Q.Two protons of equal kinetic energies enter a region of uniform magnetic field. The first proton enters normal to the field direction while the second enters at 30∘ to the field direction. Name the trajectories followed by them.
Uttarakhand UbseCBSE Class XII Board 2018Subjective· 1mImportance★★★★★
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 magnetic force qv×B acts only on the velocity component perpendicular to B; the parallel component is unaffected. A proton entering normal (90∘) has no parallel component, while one at 30∘ has bo …
Concept. The Lorentz force F=qv×B is always perpendicular to v, so it changes direction but not speed. Only the velocity component ⊥B feels a force; the component ∥B moves uniformly.
Why this trajectory.
For the first proton (θ=90∘): v∥=vcos90∘=0, so the whole velocity is perpendicular. The force provides centripetal force and the path is a circle of radius r=qBmv. …