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: …
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