Q.In a chamber, a uniform magnetic field of 6.5G (1G=10−4T) is maintained. An electron is shot into the field with a speed of 4.8×106m s−1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e=1.5×10−19C, me=9.1×10−31kg).
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).
Concept: charged particle in a ⊥ magnetic field — circular motion.
Why a circle: the force F=q(v×B) is always perpendicular to v, so it does no work — the speed is constant. With v⊥B this constant-magnitude force always points to one centre, i.e. it is centripetal, giving uniform circular motion.
Radius: from evB=rmev2, r=eBmev. With B=6.5G=6.5×10−4T and the stated e=1.5×10−19C: …
The magnetic force is perpendicular to the velocity, does no work and acts as a centripetal force, so the electron moves in a circle of radius r=eBmev≈4.48×10−2m (using the stated e=1.5×10−19C).
Why the path is a circle
The electron feels only the magnetic force F=q(v×B), which is always perpendicular to the velocity. A force perpendicular to v does no work, so the kinetic energy — and hence the speed — never changes. Since the electron enters normal to B, this force has constant magnitude evB and always points toward one fixed centre: exactly the condition for uniform circular motion.
Determining the radius
The magnetic force provides the centripetal force:
evB=rmev2⇒r=eBmev.
Convert the field: B=6.5G=6.5×10−4T. Substituting the stated data (me=9.1×10−31kg, v=4.8×106m s−1, e=1.5×10−19C):
r=(1.5×10−19)(6.5×10−4)(9.1×10−31)(4.8×106).
Numerator: 9.1×4.8=43.68, so 43.68×10−25. Denominator: 1.5×6.5=9.75, so 9.75×10−23. Hence …
The error: Students plug B=6.5G directly into formulas, forgetting the conversion factor.
Why it happens: The problem gives B in Gauss but all standard formulas use Tesla. The conversion hint (1G=10−4T) is easy to miss under time pressure.
How to avoid: Always write the conversion step explicitly:
B=6.5G=6.5×10−4T
Pro tip: Circle or underline the conversion factor in the question before starting calculations.
Mistake 2: Using Wrong Charge Value
The error: Using e=1.6×10−19C (the standard value) instead of the given e=1.5×10−19C.
Why it happens: Students memorize the standard electron charge and automatically substitute it without checking the problem's data.
How to avoid:Always use the values provided in the question, even if they differ from standard textbook values. The problem deliberately gives 1.5×10−19C — use it.
Mistake 3: Confusing the Reason for Circular Motion
The error: Saying "the electron moves in a circle because the magnetic force is perpendicular to velocity" — but not explaining why this produces a circle.
Why it happens: Students memorize the result without understanding the mechanism.
How to avoid: Explain step-by-step:
Magnetic force F=q(v×B) acts perpendicular to both v and B
Since v⊥B, the force magnitude is F=qvB
This perpendicular force provides centripetal accelerationac=v2/r
The force changes only the direction of velocity, not its magnitude
Result: uniform circular motion
Mistake 4: Sign Errors in Force Direction
The error: Forgetting that the electron has negative charge, so the force direction is opposite to that for a positive charge.
Why it happens: Students apply the right-hand rule for positive charges without flipping the direction for electrons.
How to avoid: Remember: For electrons, use left-hand rule or apply the right-hand rule and then reverse the direction. The magnitude calculation is unaffected, but conceptual questions about direction will be wrong.
Mistake 5: Formula Confusion — Radius Expression
The error: Writing r=qBmv incorrectly as r=qBmv2 or r=mvqB.
Why it happens: Mixing up centripetal force (mv2/r) with magnetic force (qvB).