Velocity selector and the cyclotron

Crossed fields: the velocity selector

Narrated video. Transcript:

This is an extra topic. It is beyond the current NCERT syllabus, but it is useful for competitive exams, and it puts the Lorentz force to work.

A charge q moves with velocity v through an electric field E and a magnetic field B together. The total force on it is the Lorentz force: F equals q, E plus v cross B.

Now arrange the fields at right angles to each other and to v. A proton moves to the right. E points down, between two plates. B points into the screen.

The electric force q E pushes the proton down. The magnetic force, q v cross B, pushes it up, with size q v B.

If the two are equal, q E equals q v B, the forces cancel and the proton goes straight through. That happens only for one speed: v equals E over B.

A faster proton feels a bigger magnetic force and bends up. A slower one bends down. Put a slit at the far end, and only protons of speed E over B get out.

That is why it is called a velocity selector. It works for any charge and any mass, because q cancels out of q E equals q v B.

Example: E equals 1 × 10⁵ volt per metre and B equals 0.1 tesla. The selected speed is 1 × 10⁵ over 0.1, which is 1 × 10⁶ metres per second.

J. J. Thomson used crossed fields like these in 1897 to measure the ratio of charge to mass of the electron.

The cyclotron

Narrated video. Transcript:

The cyclotron, invented by Lawrence and Livingston in 1934, speeds up protons and other ions to high energies. It uses an electric field to speed them up and a magnetic field to bring them back again and again.

It has two hollow D-shaped metal boxes, called dees, with a narrow gap between them. A strong magnetic field acts perpendicular to the dees. An oscillator puts an alternating voltage across the gap.

Inside a dee there is no electric field, so the proton moves in a semicircle under the magnetic force alone. Each time it crosses the gap, the voltage pushes it and it speeds up.

The key fact: the time for a half-circle does not depend on the speed. The frequency of revolution is ν c equals q B over 2πm, the cyclotron frequency.

So if the oscillator frequency ν a equals ν c, the voltage reverses just as the proton reaches the gap each time. This is the resonance condition, and the proton gains energy on every crossing.

As it speeds up, its radius r equals m v over q B grows, and the path spirals outwards. At the edge, of radius R, it leaves with speed q B R over m.

So its kinetic energy is ½ m v², which is q² B² R² over 2m. The final energy depends on B and R, not on the gap voltage. A bigger voltage just gets it there in fewer turns.

Example: an oscillator of 10 megahertz accelerates protons. B equals 2π m ν over q: 2π times 1.67 × 10⁻²⁷ times 10⁷, over 1.6 × 10⁻¹⁹. That is about 0.66 tesla.

With dees of radius 60 centimetres, v equals q B R over m: about 3.8 × 10⁷ metres per second. The energy is 1.2 × 10⁻¹² joule, about 7.4 mega electron volts.

Try it: selector and cyclotron

Interactive simulation. Things to try:

• Velocity selector: change the speed until the protons pass through the slit. Check that v = E/B.
• Compare the amber arrow qE with the blue arrow qvB: they are equal only at the right speed.
• Cyclotron: the oscillator starts in tune, and the proton spirals out to the edge. Change B and watch it fall out of step.
• Change the gap voltage: the number of crossings changes, but the final energy does not.

Which speed gets through?

In a velocity selector, E = 1.2 × 10⁵ V/m and B = 0.08 T, at right angles to each other and to the beam. Which protons pass through undeflected?

1. Those with v = 9.6 × 10³ m/s
2. Those with v = 1.5 × 10⁶ m/s
3. Those with v = 6.7 × 10⁻⁷ m/s
4. All of them, whatever their speed

Tuning the cyclotron

A cyclotron’s oscillator runs at 10 MHz. What magnetic field is needed to accelerate protons (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C)?

1. 0.10 T
2. 0.66 T
3. 1.3 T
4. 4.1 T

Summary

• Lorentz force: F = q [E + v × B]
• Velocity selector: qE = qvB ⇒ v = E/B
• Cyclotron frequency: ν_c = qB / 2πm
• Energy at the edge: K = q²B²R² / 2m

• Crossed E and B fields let through only particles of speed v = E/B, whatever their charge or mass.
• In a cyclotron the time for a half-turn is independent of speed, so a fixed-frequency oscillator can keep accelerating the ions.
• Resonance condition: oscillator frequency = cyclotron frequency, ν_a = ν_c.
• The final energy depends on B and the radius of the dees, not on the gap voltage.