Is the motor effect formula on the HSC data sheet?

Yes. The 2019 data sheet gives F = lI⊥B = lIB sin θ, the parallel-wire formula F/l = (μ0/2π)(I1I2/r) and the coil torque τ = nIA⊥B = nIAB sin θ, plus μ0 = 4π × 10^−7 N A^−2. The back-emf relation V − ε = IR is not printed; it is Ohm's law applied to the leftover voltage.

What is the angle θ in τ = nIAB sin θ?

The angle between the magnetic field and the normal to the coil. The torque is greatest (sin θ = 1) when the plane of the coil is parallel to the field, and zero when the plane is at right angles to it.

What is the difference between the motor effect and electromagnetic induction?

In the motor effect a current in a field produces a force (electrical energy to kinetic energy). In induction, motion or a changing field produces an emf (kinetic energy to electrical energy). A motor does both at once: the induced part is its back emf.

Why does a motor draw a big current when it starts?

At switch-on the coil is not turning, so there is no back emf. The whole supply voltage drives current through the coil's small resistance. As it speeds up, the back emf grows and the current falls.

Do parallel wires with currents in the same direction attract or repel?

They attract. Currents in opposite directions repel. In both cases the two forces are equal in size and opposite in direction.

A 0.25 m length of wire carries 3.0 A in a 0.40 T field. The wire makes an angle of 30° with the field. What is the size of the force on it?

0 N, because the wire is not at right angles to the field

Two long parallel wires carry 2.0 A and 6.0 A in opposite directions. Which statement about the forces on them is correct?

They repel, and the forces on the two wires are equal in size.

They attract, and the 6.0 A wire feels the larger force.

They repel, and the 6.0 A wire feels the larger force.

They attract, and the forces on the two wires are equal in size.

What does the split-ring commutator do in a simple DC motor?

It reverses the current in the coil every half turn, so the torque keeps the same direction.

It reverses the direction of the magnetic field every half turn.

It changes the DC supply into AC before it reaches the brushes.

It keeps the current in the coil flowing the same way at all times.

A DC motor running at full speed is suddenly loaded so that it slows down a lot. What happens to the current in its coil?

It increases, because the back emf decreases.

It decreases, because the coil is turning more slowly.

It stays the same, because the supply voltage has not changed.

It increases, because the back emf increases.

HSC Physics · Module 6 · Electromagnetism · 16 min read

The motor effect

A wire carrying a current in a magnetic field feels a force. That is the motor effect, and it is how every electric motor, loudspeaker and current balance works. In HSC Physics it covers the force on a wire, the force between two parallel wires, and the torque that turns the coil of a DC motor.

This guide covers all three in a few minutes: the data-sheet equations and when to use them, how a DC motor's parts work together, why a spinning motor draws less current (back emf), the mistakes that cost marks, a quick quiz and a full worked example.

What you need to know

Investigate the force F = lI_⊥B = lIB sin θ that a uniform magnetic field exerts on a current-carrying wire: when the force is greatest, how the directions of current, field and force are related, and when there is no force.
Investigate, with measurements, how two parallel wires carrying currents act on each other.
Analyse the force per unit length between parallel wires, F/l = (μ₀/2π)(I₁I₂/r), and link Newton's third law to the way the SI unit, the ampere, is defined.
Investigate a simple DC motor at work: what each part does, how it produces a torque τ = nIA_⊥B = nIAB sin θ, and the effects of back emf.
Analyse how AC induction motors and simple DC and AC generators work.

The equations (from the 2019 HSC data sheet unless marked):

F = lI_⊥B = lIB sin θForce on a length l of wire inside the field. θ is the angle between the wire (the current) and the field.

F/l = (μ₀/2π) × I₁I₂/rForce per metre on each of two parallel wires a distance r apart. μ₀ = 4π × 10⁻⁷ N A⁻² (data sheet).

B = μ₀I/2πrThe field a distance r from one long straight wire: the field the other wire sits in.

τ = nIA_⊥B = nIAB sin θTorque on a coil of n turns and area A. θ is the angle between the field and the normal to the coil.

τ = r_⊥F = rF sin θTorque from one force: use it when only part of a coil is in the field.

V − ε_back = IR not on the data sheetCurrent in a motor coil of resistance R: the supply voltage minus the back emf drives the current.

Dot points: PH-M6-05, PH-M6-06, PH-M6-07, PH-M6-13, PH-M6-14 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

The force on a current-carrying wire

Moving charges in a magnetic field feel a force, so a wire full of them does too. The size of the force is F = lIB sin θ, where l is the length of wire inside the field and θ is the angle between the wire and the field.

The force is greatest when the wire is at right angles to the field (sin 90° = 1) and zero when the wire runs parallel to the field. It always acts at right angles to both the wire and the field.

Find the direction with the right-hand palm rule: fingers point along the field (N to S), thumb along the conventional current, and the palm pushes in the direction of the force.

Field left to right, current into the page (⊗): the right-hand palm rule gives a force down the page. Reverse either the current or the field and the force reverses.

Two parallel wires and the ampere

Each wire sits in the magnetic field of the other, so two parallel currents push or pull on each other. Currents in the same direction attract; currents in opposite directions repel. Get this from the rules, not memory: the right-hand grip rule gives wire 1's field at wire 2, then the palm rule gives the force on wire 2.

The force per metre is F/l = (μ₀/2π)(I₁I₂/r). Both currents appear in it, so both wires feel the same size force even when one current is much bigger. That is Newton's third law: the forces are an action–reaction pair. If the wires move differently, it is because of their masses or how they are held, not the force.

The HSC course links this to the ampere: 1 A is the current that, flowing in two long parallel wires 1 m apart in a vacuum, gives a force of 2 × 10⁻⁷ N on each metre of each wire. (Since 2019 the SI defines the ampere through the charge on the electron instead, but the 2 × 10⁻⁷ N m⁻¹ result is unchanged.)

Currents I₁ and I₂ in the same direction: equal and opposite forces pull the wires together. Example: 5.0 A and 8.0 A, 2.0 cm apart, gives F/l = 2 × 10⁻⁷ × 5.0 × 8.0 ÷ 0.020 = 4.0 × 10⁻⁴ N m⁻¹ on each wire.

Torque on a coil

Put a rectangular coil in the field and the two sides that cross the field feel equal and opposite forces. The forces do not cancel as a turn: they act on opposite sides of the axle and make a torque, a turning effect. For n turns, τ = nIAB sin θ.

On the data sheet θ is the angle between the field and the normal to the coil (a line at right angles to its face). So the torque is greatest when the plane of the coil is parallel to the field (θ = 90°) and zero when the plane is at right angles to the field (θ = 0°). If a question gives the angle between the field and the plane of the coil, subtract it from 90° first.

The forces on the sides stay the same size as the coil turns (each side stays at right angles to the field). What changes is their lever arm, which is why the torque changes with angle.

Edge-on view. Equal and opposite forces F on the two sides turn the coil clockwise here. θ is measured from the field to the normal (green), not to the coil.

How a simple DC motor works

A simple DC motor has four main parts: the coil (armature) that turns, the magnets that give the field, the split-ring commutator that turns with the coil, and the brushes that press on the commutator and carry current in from the supply.

Without the commutator the coil would swing to the position where its plane is at right angles to the field and stop there, because past that point the torque would reverse. The split ring reverses the current in the coil every half turn, just as the coil passes that position, so the torque keeps turning it the same way. Its momentum carries it through the instant of zero torque.

To get more torque from the same motor, use more turns, a bigger current, a stronger field or a larger coil area: each is a factor in τ = nIAB sin θ.

Back emf: why a running motor draws less current

A spinning coil in a magnetic field is also a generator: the flux through it keeps changing, so an emf is induced in it. By Lenz's law this emf opposes the supply that is driving the motor, so it is called the back emf. It grows with the speed of the coil.

The current is set by the difference: I = (V − ε_back) ÷ R. At switch-on the coil is still, there is no back emf and the current is large. As the motor speeds up, the back emf rises and the current falls until the torque just balances friction and the load.

If the motor is overloaded or jammed, it slows, the back emf falls and the current rises. That is why a stalled motor overheats and can burn out. In energy terms, the back emf times the current is the electrical power turned into mechanical work: this is energy conservation applied to a motor.

Switch-on: ε_back = 0, so I = V ÷ R (largest current, largest torque).
Full speed: ε_back is close to V, so the current is small.
Loaded or jammed: slower, less back emf, more current, more heating.

AC induction motors and generators, briefly

An AC induction motor has no brushes or commutator. AC in the fixed outer coils (the stator) makes a magnetic field that rotates. The rotor, often a 'squirrel cage' of metal bars, sits in this changing field, so currents are induced in it, and by Lenz's law the rotor is dragged round after the field. It always turns a little slower than the field, otherwise the flux through it would stop changing and there would be no induced current.

Run a motor backwards and it is a generator: turn the coil and an emf is induced. Slip rings keep each end of the coil connected to the same terminal, giving AC; a split-ring commutator swaps the connections every half turn, giving a DC output that pulses but never reverses.

Common mistakes

Using the angle between the field and the plane of the coil as θ in τ = nIAB sin θ.θ is measured to the normal. A coil whose plane is 30° to the field has θ = 60°. Torque is greatest when the plane is parallel to the field.
Using the whole length of the wire (or the whole loop) as l in F = lIB sin θ.l is only the length of conductor that is inside the field. A wire passing through a 5 cm wide magnet has l = 0.05 m, however long it is.
Saying the wire with the bigger current feels the bigger force.Both wires feel the same size force (Newton's third law). F/l = μ₀I₁I₂/2πr has both currents in it, for both wires.
Saying the split-ring commutator reverses the supply or the magnetic field.It reverses the current in the coil every half turn, so the torque keeps acting in the same rotational direction.
Thinking a motor that slows down under load draws less current.Slower means less back emf, so more of the supply voltage drives current: I = (V − ε_back) ÷ R goes up.
Mixing up force and torque, or back emf and eddy currents.Torque is force × perpendicular distance from the axle. Back emf is the emf induced in the motor's own coil; eddy currents are induced in solid metal.

Worked example

A DC motor drives the winch on a model crane. Its coil has 200 turns, each 3.0 cm × 3.0 cm, and sits in a uniform magnetic field of 0.20 T. The motor runs from a 9.0 V supply and the coil's resistance is 1.5 Ω. To start lifting its load, the motor must produce a torque of at least 0.15 N m.

(a) The motor is switched on with the coil at rest and the plane of the coil at 40° to the magnetic field. Show whether the motor can start lifting the load from this position. (3 marks)

(b) Once the load is rising at a steady speed, the current in the coil is 1.8 A. Calculate the back emf in the coil, and explain why this current is smaller than the current at switch-on. (2 marks)

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. At switch-on the current is about 9 ÷ 1.5 = 6 A, so the biggest torque this coil can make is nIAB ≈ 200 × 6 × 0.0009 × 0.2 ≈ 0.22 N m. The normal is 50° from the field and sin 50° is about three-quarters, so I expect about 0.16 N m: just over the 0.15 N m needed. I predict it can just start. For (b), the current falls from 6 A to 1.8 A, less than a third, so less than a third of the 9.0 V is left to drive it: I predict a back emf of roughly 6 V.

K · Knowns

I sketch the coil edge-on with the field, mark the 40° between the field and the plane, and draw the normal. Then I convert the coil size to metres before I do anything else.

n = 200, B = 0.20 T, V = 9.0 V, R = 1.5 Ω
A = 0.030 m × 0.030 m = 9.0 × 10⁻⁴ m²
angle between field and plane = 40°, so θ (field to normal) = 90° − 40° = 50°
torque needed: τ_load = 0.15 N m; running current I_run = 1.8 A

Knowns on the sketch: B = 0.20 T, plane at 40° to the field (so the normal is at 50°), n = 200, A = 3.0 cm × 3.0 cm, 9.0 V, 1.5 Ω, torque needed 0.15 N m.

N · Not known

Part (a) asks me to decide, so I need the torque at switch-on in newton metres to compare with 0.15 N m. I don't have the current yet, but at switch-on the coil isn't moving, so there's no back emf and Ohm's law gives it. Part (b) wants the back emf in volts.

(a) τ = ? N m, then compare with 0.15 N m (on the way: I_start = ? A)
(b) ε_back = ? V

O · Organise

I write the equations in symbols first. The torque equation is on the data sheet with θ to the normal. The back-emf line is Ohm's law for the voltage that is left over.

I_start = V ÷ R (no back emf yet)
τ = nIAB sin θ
V − ε_back = I_runR, so ε_back = V − I_runR

W · Work it out

Current first, then the torque with θ = 50°, not 40°. Then the back emf from the running current.

I_start = 9.0 ÷ 1.5 = 6.0 A
τ = 200 × 6.0 × 9.0 × 10⁻⁴ × 0.20 × sin 50° = 0.216 × 0.7660 = 0.1655 N m
0.1655 N m > 0.15 N m, so the motor can start lifting the load from this position
ε_back = 9.0 − 1.8 × 1.5 = 9.0 − 2.7 = 6.3 V
Explain: the spinning coil is cutting the field, so the flux through it changes and an emf is induced. By Lenz's law it opposes the supply, so only V − ε_back = 2.7 V drives current through 1.5 Ω, giving 1.8 A instead of 6.0 A.

U · Units

Turns × amps × square metres × tesla gives N m, a torque. Volts minus amps × ohms is volts. The data are to two significant figures, so I give two.

(a) τ ≈ 0.17 N m, more than the 0.15 N m needed
(b) ε_back ≈ 6.3 V

D · Direction

Torque has a sense, clockwise or anticlockwise, but that depends on the current direction, which the question doesn't give, so I only state the size. The back emf does have a direction I can state: it acts against the supply. Sense check: 6.3 V is less than the 9.0 V supply, as it must be while the motor is still being driven.

The back emf of 6.3 V opposes the 9.0 V supply.

(a) The torque at switch-on is about 0.17 N m, more than the 0.15 N m needed, so the motor can start lifting the load from this position. (b) The back emf is about 6.3 V; it opposes the supply, so the running current is only 1.8 A.

Check against the prediction. I predicted about 0.16 N m, just enough, and roughly 6 V. The torque is 0.17 N m, under the 0.22 N m maximum but over 0.15 N m, and the back emf is 6.3 V. Both fit. Using 40° by mistake would have given 0.14 N m and the wrong conclusion that the motor can't start: here the angle decides the answer.

HSC advice

What markers keep saying

Know where θ is measured in the torque equation: from the field to the normal of the coil. Half a rotation is 180°, not 90°. The torque is greatest when the normal is at right angles to the field, which is when the coil's plane is parallel to the field, and zero when the plane is at right angles to it.NESA marking feedback, 2021, summarised
For the commutator, say exactly what it does: it reverses the current in the coil every half turn so the torque stays in one direction. For back emf, give the cause and effect (faster coil, larger back emf, smaller current) and don't confuse it with eddy currents.NESA marking feedback, 2023, summarised
With parallel wires, the forces on the two wires are equal and opposite even when the currents differ. Explain where the force comes from (one wire's field acting on the other wire's current) and put any difference in motion down to mass.NESA marking feedback, 2024, summarised
Use only the length of wire inside the field in F = lIB sin θ, and when asked how to change the torque by a factor, show the effect of each change separately (for example, × 4 from the field and × ½ from the current gives × 2).NESA marking feedback, 2025, summarised
Give force directions with a clear reference: into or out of the page, towards the top of the page, clockwise as seen from a named side. 'Up' or 'down' on its own is often ambiguous, and make sure you describe the force, not the torque, when the question asks for force.NESA marking feedback, 2019, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2019 · 2021 · 2023 · 2024 · 2025.

How often it is examined

The motor effect questions were worth 84 marks across the seven HSC Physics exams from 2019 to 2025 (about 12.0 marks a paper), and appeared in 7 of 7 papers.

Marks per paper (Module 6, The Motor Effect; Module 6, Applications of the Motor Effect). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Is the motor effect formula on the HSC data sheet?: Yes. The 2019 data sheet gives F = lI_⊥B = lIB sin θ, the parallel-wire formula F/l = (μ₀/2π)(I₁I₂/r) and the coil torque τ = nIA_⊥B = nIAB sin θ, plus μ₀ = 4π × 10⁻⁷ N A⁻². The back-emf relation V − ε = IR is not printed; it is Ohm's law applied to the leftover voltage.
What is the angle θ in τ = nIAB sin θ?: The angle between the magnetic field and the normal to the coil. The torque is greatest (sin θ = 1) when the plane of the coil is parallel to the field, and zero when the plane is at right angles to it.
What is the difference between the motor effect and electromagnetic induction?: In the motor effect a current in a field produces a force (electrical energy to kinetic energy). In induction, motion or a changing field produces an emf (kinetic energy to electrical energy). A motor does both at once: the induced part is its back emf.
Why does a motor draw a big current when it starts?: At switch-on the coil is not turning, so there is no back emf. The whole supply voltage drives current through the coil's small resistance. As it speeds up, the back emf grows and the current falls.
Do parallel wires with currents in the same direction attract or repel?: They attract. Currents in opposite directions repel. In both cases the two forces are equal in size and opposite in direction.

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