12 min readConcept sheetSection A MCQ (1 mark) on Fleming's left-hand rule; Section B short answer (3 marks): radius and period of the circular path; long answer (5 marks), often with a choice: the cyclotron; also a NEET favourite

Force on a moving charge: the Lorentz force, motion in fields and the cyclotron

A moving charge in a magnetic field feels F = qvB sin θ. This force bends its path into a circle, R = mv/qB, without changing its speed.

Lorentz forceF = q(**v** × **B**); size F = qvB sin θ, where θ is the angle between v and B

Radius of the circular pathR = mv/qB, when v is at right angles to B

Time periodT = 2πm/qB; it does not depend on the speed or the radius

Cyclotron frequencyν_c = qB/2πm; the oscillator must have this frequency

The Lorentz force and Fleming's left-hand rule

A charge q moving with velocity v through a magnetic field B feels a force. It is called the Lorentz force: F = q(v × B). Its size is F = qvB sin θ, where θ is the angle between v and B. Note these points. The force is zero if the charge is at rest, or if it moves along B. It is largest, qvB, when the charge moves at right angles to B. It always acts at right angles to both v and B. For a negative charge, such as an electron, it acts the opposite way. Find its direction with Fleming's left-hand rule. Stretch the thumb, the first finger and the middle finger of your left hand at right angles to each other. Point the first finger along the field and the middle finger along the motion of a positive charge (the current). The thumb then gives the force.

Derivation: radius and time period of the circular path

A charge q of mass m enters a uniform field B with speed v at right angles to B.

1
The force qvB is always at right angles to v. So it does no work, and the speed and kinetic energy stay the same.
2
A force of constant size, always at right angles to the motion, makes a body move in a circle. So qvB is the centripetal force.
3
Centripetal force = magnetic force: mv²/R = qvB.
4★
So, R = mv/qB. More momentum gives a bigger circle; a stronger field gives a smaller one.
5★
Time for one round: T = 2πR/v = 2πm/qB.
6
So, T and the frequency ν = qB/2πm do not depend on v or R. A faster particle runs a bigger circle in the same time.

★ key step

Motion in an electric field vs a magnetic field

In a uniform electric field EThe force F = qE acts even on a charge at rest, along E (opposite to E for a negative charge). The acceleration is a = qE/m, so v = u + (qE/m)t and s = ut + ½(qE/m)t². The speed and energy change.

In a uniform magnetic field BThe force acts only on a moving charge, at right angles to v. Along B (θ = 0°): a straight line. At right angles (θ = 90°): a circle. At any other angle: a helix, a path like a spring. This is because v cos θ along B is unchanged, while v sin θ turns in a circle. The speed never changes.

The cyclotron: how it speeds up charged particles

E. O. Lawrence built the first cyclotron in 1929. It accelerates charged particles such as protons, deuterons or alpha particles to high speeds.

1
Two hollow D-shaped metal boxes, called dees, face each other across a narrow gap. They sit in a vacuum chamber.
2
A strong electromagnet gives a uniform field B at right angles to the dees.
3
An oscillator applies an alternating voltage across the dees, so there is an electric field only in the gap.
4
A positive ion leaves a source near the centre. The electric field in the gap accelerates it into one dee.
5
Inside a dee there is no electric field. The magnetic field makes the ion move in a semicircle at constant speed.
6★
A semicircle takes a time πm/qB, whatever the speed. If the voltage reverses in exactly this time, the ion is accelerated again at the gap. So the oscillator frequency must equal ν_c = qB/2πm (cyclotron resonance).
7
Each crossing of the gap adds energy, so the radius grows and the ion spirals outward.
8
At the edge of the dees, of radius R₀, the ion leaves through a window with its largest speed, v = qBR₀/m.

★ key step

Worked example: an electron's path and a proton cyclotron

Problem

(a) An electron moving at 2.0 × 10⁷ m s⁻¹ enters a uniform magnetic field of 0.10 T at right angles. Find the radius of its path. (b) A cyclotron uses a field of 1.0 T to accelerate protons. Find the oscillator frequency needed. (Electron mass 9.1 × 10⁻³¹ kg, proton mass 1.67 × 10⁻²⁷ kg, e = 1.6 × 10⁻¹⁹ C, π = 3.14)

Working

(a) Formula: R = mv/eB.
R = (9.1 × 10⁻³¹ × 2.0 × 10⁷)/(1.6 × 10⁻¹⁹ × 0.10) m = (1.82 × 10⁻²³)/(1.6 × 10⁻²⁰) m.
R ≈ 1.1 × 10⁻³ m, about 1.1 mm.
(b) Formula: ν_c = eB/2πm.
ν_c = (1.6 × 10⁻¹⁹ × 1.0)/(2 × 3.14 × 1.67 × 10⁻²⁷) Hz = (1.6 × 10⁻¹⁹)/(1.05 × 10⁻²⁶) Hz.
ν_c ≈ 1.5 × 10⁷ Hz, that is about 15 MHz.

Answer

(a) about 1.1 × 10⁻³ m; (b) about 1.5 × 10⁷ Hz (both rounded to 2 significant figures)

How to draw the cyclotron

Draw two D-shaped halves, D₁ and D₂, with their straight edges facing each other across a narrow gap.
Mark the ion source S at the centre of the gap.
Join the two dees to an oscillator (a circle with a ~ sign inside) by two wires.
Fill the dees with dots (field out of the page) and label them B: uniform magnetic field.
Draw the path as a spiral of growing semicircles from S, ending at an exit window at the edge of one dee.
Label: dees, gap, oscillator, ion source, path of the ion and exit.

In real life

Hospitals use radioactive medicines for PET scans and some cancer treatments, and many of these are made in cyclotrons. The same force bends charged particles from the Sun towards the Earth's poles, where they light up the polar skies.

Exam angle

1 marks: “A charged particle moves parallel to a uniform magnetic field. What force does the field exert on it?” Zero, because θ = 0° and F = qvB sin 0° = 0.

3 marks: “Show that a charged particle entering a uniform magnetic field at right angles moves in a circle. Obtain the radius and show that the time period does not depend on the speed.” Force qvB at right angles to v does no work and acts as the centripetal force: mv²/R = qvB, so R = mv/qB; T = 2πR/v = 2πm/qB, which has no v in it.

5 marks: “Describe the construction and working of a cyclotron with a labelled diagram. Write the expression for the cyclotron frequency.” Dees with a gap in a vacuum; uniform B at right angles to them; oscillator across them. The ion is accelerated in the gap, runs a semicircle inside a dee in time πm/qB, and meets a reversed voltage, so it gains energy at every crossing and spirals out. ν_c = qB/2πm. Labelled diagram.

Remember it

FBI on the left hand: thumb Force, first finger B (field), middle finger I (current). A magnetic force steers but never speeds up.