Electron in a Uniform Field
An electron enters a uniform magnetic field perpendicular to the field lines with velocity v. The field strength is B. Three things you need to find: the radius of the circular path, the time period of revolution, and the force acting on the electron. This guide walks through the physics behind each part, explains the derivation approach you should follow, and shows you how to present the solution properly in an exam or assignment.
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An electron enters a uniform magnetic field B with velocity v directed perpendicular to the magnetic field lines. Find: (1) the radius r of the circular path, (2) the time period T of revolution, and (3) the magnitude of the force F acting on the electron at any point in its path.
Before writing a single equation, make sure you understand what’s physically happening. The electron has a charge (negative, magnitude e = 1.6 × 10⁻¹⁹ C) and a mass (m = 9.11 × 10⁻³¹ kg). It enters a region of space where there’s a uniform magnetic field. The velocity and the field are perpendicular — that single condition is what forces the electron into a circular orbit rather than a helix or a straight line.
Three quantities are given to you: v (velocity), B (magnetic field strength). The electron’s charge e and mass m are constants you’re expected to know or look up. Everything you derive flows from these four quantities. If you’re clear on this at the start, the algebra becomes straightforward.
Verified Reference: NIST Physics Constants
For any numerical calculation in this problem, use the values published by the NIST CODATA database (physics.nist.gov) — the internationally recognised source for fundamental physical constants. The electron mass is me = 9.1093837015 × 10⁻³¹ kg and the elementary charge is e = 1.602176634 × 10⁻¹⁹ C. In assignments, always state the source of constants you use. NIST is the gold standard.
The Physics — Why Does the Electron Move in a Circle?
This is the part most students skip in their working. Don’t. Examiners at A-Level and above want to see that you understand the physical reasoning, not just that you remembered the formula.
When a charged particle moves through a magnetic field, it experiences the Lorentz force. For an electron with charge magnitude e moving with velocity v in a field B, the force magnitude is F = evB sin θ, where θ is the angle between the velocity and the field. Since the problem states velocity is perpendicular to the field, θ = 90°, so sin θ = 1, and F = evB. Simple.
Now here’s the key insight. That force is always perpendicular to the velocity. Always. At every point in the electron’s path. A force that is always perpendicular to the direction of motion does two things: it cannot do work on the particle (so kinetic energy and speed stay constant), and it continuously changes the direction of motion. A constant-magnitude force that always points toward a centre, with constant speed — that’s circular motion. The magnetic force acts as the centripetal force.
The Lorentz Force
F = qv × B (vector form). For magnitude with perpendicular v and B: F = qvB. For an electron, the charge magnitude is e = 1.6 × 10⁻¹⁹ C. Direction: use the right-hand rule for positive charge, then reverse for negative charge (the electron).
Centripetal Force Condition
For circular motion, the net inward force must equal mv²/r. The magnetic force provides this centripetal force. Setting evB = mv²/r is the single most important equation in this problem — everything else follows from rearranging it.
No Work Done — Speed Is Constant
Work = F·d·cos90° = 0. The magnetic force does no work on the electron. This means kinetic energy is constant, speed is constant throughout the orbit. The force changes direction only — not speed. This is why the orbit is circular and not a spiral.
The magnetic force is a direction-changer, not a speed-changer. It steers the electron. It cannot accelerate it. That’s the physical fact everything else in this problem rests on.
— Concept from Griffiths, Introduction to Electrodynamics (4th ed.), Chapter 5 — the standard undergraduate electromagnetism textRadius of the Circular Path — How to Derive r = mv/eB
Radius of Circular Path
Start from the centripetal force condition and equate to the Lorentz force
Here’s how to reach this. The approach is to identify two expressions for the same force, set them equal, and rearrange for r.
Write the centripetal force expression
Any object moving in a circle of radius r at speed v requires a centripetal force directed toward the centre. The magnitude of this force is:
F_centripetal = mv² / rWrite the magnetic Lorentz force expression
The electron (charge magnitude e) moving at speed v perpendicular to magnetic field B experiences a force:
F_magnetic = evB (since sin 90° = 1)Equate the two forces
The magnetic force IS the centripetal force — there’s no other horizontal force acting. Set them equal:
evB = mv² / rRearrange for r
Multiply both sides by r, divide both sides by evB. One factor of v cancels from the right-hand side:
r = mv² / evB = mv / eB| Symbol | Quantity | SI Unit |
|---|---|---|
| r | Radius of circular orbit | metres (m) |
| m | Mass of electron = 9.11 × 10⁻³¹ kg | kilograms (kg) |
| v | Speed of electron entering the field | metres per second (m s⁻¹) |
| e | Elementary charge = 1.60 × 10⁻¹⁹ C | coulombs (C) |
| B | Magnetic field strength | tesla (T) |
Physical Intuition: What Does r = mv/eB Tell You?
Read the formula as a physical statement. A faster electron (larger v) has a larger radius — it’s harder to bend. A heavier particle (larger m) also has a larger radius — more inertia. A stronger field (larger B) produces a smaller radius — the bending force is stronger. A higher charge (larger e) produces a smaller radius — more force for the same field. This kind of dimensional and physical reasoning is exactly what gets marks in extended response questions.
Time Period of Revolution — Why T Is Independent of Velocity
Time Period of Revolution
Use the radius result from Part 1 — this derivation is shorter than it looks
The approach is to use the basic definition of period — time = distance / speed — applied to one full circular orbit.
Write the period as circumference divided by speed
The electron travels one full circumference (2πr) in one period at constant speed v. So:
T = 2πr / vSubstitute the expression for r from Part 1
You already found that r = mv/eB. Substitute this into the expression for T:
T = 2π(mv/eB) / vCancel v — and notice what disappears
The v in the numerator and the v in the denominator cancel. This is the key step:
T = 2πm / eBThe Result That Surprises Most Students — T Does Not Depend on v
The time period T = 2πm/eB contains no v. Speed has cancelled completely. A faster electron has a larger orbit radius, but it travels the larger circumference in exactly the same time. This is the isochronous property of cyclotron motion — all electrons in the same field B take the same time to complete one orbit, regardless of speed. This is the operating principle behind cyclotron particle accelerators. If an examiner asks you to comment on what the result shows, this is the observation they’re looking for.
| Derived Quantity | Formula | Notes |
|---|---|---|
| Period T | T = 2πm / eB | Independent of v — depends only on m, e, B |
| Frequency f | f = eB / 2πm | Cyclotron frequency — reciprocal of T |
| Angular frequency ω | ω = eB / m | Also called gyrofrequency; ω = 2π/T |
Force Acting on the Electron — Magnitude and Direction
Force on the Electron (Lorentz Force)
This is the simplest of the three parts — but direction requires care for negative charges
The Lorentz force on a moving charge in a magnetic field is F = qvB sin θ. Since v ⊥ B (given in the problem), sin θ = sin 90° = 1. For the electron, use the charge magnitude e = 1.60 × 10⁻¹⁹ C.
Write the general Lorentz force formula
For a particle with charge q moving at velocity v in field B, the force magnitude is:
F = qvB sin θApply the perpendicular condition
The problem states v is perpendicular to B, so θ = 90° and sin 90° = 1:
F = qvB × 1 = qvBSubstitute the electron charge
For an electron, the charge magnitude is e (we use the positive value e for the magnitude of the force). The force is directed toward the centre of the circular orbit:
F = evBState the direction using the left-hand rule
For positive charges, use the right-hand rule (Fleming’s). For electrons (negative charge), reverse the direction — use the left hand (or apply right-hand rule then flip). Point fingers in the direction of v, curl toward B — the thumb gives the force direction for a positive charge; reverse for an electron. The force always points toward the centre of the circular orbit (centripetal).
Direction: centripetal — perpendicular to v, toward centre of orbitImportant: F = evB Is Both the Lorentz Force AND the Centripetal Force Here
The force in Part 3 is the same force you equated to mv²/r in Part 1. There is no separate centripetal force — the magnetic Lorentz force is the centripetal force. In exam answers, make this explicit: “The magnetic force acting on the electron is F = evB, directed perpendicular to the velocity toward the centre of the circular orbit — this force provides the centripetal acceleration required to maintain circular motion.” That sentence scores full marks in a show-that or explain question.
Electron Constants You Must Know for This Problem
Numerical questions using this problem expect you to substitute real values for e and m. These are fixed constants — not variables. Memorise them or know where to find them quickly in an exam. Quoting them to the correct number of significant figures also matters.
| Constant | Symbol | Value | Unit | Note |
|---|---|---|---|---|
| Electron mass | mₑ | 9.109 × 10⁻³¹ | kg | In problems use 9.11 × 10⁻³¹ kg (3 s.f.) |
| Elementary charge | e | 1.602 × 10⁻¹⁹ | C | In problems use 1.60 × 10⁻¹⁹ C (3 s.f.) |
| Speed of light | c | 3.00 × 10⁸ | m s⁻¹ | Needed if checking whether relativistic effects apply (v ≪ c is assumed here) |
| Electron charge-to-mass ratio | e/mₑ | 1.76 × 10¹¹ | C kg⁻¹ | Appears directly in the cyclotron frequency ω = eB/m = (e/m)B |
Non-Relativistic Assumption — When It Applies and When It Doesn’t
The formulas r = mv/eB and T = 2πm/eB assume classical (non-relativistic) mechanics. They are valid when v ≪ c — typically for electron speeds below about 10% of the speed of light (3 × 10⁷ m s⁻¹). At higher speeds, the electron’s effective mass increases (relativistic mass correction), and both r and T become larger than the classical formulas predict. Most A-Level and first-year undergraduate problems assume non-relativistic electrons. If the problem gives you a velocity close to c, flag the relativistic regime explicitly.
Common Errors in This Problem — Where Marks Are Lost
| # | ❌ Error | Why It’s Wrong | ✓ Correct Approach |
|---|---|---|---|
| 1 | Using F = ma to find force direction without specifying centripetal acceleration | F = ma is true but unhelpful if you write it without specifying a = v²/r directed toward the centre. Examiners need to see you identify the centripetal nature of the acceleration, not just quote Newton’s second law generically. | Write F = ma = mv²/r, then state explicitly that the acceleration is centripetal (directed toward the centre of the circular orbit), and this acceleration is provided by the magnetic Lorentz force F = evB. |
| 2 | Forgetting that v cancels in the period derivation | Some students substitute r = mv/eB into T = 2πr/v but fail to cancel the v factors, leaving v in the answer: T = 2πmv/eBv. This is algebraically correct but unsimplified — it will lose marks for not reaching the standard form T = 2πm/eB. | Always simplify fully. After substituting: T = 2π(mv/eB)/v = 2πm(v/v)/eB = 2πm/eB. Show the cancellation step explicitly — it demonstrates understanding, not just manipulation. |
| 3 | Stating the force increases or decreases as the electron moves | Because v is constant (magnetic force does no work) and B is uniform (given), the force F = evB is constant in magnitude throughout the orbit. Writing “the force varies as the electron accelerates” confuses magnetic acceleration (change in direction) with kinetic acceleration (change in speed). | State clearly: “The magnetic force is constant in magnitude throughout the orbit because both v and B are constant. The force direction changes continuously (always pointing toward the centre of the orbit), but its magnitude F = evB does not change.” |
| 4 | Using the wrong rule for the direction of force on a negative charge | Fleming’s right-hand rule gives the force direction for conventional (positive) current. An electron is negative — its force is in the opposite direction. Applying the right-hand rule without reversing gives the wrong direction for the electron. | Either use the left-hand rule directly for the electron, or use the right-hand rule and explicitly reverse the direction. State which approach you’re using. In a diagram, show the force arrow pointing toward the centre of the circular path — this is always correct regardless of which hand rule you used. |
| 5 | Quoting F = Bqv without stating that this applies for perpendicular v and B | The general formula is F = qvB sin θ. Writing F = qvB is only valid for θ = 90°. If you don’t state the perpendicular condition, you haven’t justified dropping the sin θ term. | Write: “Since the velocity is perpendicular to the magnetic field (θ = 90°), sin θ = 1, and the Lorentz force simplifies to F = evB.” One sentence, full justification, no marks lost. |
| 6 | Confusing the cyclotron frequency with the time period | The cyclotron frequency f = eB/2πm and the time period T = 2πm/eB are reciprocals of each other (f = 1/T). Writing T = eB/2πm is wrong — it’s the formula for f, not T. | Derive T first from T = 2πr/v, which gives T = 2πm/eB. Then if needed, f = 1/T = eB/2πm. Don’t mix them up by working from memory — derive them. |
Extension Questions You Should Also Be Ready For
This problem has a standard three-part structure, but examiners often extend it. Knowing where the question could go next helps you anticipate what concepts to connect — and shows the kind of wider physical understanding that distinguishes top-band answers.
What If v Is Not Perpendicular to B?
If v has a component along B as well as perpendicular to it, the electron follows a helix rather than a circle. The perpendicular component causes circular motion; the parallel component is unaffected by B (force along B is zero). The resulting path is a corkscrew — helical motion. The radius is calculated using only the perpendicular component of v.
What Happens If Both E and B Fields Are Present?
This is a velocity selector problem. With a crossed electric field E and magnetic field B, the electric force qE acts opposite to the magnetic force qvB for one direction of motion. At the specific velocity v = E/B, the forces balance and the electron travels in a straight line. For other velocities, the particle curves. This is how Thomson measured the electron’s charge-to-mass ratio.
Numerical Substitution — Typical Exam Values
Given v = 2.0 × 10⁶ m s⁻¹ and B = 0.050 T: r = (9.11 × 10⁻³¹ × 2.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.050) = 0.23 m. T = 2π × 9.11 × 10⁻³¹ / (1.60 × 10⁻¹⁹ × 0.050) = 7.2 × 10⁻¹⁰ s. F = 1.60 × 10⁻¹⁹ × 2.0 × 10⁶ × 0.050 = 1.6 × 10⁻¹⁴ N.
The Cyclotron Accelerator and the Isochronous Property
The fact that T = 2πm/eB is independent of v is the operating principle of the cyclotron particle accelerator. A cyclotron alternates the accelerating electric field at a frequency matching the cyclotron frequency f = eB/2πm. As the particle accelerates and gains speed, its orbit radius increases — but the time per orbit stays the same. So the alternating field stays in phase with the particle’s orbit throughout the acceleration process. This breaks down at relativistic speeds, which is why modern synchrotrons modulate the field frequency.
Drawing the Setup and Showing the Force Direction
Examiners often ask for a diagram showing the magnetic field (into or out of the page), the initial velocity, and the resulting circular path with force direction labelled. The force always points toward the centre of the circle — it is centripetal. Use the left-hand rule to confirm direction for the electron specifically. Label: B (field direction), v (initial velocity), F (toward centre), and the curved circular path. A clear, correctly labelled diagram can score 2–3 marks on its own.
How to Write Up This Problem Properly — What Examiners Want to See
Physics marks are not just for getting the right answer. They’re for showing logical, correctly sequenced reasoning with clear physical justification at each step. A correct answer with no working shown typically scores zero or one mark. Here’s the framework for full marks.
Worked Answer Framework — Magnetic Escape Challenge
How to structure each part for maximum marks at A-Level and degree level
Pre-Submission Checklist for This Problem Type
- Identified that v ⊥ B and explicitly used sin 90° = 1 before writing F = evB
- Shown the equating step evB = mv²/r — not just the rearranged result
- Cancelled v in the period derivation and shown the step explicitly
- Stated that T is independent of v and explained what this means physically
- Given the direction of the force (centripetal — toward centre of orbit)
- Used correct values for e = 1.60 × 10⁻¹⁹ C and m = 9.11 × 10⁻³¹ kg if numerical
- Checked significant figures match the data given in the problem
- Included a labelled diagram if the question requires one (B direction, v direction, force direction)
- Applied left-hand rule (not right-hand) for force direction on the negative electron
- Stated the non-relativistic assumption if the problem doesn’t specify
FAQs: Electron in a Magnetic Field
Three Results, One Elegant Principle
The Magnetic Escape Challenge is really one problem, not three. Every result comes from a single physical idea: when a charged particle moves perpendicular to a magnetic field, the Lorentz force acts as a centripetal force, causing circular motion at constant speed. That one idea gives you F = evB, r = mv/eB, and T = 2πm/eB — each following from the previous one by straightforward algebra.
What makes this problem worth taking seriously is the physics hiding in the mathematics. The independence of T from v isn’t a coincidence — it’s a deep property of electromagnetic interactions that powered the first particle accelerators and remains central to plasma physics, MRI technology, and space physics today. An electron trapped by Earth’s magnetic field, spiralling back and forth between the poles — that’s the same physics, scaled up by many orders of magnitude.
When you write up the derivation, show the reasoning at every step. Not because examiners are pedantic, but because the reasoning is the point. For expert support with physics assignments at any level, the specialists at Smart Academic Writing are ready. See the full range of academic writing services, including physics homework help and research paper and lab report writing.