Q.What is the radius of the path of an electron (mass 9×10−31 kg and charge 1.6×10−19 C) moving at a speed of 3×107 m/s in a magnetic field of 6×10−4 T perpendicular to it? What is its frequency? Calculate its energy in keV. (1 eV=1.6×10−19 J).
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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 both v and B. Its magnitude is
F=∣q∣vBsinθ
where θ is the angle between v and B.
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
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 component v⊥=vsinθ — feels the magnetic force and drives circular motion of radius r=qBmv⊥.
- Parallel component v∥=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=9.1×10−31 kg, q=1.6×10−19 C) enters a 0.02 T field at 106 m/s, perpendicular to B:
r=qBmv=(1.6×10−19)(0.02)(9.1×10−31)(106)≈2.8×10−4 m …
Why this formula?
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 product v×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).
4. General Motion: Helical Path …
Concept: charged particle in a ⊥ magnetic field — circular motion.
Radius. The magnetic force supplies the centripetal force, qvB=rmv2, so
r=qBmv=(1.6×10−19)(6×10−4)(9×10−31)(3×107)=9.6×10−2327×10−24=0.28 m.
Frequency. f=2πmqB=2π(9×10−31)(1.6×10−19)(6×10−4)=5.65×10−309.6×10−23=1.70×107 Hz. …
Moving perpendicular to B, the electron circles with r=qBmv, cyclotron frequency f=2πmqB, and kinetic energy K=21mv2. For the given data: r≈0.28 m, f≈1.70×107 Hz, K≈2.53 keV.
Why it moves in a circle
The magnetic force F=q(v×B) is always perpendicular to v, so it does no work — the speed stays constant while the direction turns. For v⊥B this force, of constant size qvB, acts as a centripetal force and the path is a circle.
1. Radius
Set the magnetic force equal to the centripetal force and cancel one v:
qvB=rmv2 ⇒ r=qBmv.
r=(1.6×10−19)(6×10−4)(9×10−31)(3×107)=9.6×10−2327×10−24=0.281 m≈0.28 m.
2. Frequency
The period is T=v2πr=qB2πm, so the frequency (independent of speed) is
f=T1=2πmqB=2π(9×10−31)(1.6×10−19)(6×10−4)=5.65×10−309.6×10−23=1.70×107 Hz.
3. Kinetic energy in keV …
Method: Lorentz Force & Circular Motion Analysis
This problem uses the Centripetal Force from Magnetic Lorentz Force method — when a charged particle enters a uniform magnetic field perpendicularly, the magnetic force provides the necessary centripetal force for circular motion.
Step 1: Find the radius of the circular path
The magnetic force on a moving charge is:
FB=qvB
For circular motion, this equals the centripetal force:
FB=rmv2
Equating them:
qvB=rmv2
Solving for radius r:
r=qBmv
Substitute the values:
- m=9×10−31 kg
- v=3×107 m/s
- q=1.6×10−19 C
- B=6×10−4 T
r=(1.6×10−19)(6×10−4)(9×10−31)(3×107)
r=9.6×10−2327×10−24
r=0.28125 m
Step 2: Find the frequency of revolution
The time period for one complete revolution:
T=v2πr
Frequency f=T1:
f=2πrv
Alternatively, using the direct formula (derived from r expression):
f=2πmqB
Substitute:
f=2π(9×10−31)(1.6×10−19)(6×10−4)
f=5.654×10−309.6×10−23
f=1.698×107 Hz
--- …
Here are the most common mistakes students make on this problem, why they happen, and how to avoid each one.
1. Forgetting the Perpendicular Condition
Mistake: Using the formula r=qBmv without checking if the velocity is perpendicular to the magnetic field.
Why it happens: Students often plug numbers into the formula without reading the phrase “perpendicular to it.”
How to avoid: Always underline the word perpendicular in the question. If the angle θ is not 90∘, you must use r=qBsinθmv. Here, it’s given as perpendicular, so sin90∘=1 — you’re safe.
2. Mixing Up Mass and Charge Values
Mistake: Using the mass of a proton (1.67×10−27 kg) or the charge of an alpha particle (3.2×10−19 C) instead of the electron’s values.
Why it happens: Many problems use similar numbers for different particles, and students rush.
How to avoid: Write down the given data clearly at the top:
- m=9×10−31 kg
- q=1.6×10−19 C
- v=3×107 m/s
- B=6×10−4 T
Then double-check each value before substituting.
3. Incorrect Unit Conversion for Energy
Mistake: Computing energy in joules and then dividing by 1.6×10−19 incorrectly, or forgetting that 1 eV=1.6×10−19 J.
Why it happens: Students either invert the conversion factor or misplace the decimal.
How to avoid: Use the conversion as a multiplication:
E(eV)=1.6×10−19E(J)
Then convert eV to keV by dividing by 1000:
E(keV)=1000E(eV)
4. Using the Wrong Formula for Frequency
Mistake: Using f=2πrv (which is for circular motion in general) but forgetting that in a magnetic field, the frequency is independent of speed.
Why it happens: Students derive frequency from radius and speed, which works but is inefficient and error-prone.
How to avoid: Use the cyclotron frequency formula directly:
f=2πmqB
This is faster and avoids carrying over errors from the radius calculation.
5. Arithmetic Errors with Powers of 10
Mistake: Adding or subtracting exponents incorrectly when multiplying or dividing numbers like 9×10−31 and 1.6×10−19.
Why it happens: Mental math under time pressure.
How to avoid: Write each step in scientific notation and separate the coefficients from the powers of 10: …
- Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 markQ.A particle carrying charge q, moving with velocity V⃗, enters a magnetic field B⃗ that is directed perpendicular to the direction of V⃗. What will be the nature of the particle's path in the magnetic field?
›Reveal solutionSolution
Since the magnetic force on a moving charge is always perpendicular to its velocity and B is perpendicular to V here, the force acts purely as a centripetal force of constant magnitude — the particle moves in a circle.
The magnetic force on the charge is
F=qV×B …
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