Egypt Secondary 3 Physics, Week 15: Review: induction, generators, transformers and motors

Term 1 review: circuits, magnetism, induction and AC

Week 15 · Term 1 review (Chapters 1–4) and Darslab’s own review tasks · 3 lessons

Review: induction, generators, transformers and motors

Physics Sec 3 Term 1, Chapters 1–4: Review

GoalCombine Faraday’s law, Lenz’s rule, mutual and self-induction, the generator, effective values, the transformer and the motor in multi-step tasks.

New words

effective value · 0.707 of the maximum: the value of the DC that heats the samepositive direction · the direction a task fixes for the flux or the current, so that the sign of an emf has one meaningefficiency · output power ÷ input power; it changes the currents, not the turns ratio

Explanation

Chapter 3 is a second toolbox. An emf appears only while the flux changes: emf = −N ΔΦ ÷ Δt, and the minus sign is Lenz’s rule, the induced field opposes the change. Say the positive direction in the task (for example the direction of the starting flux): a falling flux then gives a positive emf, which means that the induced current’s field is in the direction of the old flux and tries to keep it. A generator turns a coil in a field: emf = NBAω sin θ with ω = 2πf and θ the angle of the normal to the coil from the field; it is largest when the coil’s plane lies along the field. Meters and mains give effective values: Veff = 0.707 Vmax, so 220 V has a peak of 311 V. A transformer follows the turns for voltages (Vs ÷ Vp = Ns ÷ Np) and the efficiency for currents (Is = η·Vp·Ip ÷ Vs). Power lost in a line is I²R. A motor is a coil in a field: F = NBIℓ on each side, τ = BIAN sin θ, and the commutator reverses the current every half turn so that the torque keeps its sense. Work step by step and check units and the size of every answer.

Topic	Tool
Faraday, Lenz	emf = −N ΔΦ ÷ Δt; Φ = BA; the induced field opposes the change
M and L	emf = −M ΔI ÷ Δt; emf = −L ΔI ÷ Δt
moving wire, generator	emf = Bℓv sin θ; emf(max) = NBAω, ω = 2πf
effective values	Ieff = 0.707 Imax; Veff = 0.707 Vmax; 220 V is an effective value
transformer	Vs ÷ Vp = Ns ÷ Np; ideal: Ip ÷ Is = Ns ÷ Np; η = VsIs ÷ VpIp
line loss, motor	P = I²R; F = NBIℓ, τ = BIAN sin θ

Worked examples

A coil of N = 90 turns has its flux fall from 5.0 × 10⁻³ Wb to 2.0 × 10⁻³ Wb in 0.05 s (positive direction: the direction of the starting flux). ΔΦ = −3.0 × 10⁻³ Wb, so emf = −90 × (−3.0 × 10⁻³) ÷ 0.05 = +5.4 V. The size is 5.4 V. The positive sign means that the induced current’s field points the same way as the old flux and opposes its fall.

Quantity	Value
ω	314 rad/s
emf(max)	75.4 V
emf at θ = 30°	37.7 V
effective value	53.3 V

A generator coil has N = 160 turns, A = 0.03 m² and B = 0.05 T, and turns at 50 Hz (π = 3.14). ω = 2 × 3.14 × 50 = 314 rad/s. emf(max) = NBAω = 160 × 0.05 × 0.03 × 314 = 75.4 V. The effective value is 0.707 × 75.4 = 53.3 V. When the normal makes 30° with the field, emf = 75.4 × sin 30° = 37.7 V.

A transformer has Np = 1000 turns, Ns = 250 turns, Vp = 220 V, Ip = 0.8 A and efficiency 90 %. Vs = 220 × 250 ÷ 1000 = 55 V. Input power = 220 × 0.8 = 176 W. Output power = 0.9 × 176 = 158.4 W. Is = 158.4 ÷ 55 = 2.88 A. Power lost as heat = 176 − 158.4 = 17.6 W.

Class activity

Paper only. Make a one-page summary of Chapter 3 with the table of tools, and add for each tool a “watch out” line: the sign of an emf and the positive direction; effective or maximum value; which angle in the generator; turns for voltage and efficiency for current. Then draw the four quarters of one turn of a generator coil, showing how the coil lies and what the emf is in each (zero or maximum).

Practice

The primary current of a pair of coils with M = 0.2 H falls from 3 A to 1 A in 0.04 s. Find the size of the emf in the secondary.

A coil of L = 0.35 H has its current fall from 6 A to 2 A in 0.14 s. Find the size of the self-induced emf and say whether it acts along the old current or against it.

A wire 0.6 m long moves at 5 m/s through a field of 0.15 T. Find the emf when the velocity is at right angles to the field, and when it makes 30° with it.

Spot the error. A student says: “The supply is 220 V, so its peak is 220 V, and an ammeter reading of 5 A means a peak of 5 A.” Correct both statements.