DSE Physics: Electromagnetic Induction and Transformers, a Tutor Prep Guide

2026-11-04

Tutor lesson-prep guides · Secondary physics

Students can usually recite that induced emf follows the rate of change of flux, then lose the turns, use minutes instead of seconds, and scale the current down with the voltage. This lesson closes those three leaks.

Objectives — Students can: (1) state that an emf appears only when flux changes, and give three ways of changing it; (2) compute emf with e = N dPhi/dt and use Lenz law for direction; (3) use Vp/Vs = Np/Ns and then power conservation for the secondary current; (4) finish a 400-turn, dPhi = 0.006 Wb, 0.2 s calculation in four minutes (12 V).

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Flux	Is Phi = BA valid for any angle?	States the perpendicular component
Induction condition	A magnet at rest inside a coil: any current?	No, because flux is unchanged
Power	Does P = VI hold on both sides of an ideal transformer?	Yes, no losses

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Three-question check	Flux, induction condition, power	Answer orally
8-22 min	Cutting field lines	Rod, coil and magnet demonstrations	Say when flux changes
22-36 min	Faraday and Lenz	Substitute N, dPhi, dt step by step	Compute one alone
36-48 min	Transformers	Voltage ratio, current ratio, power	Find primary current
48-56 min	Timed practice	Two calculations, one direction question	Work unaided
56-60 min	Exit check	Three quick questions	Answer orally

03 Teaching script (read it out)

Scene 1: Ask whether flux changes before touching a formula

Every induction question starts with one sentence: is the flux changing? Since Phi = BA, there are only three ways it changes: B changes (moving magnet, changing current), A changes (coil deforms, enters or leaves the field), or the angle changes (rotating coil). The commonest error is to start calculating because the words magnet and coil both appear, when the coil is actually at rest in a uniform field and the answer is zero. Require a written sentence naming what changes before any working.

Make students write flux changes because ____ changes; if they cannot, they have not read the question.

Scene 2: Faraday law: three slots, no exceptions

e = N dPhi/dt has three slots: turns N, flux change dPhi, time dt. In this lesson N = 400, Phi rises from zero to 0.006 Wb (B = 0.4 T over A = 0.015 m2), and dt = 0.2 s, so e = 400 x 0.006 / 0.2 = 12 V. The answer format is fixed at three lines: Phi = BA with its value, then e = N dPhi/dt, then the answer with its unit. Lenz law needs one sentence only: the induced current opposes the change in flux, so a rising flux drives a current whose field points the other way.

Have students circle N, dPhi and dt in the question paper; no circles, no calculation.

Scene 3: Transformers: voltage by turns, power unchanged

Vp/Vs = Np/Ns governs voltage only. With Np = 1200, Ns = 60 and Vp = 220 V, Vs = 220 x 60/1200 = 11 V, a step-down. Current does not follow the turns ratio: an ideal transformer conserves power, so with 2.0 A in the secondary, Ip = 22 W / 220 V = 0.10 A, which is the same as Is x Ns/Np. Students who treat current as falling with voltage write the wrong answer, and a quick P = VI check on both sides exposes it immediately.

Write power is unchanged, so current rises twenty times, and make students justify it.

04 Question types and worked steps

Type	Example	Steps	Answer
Induced emf	Coil in a changing field	Phi = BA, then e = N dPhi/dt, unit V	12 V
Direction	Which way the current flows	Flux rising or falling, then oppose it	Opposes the change
Transformer voltage	Find Vs from turns	Vs = Vp x Ns/Np	11 V
Current and power	Find primary current	P = VsIs, then Ip = P/Vp	0.10 A

Sanity check (teach this): Check the numbers: Phi = 0.4 x 0.015 = 0.006 Wb, and e = 400 x 0.006 / 0.2 = 12 V. Dropping N gives 0.030 V, which is 400 times too small and an obvious marker. For the transformer, 11 V x 2.0 A = 22 W and 220 V x 0.10 A = 22 W must agree; a primary power larger than the secondary means the current direction was reversed. Units must also be consistent: area in m2 and time in s, so cm2 or ms in the question must be converted first.

05 Independent practice (3 questions)

Question	Answer	What to watch
N = 400, flux rises from 0 to 0.006 Wb in 0.2 s: find the emf	12 V	Wrote Phi = BA first
Transformer Np = 1200, Ns = 60, Vp = 220 V, secondary draws 2.0 A: find Vs and Ip	11 V and 0.10 A	Used power conservation
A magnet at rest inside a coil: what current?	Zero, flux is unchanged	Named the reason

06 Common mistakes and what to say

Mistake	Student writes	Say this
Turns dropped	e = 0.006/0.2 = 0.030 V	The formula is N dPhi/dt, so where did N go? Circle it and redo.
Flux used instead of flux change	Multiplying 0.40 by 0.015 and stopping	dPhi is the change, and it still must be divided by time.
Current scaled down with voltage	Primary current written as 0.10 A	Power is unchanged, so current rises; check with P = VI.
Units untouched	Area left in cm2	A must be in m2, and 1 cm2 is 1e-4 m2, a factor of ten thousand.

07 Homework (paper, section and question numbers)

Physics subject index — All chapters
Practice paper — Induction: questions 3, 5 and 7

08 Exit check (last 5 minutes)

Question	Answer	Decision
Induction condition	Flux must change	Cannot say it: reteach act one
Three slots	N, dPhi, dt	Missing one: redo the worked example
Power	P = VI equal on both sides	Cannot find Ip: return to act three

09 Tutor FAQ

Is the right-hand rule required?

Opposing the change is faster; the right-hand rule only makes it concrete, so use both as a check.

Is an ideal transformer 100 per cent efficient?

Only when the question says ideal; at 80 per cent, multiply the secondary power by 0.8.

Can the emf be negative?

The formula gives magnitude, and the sign carries direction, so state the size and then the direction.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.