Electric vs Magnetic Fields
In 1897, J.J. Thomson at the Cavendish Laboratory simultaneously applied perpendicular electric and magnetic fields inside his cathode ray tube. When he tuned the fields until the beam went perfectly straight, only particles with v = E/B passed through undeflected — a velocity selector. This crossed-fields geometry, unchanged since Thomson's original apparatus, is still the first stage of every mass spectrometer built today.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A positively charged particle enters a region with both a uniform electric field (pointing downward) and a uniform magnetic field (pointing into the page). The particle is initially moving to the right.
Before reading on, answer:
- Sketch the path you predict if only the electric field is present.
- Sketch the path you predict if only the magnetic field is present.
- With both fields present, do you think the particle could travel in a straight line? If so, what condition would be required?
Warm-up — which type of charge does a magnetic field exert a force on?
Know — Field Behaviours
- E-fields act on all charges; B-fields act only on moving charges
- E-fields can do work and change speed; B-fields cannot
- E-field paths are parabolic; B-field paths are circular (perpendicular case)
Understand — Combined Fields
- How crossed E and B fields create a velocity selector
- Why $v = E/B$ is the condition for straight-line motion
- When each field dominates the particle's trajectory
Can Do — Solve Crossover Problems
- Compare and contrast E-field and B-field effects quantitatively
- Derive and apply the velocity selector equation
- Predict trajectories in combined field configurations
Core Content
How electric and magnetic fields treat charges differently
Place a stationary proton between two charged plates and it accelerates immediately — the electric field acts on it even at rest. Now place that same stationary proton in a region with only a magnetic field: nothing happens. Move the proton sideways and suddenly the magnetic force appears, pushing it perpendicular to its motion. Electric fields act on any charge regardless of motion; magnetic fields are invisible to stationary charges and only reveal themselves when a charge is moving.
| Property | Electric Field | Magnetic Field |
|---|---|---|
| Acts on | All charges (moving or stationary) | Only moving charges |
| Force equation | $F = qE$ | $F = qvB\sin\theta$ |
| Direction | Parallel to field (same or opposite) | Perpendicular to both $\vec{v}$ and $\vec{B}$ |
| Does work? | Yes — can change kinetic energy | No — force always perpendicular to motion |
| Changes speed? | Yes | No — only direction |
| Path (perpendicular) | Parabola (like projectile) | Circle |
| Depends on mass? | Only for acceleration ($a = qE/m$) | Yes — radius and period depend on $m$ |
Figure 1 — Same electron, same initial velocity: E-field produces a parabola; B-field produces a circle
A student claims: "Since both electric and magnetic forces depend on charge, they will both reverse direction if you replace a proton with an electron." Is this claim fully correct? Identify any inaccuracy.
Electric field: $F = qE$, acts on all charges (moving or stationary), parallel to $\vec{E}$, does work, changes speed → parabolic path. Magnetic field: $F = qvB\sin\theta$, acts only on moving charges, perpendicular to $\vec{v}$ and $\vec{B}$, does NO work, cannot change speed → circular path.
Pause — copy the highlighted comparison table summary into your book before moving on.
A magnetic field can do work on a moving charged particle and increase its kinetic energy.
A stationary charge in a magnetic field experiences zero magnetic force.
When crossed E and B fields produce straight-line motion
We just saw that E-fields and B-fields exert forces in very different ways. That raises a question: what happens when both fields act on the same particle simultaneously? This card answers it → at exactly one speed ($v = E/B$) the forces cancel and the particle travels straight.
When both electric and magnetic fields are present and perpendicular to each other and to the particle's velocity, the particle experiences two forces. By arranging the fields so that these forces act in opposite directions, we can find a specific velocity where they exactly cancel.
- Electric force: $F_E = qE$ (direction depends on charge sign and E-field direction)
- Magnetic force: $F_B = qvB$ (direction depends on charge sign, velocity, and B-field direction)
$F_E = F_B$
$qE = qvB$
$$v = \dfrac{E}{B}$$
Only particles with this exact speed travel straight through
Notice that $q$ cancels out. This means the velocity selector works for any charged particle regardless of its mass or charge — as long as $v = E/B$, it travels straight. This is the first stage of a mass spectrometer.
Figure 2 — Velocity selector: crossed E and B fields. Only particles with v = E/B experience balanced forces and pass through the slit.
A velocity selector has E = 2000 V/m and B = 0.040 T. A proton and an alpha particle both enter at $5.0 \times 10^4$ m/s. Which particles (if any) pass through undeflected? Explain.
Velocity selector condition: set $F_E = F_B$ → $qE = qvB$ → $v = E/B$ (m/s). The charge $q$ cancels — the condition is mass- and charge-independent. Particles faster than $E/B$: magnetic force dominates → deflected toward B-force direction. Particles slower: electric force dominates.
Add the highlighted velocity selector equation and key points to your notes before the check below.
A velocity selector has $E = 3000$ V/m and $B = 0.015$ T. What speed must a particle have to pass through undeflected?
A charged particle moving parallel to a magnetic field (angle = 0°) experiences zero magnetic force.
In a velocity selector, an electron at $v = E/B$ and a proton at $v = E/B$ both travel in a straight line.
Determining trajectory in crossed E and B fields
An electron enters a region with a uniform electric field of $E = 4000\,\text{V/m}$ directed downward and a uniform magnetic field of $B = 0.020\,\text{T}$ directed out of the page. The electron enters moving horizontally to the right at $v = 2.0 \times 10^5\,\text{m/s}$.
- Calculate the electric force magnitude and direction on the electron.
- Calculate the magnetic force magnitude and direction on the electron.
- Determine whether the electron travels in a straight line or is deflected.
- What speed would be required for the electron to travel straight through?
-
Electric force:
$F_E = qE = (1.60 \times 10^{-19})(4000) = 6.40 \times 10^{-16}\,\text{N}$
Direction: E-field points downward, so the negative electron experiences force upward. -
Magnetic force:
$F_B = qvB = (1.60 \times 10^{-19})(2.0 \times 10^5)(0.020) = 6.40 \times 10^{-16}\,\text{N}$
Direction: For a negative charge, use the right-hand rule then reverse. $v$ is right, $B$ is out of page → RH rule gives force upward on positive charge. Reverse for electron: downward. -
Net force and trajectory:
$F_E$ upward = $F_B$ downward = $6.40 \times 10^{-16}\,\text{N}$.
Net force = 0. The electron travels in a straight line. -
Velocity selector speed:
$v = \dfrac{E}{B} = \dfrac{4000}{0.020} = 2.0 \times 10^5\,\text{m/s}$
This matches the given speed, confirming the result.
If the electron's speed were increased to $4.0 \times 10^5$ m/s with the same E and B fields, in which direction would it be deflected? Justify your answer with a calculation.
Combined field method: (1) draw diagram; (2) calculate $F_E = qE$ and $F_B = qvB$ separately; (3) reverse both forces for negative charges; (4) compare magnitudes. If $v = E/B$: net force = 0, straight line. If $v > E/B$: $F_B$ dominates; if $v < E/B$: $F_E$ dominates.
Pause — write the highlighted problem-solving method into your book before moving on.
In a velocity selector, a proton travels straight through at speed $v = E/B$. If the proton is replaced by an electron with the same speed, what happens?
In combined field problems, students often forget that direction depends on charge sign for both forces. The electric force reverses for negative charges (opposite to E), and the magnetic force also reverses (opposite to the right-hand rule prediction). If you reverse both forces, they still oppose each other, and the velocity selector condition $v = E/B$ remains the same regardless of charge sign. Always draw arrows and check signs before calculating.
From velocity selector to mass analysis
We just saw how crossed E and B fields select a single speed. That raises a question: once we know every particle is travelling at exactly $v = E/B$, what can we do with them? This card answers it → add a second pure B-field and the radius $r = mv/qB$ reveals the mass-to-charge ratio.
A mass spectrometer is the velocity selector in action. By combining crossed E and B fields with a pure magnetic deflection region, scientists can separate particles by mass-to-charge ratio with extraordinary precision.
- Acceleration: Particles are accelerated through a known voltage $\Delta V$, gaining kinetic energy $q\Delta V = \frac{1}{2}mv^2$.
- Velocity selection: Crossed E and B fields filter particles so only those with $v = E/B$ pass through a narrow slit.
- Mass analysis: The selected particles enter a pure magnetic field (no E-field) and follow a circular path with radius $r = \dfrac{mv}{qB}$. Since $v$ is now known and fixed, heavier particles have larger radii.
By measuring the radius of curvature, scientists determine the mass-to-charge ratio: $\dfrac{m}{q} = \dfrac{rB}{v}$. This technique is used in chemistry (identifying isotopes), biology (protein mass analysis), and forensics (drug detection).
Figure 3 — Three-stage mass spectrometer: accelerate, select velocity, then deflect by mass
In a mass spectrometer, two singly charged ions ($q = +e$) have masses $m_1 = 20\,\text{u}$ and $m_2 = 22\,\text{u}$. They both pass through the same velocity selector and enter the same magnetic field region. Which ion follows the larger circular path? By what factor is the larger radius greater than the smaller?
Mass spectrometer: accelerate ($q\Delta V = \tfrac{1}{2}mv^2$) → velocity-select ($v = E/B$) → deflect in pure B-field ($r = mv/qB$). After selection all ions share the same $v$, so $r \propto m/q$. Heavier ions have larger radius; $r_2/r_1 = m_2/m_1$ for equal charges.
Add the highlighted three-stage mass spectrometer summary to your notes before the check below.
In a mass spectrometer, ions of different mass but the same charge pass through a velocity selector and then enter a uniform magnetic field. What happens to the radius of their circular path?
Calculate and compare field effects on different particles
- A velocity selector has $E = 5000$ V/m and $B = 50$ mT. Calculate $v_{selector} = E/B$. Show all working.
- An electron enters at $v = 1.0 \times 10^5$ m/s. Calculate both the electric and magnetic forces on it and determine whether it travels straight or is deflected.
- A proton replaces the electron with the same $E$, $B$, and speed. Does the proton also travel straight? Explain why the velocity selector condition is independent of charge sign and mass.
- With the proton at the selector speed, the electric field is doubled. What speed would now be required for straight-line motion?
Apply the three-stage mass spectrometer model
Two singly charged carbon isotopes, $^{12}$C and $^{14}$C, both pass through a velocity selector where $E = 6000$ V/m and $B_1 = 0.030$ T. They then enter a second magnetic field region $B_2 = 0.50$ T.
- Calculate the speed $v$ of ions passing through the velocity selector.
- Calculate the radius of the circular path for each isotope in the second magnetic field. Use mass of $^{12}$C = $1.993 \times 10^{-26}$ kg and $^{14}$C = $2.325 \times 10^{-26}$ kg.
- What is the separation of the two ion beams at the detector?
Key Contrasts
- E-field: all charges; B-field: moving only
- E-field does work; B-field does NOT
- E: parabola; B: circle
Velocity Selector
- $qE = qvB \Rightarrow v = E/B$
- Independent of $q$ and $m$
- Stage 1 of mass spectrometer
Mass Spectrometer
- Accelerate → Select $v$ → Deflect
- $r = mv/qB$ in pure B-field
- $r \propto m$ (heavier = larger arc)
HSC Check Signs
- Draw field and velocity vectors first
- Apply RH rule then reverse for $-q$
- Both forces reverse → $v = E/B$ holds
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 5(4 marks) 1. Compare and contrast the effect of a uniform electric field and a uniform magnetic field on a moving charged particle. In your answer, address: (i) the conditions under which each field exerts a force, (ii) whether each field can change the particle's kinetic energy, and (iii) the shape of the trajectory when the field is perpendicular to the initial velocity.
1 mark: E-field acts on all charges; B-field only on moving charges · 1 mark: E-field changes KE (does work); B-field does not · 1 mark each: parabola (E) and circle (B)
AnalyseBand 6(4 marks) 2. An electron moves through a region with both E and B fields. The magnetic force is twice as large as the electric force, and both forces act in opposite directions on the electron. Describe the electron's motion and explain whether its speed changes. Justify your answer by referring to both forces.
1 mark: net force is non-zero (B dominates) so path curves · 1 mark: the net force has a component along velocity (from the E-field component of the net) · 1 mark: magnetic component does no work but electric component does → speed changes · 1 mark: this is a complex curved path with increasing speed
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (4 marks): (i) An electric field exerts a force $F = qE$ on all charged particles regardless of whether they are moving or stationary (1 mark). A magnetic field only exerts a force $F = qvB\sin\theta$ on moving charges — a stationary charge experiences zero magnetic force (1 mark). (ii) An electric field can change a particle's kinetic energy because the electric force has a component along the direction of motion (it does work). A magnetic force is always perpendicular to the velocity, so it does zero work and cannot change speed or kinetic energy (1 mark). (iii) When the field is perpendicular to the initial velocity: an electric field produces a parabolic path (analogous to projectile motion); a magnetic field produces a circular path at constant speed (1 mark).
Q2 (4 marks): The net force is non-zero (magnetic is twice the electric, so there is a net force in the direction of the magnetic force). The path therefore curves (1 mark). The magnetic component of the force is always perpendicular to the velocity — it does no work and does not change speed (1 mark). However, the electric component of the force does have a component along the velocity direction at some instants, doing work and changing the particle's kinetic energy (1 mark). The result is a complex curved path where the electron's speed increases (1 mark).
Five timed questions on electric vs magnetic fields. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaAt the start you were asked about Thomson's 1897 CRT experiment at the Cavendish Laboratory: with E = 1.5 × 10⁴ V/m and B = 3.4 × 10⁻⁴ T, what speed did the electrons have — and does it depend on their charge sign?
The answer: $v = E/B = 1.5 \times 10^4 / 3.4 \times 10^{-4} \approx 4.4 \times 10^7$ m/s. Crucially, the condition $v = E/B$ is independent of charge sign — the velocity selector passes particles of any charge as long as their speed is right. This is why Thomson could use it for electrons, and why modern mass spectrometers use it for positive and negative ions alike.
Review your Think First predictions from this lesson and earlier lessons:
- L06: Was $v = E/B$ the condition you proposed for straight-line motion?
- L01–05: How does the circular path in a B-field connect to what you learned about circular motion in a magnetic field?