IB Physics SLTopic 4 — Force FieldsPaper 1 & 2v = E ÷ B~8 min read
Charges in Combined Fields
Switch on an electric field and a magnetic field at the same time, crossed at right angles, and their forces on a moving charge point in opposite directions. Tune them just right and the two exactly cancel — the charge sails straight through, untouched. That balancing act is a velocity selector, a clever filter that picks out particles of one precise speed.
📘 What you need to know
In crossed (perpendicular) electric and magnetic fields, a moving charge feels two forces: the electric force FE = qE and the magnetic force FB = Bqv
Set up the usual way, these forces point in opposite directions, so they can be made to cancel
When they balance, the charge moves in a straight line at constant speed: qE = Bqv
The charge cancels out, leaving the selected speed v = E ÷ B — the same for any charge or mass
This is a velocity selector: only particles with v = E ÷ B pass straight; slower ones are pulled toward the electric force, faster ones toward the magnetic force
Between parallel plates the field strength is E = V ÷ d, so the plate voltage sets the electric force
Two Fields, Two Forces
Take the standard arrangement: a charge moving right, an electric field pointing up, and a magnetic field pointing out of the page — three directions all at right angles. Each field pushes on the charge in its own way. The electric force FE = qE acts along the field; the magnetic force FB = Bqv acts at right angles to both the field and the motion (Fleming’s left-hand rule).
For a positive charge in this layout, the electric force points up and the magnetic force points down — dead against each other. Flip to a negative charge and both forces reverse, so they still oppose. Either way, the two forces line up head-to-head, which is exactly what lets them cancel.
Blue arrows are the electric field, violet dots the magnetic field (out of the page). On the positive charge the electric force (orange, up) and magnetic force (red, down) are equal and opposite, so it travels the dashed straight line. Setting them equal gives v = E ÷ B.
Balancing the Forces: v = E ÷ B
Straight-line motion means the two forces are equal in size. Write that down and the algebra almost finishes itself:
FE = FB
write them out:
qE = Bqv
the q cancels →
v = E ÷ B
Speed that passes straight throughv = E ÷ B
The striking thing is what’s missing from that result: no charge q, no mass m. Because the charge cancels, the selected speed is the same for every particle, whatever its charge or mass. Give it the field strengths and the crossed fields will only let one speed through in a straight line. If you’re working from parallel plates, get the field from the voltage using E = V ÷ d.
The Velocity Selector
What about particles that arrive at the wrong speed? The electric force qE is fixed, but the magnetic force Bqv grows with speed. So a particle moving too fast feels a bigger magnetic force — it wins, and the particle bends the magnetic way. A particle moving too slow has a weak magnetic force — the electric force wins, and it bends the other way. Only the exact speed v = E ÷ B stays balanced and flies straight out through the slit.
Same entry point, three speeds. Too slow and the electric force wins (the particle bends up); too fast and the magnetic force wins (it bends down). Only v = E ÷ B stays balanced and makes it through the slit.
Quick recap: in crossed electric and magnetic fields the electric force (qE) and magnetic force (Bqv) oppose. Balance them for straight-line motion, qE = Bqv, and the charge cancels to give v = E ÷ B — a velocity selector that passes one speed for any particle.
🧭 Crossed-field problems
Name the two forces: electric FE = qE (along/against E), magnetic FB = Bqv (Fleming’s left-hand rule)
Goes straight? The forces balance — set qE = Bqv and cancel the q to get v = E ÷ B
Given plates? Find the field from the voltage with E = V ÷ d (convert d to metres)
Off the selected speed? Faster → magnetic force wins (Bqv grows with v); slower → electric force wins
Charge sign? It flips both forces together, so the balance still holds — v = E ÷ B doesn’t care about the sign
WE 1
A velocity selector uses an electric field of 4.5 × 10⁴ N C⁻¹ crossed at right angles with a magnetic field of 0.15 T. (a) Calculate the speed of the particles that pass straight through. (b) State what happens to particles that enter faster than this.
Part (a) — straight through means v = E ÷ Bv = 4.5 × 10⁴ ÷ 0.15v = 3.0 × 10⁵ m s⁻¹Part (b) — a faster particle
Bqv grows with speed, so the magnetic force now beats the electric force
it deflects in the direction of the magnetic forceNotice the answer to (a) needs no charge or mass — every particle is filtered to the same speed.
WE 2
An electron passes in a straight line at a constant 5.0 × 10⁶ m s⁻¹ through a magnetic field of 0.080 T crossed with the electric field between two parallel plates. (a) Calculate the electric field strength. (b) The plates are 2.0 cm apart — find the potential difference across them.
Part (a) — straight line, so forces balance: v = E ÷ B → E = vBE = 5.0 × 10⁶ × 0.080E = 4.0 × 10⁵ N C⁻¹Part (b) — use E = V ÷ d → V = Ed
d = 2.0 cm = 0.020 m
V = 4.0 × 10⁵ × 0.020V = 8.0 × 10³ V
💡 Top tips
v = E ÷ B has no q or m in it. The selected speed is identical for every particle — that’s the whole point of a velocity selector
Both forces flip for a negative charge, so they still oppose and the balance condition is unchanged — don’t panic over the sign
Use E = V ÷ d to turn a plate voltage into a field, and always convert the separation to metres first
Which force wins? The magnetic force Bqv depends on speed, the electric force qE doesn’t — so faster particles feel the magnetic force take over
⚠ Common mistakes
Slipping a charge or mass into v = E ÷ B — they cancel, so the formula has neither
Assuming the two forces add up — for straight-line motion they must be equal and opposite, so check their directions
Getting the “which wins” logic backwards — since Bqv ∝ v, a faster particle is dominated by the magnetic force
Leaving the plate separation in centimetres in E = V ÷ d — convert to metres
And that completes Force Fields — from the push on a single wire all the way to charges threading crossed electric and magnetic fields. The velocity selector you’ve just met is exactly the front end of J.J. Thomson’s famous experiment: pair it with the circular path r = mv ÷ BQ and you can measure a particle’s charge-to-mass ratio. Well done getting through the whole topic — revisit any page whenever you need a refresher before the exam.
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