Year 11 Physics Module 4 ⏱ ~40 min 5 MC · 3 Short Answer Lesson 15 of 17

The Motor Effect

Michael Faraday, Royal Institution, London, September 1821: just eighteen months after Oersted's compass deflected, Faraday suspended a wire so one end dipped into a pool of mercury beside a fixed magnet. When he closed the switch, the wire began to rotate continuously around the magnet, the first demonstration that a current-carrying conductor in a magnetic field experiences a mechanical force. The modern formula for that force: $F = BIL\sin\theta$. A 20 cm wire carrying 4.0 A perpendicular to a 0.30 T field feels $F = (0.30)(4.0)(0.20) = 0.24\ \text{N}$.

Today's hook: In 1821, Faraday built a simple device: a wire dipping into mercury beside a fixed magnet, wired to a battery. Close the switch, and the wire spins continuously around the magnet, the first electric motor. The force behind it: $F = BIL\sin\theta$, maximum when the wire is perpendicular to the field, zero when parallel. A 20 cm wire carrying 4.0 A perpendicular to a 0.30 T field feels 0.24 N, tiny alone, but multiplied across hundreds of coil turns, that force becomes enough torque to spin a rotor at thousands of RPM.

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Before you read, predict

A straight wire segment 15 cm long carries a current of 3.0 A. It sits in a uniform magnetic field of 0.40 T.

Predict 1: If the wire is perpendicular to the field, what force does it experience?

Predict 2: If the wire is rotated so it becomes parallel to the field (still carrying 3.0 A), what happens to the force?

The force on a current-carrying conductor in a magnetic field is greatest when the angle between the current and the field is:

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Learning Intentions

Know

Force on a current-carrying conductor: $F = BIL\sin\theta$
$\theta$ is the angle between the current direction and the field direction
Right-hand palm rule for direction of force
The split-ring commutator reverses current every half-rotation in a DC motor

Understand

Why the force is maximum at $\theta = 90°$ and zero at $\theta = 0°$
Why the force is always perpendicular to both current and field (a cross-product relationship)
Why a coil without a commutator would not rotate continuously in one direction

Can Do

Calculate $F$ for a current-carrying conductor perpendicular to a field
Determine the direction of $F$ using the right-hand palm rule
Explain how a split-ring commutator maintains continuous rotation and how torque depends on $N$, $B$, $I$ and coil area

Beyond the syllabus. This whole lesson is extension. Under the 2017 syllabus, the motor effect, the DC motor and the split-ring commutator belong to Year 12 Module 6, Electromagnetism — Year 11 Module 4 finishes at the magnetic fields produced by magnets, current-carrying wires and solenoids. Nothing in this lesson is a mastery gate for Year 11. What is still examinable in Year 11: the right-hand rules and field patterns for straight wires (Lesson 13) and solenoids (Lesson 14) — this lesson is the payoff of those rules, so treat it as a well-earned head start on Year 12.

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Key Terms

Motor effectThe mechanical force experienced by a current-carrying conductor placed in an external magnetic field, given by $F = BIL\sin\theta$.

Right-hand palm rulePoint the fingers of the right hand in the direction of the magnetic field (N to S). Point the thumb in the direction of conventional current. The palm pushes in the direction of the force.

Split-ring commutatorA rotating switch in a DC motor that reverses the direction of current through the coil every half-rotation, so the torque keeps turning the coil in the same rotational direction.

Torque (motor)The turning effect produced on a current-carrying coil in a magnetic field. It increases with the number of turns $N$, the field strength $B$, the current $I$, and the coil's area.

Cross-lesson links: L13 and L14 established that current-carrying wires create magnetic fields. L15 flips the relationship: what happens when a current-carrying wire sits inside someone else's magnetic field? Oersted's compass (L13) simply responded to a field; here, the wire itself feels a mechanical push, the motor effect, the principle behind every electric motor since Faraday's first spinning wire in 1821. L16 extends this idea to two current-carrying wires acting on each other at the same time.

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Force on a Current-Carrying Conductor

Core Content

Force on a Current-Carrying Conductor

+5 XP

Dip one end of a wire into a pool of mercury next to a fixed bar magnet, connect the other end to a battery, and close the switch. The wire begins to rotate steadily around the magnet, no motor, no gears, just a current in a magnetic field. This is what Michael Faraday demonstrated at the Royal Institution in September 1821, only eighteen months after Oersted's compass had deflected in Copenhagen. Faraday's rotating wire proved something new: a magnetic field does not just get created by a current, it can also push on a current. That push is the motor effect, and it is the reason every electric motor, from a kitchen blender to a Tesla drive unit, can turn mechanical energy out of electrical energy.

Force on a current-carrying conductor

$F = BIL\sin\theta$

$F$: force (N); $B$: magnetic field strength (T); $I$: current (A); $L$: length of conductor in the field (m); $\theta$: angle between the current direction and the field direction

Key proportionalities

$F \propto \sin\theta$: maximum force at $\theta = 90°$ (perpendicular), zero force at $\theta = 0°$ (parallel to the field).

$F \propto I$ and $F \propto B$ and $F \propto L$: doubling any one of these doubles the force.

Worked example, perpendicular wire

A 20 cm wire carries a current of 4.0 A perpendicular to a magnetic field of 0.30 T. Calculate the force.

Perpendicular means $\theta = 90°$, so $\sin\theta = 1$.
$F = BIL\sin\theta = (0.30)(4.0)(0.20)(1)$
$F = 0.24\ \text{N}$

Right-hand palm rule

Direction of F

Point the fingers of the right hand in the direction of the magnetic field (N to S). Point the thumb in the direction of conventional current. The palm pushes in the direction of the force. The force is always perpendicular to both the current and the field.

Motor-effect sequence showing field upward and current right producing force out of the page, reversed current producing force into the page, and opposite forces on a current-carrying coil producing torque. — The motor effect is a three-direction relationship. Reversing either the current or the field reverses the force; opposite forces on the sides of a coil form a couple and produce torque.

The force on a current-carrying conductor in a magnetic field is $F = BIL\sin\theta$ (N), maximum at $\theta = 90°$, zero at $\theta = 0°$. The force is always perpendicular to both the current and the field. Direction: right-hand palm rule, fingers point along the field, thumb along the current, palm pushes in the direction of the force.

Pause, copy the highlighted formula and rule into your book before the check below.

The force on a current-carrying wire is greatest when the wire is perpendicular to the magnetic field.

A current-carrying wire oriented parallel to a magnetic field experiences the maximum possible force.

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The DC Motor and the Split-Ring Commutator

+5 XP

Place a single rectangular coil of wire in a magnetic field and pass a current through it. Both sides of the coil feel a force (motor effect), and because the current flows in opposite directions on each side, the two forces act in opposite directions, and the coil begins to rotate. But there is a problem: once the coil turns past the halfway point, those same forces would start slowing it down and turning it back. Left alone, the coil would just oscillate, not spin continuously. The fix is the split-ring commutator, a rotating switch that flips the current direction through the coil at exactly the moment the coil passes the halfway point, so the torque always acts to keep the coil turning the same way.

Torque in a DC motor

Torque $\propto N \times B \times I \times A$

$N$: number of turns in the coil; $B$: field strength; $I$: current; $A$: area of the coil. Increasing the current, the field strength, the number of turns, or the coil area all increase the torque.

Why the commutator is needed

1. Current flows through the coil; each side feels a force (motor effect), producing a torque that starts the coil rotating.

2. As the coil rotates past the vertical (halfway) position, the torque direction would reverse if the current stayed the same, decelerating the coil.

3. The split-ring commutator breaks contact and reconnects with reversed polarity at exactly this point, reversing the current in the coil.

4. With the current reversed, the torque continues to act in the same rotational sense, so the coil keeps spinning in one direction.

A DC motor uses the motor effect on a current-carrying coil to produce torque, proportional to $N$, $B$, $I$ and the coil's area. A split-ring commutator reverses the current in the coil every half-rotation, so the torque always acts in the same rotational direction and the coil spins continuously rather than oscillating.

Add the torque proportionality and the commutator's role to your notes before the check below.

In a DC motor, the split-ring commutator's role is to:

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Activity 1, Calculating Motor-Effect Force

ApplyBand 3

A wire 30 cm long carries a current of 6.0 A, perpendicular to a magnetic field of 0.25 T.

Calculate the force on the wire.
If the current is reduced to 3.0 A (same length and field), recalculate the force.
If the wire is instead oriented parallel to the field (still carrying 6.0 A), what is the force now?

Three of these are correct statements about the motor effect. Pick the odd one out.

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Activity 2, Direction and the DC Motor

AnalyseBand 4

A horizontal wire carries current toward the west, in a magnetic field that points vertically upward.

Using the right-hand palm rule, determine the direction of the force on the wire.
Explain why a simple current-carrying coil, without a commutator, would not rotate continuously in one direction.
State one factor (besides current) that would increase a DC motor's torque, and explain why.

A 25 cm wire carrying 5.0 A perpendicular to a 0.40 T field experiences a force of _____ N.

A current-carrying wire experiences zero force in a magnetic field when the angle between the current and the field is:

✓

Show what you have learned

Quick recall, the motor effect

+5 XP

Short Answer, 10 marks

+5 XP

ApplyBand 3(3 marks) 3. A wire 40 cm long carries a current of 8.0 A perpendicular to a magnetic field of 0.20 T. Calculate (a) the force on the wire, (b) the force if the current is doubled, and (c) the force if the wire is instead at 30° to the field, keeping the current and length the same.

AnalyseBand 4(3 marks) 4. Two identical DC motors carry the same current in the same field, except Motor A's coil has twice the number of turns of Motor B's coil. (a) Compare the torque produced by each motor and explain why. (b) Explain the role of the split-ring commutator in maintaining continuous rotation. (c) State one change to the field that would decrease the torque of either motor.

EvaluateBand 6(4 marks) 5. Faraday's 1821 rotating wire seems primitive compared with a modern DC motor. Using your understanding of the motor effect ($F = BIL\sin\theta$) and torque, evaluate why Faraday's demonstration was nonetheless a turning point, and explain what a modern motor's split-ring commutator adds to make this force useful for continuous rotation.

Show all answers

Activity 1, Model Answers

$F = BIL\sin\theta = (0.25)(6.0)(0.30)(1) = 0.45\ \text{N}$
$F = (0.25)(3.0)(0.30)(1) = 0.225\ \text{N} \approx 0.23\ \text{N}$, halving the current halves the force.
Parallel to the field means $\theta = 0°$, so $\sin\theta = 0$ and $F = 0\ \text{N}$.

Activity 2, Model Answers

Right-hand palm rule: fingers point up (field), thumb points west (current). The palm pushes toward the south. The force is directed south.
Without a commutator, once the coil rotates past the vertical (halfway) position, the force on each side would reverse relative to the rotation, decelerating the coil and pulling it back rather than continuing to turn it. The coil would oscillate rather than spin continuously.
Increasing the magnetic field strength $B$ (or the number of turns $N$, or the coil area) increases torque, because torque $\propto N \times B \times I \times A$.

Short Answer, Model Answers

Q3 (3 marks): (a) $F = BIL\sin\theta = (0.20)(8.0)(0.40)(1) = 0.64\ \text{N}$. (b) Doubling $I$ doubles $F$: new $F = 1.28\ \text{N}$. (c) At 30°: $F = (0.20)(8.0)(0.40)\sin 30° = 0.64 \times 0.5 = 0.32\ \text{N}$.

Q4 (3 marks): (a) Torque $\propto N \times B \times I \times A$. With twice the turns and everything else equal, Motor A produces twice the torque of Motor B. (b) The split-ring commutator reverses the current in the coil every half-rotation, so the torque always acts to turn the coil in the same rotational direction rather than oscillating back and forth. (c) Decreasing the magnetic field strength $B$ would decrease the torque of either motor, since torque $\propto B$.

Q5 (4 marks): Faraday's 1821 demonstration was significant because it was the first experimental proof that a magnetic field could exert a continuous mechanical force on a current-carrying conductor, converting electrical energy directly into rotational motion. Before this, Oersted (1820) had shown current produces a field, but Faraday showed the reverse interaction: a field can push on a current. This established the motor effect as a genuine, repeatable physical principle rather than an isolated curiosity. However, Faraday's simple wire-and-mercury device only produced a modest, somewhat erratic rotation of a single wire. A modern DC motor adds a coil of many turns (increasing torque via $N$), a strong field (often from an electromagnet or permanent magnet), and crucially a split-ring commutator, which reverses the current every half-turn so that the torque never opposes the rotation. Without the commutator, any coil-based motor would only oscillate; the commutator is what converts Faraday's single push into continuous, usable rotational motion, exactly the property needed to drive real machinery.

✓

Retrieve, reflect and finish

Check what actually stuck

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How did your thinking change?

Michael Faraday, Royal Institution, London, September 1821: a wire dipping into mercury beside a fixed magnet spun continuously once the current flowed, the first electric motor. The force behind it, $F = BIL\sin\theta$, gives just 0.24 N for a 20 cm wire carrying 4.0 A perpendicular to a 0.30 T field. Alone that force is barely noticeable; multiplied across hundreds of turns in a modern motor's coil, and kept turning the same way by a split-ring commutator, it becomes enough torque to drive a car, a drill, or a washing machine drum.

Now check your Think First answers: perpendicular to the 0.40 T field, a 15 cm wire carrying 3.0 A feels $F = (0.40)(3.0)(0.15) = 0.18\ \text{N}$. Rotated parallel to the field, $\sin\theta = 0$, so the force drops to zero, even though the same current still flows.

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