Concept understanding — Charged Particle in Magnetic Field
Charged Particle in a Magnetic Field
When a charged particle moves through a magnetic field, the field grabs it sideways. Unlike an electric field, which can speed a charge up or slow it down, a magnetic field only bends the path — it never changes the particle's speed. Understanding why leads directly to circular and helical motion, the basis of cyclotrons, mass spectrometers and the aurora.
The force: always sideways
A particle of charge q moving with velocity v in a magnetic field B feels the magnetic (Lorentz) force:
F=q(v×B)
Because of the cross product, F is perpendicular to bothv and B. Its magnitude is
F=∣q∣vBsinθ
where θ is the angle between v and B.
Important
Since F⊥v, the force does no work: F⋅v=0. Therefore the kinetic energy and the speed stay constant — the field only changes the direction of motion, never the magnitude.
Case 1: velocity perpendicular to the field → a circle
If v⊥B (θ=90∘), the force F=qvB stays constant in size and always points toward one central point. That is exactly the condition for uniform circular motion, with the magnetic force acting as the centripetal force:
qvB=rmv2
Solving for the radius:
r=qBmv
The time period of one revolution is
T=v2πr=qB2πm
Note
The period T (and the frequency f=qB/2πm, the cyclotron frequency) does not depend on the speed or the radius. A faster particle traces a bigger circle but takes exactly the same time per loop. This speed-independence is what makes the cyclotron work.
Case 2: velocity at an angle → a helix
If v makes an angle θ with B, split it into two parts:
Perpendicular componentv⊥=vsinθ — feels the magnetic force and drives circular motion of radius r=qBmv⊥.
Parallel componentv∥=vcosθ — feels no force (since v∥×B=0) and carries the particle steadily along the field line.
Combining a circle with a steady drift gives a helix. The distance advanced along the field in one full turn is the pitch:
p=v∥T=vcosθ⋅qB2πm
A quick example
An electron (m=9.1×10−31kg, q=1.6×10−19C) enters a 0.02T field at 106m/s, perpendicular to B:
Charged Particle in a Magnetic Field — Why the Key Formulas Hold
Let's build this from first principles. The core idea is that a magnetic field exerts a force only on a moving charge, and that force is always perpendicular to both the velocity and the field.
1. The Fundamental Force Law: Lorentz Force
The starting point is the Lorentz force for a charge q moving with velocity v in a magnetic field B:
Fm=q(v×B)
Why this form?
Cross productv×B means the force is perpendicular to both v and B.
Magnitude: Fm=∣q∣vBsinθ, where θ is the angle between v and B.
Direction: given by the right-hand rule (for positive q).
Key insight: Because Fm⊥v, the magnetic force does no work — it changes only the direction of velocity, not its speed.
2. Circular Motion in a Uniform Magnetic Field
Consider a charge q moving with speed v perpendicular to a uniform B (so θ=90∘, sinθ=1).
Step 1: Force provides centripetal acceleration
The magnetic force is the only radial force:
Fm=qvB
This must equal the centripetal force required for circular motion:
Fc=rmv2
Step 2: Equate and solve for r
qvB=rmv2
Cancel one v (assuming v=0):
qB=rmv
Thus:
r=qBmv
This is the radius of the circular path (cyclotron radius).
Why this makes sense:
Larger mass m → harder to turn → larger r
Larger charge q or stronger B → stronger force → tighter turn → smaller r
Faster speed v → more momentum → larger r
3. Angular Frequency (Cyclotron Frequency)
From the circular motion relation:
ω=rv
Substitute r=qBmv:
ω=qBmvv=mqB
Thus:
ωc=mqB
Why this is remarkable:
ωc is independent of speed v — all particles with same q/m have the same angular frequency, regardless of how fast they move.
This is the principle behind cyclotrons (particle accelerators).
The key idea is that cyclotron frequency comes from equating the magnetic force to the centripetal force for a charged particle moving in a uniform magnetic field.
Step 1: For a particle of charge e, mass m, and speed v moving perpendicular to a field B, the magnetic force evB provides the centripetal force mv2/r.
Step 2: Equating: evB=rmv2. Cancelling v gives eB=rmv.
Step 3: Angular frequency is ω=v/r. Substituting v/r=eB/m yields ω=meB. …
The cyclotron frequency ω=eB/m has dimensions of [T]−1 because the Lorentz force law F=qvB gives [eB]=[M][T]−1 and dividing by mass [M] leaves [T]−1.
The key insight is that dimensions must match on both sides of any physical equation. For cyclotron frequency, we're checking that ω=eB/m indeed gives inverse time — the unit of frequency.
Let's work through this systematically.
Start with what we know about dimensions. Frequency ω has dimensions of [T]−1 — that's what we need to verify. The right side is eB/m, so we need the dimensions of e, B, and m.
Mass is straightforward. Mass m has dimension [M].
For charge e and magnetic field B, we need a physical relation. The Lorentz force gives us the link: a charge q moving with velocity v in a magnetic field B experiences force F=qvB.
Write this dimensionally. Force has dimensions [M][L][T]−2. Velocity has [L][T]−1. So:
[F]=[q][v][B]
[M][L][T]−2=[e][L][T]−1[B]
Solve for [eB]. Multiply both sides by [T]:
[M][L][T]−1=[e][L][B]
Cancel [L]:
[M][T]−1=[e][B]
Note
This is a neat result: the product eB has dimensions [M][T]−1 — mass per unit time.
Now divide by mass. The cyclotron frequency is:
ω=meB
Dimensionally:
[ω]=[m][eB]=[M][M][T]−1=[T]−1
This matches exactly what we expect for frequency. …