IB Physics HLTopic 4 — Force FieldsPaper 1 & 2v = E/B~16 min read
Charged Particles in Combined Fields
You have met the two fields on their own. A magnetic field bends a charge into a circle and never changes its speed. An electric field pushes it into a parabola and speeds it up. Now switch both on at once, and aim them so their forces fight each other. Turn the dials just right and the two forces cancel dead — the charge sails straight through, as if there were no fields at all. But here is the clever part: it only balances at one special speed. Every other speed loses. That single idea is a velocity selector, and it is how the electron was first weighed.
📘 What you need to know
In crossed (perpendicular) electric and magnetic fields, a charge feels two forces at once
The electric force is FE = qE; the magnetic force is FB = Bqv
The fields are arranged so these forces point in opposite directions
The particle goes straight only when the forces balance: qE = Bqv
The q cancels, giving the selected speedv = E/B — it does not depend on charge or mass
This device is a velocity selector: only particles at exactly v = E/B pass undeflected
The charge-to-mass ratio is q/m, and for an electron equals 1.76 × 1011 C kg−1
J.J. Thomson used crossed fields, then the magnetic field alone, to measure q/m for the electron
Switch the electric field off and only the magnetic force remains, so the beam curves into a circle
Two forces, pulling opposite ways
Send a charge through a region where an electric field and a magnetic field are set at right angles to each other and to the motion. The charge now feels both forces together.
The two forces on the chargeelectric: FE = qE • magnetic: FB = Bqv
The fields are lined up so the electric force pushes the charge one way and the magnetic force pushes it the other way. Whether the charge deflects — and which way — comes down to which force is winning.
The electric force (green, up) and the magnetic force (purple, down) are drawn equal and opposite. When they balance exactly, the charge feels no net force and flies straight.
The balance condition
Set the two forces equal and something wonderful happens.
Step 1 — forces balance for a straight pathFE = FB → qE = Bqv
Step 2 — the charge q cancels from both sidesE = Bv
Step 3 — rearrange for the selected speedv = E / Bthe one speed that passes straight through — no q, no m in sight
Electric force qE
balances the
Magnetic force Bqv
cancel q
v = E/B
Look at what dropped out. The charge q cancelled, and the mass m was never there. So the selected speed v = E/B is the same for every particle — an electron, a proton, a dust speck, whatever. That is exactly why it’s called a velocity selector: it sorts purely by speed and ignores everything else. Say that in the exam and the marks are yours.
Why only one speed survives
The trick is that the two forces respond differently to speed. The electric force qE does not care how fast the charge goes — it’s fixed. But the magnetic force Bqv grows with speed. So:
Too slow: the magnetic force is weak, the electric force wins, and the charge deflects toward the electric-force side
Just right (v = E/B): the two are equal, and the charge goes straight
Too fast: the magnetic force is strong, it wins, and the charge deflects the other way
Slow particles bend toward the electric force; fast ones bend toward the magnetic force. Put a slit at the far end and only the one speed v = E/B makes it out — a clean beam of a single, known speed.
WE 1
A velocity selector uses a magnetic field of flux density 0.040 T, crossed with an electric field produced by two plates 5.0 cm apart with 1200 V across them. (a) Calculate the electric field strength. (b) Calculate the speed of the particles that pass through undeflected. (c) State whether a proton and an electron at this speed would both pass straight through.
(a) Step 1 — field between the platesE = V/d = 1200 / 0.050E = 2.4 × 10⁴ N C⁻¹(b) Step 2 — the selected speedbalance: qE = Bqv → v = E/Bv = (2.4 × 10⁴) / 0.040v = 6.0 × 10⁵ m s⁻¹(c) proton and electron?
v = E/B has no q and no m in it.
yes — both pass straight through, at any charge or massThe whole point: the selector filters by speed alone. A proton, an electron, an ionised atom — if it’s doing 6.0 × 10⁵ m s⁻¹ it sails through, and if it isn’t, it gets bent into a wall. Charge and mass simply don’t enter the balance.
Weighing the electron: charge-to-mass ratio
You cannot put an electron on a balance. But you can measure the ratio of its charge to its mass, q/m — and that ratio is a fingerprint of the particle.
Charge-to-mass ratio (a definition)q / m = charge / masselectron: 1.76 × 1011 C kg−1 • proton: 9.58 × 107 C kg−1
J.J. Thomson found this ratio for the electron with a beautiful two-step experiment, using exactly the fields on this page.
Step 1: balance both fields to fix the speed, v = E/B. Step 2: switch the electric field off, so only the magnetic force is left and the beam curls into a circle whose radius reveals q/m.
Putting it together
With the electric field off, the magnetic force alone provides the centripetal force, exactly as on the “Charged Particles in Magnetic Fields” page:
Magnetic force provides the centripetal forceBqv = mv2/r → q/m = v / rB
Now substitute the speed found in step 1, v = E/B = V/(dB), and everything on the right is measurable:
Charge-to-mass ratio from measurementsq/m = V / (rB2d)just four measured quantities: V, d, r and B
WE 2
Show that the charge-to-mass ratio of an electron is about 1.8 × 1011 C kg−1, and that a proton’s is far smaller. (e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg, mp = 1.67 × 10−27 kg)
electrone/mₕ = (1.60 × 10⁻¹⁹) / (9.11 × 10⁻³¹)= 1.76 × 10¹¹ C kg⁻¹protone/mₚ = (1.60 × 10⁻¹⁹) / (1.67 × 10⁻²⁷)= 9.58 × 10⁷ C kg⁻¹The electron’s ratio is about 1840× bigger than the proton’s — same size of charge, but the electron is 1840× lighter. That huge ratio is why an electron whips round a tiny circle while a proton needs a far bigger one in the same field.
WE 3
In a Thomson-style tube, a charged particle first passes undeflected through crossed fields with plates 5.0 cm apart, 1800 V across them, and a magnetic flux density of 1.5 mT. The electric field is then switched off and the particle curves with a radius of 9.1 cm. (a) Find the speed of the particle. (b) Determine its charge-to-mass ratio. (c) Suggest what the particle is.
(a) Step 1 — speed from the balanced fieldsE = V/d = 1800 / 0.050 = 3.6 × 10⁴ N C⁻¹v = E/B = (3.6 × 10⁴) / (1.5 × 10⁻³)v = 2.4 × 10⁷ m s⁻¹(b) Step 2 — ratio from the circular pathq/m = v / (rB) = (2.4 × 10⁷) / [(0.091)(1.5 × 10⁻³)]q/m = 1.76 × 10¹¹ C kg⁻¹(c) what is it?
This matches the known ratio for an electron.
the particle is an electronNotice how the speed you found in part (a) feeds straight into part (b) — that’s the whole method in miniature. You could also go in one line with q/m = V/(rB²d); it gives the same 1.76 × 10¹¹. Getting the electron’s fingerprint out of four ordinary measurements is exactly what made Thomson famous.
🧲 Working a combined-fields question
Straight through? The forces balance: qE = Bqv, so v = E/B.
Given plates? Get the field first: E = V/d, with d in metres.
Selected speed has no q or m — it’s the same for every particle.
Charge-to-mass? Switch E off, use the circle: q/m = v/(rB).
All four measured? Combine in one step: q/m = V/(rB2d).
Identify a particle by comparing its q/m to the electron’s 1.76 × 1011.
💡 Top tips
Straight-line path ⇒ write qE = Bqv at once, then cancel q to get v = E/B.
Say the magic line: the selected speed is independent of charge and mass.
Convert plate separation to metres before E = V/d.
For q/m, the field is switched off and the path becomes a circle.
The one-line combined formula is q/m = V/(rB2d) — note B is squared.
Electron q/m = 1.76 × 1011; proton = 9.58 × 107. The electron’s is ~1840× bigger.
⚠ Common mistakes
Adding the two forces instead of balancing them. For a straight path they’re equal and opposite
Thinking the selected speed depends on the particle. v = E/B has no q or m
Leaving d in cm in E = V/d
Forgetting to square B in q/m = V/(rB2d)
Using the circle formula with both fields on. The circle only appears once E is off
Mixing up the forces: FE = qE is fixed, but FB = Bqvgrows with speed
Claiming Thomson measured the electron’s charge or mass. He measured their ratio
Quick recap: In crossed electric and magnetic fields a charge feels two opposing forces, FE = qE and FB = Bqv. It travels straight only when they balance, and cancelling q gives the selected speed v = E/B — the same for every particle, so the device is a velocity selector. Switch the electric field off and the magnetic force curls the beam into a circle, from which q/m = v/(rB) = V/(rB2d). That is how J.J. Thomson measured the electron’s charge-to-mass ratio, 1.76 × 1011 C kg−1.
That wraps up Motion in Electromagnetic Fields — you can now handle a charge in a magnetic field, an electric field, and both at once. The natural next step is to stop pushing charges and start asking what a field is: how strong it is, which way it points, and how we picture it with field lines. Next up: Electric Fields & Field Strength.
Velocity selector making your head spin?
Book a free meeting and we’ll nail the qE = Bqv balance, the v = E/B derivation, and Thomson’s two-step q/m method.
