Magnetic Field and Magnetic Forces

Lessons

Senior 5 · Physics

Magnetic Field and Magnetic Forces

4 lessons

Lesson 3 of 4

Charged Particles in Magnetic Fields and Their Applications

25 min

Glowing circles

In a darkened laboratory, a glass bulb filled with a trace of hydrogen gas glows with a perfect violet ring. Inside, an electron gun fires electrons sideways, and a pair of large coils outside the bulb makes a uniform magnetic field. The electrons, bent continuously by the field, travel in a circle and make the gas glow along their path. Measure the circle and you can find the ratio of the electron's charge to its mass, as J. J. Thomson did in 1897 by a related method. The same physics separates isotopes in a mass spectrometer, accelerates particles in a cyclotron and traps the charged particles that make the polar aurorae.

Circular motion in a uniform field

A charged particle moving at right angles to a uniform field feels a force $q v B$ that is always perpendicular to its velocity. A force of constant size that is always at right angles to the velocity produces uniform circular motion, so the magnetic force provides the centripetal force:

qvB = \frac{mv^2}{r} \quad \Rightarrow \quad r = \frac{mv}{qB} = \frac{p}{qB}

The time for one revolution is

T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}, \qquad f = \frac{qB}{2\pi m}

The period does not depend on the speed or the radius: faster particles move in bigger circles but take the same time to go round. This cyclotron frequency is the basis of the cyclotron accelerator.

If the velocity makes an angle $\theta$ with the field, split it into $v\sin\theta$ across the field, which gives circular motion of radius $\frac{mv\sin\theta}{qB}$ , and $v\cos\theta$ along the field, which is unaffected. The particle moves in a helix whose pitch (distance moved along the field per turn) is $v\cos\theta \times T$ . Charged particles from the Sun spiral like this along the Earth's field lines towards the poles, where they make the aurorae.

Worked example 1: an electron in a field

An electron moving at $3.0 \times 10^{7}$ m s⁻¹ enters a uniform 2.0 mT field at right angles. Find (a) the radius of its path, (b) its period and frequency. (c) What would the radius and pitch be if it entered at 30° to the field?

Solution

(a) $r = \frac{mv}{qB} = \frac{(9.11 \times 10^{-31})(3.0 \times 10^{7})}{(1.60 \times 10^{-19})(2.0 \times 10^{-3})} = 0.085$ m.

(b) $T = \frac{2\pi m}{qB} = \frac{2\pi(9.11 \times 10^{-31})}{(1.60 \times 10^{-19})(2.0 \times 10^{-3})} = 1.8 \times 10^{-8}$ s; $f = 5.6 \times 10^{7}$ Hz.

(c) Radius uses $v\sin 30^\circ = 1.5 \times 10^{7}$ m s⁻¹, giving 0.043 m. Pitch $= v\cos 30^\circ \times T = (2.6 \times 10^{7})(1.79 \times 10^{-8}) = 0.46$ m.

The velocity selector

If a particle crosses a region where an electric field $E$ and a magnetic field $B$ are at right angles to each other and to the beam, the two forces act in opposite directions. They balance when $q E = q v B$ , that is, for one speed only:

v = \frac{E}{B}

Particles of exactly this speed pass straight through, whatever their charge or mass; faster ones are bent one way and slower ones the other, and are stopped by a slit. Such crossed fields form a velocity selector.

Thomson's measurement of $\frac{e}{m}$

In 1897 J. J. Thomson used crossed fields to measure the charge-to-mass ratio of cathode rays. He adjusted $E$ and $B$ until the beam was undeflected, which gave the speed $v = \frac{E}{B}$ . Then, using the magnetic field alone, he measured the radius $r$ of the beam's curve. From $r = \frac{mv}{eB}$ ,

\frac{e}{m} = \frac{v}{rB} = \frac{E}{rB^2}

Alternatively, if the electrons are accelerated through a potential difference $V$ , then $eV = \tfrac{1}{2}mv^2$ , and combining with $r = \frac{mv}{eB}$ gives $\frac{e}{m} = \frac{2V}{B^2r^2}$ . Thomson found a ratio about 1800 times that of the hydrogen ion, the same for every cathode material: cathode rays were a universal particle, much lighter than any atom, the electron. The modern value is $\frac{e}{m_e} = 1.76 \times 10^{11}\,\mathrm{C\,kg^{-1}}$ .

Worked example 2: a Thomson-type experiment

A beam of electrons passes undeflected through crossed fields $E = 1.5 \times 10^{4}\,\mathrm{V\,m^{-1}}$ and $B = 1.0$ mT. When the electric field is switched off, the beam moves in a circle of radius 8.53 cm. Find (a) the speed of the electrons, (b) $\frac{e}{m}$ .

Solution

(a) $v = \frac{E}{B} = \frac{1.5 \times 10^{4}}{1.0 \times 10^{-3}} = 1.5 \times 10^{7}\,\mathrm{m\,s^{-1}}$ .

(b) $\frac{e}{m} = \frac{v}{rB} = \frac{1.5 \times 10^{7}}{(0.0853)(1.0 \times 10^{-3})} = 1.76 \times 10^{11}\,\mathrm{C\,kg^{-1}}$ .

The mass spectrometer

A mass spectrometer separates ions by mass. In Bainbridge's design, ions from a source pass through a velocity selector, so they all have the same speed $v = \frac{E}{B}$ . They then enter a second uniform field $B^{'}$ , where they travel in semicircles of radius $r = \frac{mv}{qB'}$ and strike a detector at a distance $2 r$ from the entrance. For ions of the same charge, the radius is proportional to the mass:

m = \frac{qB'r}{v} = \frac{qBB'r}{E}

Mass spectrometers are used to identify isotopes and measure their abundances, to date ancient objects by counting carbon-14 atoms, to test athletes and food for traces of drugs and pesticides, and to analyse the gases from volcanoes.

Definition

A velocity selector is a region of crossed electric and magnetic fields that lets through undeflected only particles of speed $v = \frac{E}{B}$ . A mass spectrometer is an instrument that separates ions according to their mass (strictly, their mass-to-charge ratio) by bending them in a magnetic field. Example: with $E = 4.0 \times 10^{4}\,\mathrm{V\,m^{-1}}$ and $B = 0.080$ T, only ions moving at $5.0 \times 10^{5}\,\mathrm{m\,s^{-1}}$ pass.

Worked example 3: separating the isotopes of neon

Singly charged neon ions leave a velocity selector with $E = 4.0 \times 10^{4}\,\mathrm{V\,m^{-1}}$ and $B = 0.080$ T, and enter a field $B^{'} = 0.50$ T. Find the radii for $^{20}\mathrm{Ne}^{+}$ (mass 20 u) and $^{22}\mathrm{Ne}^{+}$ (mass 22 u), and the distance between their marks on the detector. ( $1\,\mathrm{u} = 1.66 \times 10^{-27}$ kg.)

Solution

$v = \frac{E}{B} = 5.0 \times 10^{5}\,\mathrm{m\,s^{-1}}$ .

$r_{20} = \frac{(20 \times 1.66 \times 10^{-27})(5.0 \times 10^{5})}{(1.60 \times 10^{-19})(0.50)} = 0.208$ m; $r_{22} = \frac{22}{20} \times 0.2075 = 0.228$ m.

The marks are at $2 r$ from the entrance, so they are $2 (0.228 - 0.208) = 0.042$ m, 4.2 cm, apart, as in the diagram. In 1913 this kind of experiment showed that neon has more than one isotope.

Warning

Direction depends on the sign of the charge. An electron curves the opposite way to a proton in the same field. And $r = \frac{mv}{qB}$ uses the momentum: doubling the speed doubles the radius, but leaves the period unchanged.

Tip

One line does it: magnetic force = centripetal force, $qvB = \frac{mv^2}{r}$ . Radius, period, $\frac{e}{m}$ and the mass spectrometer all follow from it.

e/m Equipment Tutorial, SFU Physics. Watch the electron beam bend into a circle as the coil current is raised, and how the radius is measured to find e/m.

✦

Practice

Show that the period of a charged particle in a uniform magnetic field does not depend on its speed.
A proton moves at $4.0 \times 10^{6}$ m s⁻¹ perpendicular to a 0.25 T field. Find the radius of its path ( $m_p = 1.67 \times 10^{-27}$ kg).
Find the cyclotron frequency of a proton in a 1.5 T field.
In a velocity selector, $B = 0.040$ T. What electric field passes ions of speed $2.0 \times 10^{5}$ m s⁻¹?
Electrons accelerated through 200 V move in a circle of radius 5.97 cm in a field of 0.80 mT. Use $\frac{e}{m} = \frac{2V}{B^2r^2}$ to find $\frac{e}{m}$ .
In a mass spectrometer, $^{12}\mathrm{C}^{+}$ ions follow a path of radius 12.0 cm. What is the radius for $^{14}\mathrm{C}^{+}$ ions of the same speed?
An alpha particle (charge $+ 2 e$ , mass 4 u) and a proton (charge $+ e$ , mass 1 u) enter the same field with the same speed. Compare their radii.
A student says the magnetic field in a cyclotron speeds up the protons. Evaluate this claim.

Show answersHide answers

$r = \frac{mv}{qB}$ and $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$ : $v$ cancels.
$r = \frac{(1.67 \times 10^{-27})(4.0 \times 10^{6})}{(1.60 \times 10^{-19})(0.25)} = 0.17$ m.
$f = \frac{qB}{2\pi m} = \frac{(1.60 \times 10^{-19})(1.5)}{2\pi(1.67 \times 10^{-27})} = 2.3 \times 10^{7}$ Hz.
$E = vB = (2.0 \times 10^{5})(0.040) = 8.0 \times 10^{3}\,\mathrm{V\,m^{-1}}$ .
$\frac{2(200)}{(0.80 \times 10^{-3})^2(0.0597)^2} = 1.75 \times 10^{11}\,\mathrm{C\,kg^{-1}}$ .
$r \propto m$ : $12.0 \times \frac{14}{12} = 14.0$ cm.
$r \propto \frac{m}{q}$ : alpha $\frac{4}{2} = 2$ , proton $\frac{1}{1} = 1$ ; the alpha particle's radius is twice the proton's.
Incorrect. A magnetic force is perpendicular to the velocity and does no work, so it cannot change the speed. In a cyclotron the magnetic field only bends the protons round; they are speeded up by the alternating electric field across the gap between the two halves (dees), which they cross twice per revolution.

Key points

Perpendicular to a uniform field: circle of radius $r = \frac{mv}{qB}$ , period $T = \frac{2\pi m}{qB}$ independent of speed; at an angle: a helix.
Velocity selector: crossed fields pass only $v = \frac{E}{B}$ .
Thomson: $\frac{e}{m} = \frac{E}{rB^2}$ or $\frac{2V}{B^2r^2}$ ; $\frac{e}{m_e} = 1.76 \times 10^{11}\,\mathrm{C\,kg^{-1}}$ .
Mass spectrometer: $m = \frac{qB'r}{v}$ ; ions of the same charge separate by mass.

Sources and credits

Photo: Cyclotron motion by Marcin Białek, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: e/m Equipment Tutorial, SFU Physics, embedded from YouTube.

Written by Ganzaa. The two diagrams (the mass spectrometer is drawn to scale for Worked example 3), the worked examples and the practice exercises are by Ganzaa. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Glowing circles

2Circular motion in a uniform field

3The velocity selector

4Thomson's measurement of em \frac{e}{m} me​

5The mass spectrometer

Glowing circles

Circular motion in a uniform field

The velocity selector

Thomson's measurement of $\frac{e}{m}$

The mass spectrometer