A wire of length 0.20\ m carrying 3.0\ A sits at right angles to a 0.40\ T field. Find the force. [2 marks]

OCR OCR 2018-style practice: A straight wire of length 0.25\ m carries a current of 4.0\ A at right angles to a uniform magnetic field of flux density 0.30\ T. Calculate the force on the wire and state how the force changes if the wire is rotated to lie along the field.

Force on a current-carrying conductor: F = BILsinθ, with θ = 90^ so sinθ = 1. F = (0.30)(4.0)(0.25) = 0.30\ N. When the wire lies along the field, θ = 0 and sinθ = 0, so the force is zero. Markers reward F = BIL giving 0.30\ N, and recognising that the force falls to zero when the current is parallel to the field.

OCR OCR 2021-style practice: An electron travels at 2.0 × 10^7\ m s^-1 at right angles to a magnetic field of flux density 0.050\ T. Calculate the force on the electron and the radius of its circular path. Take e = 1.6 × 10^-19\ C and m = 9.11 × 10^-31\ kg.

Force on a moving charge: F = BQv = (0.050)(1.6 × 10^-19)(2.0 × 10^7) = 1.6 × 10^-13\ N. For circular motion this force is centripetal: BQv = (mv^2)/(r), so r = (mv)/(BQ). r = (9.11 × 10^-31)(2.0 × 10^7)(0.050)(1.6 × 10^-19) = 1.82 × 10^-238.0 × 10^-21 = 2.3 × 10^-3\ m. Markers reward F = BQv about 1.6 × 10^-13\ N, equating it to (mv^2)/(r), and the radius about 2.3\ mm.

EnglandPhysicsSyllabus dot point

What force does a magnetic field exert on a current and on a moving charge?

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

A focused answer to the OCR H556 content on magnetic fields and the motor effect, covering magnetic flux density and its definition, the force on a current-carrying conductor with Fleming's left-hand rule, the force on a moving charge, and the circular motion of charged particles in a uniform magnetic field.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

OCR wants you to define magnetic flux density, calculate the force on a current-carrying conductor in a magnetic field and find its direction with Fleming's left-hand rule, calculate the force on a moving charge, and analyse the circular motion of charged particles in a uniform magnetic field.

The answer

Magnetic flux density and the motor effect

Force on a current-carrying conductor

Force on a moving charge

Circular motion of charged particles

Force per unit length on a power cable

A horizontal cable carries a current of $150\ \text{A}$ from east to west through the Earth's magnetic field, which has a horizontal flux density of $1.8 \times 10^{-5}\ \text{T}$ pointing north. Find the force per metre on the cable.

step 1: Identify the angle

The current (east-west) is perpendicular to the field (north), so $\theta = 90^{\circ}$ and $\sin\theta = 1$ .

step 2: Force per unit length

$\frac{F}{L} = BI = (1.8 \times 10^{-5})(150) = 2.7 \times 10^{-3}\ \text{N m}^{-1}$ .

step 3: Interpret

The force is small but acts vertically (by Fleming's left-hand rule), showing why everyday currents in the Earth's weak field produce only tiny forces.

Examples in context

The motor effect drives every electric motor and loudspeaker, where a current in a magnetic field produces a turning force or vibration. The circular motion of charges in a field is used in the mass spectrometer to separate ions by mass-to-charge ratio, in the cyclotron to accelerate particles, and in the bending magnets of particle accelerators. The same force confines hot plasma in fusion reactors and shapes the auroras as charged particles spiral along the Earth's field lines.

Try this

Q1. State the equation for the force on a current-carrying conductor and define each term. [2 marks]

Cue. $F = BIL\sin\theta$ : flux density, current, length and the angle between current and field.

Q2. A wire of length $0.20\ \text{m}$ carrying $3.0\ \text{A}$ sits at right angles to a $0.40\ \text{T}$ field. Find the force. [2 marks]

Cue. $F = BIL = (0.40)(3.0)(0.20) = 0.24\ \text{N}$ .

Q3. Explain why a magnetic field does no work on a moving charge. [2 marks]

Cue. The force is always perpendicular to the velocity, so it changes only the direction of motion, not the speed, and does no work.

Exam-style practice questions

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

OCR 20183 marksA straight wire of length

0.25\ \text{m}

carries a current of

4.0\ \text{A}

at right angles to a uniform magnetic field of flux density

0.30\ \text{T}

. Calculate the force on the wire and state how the force changes if the wire is rotated to lie along the field.

Show worked answer →

Force on a current-carrying conductor: $F = BIL\sin\theta$ , with $\theta = 90^{\circ}$ so $\sin\theta = 1$ .

$F = (0.30)(4.0)(0.25) = 0.30\ \text{N}$ .

When the wire lies along the field, $\theta = 0$ and $\sin\theta = 0$ , so the force is zero.

Markers reward $F = BIL$ giving $0.30\ \text{N}$ , and recognising that the force falls to zero when the current is parallel to the field.

OCR 20214 marksAn electron travels at

2.0 \times 10^{7}\ \text{m s}^{-1}

at right angles to a magnetic field of flux density

0.050\ \text{T}

. Calculate the force on the electron and the radius of its circular path. Take

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

and

m = 9.11 \times 10^{-31}\ \text{kg}

Show worked answer →

Force on a moving charge: $F = BQv = (0.050)(1.6 \times 10^{-19})(2.0 \times 10^{7}) = 1.6 \times 10^{-13}\ \text{N}$ .

For circular motion this force is centripetal: $BQv = \frac{mv^2}{r}$ , so $r = \frac{mv}{BQ}$ .

$r = \frac{(9.11 \times 10^{-31})(2.0 \times 10^{7})}{(0.050)(1.6 \times 10^{-19})} = \frac{1.82 \times 10^{-23}}{8.0 \times 10^{-21}} = 2.3 \times 10^{-3}\ \text{m}$ .

Markers reward $F = BQv$ about $1.6 \times 10^{-13}\ \text{N}$ , equating it to $\frac{mv^2}{r}$ , and the radius about $2.3\ \text{mm}$ .

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)