Charged Particles in Electric and Magnetic Fields Explained

Steering Electrons and Ions

Old television tubes painted pictures by sweeping a beam of electrons across a screen. Mass spectrometers identify drugs, pollutants and the ages of rocks by bending beams of ions. Particle accelerators and hospital cyclotrons whirl protons round in circles. All of these work by pushing charged particles with electric and magnetic fields. The same ideas allowed J. J. Thomson to discover the electron and Robert Millikan to measure its charge. This tutorial covers charged particles in electric and magnetic fields as taught in PHY 102 and university physics: acceleration through a potential difference, deflection between charged plates, circular motion in a magnetic field, the velocity selector, Thomson's e/m experiment, Millikan's oil-drop experiment, the mass spectrometer and the cyclotron, with worked examples. It builds on the tutorials on electric fields, magnetic fields and circular motion.

Useful Constants

These constants appear in almost every calculation involving charged particles:

Electron charge: e = 1.60 × 10⁻¹⁹ C
Electron mass: m(e) = 9.11 × 10⁻³¹ kg
Proton mass: m(p) = 1.67 × 10⁻²⁷ kg
Atomic mass unit: u = 1.66 × 10⁻²⁷ kg

Accelerating a Charge Through a Potential Difference

When a charge q moves through a potential difference V, the electric field does work qV on it. If the charge starts from rest, all of that work becomes kinetic energy:

½mv² = qV, so v = √(2qV ÷ m)

This is how an electron gun works: electrons released from a hot cathode by thermionic emission are accelerated towards a positive anode. The energy gained by one electron accelerated through 1 V is 1 electronvolt (1 eV = 1.60 × 10⁻¹⁹ J).

Worked example 1

An electron is accelerated from rest through 2000 V. Find its kinetic energy and its speed.

KE = eV = 1.60 × 10⁻¹⁹ × 2000 = 3.2 × 10⁻¹⁶ J (that is, 2000 eV).

v = √(2 × 3.2 × 10⁻¹⁶ ÷ 9.11 × 10⁻³¹) ≈ 2.65 × 10⁷ m/s, nearly a tenth of the speed of light.

At accelerating voltages above a few tens of kilovolts, electrons approach the speed of light, and the formula must be corrected using special relativity.

Deflection Between Parallel Plates

Between two parallel plates with p.d. V and separation d, the electric field is uniform: E = V ÷ d. A charge q feels a constant force F = qE towards one plate, and so a constant acceleration a = qE ÷ m.

A particle that enters the field moving parallel to the plates behaves just like a projectile under gravity:

Parallel to the plates, there is no force, so its velocity stays constant.
Perpendicular to the plates, it accelerates steadily, so it follows a parabolic path.

Worked example 2

The electrons from example 1 (speed 2.65 × 10⁷ m/s) pass between deflecting plates 5.0 cm long and 2.0 cm apart, with a p.d. of 200 V between them. Find how far they are deflected by the time they leave the plates.

Field: E = 200 ÷ 0.020 = 1.0 × 10⁴ V/m.
Acceleration: a = eE ÷ m = 1.60 × 10⁻¹⁹ × 10⁴ ÷ 9.11 × 10⁻³¹ ≈ 1.76 × 10¹⁵ m/s².
Time between the plates: t = 0.050 ÷ 2.65 × 10⁷ ≈ 1.89 × 10⁻⁹ s.
Deflection: y = ½at² = ½ × 1.76 × 10¹⁵ × (1.89 × 10⁻⁹)² ≈ 3.1 mm.

Interestingly, combining the formulas gives y = V(plates) L² ÷ (4dV(accelerating)) = 200 × 0.05² ÷ (4 × 0.02 × 2000) = 3.1 mm, which does not depend on the particle's charge or mass at all. This is how the cathode ray oscilloscope works: the spot's deflection on the screen is proportional to the p.d. applied to the plates.

Motion in a Magnetic Field

A charged particle of charge q moving with velocity v at right angles to a magnetic field B feels a force:

F = qvB

Its direction is given by Fleming's left-hand rule (for a negative charge such as an electron, the current direction is opposite to its motion). The force is always perpendicular to the velocity, so it changes the particle's direction but never its speed, and it does no work. The particle therefore moves in a circle, with the magnetic force providing the centripetal force:

qvB = mv² ÷ r, so r = mv ÷ (qB)

Faster or heavier particles move in larger circles, and a stronger field or a larger charge makes the circles tighter.

Worked example 3

Electrons moving at 2.65 × 10⁷ m/s enter a uniform magnetic field of 1.0 mT at right angles. Find the radius of their path.

r = 9.11 × 10⁻³¹ × 2.65 × 10⁷ ÷ (1.60 × 10⁻¹⁹ × 1.0 × 10⁻³) ≈ 0.15 m.

Period and frequency

The time for one revolution is T = 2πr ÷ v = 2πm ÷ (qB). Remarkably, it does not depend on the speed: faster particles travel in bigger circles, but take exactly the same time to go round. This is the key to the cyclotron.

The Velocity Selector

If a particle passes through crossed electric and magnetic fields, at right angles to each other and to the beam, the electric force qE and the magnetic force qvB act in opposite directions. Only particles with exactly the right speed have balanced forces and pass straight through undeflected:

qE = qvB, so v = E ÷ B

Slower particles are deflected one way, and faster ones the other way, so a narrow slit at the exit lets through only particles of one speed.

Worked example 4

A velocity selector has E = 1.0 × 10⁴ V/m and B = 2.0 × 10⁻³ T. What speed of particle passes straight through?

v = 1.0 × 10⁴ ÷ 2.0 × 10⁻³ = 5.0 × 10⁶ m/s, whatever the particle's charge or mass.

Thomson's Measurement of e/m (1897)

J. J. Thomson showed that cathode rays are streams of tiny negative particles, which we now call electrons. He used crossed fields to find the speed of the rays (v = E/B), then switched off the electric field and measured the radius of their circular path in the magnetic field alone. Since r = mv/(eB), the specific charge is:

e ÷ m = v ÷ (rB) = E ÷ (rB²)

He found e/m ≈ 1.76 × 10¹¹ C/kg, about 1800 times larger than for a hydrogen ion. This meant that either the particles carried an enormous charge or, as turned out to be true, they were far lighter than any atom. The electron was the first subatomic particle to be discovered.

Millikan's Oil-Drop Experiment (1909)

Robert Millikan measured the charge on tiny charged oil drops between horizontal plates. By adjusting the voltage until a drop hung stationary, he balanced the upward electric force against the drop's weight:

qE = mg, so q = mgd ÷ V

The drop's mass was found from its terminal velocity when falling with the field switched off, using Stokes' law.

Worked example 5

An oil drop of mass 3.26 × 10⁻¹⁵ kg is held stationary between plates 1.0 cm apart when the p.d. is 1000 V. Find its charge, and how many electrons this represents. (g = 9.81 m/s²)

q = 3.26 × 10⁻¹⁵ × 9.81 × 0.010 ÷ 1000 ≈ 3.2 × 10⁻¹⁹ C, which is 3.2 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 2 electron charges.

Millikan found that every charge was a whole-number multiple of 1.60 × 10⁻¹⁹ C. Charge is quantised: it comes in units of e. Combined with Thomson's e/m, this gave the mass of the electron, about 9.1 × 10⁻³¹ kg.

The Mass Spectrometer

A mass spectrometer separates ions by mass, so it can identify substances and measure the proportions of isotopes.

Atoms are ionised, usually by knocking out an electron, giving each a charge of +e.
The ions are accelerated by an electric field.
A velocity selector lets through only ions of one speed.
In a uniform magnetic field, the ions move in semicircles of radius r = mv/(qB), so heavier ions travel in larger circles.
A detector records where the ions land and how many arrive, giving a spectrum of masses.

Worked example 6

Neon-20 and neon-22 ions, each with charge +e, enter a magnetic field of 0.50 T at 1.0 × 10⁵ m/s. Find the radius of each path and the distance between the points where they land after a semicircle.

r(20) = 20 × 1.66 × 10⁻²⁷ × 1.0 × 10⁵ ÷ (1.60 × 10⁻¹⁹ × 0.50) ≈ 4.15 cm.

r(22) = 22 × 1.66 × 10⁻²⁷ × 1.0 × 10⁵ ÷ (1.60 × 10⁻¹⁹ × 0.50) ≈ 4.56 cm.

After a semicircle, each ion lands a distance 2r from its entry point, so the separation is 2 × (4.56 − 4.15) ≈ 0.83 cm. The two isotopes are clearly separated. It was with an early mass spectrometer that neon was found to have more than one isotope.

The Cyclotron

A cyclotron accelerates protons or other ions to high energies in a compact space. Two hollow D-shaped electrodes ("dees") sit in a strong magnetic field, with an alternating p.d. across the gap between them.

Inside a dee, there is no electric field, and the ion moves in a semicircle.
Each time it crosses the gap, the alternating voltage has reversed, so the ion is accelerated again.
It speeds up and spirals outwards, and because the period of revolution does not depend on speed, a constant frequency stays in step with it.

The required frequency is f = qB ÷ (2πm).

Worked example 7

Find the frequency of the alternating voltage needed to accelerate protons in a cyclotron with a magnetic field of 1.5 T.

f = 1.60 × 10⁻¹⁹ × 1.5 ÷ (2π × 1.67 × 10⁻²⁷) ≈ 2.3 × 10⁷ Hz (23 MHz), a radio frequency.

Hospitals use cyclotrons to produce short-lived radioactive isotopes for medical scans, and to provide proton beams for cancer therapy.

Common Mistakes with Charged Particles in Fields

Thinking a magnetic field speeds particles up. The magnetic force is perpendicular to the motion, so it changes only the direction.
Using Fleming's left-hand rule directly for electrons. Point the "current" finger opposite to the electrons' motion.
Forgetting to convert units: cm to m, mT to T, and the mass in u to kg.
Using v = √(2qV/m) when the particle does not start from rest. Then ½mv² − ½mu² = qV.
Treating the path between the plates as circular. In a uniform electric field, the path is a parabola. Circles occur in magnetic fields.

Charged Particles in Fields: Practice Questions

Find the speed of an electron accelerated from rest through 500 V.
A proton moves at 2.0 × 10⁶ m/s at right angles to a 0.20 T field. Find the radius of its path.
In a velocity selector, B = 0.10 T. What electric field is needed to select particles moving at 3.0 × 10⁵ m/s?
An oil drop carrying three electron charges is held stationary between plates 5.0 mm apart. If the drop's mass is 4.9 × 10⁻¹⁵ kg, find the p.d. needed. (g = 9.8 m/s²)
Explain why the period of revolution of a charged particle in a uniform magnetic field does not depend on its speed.

Answers: 1. v = √(2 × 1.60 × 10⁻¹⁹ × 500 ÷ 9.11 × 10⁻³¹) ≈ 1.33 × 10⁷ m/s. 2. r = 1.67 × 10⁻²⁷ × 2.0 × 10⁶ ÷ (1.60 × 10⁻¹⁹ × 0.20) ≈ 0.104 m. 3. E = vB = 3.0 × 10⁴ V/m. 4. V = mgd ÷ q = 4.9 × 10⁻¹⁵ × 9.8 × 0.005 ÷ (4.8 × 10⁻¹⁹) = 500 V. 5. T = 2πr/v and r = mv/(qB), so T = 2πm/(qB): a faster particle moves in a proportionally larger circle, so the time per revolution stays the same.

Conclusion

Electric fields accelerate charged particles, giving ½mv² = qV, and deflect them in parabolic paths between plates, as in the oscilloscope. Magnetic fields exert a force qvB perpendicular to the motion, bending particles into circles of radius r = mv/(qB) with a period that does not depend on speed. Crossed fields select a single speed v = E/B. These principles let Thomson measure e/m and discover the electron, let Millikan show that charge is quantised in units of 1.60 × 10⁻¹⁹ C, and drive mass spectrometers and cyclotrons in laboratories and hospitals today.