A wire of length 0.30 m carries 4.0 A at right angles to a 0.50 T field. Calculate the force. [2 marks]

AQA AQA 2019-style practice: A proton (m = 1.67 × 10^-27 kg, q = 1.60 × 10^-19 C) moves at 2.0 × 10^6 m s^-1 perpendicular to a uniform magnetic field of flux density 0.15 T. Calculate the radius of its circular path.

The magnetic force provides the centripetal force, so Bqv = mv^2r, which rearranges to r = mvBq. r = (1.67 × 10^-27)(2.0 × 10^6)(0.15)(1.60 × 10^-19). The denominator is 2.4 × 10^-20, so r = 3.34 × 10^-212.4 × 10^-20 = 0.14 m. Markers reward equating the magnetic and centripetal forces, the correct rearrangement, and substitution to give the radius.

EnglandPhysicsSyllabus dot point

What force does a magnetic field exert on a current or a moving charge, and why does it cause circular motion?

Magnetic flux density, the force on a current-carrying conductor, the force on a moving charge, Fleming's left hand rule, and the circular motion of charged particles.

A focused answer to AQA A-Level Physics 3.7.5.1, covering magnetic flux density, the force on a current-carrying conductor, Fleming's left hand rule, the force on a moving charge, and the resulting circular motion of charged particles.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Magnetic flux density
Force on a current-carrying conductor
Force on a moving charge
Circular motion of charged particles
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What this dot point is asking

AQA specification point 3.7.5.1 wants you to define magnetic flux density, calculate the force on a current-carrying conductor and on a moving charge, use Fleming's left hand rule, and explain the circular motion of charged particles in a magnetic field.

Magnetic flux density

Flux density is a measure of the strength of the field, distinct from the total flux $\Phi = BA$ . One tesla is a strong field; the Earth's surface field is only about $5 \times 10^{-5} \text{ T}$ , while an MRI scanner uses a few tesla.

Force on a current-carrying conductor

This is the motor effect: the interaction between the field of the magnet and the field of the current produces a force. If the conductor makes an angle $\theta$ with the field, only the perpendicular component contributes, so the force becomes $F = BIl\sin\theta$ . The direction of the force is given by Fleming's left hand rule: with the left hand, the thumb points along the force (motion), the first finger along the field (from north to south), and the second finger along the conventional current.

Force on a moving charge

A single charge moving through the field is effectively a tiny current, so it experiences a force:

This follows from $F = BIl$ because a charge $q$ crossing a length $l$ in time $t$ constitutes a current $I = \dfrac{q}{t}$ at speed $v = \dfrac{l}{t}$ , giving $BIl = Bq\dfrac{l}{t} = Bqv$ .

Circular motion of charged particles

This principle underlies mass spectrometers (heavier ions follow larger circles) and cyclotrons and synchrotrons (which use magnetic fields to bend charged particles around a ring).

Finding the radius of an electron's circular path

An electron ( $m = 9.11 \times 10^{-31} \text{ kg}$ , $q = 1.60 \times 10^{-19} \text{ C}$ ) moves at $3.0 \times 10^7 \text{ m s}^{-1}$ perpendicular to a $0.020 \text{ T}$ field. Find the radius of its path.

step 1: Equate magnetic and centripetal forces

$Bqv = \dfrac{mv^2}{r}$ , which rearranges to $r = \dfrac{mv}{Bq}$ .

step 2: Substitute the values

$r = \dfrac{(9.11 \times 10^{-31})(3.0 \times 10^7)}{(0.020)(1.60 \times 10^{-19})}$ .

step 3: Evaluate the denominator

$(0.020)(1.60 \times 10^{-19}) = 3.2 \times 10^{-21}$ .

step 4: Compute the radius

$r = \dfrac{2.73 \times 10^{-23}}{3.2 \times 10^{-21}} = 8.5 \times 10^{-3} \text{ m}$ , about $8.5 \text{ mm}$ .

Try this

Q1. State the unit of magnetic flux density and define it. [2 marks]

Cue. The tesla; the flux density giving a force of one newton per metre on a conductor carrying one amp perpendicular to the field.

Q2. A wire of length $0.30 \text{ m}$ carries $4.0 \text{ A}$ at right angles to a $0.50 \text{ T}$ field. Calculate the force. [2 marks]

Cue. $F = BIl = 0.50 \times 4.0 \times 0.30 = 0.60 \text{ N}$ .

Q3. State why the magnetic force does no work on a moving charge. [1 mark]

Cue. It is always perpendicular to the velocity.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksA proton (

m = 1.67 \times 10^{-27} \text{ kg}

q = 1.60 \times 10^{-19} \text{ C}

) moves at

2.0 \times 10^6 \text{ m s}^{-1}

perpendicular to a uniform magnetic field of flux density

0.15 \text{ T}

. Calculate the radius of its circular path.

Show worked answer →

The magnetic force provides the centripetal force, so $Bqv = \dfrac{mv^2}{r}$ , which rearranges to $r = \dfrac{mv}{Bq}$ .

$r = \dfrac{(1.67 \times 10^{-27})(2.0 \times 10^6)}{(0.15)(1.60 \times 10^{-19})}$ .

The denominator is $2.4 \times 10^{-20}$ , so $r = \dfrac{3.34 \times 10^{-21}}{2.4 \times 10^{-20}} = 0.14 \text{ m}$ .

Markers reward equating the magnetic and centripetal forces, the correct rearrangement, and substitution to give the radius.

AQA 20223 marksExplain why a charged particle moving at constant speed perpendicular to a uniform magnetic field follows a circular path, and state why its speed does not change.

Show worked answer →

The magnetic force on the moving charge, $F = Bqv$ , is always perpendicular to the velocity (Fleming's left hand rule), so it constantly changes the direction of motion but not the speed.

A constant-magnitude force always directed perpendicular to the velocity is exactly a centripetal force, so the particle moves in a circle of constant radius.

Because the force is perpendicular to the displacement at every instant, it does no work, so the kinetic energy and therefore the speed remain constant.

Markers reward the perpendicular force, identifying it as centripetal, and explaining no work is done so the speed is unchanged.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)