What are electromagnetic effects in the context of Cambridge IGCSE Physics?

Electromagnetic effects refer to the phenomena that occur when electric currents interact with magnetic fields. This includes the generation of magnetic fields by electric currents, the force exerted on a current-carrying conductor in a magnetic field, and electromagnetic induction. These concepts are fundamental in the study of Electricity and Magnetism in the Cambridge IGCSE Physics curriculum.

How does electromagnetic induction work in simple terms?

Electromagnetic induction is the process by which a changing magnetic field within a coil of wire induces an electromotive force (EMF) or voltage in the wire. This can occur when a magnet is moved in and out of a coil or when the coil itself is moved through a magnetic field. This principle is the basis for many electrical generators and transformers, which are key topics in the Cambridge IGCSE Physics syllabus.

What is the right-hand rule in the context of electromagnetic effects?

The right-hand rule is a mnemonic used to determine the direction of the magnetic field around a current-carrying conductor. According to this rule, if you point the thumb of your right hand in the direction of the current, the curl of your fingers shows the direction of the magnetic field lines. This concept helps students visualize the orientation of magnetic fields in the study of Electricity and Magnetism for Cambridge IGCSE Physics.

Why is the study of electromagnetic effects important in everyday life?

Understanding electromagnetic effects is crucial because they underpin many technologies we use daily, such as electric motors, transformers, and generators. These devices convert electrical energy into mechanical energy and vice versa, making them essential for power generation, transportation, and various electronic devices. The Cambridge IGCSE Physics curriculum emphasizes these applications to highlight the practical significance of electromagnetic principles.

What is the difference between a solenoid and a bar magnet in terms of magnetic fields?

A solenoid is a coil of wire that generates a magnetic field when an electric current passes through it, resembling the field of a bar magnet. However, unlike a permanent bar magnet, the magnetic field of a solenoid can be turned on and off and its strength can be adjusted by changing the current. This distinction is important in the study of electromagnetic effects within the Cambridge IGCSE Physics framework, as it highlights the versatility of electromagnets in various applications.

Why is "Using the right hand for the motor effect (or left for induction)" a common mistake in Electromagnetic Effects?

Why it happens: Mixing the rules. How to avoid it: MOTOR effect → LEFT hand. GENERATOR (induction) → RIGHT hand. Source: 0625/42 — recurring

Why is "Inverting the transformer ratio" a common mistake in Electromagnetic Effects?

Why it happens: Memory slip. How to avoid it: V is proportional to N. Ratio with V on top mirrors N on top.

Why is "Saying transformers work on DC" a common mistake in Electromagnetic Effects?

Why it happens: Confusing AC and DC. How to avoid it: Transformers need a CHANGING magnetic flux. DC produces no change → no induced e.m.f. (only AC works).

Why is "Stating 'a magnet near a coil' produces e.m.f. (without motion)" a common mistake in Electromagnetic Effects?

Why it happens: Forgetting the change requirement. How to avoid it: Induction needs a CHANGE in flux — relative motion or a changing field.

Electricity and MagnetismCambridge IGCSE Physics 0625 Extended

Electromagnetic Effects

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Electromagnetic Effects — Cambridge IGCSE 0625 Physics Extended (2026)

Solenoids, the motor effect, electromagnetic induction, generators and transformers. The biggest ideas in physics — they power everything.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

4.5 — Describe the magnetic field of a current-carrying wire and a solenoid.
4.5 — Apply Fleming's left-hand rule to predict force direction.
4.5 — Describe electromagnetic induction in a coil.
4.5 — Describe construction and use of an AC generator.
4.5 — Recall and use the transformer equation.

Current creates a magnetic field

A wire with current has a circular field around it. A coil concentrates the field; a solenoid acts like a bar magnet.

Straight wire. Current produces concentric circular field lines around the wire.

Direction by right-hand grip rule: thumb points in current direction; fingers curl in field direction.

Coil / solenoid. A coil of wire carrying current has:

Strong, roughly uniform field INSIDE the coil.
Field lines outside resemble those of a bar magnet — with N and S poles at the ends.
Direction: use the right-hand grip on the WIRE LOOP. Or: looking at one end, if current flows ANTICLOCKWISE → that end is N.

A current-carrying coil behaves like a bar magnet — uniform field inside, N and S poles at the ends.

Strengthening a solenoid's field.

More turns per metre.
Higher current.
A SOFT IRON CORE inside (greatly amplifies the field).

Electromagnet. A solenoid with a soft iron core. Switch the current on → magnet on. Switch off → magnet off (instant). Used in scrap-yard cranes, relays, electric bells.

Using the magnetic effect of a current — relays and loudspeakers.

Relay. A small current in a coil makes an electromagnet that attracts a soft-iron armature. The armature pivots and closes (or opens) the contacts of a separate circuit. This lets a small, safe current switch a much larger current, with the two circuits electrically isolated.
Loudspeaker. An alternating current (the audio signal) flows through a coil sitting in the field of a permanent magnet. The motor effect makes the coil move back and forth as the current changes; the coil is attached to a paper cone, so the cone vibrates and produces sound waves.

Current → field. Right-hand grip rule.
Solenoid: bar-magnet-like field.
Strengthen: more turns, more current, iron core.
Electromagnet: solenoid + iron core; switchable.
Magnetic effect of a current is used in relays and loudspeakers.

The motor effect

Current-carrying wire in a magnetic field experiences a force. Fleming's left-hand rule.

Motor effect. When a current-carrying wire is placed in an EXTERNAL magnetic field, the wire experiences a force PERPENDICULAR to BOTH current and field.

Fleming's left-hand rule (predicts the force direction).

Thumb: Force ( $F$ ).
First finger: B-field.
Second finger: I (current). Hold them mutually perpendicular on your LEFT hand.

LEFT hand for motors: thumb = Force, first finger = B-field, second finger = Current.

Reverse the current OR the field: force reverses too.

Force size depends on:

Current size.
Magnetic field strength.
Length of wire in the field.
Angle (max when wire is at 90° to the field).

DC motor. A coil in a magnetic field, with a split-ring commutator. Current in the coil creates forces on opposite sides → torque → rotation. Commutator reverses current direction every half-turn so the torque keeps pushing the same way → continuous rotation.

Worked qualitative. Why does a stronger magnet make the motor spin faster? Larger force on each side of the coil → larger torque → faster acceleration → higher final speed (limited by friction and back-EMF).

Force = current × field × length.
Fleming's LEFT hand for motors.
Reverse $I$ or $B$ → reverse $F$ .
DC motor: coil + commutator.

Electromagnetic induction

Changing flux through a coil induces an EMF. Move a magnet, OR change the current in a nearby coil.

The induction effect (qualitative). When the magnetic field linking a conductor or coil changes, an e.m.f. is induced in it. (The formal name "Faraday's law" and the magnetic-flux symbol $\Phi$ are AS/A-level terminology — see Going deeper — but the underlying idea is fully in scope for 0625.)

Lenz's law. The induced current opposes the change that caused it.

Ways to induce current.

Move a magnet into / out of a coil.
Move the coil instead.
Change the current in a NEARBY coil (mutual induction; basis of transformers).

Size of induced EMF depends on:

Speed of the change (faster = larger EMF).
Number of turns on the coil.
Strength of the magnet.
Area of the coil.

Direction (Lenz's law illustrated). Push the N pole of a magnet INTO a coil. The induced current creates a magnetic field that REPELS the incoming N pole — i.e. the coil end facing the magnet acts as N. Pull the magnet OUT and the coil end becomes S, attracting the magnet back.

Worked qualitative. Why does dropping a strong magnet down a copper tube fall slowly? As the magnet falls, the changing flux in the tube induces eddy currents that oppose the motion (Lenz). The drag slows the magnet — it "floats" through the tube.

Changing magnetic field → induced EMF.
Lenz: induced current opposes the change.
Faster change → larger EMF.
Right-hand rule for direction (or apply Lenz).

Generators and transformers

Generator: rotation → AC. Transformer: AC voltage step-up/step-down via two coils on an iron core.

AC generator. A rotating coil in a magnetic field. As the coil turns, the flux through it changes sinusoidally → induced EMF is sinusoidal → AC output. Slip rings (NOT split rings) connect the rotating coil to the external circuit.

Output: peak EMF when the coil sides are moving fastest across the field (parallel to it); zero when coil is parallel to the field.

Transformer. Two coils (primary $N_{p}$ , secondary $N_{s}$ ) sharing a soft iron core. AC in the primary creates a changing flux in the core → induces an AC EMF in the secondary.

Equation. $\dfrac{V_{p}}{V_{s}} = \dfrac{N_{p}}{N_{s}}.$

More turns on the secondary steps the voltage up: Vp ÷ Vs = Np ÷ Ns.

Step-up: $N_{s} > N_{p}$ , $V_{s} > V_{p}$ . Used to raise voltage for transmission. Step-down: $N_{s} < N_{p}$ , $V_{s} < V_{p}$ . Used near homes to bring it back to safe levels.

Ideal transformer (100% efficient): $V_{p}I_{p} = V_{s}I_{s}$ . Stepping voltage UP steps current DOWN by the same factor.

Worked. Step-up transformer: $N_{p} = 200$ , $N_{s} = 1000$ , $V_{p} = 230\,\text{V}$ .

$V_{s} = 230 \times 1000/200 = 1150\,\text{V}$ .

Why high voltage for transmission? Power lost in cables = $I^{2}R$ . Step UP voltage → step DOWN current → losses cut by current squared. National grid uses up to $400\,\text{kV}$ .

Generator: rotating coil → AC EMF (slip rings).
Transformer: $V_{p}/V_{s} = N_{p}/N_{s}$ .
Step-up = more secondary turns.
Ideal: $V_{p}I_{p} = V_{s}I_{s}$ .
High voltage transmission cuts $I^{2}R$ losses.

How it’s examined

Electromagnetic effects appear on every Paper 4 (10-12 marks) — typically a transformer calculation, a Fleming's rule diagram, and a generator description. Examiner reports flag two errors: confusing left and right hand rules, and forgetting that transformers only work with AC.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (4.5); 0625/42 Oct/Nov 2024 — Q18 (transformer + induction); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

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Step-by-step worked examples — Electromagnetic Effects

Step-by-step solutions to past-paper-style questions on electromagnetic effects, written exactly the way a tutor would explain them at the board.

1Force on a current-carrying wire
Extended• motor
Question
A wire carries current to the right. The magnetic field points into the page. Use Fleming's left-hand rule to find the force direction.
Step-by-step solution
1. Step 1
  First finger = field (into page). Second finger = current (right). Thumb gives motion.
2. Step 2
  Result: force points UPWARDS.
Answer
Force is directed upward (out of the plane of the wire).
Examiner tip
Use the LEFT hand for the motor effect (Force, Field, Current = thuMb, First, seCond fingers).
2Transformer ratio
Extended• Adapted from 0625/42 May/Jun 2024 Q18• transformer
Question
A transformer has $200$ primary turns and $50$ secondary turns. The primary input is $230\,\text{V}$ . Find the secondary voltage.
Step-by-step solution
1. Step 1
  $\dfrac{V_{p}}{V_{s}} = \dfrac{N_{p}}{N_{s}}$ .
  $\dfrac{230}{V_{s}} = \dfrac{200}{50}$
2. Step 2
  Solve.
  $V_{s} = 230 \times \dfrac{50}{200} = 57.5\,\text{V}$
Answer
$57.5\,\text{V}$ (step-down).
3Induced e.m.f.
Extended• induction
Question
State two ways to increase the e.m.f. induced by a magnet moved through a coil.
Step-by-step solution
1. Step 1
  Move the magnet faster (higher rate of change of flux).
2. Step 2
  Use a coil with more turns; or a stronger magnet.
Answer
(1) Move the magnet faster. (2) Increase the number of coil turns (or use a stronger magnet).
4Why national grid uses high voltage
Extended• grid
Question
Explain why the National Grid transmits power at high voltage.
Step-by-step solution
1. Step 1
  $P = VI$ — for a given $P$ , increasing $V$ reduces $I$ .
2. Step 2
  Power lost in the cables is $P_{\text{lost}} = I^{2}R$ — reducing $I$ massively reduces losses.
Answer
Higher $V$ → lower $I$ → less $I^{2}R$ heating loss in transmission cables.
5Field around a long straight wire
Core• right-hand grip
Question
A long straight wire carries a current vertically UPWARDS. Describe the direction of the magnetic field around it.
Step-by-step solution
1. Step 1
  Use the right-hand grip rule: thumb points in the direction of conventional current (upwards); fingers curl in the direction of the field.
2. Step 2
  Field lines form concentric horizontal circles around the wire. Looking DOWN at the wire from above, the field circles run ANTICLOCKWISE.
Answer
Concentric circles around the wire, anticlockwise when viewed from above.
6The magnetic effect of a current — the relay
Core• relay, electromagnet
Question
A relay uses the magnetic effect of a current. (a) Explain how a relay works. (b) State one reason a relay is useful.
Step-by-step solution
1. Step 1
  (a) A small current in the relay coil makes it an electromagnet.
2. Step 2
  The electromagnet attracts a soft-iron armature, which pivots and closes (or opens) a separate pair of contacts in a second circuit.
3. Step 3
  (b) A relay lets a small, safe current switch on a much larger current in a separate circuit — the two circuits are kept electrically isolated.
Answer
(a) A small current makes the coil an electromagnet; it attracts an iron armature that closes the contacts of a second circuit. (b) A small/safe current can switch a large current, with the two circuits isolated.
7The magnetic effect of a current — the loudspeaker
Core• loudspeaker, motor effect
Question
A loudspeaker uses the magnetic effect of a current. Explain how it converts an electrical signal into sound.
Step-by-step solution
1. Step 1
  An alternating current (the audio signal) flows through a coil that sits in the field of a permanent magnet.
2. Step 2
  Because a current-carrying coil in a magnetic field experiences a force (the motor effect), the coil is pushed back and forth as the current changes size and direction.
3. Step 3
  The coil is attached to a paper cone, so the cone vibrates. The vibrating cone makes the air vibrate, producing sound waves that match the original signal.
Answer
An alternating current in a coil placed in a magnetic field makes the coil (and an attached cone) vibrate due to the motor effect; the vibrating cone produces sound waves.
8Beyond 0625 — Force on a current-carrying wire ( $F = BIL$ )
Extended• Adapted from 0625/42 Oct/Nov 2022 Q10• F=BIL, motor
Question
A wire of length $0.20\,\text{m}$ carries a current of $4.0\,\text{A}$ at right angles to a magnetic field of flux density $0.50\,\text{T}$ . Find the force on the wire.
Step-by-step solution
1. Step 1
  Use $F = BIL$ (current perpendicular to field).
2. Step 2
  Substitute.
  $F = 0.50 \times 4.0 \times 0.20 = 0.40\,\text{N}$
Answer
$0.40\,\text{N}$
Examiner tip
Enrichment beyond Cambridge IGCSE 0625 — the equation F = BIL, magnetic flux density and the unit tesla (T) are NOT in the 0625 syllabus. 0625 §4.5.4 is qualitative: it only requires the relative directions of force, field and current (Fleming's left-hand rule), not a force calculation. This belongs to AS/A-level physics and will not be examined on 0625.
9EMF in a rotating coil
Extended• generator, coil
Question
A flat coil is rotated steadily in a magnetic field, producing an alternating e.m.f. State two ways to INCREASE the peak e.m.f. and explain each.
Step-by-step solution
1. Step 1
  Rotate the coil FASTER — the rate of change of flux linkage rises, so the induced e.m.f. rises.
2. Step 2
  Increase the NUMBER OF TURNS on the coil — every turn adds its own induced e.m.f., so more turns gives a larger total e.m.f. (The named Faraday's law with magnetic flux $\Phi$ — $\varepsilon \propto N \,\Delta \Phi / \Delta t$ — is AS/A-level content beyond 0625; 0625 only requires this qualitatively.)
3. Step 3
  (Other valid changes: stronger magnet, larger coil area, soft-iron core to increase flux density.)
Answer
Faster rotation, more turns (or stronger magnet / larger area / soft-iron core).
10Rate of change of flux — qualitative
Extended• Faraday, induction
Question
A bar magnet is dropped vertically through a flat coil. Describe how the induced e.m.f. varies as the magnet enters, passes through, and exits the coil.
Step-by-step solution
1. Step 1
  Entering: the magnetic flux through the coil INCREASES rapidly → induced e.m.f. in one direction (Lenz: opposes the increase).
2. Step 2
  Centred in the coil: rate of change of flux is at a MINIMUM (flux is at a maximum), so the induced e.m.f. briefly DROPS to near zero.
3. Step 3
  Leaving: flux DECREASES → induced e.m.f. REVERSES direction (Lenz: opposes the decrease).
4. Step 4
  Because the magnet has accelerated under gravity, the exit pulse is BIGGER and BRIEFER than the entry pulse.
Answer
A positive pulse on entry, near-zero at the centre, a larger negative pulse on exit. Magnitude of e.m.f. is proportional to the RATE of flux change.
Examiner tip
The examiner report flags candidates often saying 'no e.m.f. is induced' when the magnet is inside the coil — they confuse maximum flux with maximum rate of change of flux.
11DC motor commutator vs AC slip rings
Extended• commutator, slip rings
Question
A DC motor uses a split-ring commutator. An AC generator uses slip rings. Explain the function of each.
Step-by-step solution
1. Step 1
  DC motor: as the coil rotates past the vertical, the force on each side reverses. The split-ring (commutator) swaps the connections every half-turn so that current in each side of the coil is always in the direction needed to keep the coil rotating the same way.
2. Step 2
  AC generator: slip rings keep the same end of the coil permanently connected to the same external terminal. As the coil rotates, the induced e.m.f. naturally reverses every half-turn → the output is AC.
Answer
Commutator REVERSES connections every half-turn (DC motor — gives continuous rotation). Slip rings keep connections the SAME (AC generator — output naturally alternates).
12Step-down transformer current
Extended• transformer, current
Question
An ideal transformer steps $230\,\text{V}$ down to $12\,\text{V}$ . The secondary delivers $5.0\,\text{A}$ to a load. Find the primary current.
Step-by-step solution
1. Step 1
  Ideal transformer: $V_{p} I_{p} = V_{s} I_{s}$ (power conserved).
2. Step 2
  Rearrange and substitute.
  $I_{p} = \dfrac{V_{s} I_{s}}{V_{p}} = \dfrac{12 \times 5.0}{230} \approx 0.26\,\text{A}$
Answer
$I_{p} \approx 0.26\,\text{A}$
Examiner tip
The examiner report flags candidates often using the TURNS RATIO for current the same way as for voltage. For an ideal transformer, voltage is in the ratio $N_{p}/N_{s}$ but current is in the INVERSE ratio $N_{s}/N_{p}$ .
13Transmission power loss — synoptic
Challenge• Adapted from 0625/42 May/Jun 2023 Q9• transformer, I^2R, synoptic
Question
A power station produces $100\,\text{kW}$ . It is transmitted through cables of total resistance $4\,\Omega$ . Compare the power lost in the cables if the transmission voltage is (a) $1000\,\text{V}$ or (b) $100\,\text{kV}$ .
Step-by-step solution
1. Step 1
  (a) Transmission current: $I = P/V$ .
  $I_{a} = \dfrac{100\,000}{1000} = 100\,\text{A}$
2. Step 2
  Power lost: $P_{\text{loss}} = I^{2}R$ .
  $P_{\text{a, loss}} = 100^{2} \times 4 = 40\,000\,\text{W} = 40\,\text{kW}$
3. Step 3
  (b) New current at $100\,\text{kV} = 100\,000\,\text{V}$ :
  $I_{b} = \dfrac{100\,000}{100\,000} = 1.0\,\text{A}$
4. Step 4
  Power lost at high voltage:
  $P_{\text{b, loss}} = (1.0)^{2} \times 4 = 4\,\text{W}$
5. Step 5
  Ratio: $40\,000 / 4 = 10\,000$ — increasing $V$ by $\times 100$ reduces loss by $\times 10\,000$ (because $P_{\text{loss}} \propto 1/V^{2}$ ).
Answer
At $1\,\text{kV}$ : $40\,\text{kW}$ lost ( $40\%$ ). At $100\,\text{kV}$ : only $4\,\text{W}$ lost. Power loss falls as the square of voltage.
14Beyond 0625 — EMF induced in a moving rod ( $\varepsilon = BvL$ )
Challenge• BvL, induction
Question
A horizontal metal rod of length $L = 0.50\,\text{m}$ moves at speed $v = 4.0\,\text{m/s}$ perpendicular to a vertical magnetic field of flux density $B = 0.20\,\text{T}$ . Find the e.m.f. induced across the ends of the rod.
Step-by-step solution
1. Step 1
  Each free electron in the rod experiences a force $F = Bev$ perpendicular to $v$ and $B$ . This separates charges along the rod until the electric field balances the magnetic force.
2. Step 2
  At equilibrium, the induced e.m.f. is $\varepsilon = BvL$ .
3. Step 3
  Substitute.
  $\varepsilon = 0.20 \times 4.0 \times 0.50 = 0.40\,\text{V}$
Answer
$\varepsilon = 0.40\,\text{V}$
Examiner tip
Enrichment beyond Cambridge IGCSE 0625 — the equation ε = BvL, magnetic flux density and the unit tesla (T) are NOT in the 0625 syllabus. 0625 §4.5.1 is qualitative: it only requires the factors affecting the magnitude of an induced e.m.f. (speed of the change, number of turns, field strength), not a quantitative e.m.f. calculation. This belongs to AS/A-level physics and will not be examined on 0625.

Key Formulae — Electromagnetic Effects

The formulae you need to memorise for electromagnetic effects on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Transformer (turns) ratio
$\dfrac{V_{p}}{V_{s}} = \dfrac{N_{p}}{N_{s}}$
$N_{p}, N_{s}$
primary / secondary turns
When to use
Ideal transformer.
Ideal transformer power
$V_{p} I_{p} = V_{s} I_{s}$
When to use
Conservation of power for a 100%-efficient transformer.
Fleming's left-hand rule
$\text{Thumb: motion (F) | First: field (B) | Second: current (I)}$
When to use
Force on a current in a magnetic field.

Key Definitions and Keywords — Electromagnetic Effects

Definitions to memorise and the exact keywords mark schemes credit for electromagnetic effects answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Electromagnetic induction
Examiner keyword
Generation of an e.m.f. in a conductor when the magnetic flux through it changes.
Motor effect
Examiner keyword
A current-carrying wire in a magnetic field experiences a force.
Transformer
Examiner keyword
Device with two coils on a soft-iron core that uses electromagnetic induction to step voltage up or down.
Step-up vs step-down
Examiner keyword
Step-up: $N_{s} > N_{p}$ , $V_{s} > V_{p}$ . Step-down: $N_{s} < N_{p}$ .

Common Mistakes and Misconceptions — Electromagnetic Effects

The traps other students keep falling into on electromagnetic effects questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Using the right hand for the motor effect (or left for induction)
0625/42 — recurring
Why it happens
Mixing the rules.
How to avoid it
MOTOR effect → LEFT hand. GENERATOR (induction) → RIGHT hand.
✕Inverting the transformer ratio
Why it happens
Memory slip.
How to avoid it
$V$ is proportional to $N$ . Ratio with $V$ on top mirrors $N$ on top.
✕Saying transformers work on DC
Why it happens
Confusing AC and DC.
How to avoid it
Transformers need a CHANGING magnetic flux. DC produces no change → no induced e.m.f. (only AC works).
✕Stating 'a magnet near a coil' produces e.m.f. (without motion)
Why it happens
Forgetting the change requirement.
How to avoid it
Induction needs a CHANGE in flux — relative motion or a changing field.

Practice questions

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Electromagnetic Effects

Short Study Notes in the form of Slides

Short Study Notes — Electromagnetic Effects

Detailed Notes

What you’ll learn

How it’s examined

1Force on a current-carrying wire

2Transformer ratio

3Induced e.m.f.

4Why national grid uses high voltage

5Field around a long straight wire

6The magnetic effect of a current — the relay

7The magnetic effect of a current — the loudspeaker

8Beyond 0625 — Force on a current-carrying wire (F=BILF = BILF=BIL)

9EMF in a rotating coil

10Rate of change of flux — qualitative

11DC motor commutator vs AC slip rings

12Step-down transformer current

13Transmission power loss — synoptic

14Beyond 0625 — EMF induced in a moving rod (ε=BvL\varepsilon = BvLε=BvL)

Transformer (turns) ratio

Ideal transformer power

Fleming's left-hand rule

Electromagnetic induction

Motor effect

Transformer

Step-up vs step-down

✕Using the right hand for the motor effect (or left for induction)

✕Inverting the transformer ratio

✕Saying transformers work on DC

✕Stating 'a magnet near a coil' produces e.m.f. (without motion)

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Electromagnetic Effects — frequently asked questions

8Beyond 0625 — Force on a current-carrying wire ( $F = BIL$ )

14Beyond 0625 — EMF induced in a moving rod ( $\varepsilon = BvL$ )