A charge of 3.2 × 10^-19\ C moves at 5.0 × 10^6\ m s^-1 perpendicular to a field of 0.40\ T. Find the force on it. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A straight wire of length 0.080\ m carries a current of 4.5\ A at right angles to a uniform magnetic field of flux density 0.25\ T. Calculate the force on the wire.

The wire is perpendicular to the field, so the full force law applies: F = BIL. F = 0.25 × 4.5 × 0.080 = 0.090\ N. Markers reward F = BIL for a wire perpendicular to the field, and the force 0.090\ N.

WJEC Eduqas Eduqas 2021-style practice: A proton enters a uniform magnetic field of flux density 0.30\ T at right angles, moving at 2.0 × 10^6\ m s^-1. Calculate the radius of its circular path. Take the proton mass m = 1.67 × 10^-27\ kg and charge e = 1.6 × 10^-19\ C.

The magnetic force provides the centripetal force: Bqv = mv^2r, so r = mvBq. r = (1.67 × 10^-27)(2.0 × 10^6)(0.30)(1.6 × 10^-19) = 3.34 × 10^-214.8 × 10^-20 = 0.070\ m. Markers reward equating Bqv to the centripetal force giving r = (mv)/(Bq), correct substitution, and the radius about 0.070\ m (7.0 cm).

EnglandPhysicsSyllabus dot point

What force does a magnetic field exert on currents and moving charges, and how is flux density defined?

Magnetic fields: magnetic flux density, the force on a current-carrying conductor F = BIL, the force on a moving charge F = Bqv, and the circular motion of charged particles in a field.

A focused answer to the Eduqas A-Level Physics Component 3 magnetic fields content, covering magnetic flux density, the force on a current-carrying conductor F = BIL, the force on a moving charge F = Bqv, and the circular motion of charged particles moving through a magnetic field.

Generated by Claude Opus 4.812 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
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What this dot point is asking

Eduqas wants you to define magnetic flux density, use the force on a current-carrying conductor $F = BIL$ , use the force on a moving charge $F = Bqv$ , apply Fleming's left-hand rule, and analyse the circular motion of a charged particle moving through a magnetic field.

The answer

Magnetic flux density and the force on a conductor

Fleming's left-hand rule

The force on a moving charge

Circular motion of charged particles

Speed of an electron from its circular radius

An electron moves in a circle of radius $4.0\ \text{cm}$ in a magnetic field of flux density $1.5 \times 10^{-3}\ \text{T}$ . Find its speed. Take $m_e = 9.11 \times 10^{-31}\ \text{kg}$ and $e = 1.6 \times 10^{-19}\ \text{C}$ .

step 1: Rearrange the circular-motion result

From $r = \dfrac{mv}{Bq}$ , the speed is $v = \dfrac{Bqr}{m}$ .

step 2: Substitute

$v = \dfrac{(1.5 \times 10^{-3})(1.6 \times 10^{-19})(0.040)}{9.11 \times 10^{-31}} = \dfrac{9.6 \times 10^{-24}}{9.11 \times 10^{-31}}$ .

step 3: Evaluate

$v = 1.05 \times 10^{7}\ \text{m s}^{-1}$ , about $1.1 \times 10^{7}\ \text{m s}^{-1}$ .

Examples in context

The motor effect drives electric motors, loudspeakers and the read heads of older hard drives. The force on a moving charge steers electron beams in cathode-ray tubes and bends particles in mass spectrometers (to separate isotopes by mass) and in accelerators such as the cyclotron and synchrotron. The Earth's magnetic field deflects charged cosmic rays and solar-wind particles, channelling them towards the poles to create the auroras.

Try this

Q1. State the equation for the force on a current-carrying conductor perpendicular to a magnetic field. [1 mark]

Cue. $F = BIL$ .

Q2. A charge of $3.2 \times 10^{-19}\ \text{C}$ moves at $5.0 \times 10^{6}\ \text{m s}^{-1}$ perpendicular to a field of $0.40\ \text{T}$ . Find the force on it. [2 marks]

Cue. $F = Bqv = 0.40 \times 3.2 \times 10^{-19} \times 5.0 \times 10^{6} = 6.4 \times 10^{-13}\ \text{N}$ .

Q3. Explain why the magnetic force makes a charged particle move in a circle. [2 marks]

Cue. The force is always perpendicular to the velocity, so it provides a centripetal force, changing direction but not speed.

Exam-style practice questions

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

Eduqas 20193 marksA straight wire of length

0.080\ \text{m}

carries a current of

4.5\ \text{A}

at right angles to a uniform magnetic field of flux density

0.25\ \text{T}

. Calculate the force on the wire.

Show worked answer →

The wire is perpendicular to the field, so the full force law applies: $F = BIL$ .

$F = 0.25 \times 4.5 \times 0.080 = 0.090\ \text{N}$ .

Markers reward $F = BIL$ for a wire perpendicular to the field, and the force $0.090\ \text{N}$ .

Eduqas 20215 marksA proton enters a uniform magnetic field of flux density

0.30\ \text{T}

at right angles, moving at

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

. Calculate the radius of its circular path. Take the proton mass

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

and charge

e = 1.6 \times 10^{-19}\ \text{C}

Show worked answer →

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

$r = \dfrac{(1.67 \times 10^{-27})(2.0 \times 10^{6})}{(0.30)(1.6 \times 10^{-19})} = \dfrac{3.34 \times 10^{-21}}{4.8 \times 10^{-20}} = 0.070\ \text{m}$ .

Markers reward equating $Bqv$ to the centripetal force giving $r = \frac{mv}{Bq}$ , correct substitution, and the radius about $0.070\ \text{m}$ (7.0 cm).

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)