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4. Electromagnetism & Induction

Curiominds Electricity Lab · Mr Lau Chin Hang / @curiominds.inc · Syllabus: 6091 topics 17–19 · IGCSE 4.1, 4.5

How the sim works. Each chapter runs Explain → Predict → Test → Challenge, then a set of exam-style questions. A free-build Sandbox tab is always available. The toolbar toggles Symbol view (exam-style symbols) and Real view (illustrated parts), and conventional current or electron flow; the label always shows which is active. Challenges are checked automatically. Nothing is recorded or sent anywhere; progress is kept in the browser only if it allows local storage.

Syllabus check. Learning outcomes were checked on 13 October 2026 against the published SEAB O-Level Physics 6091 syllabus for 2026 (topics 14–19). Where O-Level and IGCSE wording differ, the sim follows O-Level and labels IGCSE-only material (for example resistivity ρ).

Using it in SLS or offline. The page is a single self-contained HTML file: it needs no internet connection, makes no network requests and opens no pop-ups. Upload it to SLS as a file resource or link to the hosted copy; it also runs from a USB stick. Deep links open a chapter directly, e.g. electromagnetism.html#fields or #fields/predict.

Chapters at a glance

Chapter	Syllabus	Predict-then-test answer
a. Magnetic fields	6091 17(a)–(e), 18(a), (b) · IGCSE 4.1, 4.5.2	The needle turns through 180°.
b. Force on a current in a field	6091 18(c), (d) · IGCSE 4.5.4	Down.
c. The d.c. motor	6091 18(e), (f) · IGCSE 4.5.5	It rocks back and forth, then stops with the coil vertical.
d. Electromagnetic induction	6091 19(a) · IGCSE 4.5.1	Returns to zero.
e. The a.c. generator	6091 19(b), (c) · IGCSE 4.5.1	Double the peak and double the frequency.
f. Transformers	6091 19(d)–(f) · IGCSE 4.5.6	11.5 V.

a. Magnetic fields

6091 17(a)–(e), 18(a), (b) · IGCSE 4.1, 4.5.2

Lesson flow

Field model: each magnet face is three point “poles” (1/r² field); wires give a 1/r tangential field; the solenoid is a row of wire pairs. Patterns are qualitative. Start with the compass alone and Mark, then reveal the lines. Use N facing N for the neutral point.

Predict-then-test

Prompt: A compass sits beside a wire carrying current out of the page. Predict what the needle does when the current is reversed.
Outcome: The needle turns through 180°.

Challenge (auto-checked)

Task: Two N poles face each other. Find the neutral point, where the two fields cancel and the compass is not pushed either way. Drag the compass there.

Answer: Neutral point at the midpoint between the two N poles (the origin).

Exam-style questions and answers

Which material is best for the core of an electromagnet that must be switched on and off? [1]
A Steel
B Soft iron
C Copper
D Aluminium
B (Soft iron). Soft iron is a temporary magnet: it magnetises and demagnetises easily.
In which direction do magnetic field lines point outside a bar magnet? [1]
A From S to N
B From N to S
C From the middle outwards
D Towards the middle
B (From N to S).
Current flows into the page in a straight wire. Viewed from above the page, the field lines are… [1]
A clockwise circles
B anticlockwise circles
C straight lines parallel to the wire
D straight lines radiating out
A (clockwise circles). Right-hand grip rule: thumb into the page, fingers curl clockwise.
How can the magnetic field of a solenoid be made stronger? [1]
A Reverse the current
B Increase the current or the number of turns
C Use a copper core
D Make the coil longer with the same number of turns
B (Increase the current or the number of turns).
Describe how a plotting compass can be used to draw the field lines around a bar magnet. [3]
Place the magnet on paper. Put the compass near the N pole and mark the position of the N end of the needle with a dot. Move the compass so the other end of the needle is at that dot, and mark again. Repeat until reaching the S pole, then join the dots with a smooth line and add an arrow from N to S. Repeat from other starting points.

b. Force on a current in a field

6091 18(c), (d) · IGCSE 4.5.4

Lesson flow

Use the catapult-field toggle to link the rule to field superposition. The beam mode covers the “beam of charged particles” outcome: remind students that conventional current is opposite to electron motion.

Predict-then-test

Prompt: Field left → right, current into the page. Predict the direction of the force.
Outcome: Down.

Challenge (auto-checked)

Task: Find two different settings of current direction and field direction that push the wire upwards.

Answer: Current out of page with field left → right; or current into page with field right → left.

Exam-style questions and answers

A wire carries current into the page between a N pole on the left and a S pole on the right. Which way is the force on the wire? [1]
A Up the page
B Down the page
C Towards N
D Towards S
B (Down the page).
The current in the wire and the magnetic field are both reversed. What happens to the force? [1]
A It reverses
B It is unchanged
C It becomes zero
D It doubles
B (It is unchanged).
Which change does NOT increase the force on a current-carrying wire in a magnetic field? [1]
A Increasing the current
B Using a stronger magnet
C Reversing the current
D Making a longer length of wire lie in the field
C (Reversing the current).
A beam of electrons moves left to right across the page through a field directed into the page. Which way is it deflected? [1]
A Up the page
B Down the page
C Into the page
D It is not deflected
B (Down the page). Conventional current is right to left. Field into page, current to the left: thumb points down.
State Fleming’s left-hand rule and explain how it shows that the force on a current-carrying conductor reverses when the current reverses. [3]
Hold the thumb, first and second fingers of the left hand mutually at right angles: first finger along the field, second finger along the (conventional) current; the thumb gives the direction of the force (motion). Reversing the current means the second finger points the opposite way; keeping the first finger fixed, the thumb then points the opposite way, so the force reverses.

c. The d.c. motor

6091 18(e), (f) · IGCSE 4.5.5

Lesson flow

Model: torque = k N B I cos φ (× 2 with soft-iron core), commutator reverses the coil current when cos φ changes sign, viscous friction gives a steady speed proportional to N B I. Above ~240 rpm the coil is drawn as a blur. Use Predict to show what the commutator is for.

Predict-then-test

Prompt: Predict what the coil does if the split-ring commutator is replaced by slip rings.
Outcome: It rocks back and forth, then stops with the coil vertical.

Challenge (auto-checked)

Task: Make the motor turn clockwise at more than 150 rpm.

Answer: Reverse the current (or the magnet) and raise N, I, B or add the core, e.g. 20 turns, 2.0 A, iron core gives about 320 rpm clockwise.

Exam-style questions and answers

What is the purpose of the split-ring commutator in a d.c. motor? [1]
A To make the current larger
B To reverse the current in the coil every half-turn so the coil keeps turning the same way
C To make the magnet stronger
D To change d.c. into a.c. in the supply
B (To reverse the current in the coil every half-turn so the coil keeps turning the same way).
Which change would make a d.c. motor turn in the opposite direction? [1]
A Doubling the current
B Reversing the battery
C Adding more turns
D Adding a soft-iron core
B (Reversing the battery).
In which position of the coil is the turning effect zero? [1]
A Coil plane parallel to the field
B Coil plane at right angles to the field
C At every position
D Coil at 45° to the field
B (Coil plane at right angles to the field). When the coil is vertical, the forces on the sides act through the axle.
State three ways of increasing the speed of a d.c. motor. [3]
Any three: increase the current; increase the number of turns on the coil; use a stronger magnet; wind the coil on a soft-iron core.
Explain why the coil of a simple d.c. motor turns. [3]
The current in the two sides of the coil is in opposite directions, and both sides are at right angles to the magnetic field. By Fleming’s left-hand rule one side experiences an upward force and the other a downward force. These forces form a couple that turns the coil about its axle; the commutator reverses the current every half-turn so the rotation continues in the same direction.

d. Electromagnetic induction

6091 19(a) · IGCSE 4.5.1

Lesson flow

Flux model: on-axis dipole profile 1/(1+(d/a)²)^1.5 averaged over the coil length; e.m.f. = −N dΦ/dt, scaled to mV. The induced pole is shown in magenta so Lenz’s law is visible. Record rows to compare speed, turns and strength.

Predict-then-test

Prompt: Pushing N in gives a right deflection. Predict the needle when the magnet is held still inside the coil.
Outcome: Returns to zero.

Challenge (auto-checked)

Task: Make the galvanometer needle swing to the left by at least 4.0 mV.

Answer: Pull the N pole out fast (or push a S pole in fast) with more turns: e.g. 400 turns at 4 units/s gives about −8 mV.

Exam-style questions and answers

A magnet rests inside a coil connected to a galvanometer. What does the galvanometer read? [1]
A A steady current
B Zero
C A current that keeps reversing
D A current that slowly increases
B (Zero).
Which change does NOT increase the e.m.f. induced when a magnet is pushed into a coil? [1]
A Pushing it in faster
B Using more turns
C Using a stronger magnet
D Using a thicker wire with the same number of turns
D (Using a thicker wire with the same number of turns).
The N pole of a magnet is pulled out of a coil. What pole is induced at the end of the coil nearest the magnet? [1]
A N, to repel the magnet
B S, to attract the magnet back
C No pole
D Both N and S
B (S, to attract the magnet back). Lenz’s law: the induced pole opposes the change (the magnet leaving).
State Faraday’s law of electromagnetic induction and give three factors that affect the size of the induced e.m.f. [4]
An e.m.f. is induced in a conductor whenever the magnetic field linking it changes; its size is proportional to the rate of change. Factors: speed of relative movement; strength of the magnet; number of turns on the coil (also the area of the coil).
Use Lenz’s law to explain why work must be done to push a magnet into a coil that is part of a complete circuit. [3]
The induced current produces a magnetic field that opposes the change causing it: the end of the coil facing the approaching N pole becomes a N pole and repels the magnet. A force must be applied against this repulsion, so work is done; this work is transferred to electrical energy in the circuit.

e. The a.c. generator

6091 19(b), (c) · IGCSE 4.5.1

Lesson flow

Output = 0.012 V × N × f × sin φ (slip rings) or |sin φ| (commutator), with φ = 0 when the coil is at right angles to the field. The graph shows true time over 1 s; the coil animation is slowed ×10 so it can be followed. Compare with the d.c. motor: same parts, opposite energy transfer.

Predict-then-test

Prompt: Generator at 2 rev/s gives 2.4 V peak at 2 Hz. Predict the output at 4 rev/s.
Outcome: Double the peak and double the frequency.

Challenge (auto-checked)

Task: Set up the generator to give an a.c. output with a peak of 6.0 V and a frequency of 5 Hz.

Answer: 100 turns, 5 rev/s, slip rings: 0.012 × 100 × 5 = 6.0 V.

Exam-style questions and answers

In a simple a.c. generator, when is the induced e.m.f. zero? [1]
A When the plane of the coil is parallel to the field
B When the plane of the coil is at right angles to the field
C When the coil turns fastest
D Never
B (When the plane of the coil is at right angles to the field).
What is the job of the slip rings in an a.c. generator? [1]
A To reverse the output every half-turn
B To keep each end of the coil connected to the same brush
C To make the field stronger
D To make the output d.c.
B (To keep each end of the coil connected to the same brush).
A generator’s output has a period of 0.020 s. What is its frequency? [1]
A 0.020 Hz
B 20 Hz
C 50 Hz
D 200 Hz
C (50 Hz).
Sketch the output p.d. against time for a simple a.c. generator over two turns, and show on the sketch the effect of turning it at twice the speed. [3]
A sine-shaped curve crossing zero twice per turn (two full cycles for two turns). At twice the speed the peak is twice as high and the cycles are half as long (twice the frequency).
Explain why an e.m.f. is induced in the coil of a generator. [2]
As the coil rotates, the magnetic field (flux) through it changes continuously; by Faraday’s law a changing magnetic field linking the coil induces an e.m.f. in it.

f. Transformers

6091 19(d)–(f) · IGCSE 4.5.6

Lesson flow

Added in v1 at the owner’s request (brief §12 open item); kept light, with no separate transformer calculator. The model is ideal (100% efficient). Use the d.c. option to show why transformers need a.c. The transmission view uses 2.0 MW and 4.0 Ω cables; 230 V shows that the loss would exceed the power sent.

Predict-then-test

Prompt: Np = 1000, 230 V a.c. on the primary. Predict Vs when Ns = 50.
Outcome: 11.5 V.

Challenge (auto-checked)

Task: Design a transformer that gives 12 V a.c. from the 230 V mains (for a model railway).

Answer: Any turns ratio Ns/Np = 12/230, e.g. Np = 920, Ns = 48, or Np = 1150, Ns = 60.

Exam-style questions and answers

A transformer has 2000 turns on its primary and 100 turns on its secondary. The primary is connected to 230 V a.c. Calculate the secondary p.d. [2]
11.5 V.
An ideal transformer steps 230 V down to 6.0 V. The secondary current is 2.3 A. Calculate the primary current. [2]
0.06 A. Ip = VsIs / Vp = 6.0 × 2.3 / 230 = 0.060 A.
Why will a transformer not work with a steady d.c. supply? [1]
A d.c. is too small
B A steady current gives a magnetic field that does not change, so no e.m.f. is induced
C The core would melt
D d.c. cannot flow in a coil
B (A steady current gives a magnetic field that does not change, so no e.m.f. is induced).
Electricity is transmitted at 400 kV rather than 11 kV for the same power. The heating loss in the cables is… [1]
A about 36 times smaller
B about 1300 times smaller
C the same
D larger
B (about 1300 times smaller). Current falls by 400/11 ≈ 36; loss ∝ I², so about 1300 times smaller.
Explain why electrical energy is transmitted over long distances at a very high p.d. [3]
For a given power P = VI, a higher p.d. means a smaller current in the cables. The power wasted as heat in the cables is I²R, so a smaller current greatly reduces the energy lost. Thinner, cheaper cables can also be used. Transformers step the p.d. up at the power station and down again near consumers.

Physics models

Every part is ideal unless stated. The models are chosen to give O-Level-correct behaviour, not to imitate real equipment exactly.

Model	Description
Field patterns	Qualitative 2-D model. Bar-magnet faces are three point “poles” each (2-D 1/r field, so lines run from N round to S); straight wires give a tangential field ∝ I/r; a solenoid is a row of wire pairs. Field lines are traced with a Runge–Kutta step and arrows show the direction of B.
Motor	Torque = k N B I cos φ (doubled with a soft-iron core); the split-ring commutator reverses the coil current whenever cos φ changes sign; viscous friction gives a steady speed ∝ N B I (about 40 rpm with 10 turns, 1.0 A).
Induction	Flux through the coil uses an on-axis dipole profile, Φ ∝ 1/(1 + (d/a)²)^3/2, averaged over the coil; e.m.f. = −N dΦ/dt scaled to millivolts. The induced pole (Lenz’s law) is drawn on the coil.
Generator	Output = 0.012 V × N × f × sin φ with slip rings, or \|sin φ\| with a commutator, where φ = 0 when the coil is at right angles to the field (flux largest, e.m.f. zero). The graph uses true time; the animation is slowed ×10.
Transformer	Ideal: V_s = V_p N_s/N_p and V_pI_p = V_sI_s; no output from steady d.c. Transmission: 2.0 MW through 4.0 Ω of cable, loss = I²R.

Transformers (chapter f) were added to v1 at the owner’s request. There is no separate transformer calculator.

Induced magnetism and temporary versus permanent magnets (6091 17(b), (c)) are covered in the Explain text and questions of chapter a rather than as an interactive.