IB Physics SLTopic 4 — Force FieldsPaper 1 & 2r = mv ÷ BQ~8 min read
Charges in Magnetic Fields
Last page we found that a magnetic field always pushes a moving charge sideways — at right angles to its motion. Follow that through and something neat happens: a charge fired into a uniform field doesn’t just bend, it loops all the way round into a perfect circle. This is the physics behind mass spectrometers, particle accelerators, and the aurora dancing over the poles.
📘 What you need to know
A charged particle moving through a uniform magnetic field follows a circular path, because the magnetic force stays perpendicular to the velocity and always points to the centre
That magnetic force provides the centripetal force: BQv = mv2 ÷ r
Rearranging gives the radius of the path: r = mv ÷ BQ
So the circle is wider for a faster (↑v) or heavier (↑m) particle, and tighter for a bigger charge (↑q) or stronger field (↑B)
The magnetic force does no work — it changes direction, never speed — so the particle circles at constant speed (uniform circular motion)
It’s the same centripetal machinery as any circular motion, so the acceleration also points to the centre and F = ma still applies
Why the Path Is a Circle
Here’s the chain of reasoning. The magnetic force on the moving charge is always perpendicular to its velocity (that’s just how F = Bqv works). A force that stays at right angles to the motion can’t speed the particle up or slow it down — it can only swing the direction round. And a constant-size force forever pointing “inward”, square to the motion, is exactly the recipe for uniform circular motion. So the particle traces a circle at steady speed, with the magnetic force acting as the centripetal force.
The velocity (teal) is always tangent to the circle, while the magnetic force (red) always points to the centre. A centre-seeking force of constant size is exactly what produces steady circular motion — the particle’s speed never changes, only its direction.
The Radius of the Path
Because the magnetic force is doing the job of the centripetal force, we can set the two equal and solve for the radius. The magnetic force is BQv; the centripetal force needed is mv2 ÷ r:
magnetic force = centripetal force
set equal:
BQv = mv2 ÷ r
cancel a v, rearrange →
r = mv ÷ BQ
Radius of a charge’s circular pathr = mv ÷ BQ
where r is the radius (m), m the mass (kg), v the speed (m s−1), B the flux density (T), and Q the charge (C). Being able to derive this from the force balance is a favourite exam ask, so practise the two lines above until they’re automatic.
What makes the circle bigger or smaller
Reading the formula tells you everything about the size of the loop. Mass and speed sit on top, so more of either widens the circle; charge and field sit underneath, so more of either tightens it.
Same field and charge, different speeds: the faster particle sweeps a wider circle (r ∝ v). Mass behaves the same way; charge and field do the opposite, pulling the radius down.
Quick recap: the magnetic force on a moving charge is always perpendicular to its velocity, so it acts as a centripetal force and the charge circles at constant speed. Setting BQv = mv2 ÷ r gives r = mv ÷ BQ — wider for fast, heavy particles; tighter in strong fields or for big charges.
🧭 Radius-of-path problems
State the force balance: the magnetic force is the centripetal force, so BQv = mv2 ÷ r
Rearrange to r = mv ÷ BQ — do the algebra yourself; examiners often ask for the derivation
List your quantities in SI units (mass in kg, speed in m s−1, charge in C), converting any mT or mm first
For a “compare two particles” question, use the proportionalities — r ∝ m ÷ Q at fixed speed and field — rather than crunching both from scratch
Sanity check: a stronger field or a bigger charge should always give you a smaller radius
WE 1
A proton (mass 1.67 × 10⁻²⁷ kg, charge 1.60 × 10⁻¹⁹ C) enters a uniform magnetic field of flux density 0.48 T at right angles, moving at 2.4 × 10⁶ m s⁻¹. (a) Calculate the radius of its circular path. (b) State the new radius if the field is increased to 0.96 T at the same speed.
Part (a) — use r = mv ÷ BQ
m = 1.67 × 10⁻²⁷ kg, v = 2.4 × 10⁶ m s⁻¹
B = 0.48 T, Q = 1.60 × 10⁻¹⁹ C
r = (1.67 × 10⁻²⁷ × 2.4 × 10⁶) ÷ (0.48 × 1.60 × 10⁻¹⁹)r ≈ 0.052 m (5.2 cm)Part (b) — field doubled (0.48 → 0.96 T)
r ∝ 1 ÷ B, so doubling B halves r
r ≈ 0.026 m (2.6 cm)
WE 2
A proton and an alpha particle enter the same uniform magnetic field at right angles, moving at the same speed. The alpha particle has 4 times the mass of a proton and twice the charge. Determine how the radius of the alpha particle’s path compares with the proton’s.
Use the proportionality
At the same v and B: r ∝ m ÷ Q
Take the ratio (alpha ÷ proton)rα ÷ rp = (mα ÷ mp) ÷ (Qα ÷ Qp) = 4 ÷ 2the alpha’s radius is 2 × the proton’sMore mass widens the circle, more charge tightens it — here the 4× mass wins over the 2× charge, so the alpha loops twice as wide.
💡 Top tips
Know the derivation. Start from “magnetic force = centripetal force”, write BQv = mv2 ÷ r, cancel a v, and read off r = mv ÷ BQ
The magnetic force does no work. It’s perpendicular to the motion, so the speed stays constant — only the direction turns
Compare with proportions. For two-particle questions, r ∝ m ÷ Q (same field and speed) beats plugging numbers twice
It’s ordinary circular motion. The same F = mv2 ÷ r you met for orbits applies — the magnetic force is just what supplies it
⚠ Common mistakes
Equating the centripetal force to BIL instead of BQv — this page is about a single charge, so it’s Bqv
Getting the proportionalities backwards — a stronger field gives a smaller circle, not a bigger one
Claiming the magnetic force speeds the particle up; it only changes direction, so the speed is constant
Slipping in the algebra — remember one factor of v cancels, leaving r = mv ÷ BQ, not mv2 ÷ BQ
Up next: we swap the magnetic field for an electric one. A charge fired between two charged parallel plates doesn’t loop in a circle — it curves in a parabola, just like a ball thrown sideways under gravity. Next page maps that path out.
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