IB Physics HLCharges Moving in FieldsPaper 1 & 2r = mv/Bq~16 min read
Charges in Magnetic Fields
A force of constant size, always at right angles to the velocity, never changing the speed. You have met that before. It is a centripetal force, and it means a charged particle fired into a uniform magnetic field does not fly off in some new direction — it goes round in a circle, forever. Set the magnetic force equal to the centripetal force, cancel one v, and the radius of that circle drops straight out.
📘 What you need to know
A charged particle entering a uniform magnetic field at right angles travels in a circular path
Because F is always perpendicular to v and always directed towards the centre of the path
The magnetic force provides the centripetal force: Bqv = mv2/r
Cancelling one v and rearranging gives r = mv / Bq
r ∝ v — faster particles sweep larger circles
r ∝ m — heavier particles sweep larger circles
r ∝ 1/q and r ∝ 1/B — more charge, or a stronger field, means tighter circles
The magnetic field does no work: the speed and the kinetic energy are constant
The centripetal acceleration is towards the centre, and F = ma still applies
Positive and negative charges circle in opposite senses
Why a circle?
Three facts, and the conclusion is forced.
The magnitude of the force is constant, since B, q and v never change
The force is always perpendicular to the velocity, so it can never speed the particle up or slow it down
A constant force, always at right angles to the motion, is the definition of a centripetal force
Every red arrow is exactly 90° from its green one, at every point. That is why the kinetic energy is untouchable — a force can only do work along the direction you move.
Deriving the radius
The magnetic force is the centripetal force. Write both down and set them equal.
Step 1 — magnetic force provides the centripetal forceBqv = mv2 / r
Step 2 — one v cancels from each sideBq = mv / r
Step 3 — rearrange for the radiusr = mv / Bqr = radius (m) • m = mass (kg) • v = speed (m s−1) • B = flux density (T) • q = charge (C)
Magnetic force Bqv
provides the
Centripetal force mv2/r
cancel one v
r = mv/Bq
Examiners ask you to derive this, not just quote it. Three lines: write Bqv = mv2/r, say “the magnetic force provides the centripetal force”, cancel a v, rearrange. That is the whole thing, and it is worth three marks that most students throw away by starting from the answer.
What the equation tells you
Triple the speed and you triple the radius. Flip the sign of the charge and the particle curls the other way — which is exactly how a mass spectrometer sorts ions.
Increase…
Effect on the radius
Why
Speed v
Bigger circle, r ∝ v
More momentum to turn
Mass m
Bigger circle, r ∝ m
More inertia to turn
Charge q
Smaller circle, r ∝ 1/q
A bigger force does the turning
Flux density B
Smaller circle, r ∝ 1/B
A bigger force does the turning
Two of the four proportionalities are direct (v and m) and two are inverse (q and B). The equation r = mv/Bq tells you which is which at a glance.
Here is a free extra, one line from what you already have. The period of the circular motion is T = 2πr/v. Substitute r = mv/Bq and the v cancels, leaving T = 2πm/Bq. The time to go round does not depend on the speed. A fast particle sweeps a bigger circle but covers it in exactly the same time. That single fact is what makes a cyclotron possible.
🌀 Working a circular-motion question
State the physics first: “the magnetic force provides the centripetal force”, then Bqv = mv2/r.
Cancel one v and rearrange. Do not quote r = mv/Bq from memory if asked to derive.
Convert. mT → T. Use the magnitude of q.
Given e/me instead of the mass? Then r = v / [(e/me)B].
Comparing two particles?r ∝ mv/Bq. Cancel whatever they share.
Speed or energy? They never change. The field does no work.
WE 1
A proton enters a uniform magnetic field of flux density 0.35 T at right angles, with a speed of 2.4 × 106 m s−1. (a) Calculate the radius of its circular path. (b) Calculate its centripetal acceleration. (c) State what happens to the radius and to the speed if the proton entered twice as fast. (mp = 1.67 × 10−27 kg, q = 1.60 × 10−19 C)
(a) Step 1 — the magnetic force provides the centripetal forceBqv = mv²/r → r = mv / Bqr = (1.67 × 10⁻²⁷)(2.4 × 10⁶) / [(0.35)(1.60 × 10⁻¹⁹)]r = (4.008 × 10⁻²¹) / (5.6 × 10⁻²⁰)r = 7.2 × 10⁻² m = 7.2 cm(b) Step 2 — centripetal accelerationa = v² / r = (2.4 × 10⁶)² / 0.0716a = 8.0 × 10¹³ m s⁻²(c) Step 3 — double the speedr ∝ v, so the radius doublesr = 14 cm, and the speed stays 4.8 × 10⁶ m s⁻¹ foreverCheck (b) another way: a = F/m = Bqv/m = (0.35)(1.60 × 10⁻¹⁹)(2.4 × 10⁶) / (1.67 × 10⁻²⁷) = 8.0 × 10¹³ m s⁻². Same answer. That is a colossal acceleration — yet the speed never budges, because it is all sideways.
WE 2
An alpha particle and a proton enter the same uniform magnetic field at right angles, with the same speed. (a) Derive an expression for the radius of a charged particle’s path. (b) Determine the ratio of the alpha particle’s radius to the proton’s. (c) State, with a reason, whether the two particles curve the same way. (Alpha particle: mass 4u, charge +2e)
(a) the derivation
The magnetic force provides the centripetal force:
Bqv = mv² / r
One v cancels from each side:
Bq = mv / rr = mv / Bq(b) B and v are the same for both, so r ∝ m/qalpha: m/q = 4u / 2e = 2 (u/e)proton: m/q = 1u / 1e = 1 (u/e)ralpha / rproton = 2(c) which way do they curve?
Both particles are positively charged.
yes — they curve the same way, the alpha simply on a circle twice as wideThe alpha has four times the mass but only twice the charge, so the mass wins by a factor of two. Double the charge would have halved the radius; four times the mass doubles it twice over.
WE 3
An electron travels at right angles to a uniform magnetic field of flux density 8.0 mT with a speed of 5.0 × 106 m s−1. (a) Calculate the radius of its path, using the charge-to-mass ratio e/me = 1.76 × 1011 C kg−1. (b) A proton enters the same field at the same speed. Determine how many times larger its radius is. (mp/me = 1840)
(a) Step 1 — rewrite r in terms of e/mr = mev / eB = v / [(e/me) B]Step 2 — convert the field, then substituteB = 8.0 mT = 8.0 × 10⁻³ Tr = (5.0 × 10⁶) / [(1.76 × 10¹¹)(8.0 × 10⁻³)] = (5.0 × 10⁶) / (1.408 × 10⁹)r = 3.6 × 10⁻³ m = 3.6 mm(b) Step 3 — same B, same v, same magnitude of charger ∝ m, so rp / re = mp / me1840 times larger, giving about 6.5 mThree millimetres against six and a half metres. That enormous gap is exactly why an electron is easy to bend round a small tube in a laboratory, while bending protons needs a machine the size of a town.
💡 Top tips
Always say the sentence: “the magnetic force provides the centripetal force”. It is a marked statement.
Derive, don’t quote: Bqv = mv2/r, cancel a v, rearrange.
The speed and the kinetic energy never change. The field does no work.
Direct: r ∝ v and r ∝ m. Inverse: r ∝ 1/q and r ∝ 1/B.
Given e/me? Use r = v / [(e/me)B].
Comparing particles? Cancel what they share and compare m/q.
Positive and negative charges circle opposite ways in the same field.
⚠ Common mistakes
Forgetting to cancel a v, and ending up with r = mv2/Bq
Saying the magnetic field accelerates the particle in the sense of speeding it up. It only turns it
Claiming the kinetic energy changes. The force is perpendicular, so no work is done
Getting the proportionalities inverted — a stronger field gives a tighter circle
Leaving B in mT, or using a negative q in the arithmetic
Quoting r = mv/Bq when the question says “derive“. That’s zero marks for the derivation
Assuming a heavier particle curves more. More inertia means it curves less
Quick recap: A charge entering a uniform magnetic field at right angles moves in a circle, because the magnetic force is constant in size and always perpendicular to the velocity. The magnetic force provides the centripetal force: Bqv = mv2/r, and cancelling one v gives r = mv/Bq. So r ∝ v and r ∝ m, while r ∝ 1/q and r ∝ 1/B. The field does no work, so the speed and kinetic energy are constant, and positive and negative charges circle in opposite senses.
Now switch the magnetic field off and switch an electric field on. Everything changes. The electric force is parallel to the field, not perpendicular to the velocity, so it does do work — the particle speeds up, its kinetic energy rises, and it no longer curls into a circle. It behaves exactly like a ball thrown horizontally in gravity: constant velocity one way, constant acceleration the other. The path is a parabola. Next page: Charged Particles in Electric Fields.
Can’t get the derivation to come out?
Book a free meeting and we’ll drill Bqv = mv2/r, the four proportionalities and the “no work done” argument.
