Physics › Magnetic fields › Force on a moving charge

Force on a moving charge

The same rule applies to a single moving charge as to a current-carrying wire: F = BQv sin θ, at right angles to both the velocity and the field. A force always perpendicular to the motion does no work, so it changes direction without changing speed, which sends the charge in a circle. That is the principle of the cyclotron.

Year 13AQA 3.7.5.2 CIE 20.3.1, 20.3.2, 20.3.3, 20.3.4, 20.3.5, 20.3.6 OCR A 6.3.2(a), 6.3.2(b), 6.3.2(c)EDEXCEL 122, 133, 134

Builds on Magnetic flux density and the force on a wire and Circular motion.

IN THIS TOPIC

Use F = BQv for a charge moving perpendicular to the field, with the correct direction for either sign.
Explain why a magnetic force can do no work on a moving charge.
Explain why the path is a circle, and derive r = mv/BQ.
Describe the cyclotron, with magnetic steering set against electric acceleration.

COMMON MISCONCEPTION

Magnetic fields can speed particles up.

One charge, same rule

A current is charge in motion, so the wire's force law has a single-particle version. In general, a particle of charge magnitude Q moving at speed v at an angle θ to a field of flux density B feels a force of magnitude BQv sin θ, at right angles to both the velocity and the field. Every board starts from the perpendicular case, θ = 90°, where a charge moving at right angles to the field feels

F = BQv

ON THE AQA DATA SHEET

FIG. 1Same field, same velocity: the force on a negative charge is exactly opposite to the force on a positive one.

with the direction from Fleming's left hand once more, remembering that the seCond finger follows conventional current. A negative charge moving right counts as conventional current moving left, so electrons deflect exactly opposite to protons in the same field. A stationary charge, with v = 0, feels nothing at all.

The general form above is the wire's in-between case with a single-particle twin, and CIE and Edexcel set it as an equation, with θ the angle between the velocity and the field,

F = BQv \sin\theta

and Edexcel prints it while CIE does not, so it is a recall item on 9702 alone. It says nothing new. B sin θ is the component of the field across the velocity, so θ = 90° gives sin θ = 1 and hands back F = BQv, while θ = 0 gives zero for a charge flying straight along the field. Everything between the two scales by the sine. Take 2.0 nC crossing a 0.40 T field at 5.0 × 10⁴ m s⁻¹. Square on it feels 0.40 × 2.0 × 10⁻⁹ × 5.0 × 10⁴ = 4.0 × 10⁻⁵ N, and at 30° to the field exactly half of that, 2.0 × 10⁻⁵ N. The other two boards examine the perpendicular case alone, and it is the bare F = BQv that their data sheets print.

Circles, because no work is done

FIG. 2A charge circling in a uniform field. The magnetic force, always at right angles to the motion, can swing the cyan velocity arrow round but never stretch it: direction changes, speed does not. The coral arrow is the resulting acceleration, pointing at the centre at every instant, with F = BQv filling the centripetal role that gravity fills for a satellite.

Because the force stays perpendicular to the velocity, it can never do work on the charge. Speed never changes, only direction, and that disposes of the opening misconception outright. A constant-magnitude force held forever at right angles to the motion is precisely the condition for circular motion, with F = BQv in the centripetal seat. Set the two faces equal, BQv = mv²/r, and the radius follows,

r = \frac{mv}{BQ}

NOT ON THE AQA DATA SHEET: LEARN IT

so faster or heavier particles sweep wider circles, while stronger fields and bigger charges bend tighter. This one relation is how bubble-chamber photographs are read. Curvature of a track hands over the particle's momentum, and direction of curl hands over its sign.

WORKED EXAMPLE

A proton on a curve

A proton (m = 1.67 × 10⁻²⁷ kg) moves at 4.0 × 10⁶ m s⁻¹ at right angles to a 0.30 T field. Find the force on it and the radius of its circle.

F = BQv = 0.30 × 1.60 × 10⁻¹⁹ × 4.0 × 10⁶ = 1.9 × 10⁻¹³ N, at right angles to the motion.

r = mv/BQ = (1.67 × 10⁻²⁷ × 4.0 × 10⁶)/(0.30 × 1.60 × 10⁻¹⁹) = 0.14 m.

The speed never changes, only the direction. The force does no work at all, and that fourteen-centimetre circle is momentum written as curvature.

Crossed fields, a CIE extension

Neither the Hall effect nor the velocity selector appears anywhere in AQA 7408. CIE asks for both, so read this section only if that is your board.

Put a magnetic force and an electric force at right angles across the same moving charge and each one can measure the other. In a current-carrying slice of semiconductor sitting in a field, the magnetic force pushes the drifting charges to one face until the charge piling up there supplies an electric force that balances it, so eE = Bev. The steady pd across the faces is the Hall voltage,

V_{H} = \frac{BI}{ntq}

with n the charge-carrier density and t the slice's thickness in the field direction. Since V_H is proportional to B, a calibrated slice becomes a magnetometer, the Hall probe, and that is the standard instrument for measuring flux density.

The same balance with its geometry rearranged is the velocity selector. A beam crossing perpendicular E and B fields passes undeflected only where the electric and magnetic forces cancel, at v = E/B, and every other speed bends away, so the pair of fields skims one velocity out of a mixed beam.

Accelerators, the cyclotron and the linac

FIG. 3Each time the particle crosses the gap the alternating voltage kicks it, so every half-circle is wider and faster than the last. But watch the gap flashes: perfectly even. Radius grows in step with speed, so at non-relativistic speeds the half-turn time never changes, and that is the machine's trick: one fixed frequency can keep kicking a particle that is forever speeding up.

The cyclotron is the named application here. Two hollow D-shaped electrodes sit in a uniform magnetic field, with an alternating pd across the gap between them. Inside each dee the magnetic field steers the particle round a half-circle. At every gap crossing the electric field does the accelerating, timed to push whichever way the particle happens to be going. Each crossing raises v, and by r = mv/BQ each half-circle comes out wider than the last, so the path is an outward spiral and the particle leaves at the rim carrying the energy of many small kicks. That division of labour is the design. Magnetic fields steer for free, and electric fields do the work.

GUIDED PRACTICE

The cyclotron's fixed drumbeat

Show that the proton above (0.30 T) completes each half-circle in a time that does not depend on its speed, and find its orbital frequency.

Show the working

The full turn takes T = 2πr/v, and r = mv/BQ, so T = 2πm/BQ and each half-circle takes πm/BQ, with v cancelled clean out. Faster protons ride bigger circles in exactly the same time.

f = BQ/2πm = (0.30 × 1.60 × 10⁻¹⁹)/(2π × 1.67 × 10⁻²⁷) = 4.6 MHz.

That cancellation holds under one condition, and it is why bigger machines stop looking like cyclotrons. Treating T = 2πm/BQ as constant is the non-relativistic case, speeds well below c. Push a particle close to light speed and its momentum grows as p = γmv while the mass itself stays the invariant rest mass, so the turns take longer and a fixed drive frequency slips out of step. (AQA's Turning points option keeps the older relativistic-mass wording, m = γm₀, for the same physics.) Real high-energy machines answer that by sweeping the frequency or ramping the field as the beam gains energy.

The linac takes the other route, with no magnetic field doing the steering, only a line of tube electrodes with an alternating pd between neighbours. The particle is accelerated in each gap and coasts inside each tube while the polarity flips, so successive tubes must be longer as it speeds up if the flip timing is to stay in step. Edexcel asks for both machines. The cyclotron spirals inside a fixed field, and the linac runs straight and simply grows.

Working linacs are not magnet-free, mind, and the exam answer should not say so. They carry magnets to focus the beam and to nudge it back on axis, and a medical linac bends its beam onto the patient with one. The difference from the cyclotron is what the magnets are for. In a linac no field curves the path into the circle the acceleration depends on.

ASSESSMENT FOCUS

F = BQv needs the perpendicular condition stated, and it gives zero for a charge moving along the field or standing still.
CIE and Edexcel both set the general F = BQv sin θ, with θ measured between the velocity and the field. Edexcel prints it on its sheet; CIE makes you recall it. AQA and OCR examine only the perpendicular form, which is on the sheet, so do not spend memory on the angle unless the angle is your board's.
Handle the sign first. Convert the particle's motion into conventional current before the left hand comes out, because electrons curl opposite to protons.
The no-work argument earns its marks almost verbatim. The force is perpendicular to the velocity, so no work is done and the speed stays constant, while only the direction changes.
Derive r = mv/BQ by equating BQv to mv²/r. It is quick, it is asked often, and it is not printed with the magnetism equations. The Turning points option page carries an electron-only r = mv/Be, but do not bank on spotting that under pressure, so the derivation doubles as your recall.
For the cyclotron, split the two jobs cleanly. The magnetic field does the circular steering, and the alternating electric field between the dees supplies the energy. Mixing them up loses every explanation mark on offer.
T = 2πm/BQ is speed-independent only while the speed is well below c, so write the non-relativistic condition into the “show that” answer. The linac's contrast is that no magnetic field bends its beam, not that it carries no magnets.
Watch the trap in a question with two fields at once. Only the magnetic part is guaranteed to do no work, so a charge crossing an electric field as well can and does change speed.

CHECK YOURSELF

An electron moves at 2.0 × 10⁷ m s⁻¹ at right angles to a field of 0.50 mT. Find the force on it and the radius of its circular path.

Show a hint

BQv for the force; then let it be the centripetal force.

Show the answer

$F = BQv$ = 5.0 × 10⁻⁴ × 1.60 × 10⁻¹⁹ × 2.0 × 10⁷ = 1.6 × 10⁻¹⁵ N.

$r = mv/BQ$ = (9.11 × 10⁻³¹ × 2.0 × 10⁷) / (5.0 × 10⁻⁴ × 1.60 × 10⁻¹⁹) = 0.23 m.

The speed stays 2.0 × 10⁷ m s⁻¹ all the way round. The field steered the electron without handing it a single joule.

F = BQv steers and never works, so the speed holds and the path curls.

Radius mv/BQ is momentum written as curvature.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the force on a moving charge questions page.

7 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Trigonometry and resolving vectors is the maths this lesson leans on, worked through from GCSE.

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Use F = BQv for a charge moving perpendicular to the field, with the correct direction for either sign.
Explain why a magnetic force can do no work on a moving charge.
Explain why the path is a circle, and derive r = mv/BQ.
Describe the cyclotron, with magnetic steering set against electric acceleration.

Open the full revision checklist to track your progress across the whole unit.