lambda \gtrsim 1\,\mathrm{km}$ , i.e. $p \lesssim 10^5\ \mathrm{Pa} \times 7\times10^{-8}/10^3 = 7 \times 10^{-6}\,\mathrm{Pa}$ — a high vacuum.

12. $\gamma - 1 = 27/938 = 0.029$ ; the period grows by $2.9\%$ . Over $270$ turns the slip accumulates to several periods — the last turns would be out of phase; in practice $B$ and the injection phase are trimmed, and $27\,\mathrm{MeV}$ is near the limit of the simple machine.

13. $E_k = (eBR)^2/2m$ : $BR = \sqrt{2mE_k}/e = \sqrt{2 \times 1.67\times 10^{-27} \times 3.68\times10^{-11}}/1.6\times10^{-19} = 2.2\,\mathrm{T}\,\mathrm{m}$ ; $R = 1.1\,\mathrm{m}$ at $2\,\mathrm{T}$ .

14. $T = 2\pi\gamma m/eB$ grows by $24\%$ from center to rim: a fixed frequency falls hopelessly out of step.

15. Synchrocyclotron: lower the frequency as the bunch gains energy — only one bunch at a time, low average current. Isochronous: shape $B(R) \propto \gamma$ so the period stays constant — continuous beam, the modern choice.

16. $p = \gamma mv = 1.245 \times 1.67\times10^{-27} \times 1.8\times10^8 = 3.7 \times 10^{-19}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$ ; $R = p/eB = 1.2\,\mathrm{m}$ , a little more than the non-relativistic $1.1\,\mathrm{m}$ .

17. Transverse impulse $eEL/v = 1.6\times10^{-19} \times 10^5 \times 1.0/1.8\times10^8 = 8.9 \times 10^{-23}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$ ; angle $\approx 8.9\times10^{-23}/ 3.7\times10^{-19} = 2.4 \times 10^{-4}\,\mathrm{rad}$ : hopeless for steering.

18. Magnetic: $R = p/eB = 2.3\,\mathrm{m}$ , angle $L/R = 0.43\,\mathrm{rad} \approx 25^\circ$ : a thousand times more — magnets it is.

19. Energy per proton $3.68 \times 10^{-11}\,\mathrm{J}$ ; dose $2.0 \times 0.50 = 1.0\,\mathrm{J}$ : $2.7\times10^{10}$ protons.

20. $10^{-9}/1.6\times10^{-19} = 6.2 \times 10^{9}\,\mathrm{protons}/\mathrm{s}$ : about $4.4\,\mathrm{s}$ .

21. $P = 230\,\mathrm{MeV} \times 6.2 \times 10^{9}\,\mathrm{s}^{-1} = 0.23\,\mathrm{W}$ : a night light, delivered exactly where wanted.

22. $58\,\mathrm{MeV}$ in $1\,\mathrm{cm}$ : about $6\,\mathrm{keV}/\text{µ}\mathrm{m}$ — dense ionization along the last track.

23. $230\times10^6/30 \approx 8 \times 10^{6}$ ionizations per proton — the molecular damage that kills the cell.

24. X-rays deposit energy all along their path, most near the entrance; protons deposit most of theirs at the end of their range, sparing the tissue in front and leaving nothing behind.

25. $R = mv/eB$ and $T = 2\pi m/eB$ (the circle), $E_k = (eBR)^2/2m$ (the energy), $E_k/2eU$ (the turns), dose $=$ (protons $\times$ energy)/mass and current $= e \times$ rate (the time).

Terms defined in this chapter

See all 393 terms in the glossary