12 min readConcept sheet1-mark MCQ and assertion-reason, a 3-mark derivation of the radius and frequency of circular motion, and numericals; also a NEET and JEE favourite

Force on a moving charge: the Lorentz force and motion in a magnetic field

Lorentz force F = q(E + v × B). Its magnetic part does no work; it only bends the path into a circle or helix.

Lorentz force**F** = q(**E** + **v** × **B**); magnetic part F = qvB sin θ

Radius of the circular pathr = mv/qB, when v is perpendicular to B

Frequency of revolutionν = qB/2πm, the same for all speeds; ω = qB/m

Pitch of the helixp = v∥ × T = 2πm v∥/qB, where v∥ is the part of v along B

The Lorentz force

Let a point charge q move with velocity v where there is an electric field E and a magnetic field B. The total force on it is the Lorentz force: F = q(E + v × B). The electric part qE acts whether the charge moves or not, along E. The magnetic part q(v × B) acts only on a moving charge. Its size is qvB sin θ, where θ is the angle between v and B. It is zero if the charge is at rest or moves along B. It is largest (qvB) when v is perpendicular to B. It is always perpendicular to both v and B. For a negative charge, such as an electron, the force is opposite to that on a positive charge. Find its direction with the right-hand rule for a cross product, or with Fleming's left-hand rule, taking the current along the motion of a positive charge.

Electric force vs magnetic force on a charge

Electric force qEActs on a charge at rest or moving, along E (opposite to E for a negative charge). It can do work, so it can change the speed and kinetic energy.

Magnetic force q(v × B)Acts only on a moving charge and is always perpendicular to v. It does no work, so the speed and kinetic energy stay the same; only the direction of motion changes.

Derivation: circular motion in a uniform magnetic field

A charge q of mass m enters a uniform magnetic field B with speed v perpendicular to B.

1
The magnetic force qvB is perpendicular to v at every instant, so it changes only the direction of v.
2
A force of constant size that is always perpendicular to the velocity is a centripetal force. So the path is a circle.
3
Centripetal force = magnetic force: mv²/r = qvB.
4★
So, the radius is r = mv/qB.
5
Time for one revolution: T = 2πr/v = 2πm/qB.
6★
Frequency ν = 1/T = qB/2πm; angular frequency ω = 2πν = qB/m.
7
So, T and ν do not depend on v or r. A faster particle moves in a bigger circle but takes the same time.

★ key step

When v makes an angle with B: the helix

Split v into v∥ = v cos θ along B and v⊥ = v sin θ perpendicular to B.
v∥ feels no force, so the particle keeps drifting along B at a steady speed.
v⊥ makes the particle go round in a circle of radius r = mv⊥/qB.
Together these give a helix, a path like the coils of a spring.
The pitch is the distance moved along B in one turn: p = v∥T = 2πm v∥/qB.
If v is exactly along B (θ = 0°), there is no force and the path is a straight line.

Worked example: an electron in a magnetic field

Problem

An electron moving at 2.0 × 10⁶ m/s enters a uniform magnetic field of 1.0 × 10⁻³ T at right angles to it. Find (a) the radius of its path and (b) its frequency of revolution. (m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)

Working

(a) Formula: r = mv/eB.
r = (9.1 × 10⁻³¹ × 2.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 1.0 × 10⁻³) m.
r = (18.2 × 10⁻²⁵)/(1.6 × 10⁻²²) m = 11.4 × 10⁻³ m ≈ 1.1 cm.
(b) Formula: ν = eB/2πm.
ν = (1.6 × 10⁻¹⁹ × 1.0 × 10⁻³)/(2 × 3.14 × 9.1 × 10⁻³¹) Hz.
ν = (1.6 × 10⁻²²)/(5.71 × 10⁻³⁰) Hz ≈ 2.8 × 10⁷ Hz.

Answer

(a) about 1.1 cm; (b) about 2.8 × 10⁷ Hz (both rounded to 2 significant figures)

Worked example: size and direction of the magnetic force

Problem

A proton moves along +x at 2.0 × 10⁶ m/s through a uniform magnetic field of 0.020 T along +y. Find the size and direction of the magnetic force on it. What changes for an electron moving the same way? (e = 1.6 × 10⁻¹⁹ C)

Working

Formula: F = qvB sin θ, with θ = 90° between v (along +x) and B (along +y).
F = 1.6 × 10⁻¹⁹ C × 2.0 × 10⁶ m/s × 0.020 T = 6.4 × 10⁻¹⁵ N.
Direction: v × B is along x̂ × ŷ = ẑ, so the force on the proton is along +z.
The electron has charge −e, so its force has the same size but points along −z.

Answer

6.4 × 10⁻¹⁵ N along +z on the proton; the same size along −z on the electron

In real life

Charged particles from the Sun spiral along Earth's magnetic field lines, and many are guided towards the poles. ISRO's Aditya-L1 spacecraft carries instruments that measure such particles and the magnetic field in space.

Exam angle

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

3 marks: “Show that a charged particle entering a uniform magnetic field at right angles moves in a circle. Derive its radius and show that its frequency does not depend on its speed.” Force qvB ⊥ v acts as a centripetal force; mv²/r = qvB gives r = mv/qB; T = 2πr/v = 2πm/qB; ν = qB/2πm, which has no v in it.

2 marks: “A proton and an alpha particle enter the same uniform magnetic field at right angles, with the same speed. Find the ratio of the radii of their paths.” r = mv/qB. For the alpha particle m = 4m_p and q = 2e, so r_α/r_p = 4/2 = 2. Ratio r_p : r_α = 1 : 2.

Remember it

A magnetic force steers but never speeds up. r = mv/qB, while ν = qB/2πm has no v in it.