32  Electromagnetic Induction

32.1 Syllabus inquiry question

  • How are electric and magnetic fields related?
Feynman Insight

From The Feynman Lectures on Physics, Vol II, Chapter 17:

Faraday discovered that a changing magnetic field creates an electric field. This seemingly simple observation is one of the great unifying principles of physics—it connects electricity and magnetism into a single phenomenon.

32.2 Learning Objectives

  • Calculate magnetic flux: \(\Phi = BA\cos\theta\)
  • Apply Faraday’s law: \(\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\)
  • Explain Lenz’s law and its connection to energy conservation
  • Analyse transformer operation: turns ratio and power conservation
  • Evaluate real transformer efficiency and losses

32.3 Content

32.3.1 Magnetic Flux

Magnetic flux measures how much magnetic field passes through a surface:

\[\Phi = BA\cos\theta\]

where: - \(\Phi\) = magnetic flux (Wb, Weber) - \(B\) = magnetic field strength (T) - \(A\) = area of surface (m²) - \(\theta\) = angle between field and normal to surface

Understanding θ
  • \(\theta = 0°\): Field perpendicular to surface → maximum flux (\(\Phi = BA\))
  • \(\theta = 90°\): Field parallel to surface → zero flux (\(\Phi = 0\))

32.3.2 Interactive: Magnetic Flux Visualisation

32.3.3 Faraday’s Law

A changing magnetic flux induces an electromotive force (EMF):

\[\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\]

where: - \(\varepsilon\) = induced EMF (V) - \(N\) = number of turns in coil - \(\Delta\Phi\) = change in flux (Wb) - \(\Delta t\) = time interval (s)

Ways to Change Flux
  1. Change the magnetic field strength (B)
  2. Change the area of the loop (A)
  3. Change the orientation angle (θ)
  4. Move the loop in/out of the field

32.3.4 Lenz’s Law

The negative sign in Faraday’s law represents Lenz’s Law:

The induced current flows in a direction that opposes the change in flux that caused it.

Energy Conservation

Lenz’s law is a consequence of energy conservation. If the induced current aided the change: - It would create a runaway effect - Energy would appear from nowhere - This violates conservation of energy

32.3.5 Interactive: Faraday and Lenz

32.3.6 Motional EMF

When a conductor moves through a magnetic field, an EMF is induced:

\[\varepsilon = Blv\]

where: - \(B\) = magnetic field strength (T) - \(l\) = length of conductor in field (m) - \(v\) = velocity perpendicular to field (m/s)

This is equivalent to Faraday’s law: the conductor sweeps out an area, changing the flux.

32.3.7 Transformers

A transformer transfers electrical energy between circuits through electromagnetic induction:

Turns Ratio: \[\frac{V_p}{V_s} = \frac{N_p}{N_s}\]

  • Step-up: \(N_s > N_p\) → voltage increases
  • Step-down: \(N_s < N_p\) → voltage decreases

Power Conservation (ideal): \[V_p I_p = V_s I_s\]

32.3.8 Interactive: Transformer Operation

32.3.9 Transformer Efficiency

Real transformers have energy losses:

Loss Type Cause Mitigation
Resistive (I²R) Current through wire resistance Thicker wires
Eddy currents Induced currents in core Laminated core
Hysteresis Magnetisation cycles in core Soft iron core
Flux leakage Not all flux links both coils Tight coil winding

Efficiency: \[\eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{V_p I_p}\]

Large power transformers achieve >99% efficiency.

32.4 Worked Examples

32.4.1 Example 1: Magnetic flux calculation

A rectangular coil (0.15 m × 0.10 m) is in a 0.40 T field at 60° to the normal. Find the flux through it.

Solution:

  1. Use \(\Phi = BA\cos\theta\)

  2. \(A = 0.15 \times 0.10 = 0.015\ \text{m}^2\)

  3. \(\Phi = 0.40 \times 0.015 \times \cos(60°) = 0.006 \times 0.5 = 0.0030\ \text{Wb}\)

32.4.2 Example 2: Induced EMF from flux change

A 200-turn coil experiences a flux change from 0.008 Wb to 0.002 Wb in 0.025 s. Find the induced EMF.

Solution:

  1. \(\Delta\Phi = 0.002 - 0.008 = -0.006\ \text{Wb}\)

  2. Use \(\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\)

  3. \(\varepsilon = -200 \times \frac{-0.006}{0.025} = -200 \times (-0.24) = 48\ \text{V}\)

32.4.3 Example 3: Motional EMF

A 0.20 m conductor moves at 3.0 m/s perpendicular to a 0.50 T field. Find the induced EMF.

Solution:

  1. Use \(\varepsilon = Blv\)

  2. \(\varepsilon = 0.50 \times 0.20 \times 3.0\)

  3. \(\varepsilon = 0.30\ \text{V}\)

32.4.4 Example 4: Transformer calculations

A transformer has 800 primary turns and 200 secondary turns. If the primary voltage is 240 V and the secondary current is 2.0 A, find the secondary voltage and primary current (assuming ideal).

Solution:

  1. Secondary voltage: \(V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{200}{800} = 60\ \text{V}\)

  2. Primary current (power conservation): \(I_p = \frac{V_s I_s}{V_p} = \frac{60 \times 2.0}{240} = 0.50\ \text{A}\)

32.4.5 Example 5: Transformer efficiency

A transformer has input power 500 W and output power 475 W. Calculate its efficiency and the power lost.

Solution:

  1. Efficiency: \(\eta = \frac{P_{out}}{P_{in}} = \frac{475}{500} = 0.95 = 95\%\)

  2. Power lost: \(P_{lost} = P_{in} - P_{out} = 500 - 475 = 25\ \text{W}\)

32.5 Common Misconceptions

Common Misconceptions

  • Misconception: A steady magnetic field induces an EMF. Correction: Only a changing flux induces an EMF. A constant field through a stationary loop produces zero EMF.

  • Misconception: The negative sign in Faraday’s law means EMF is always negative. Correction: The sign indicates the induced EMF opposes the change (Lenz’s law), not that the value is negative.

  • Misconception: Transformers work with DC. Correction: Transformers require AC. DC produces no flux change, so no EMF is induced in the secondary.

  • Misconception: Step-up transformers create energy. Correction: Power is conserved: higher voltage means lower current. Total power out ≤ power in.

  • Misconception: Laminating the core increases flux. Correction: Lamination reduces eddy currents by increasing core resistance, reducing energy loss.

32.6 Practice Questions

32.6.1 Easy (2 marks)

Calculate the magnetic flux through a 0.04 m² coil in a 0.25 T field when the field is perpendicular to the coil.

  • Use \(\Phi = BA\cos(0°) = BA\) (1)
  • \(\Phi = 0.25 \times 0.04 = 0.010\) Wb (1)

Answer: 0.010 Wb (or 10 mWb)

32.6.2 Medium (4 marks)

A 150-turn coil has its flux reduced from 0.012 Wb to zero in 0.030 s. Calculate the average induced EMF and explain, using Lenz’s law, the direction of the induced current.

  • \(\Delta\Phi = 0 - 0.012 = -0.012\) Wb (1)
  • \(\varepsilon = -N\Delta\Phi/\Delta t = -150 \times (-0.012)/0.030 = 60\) V (1)
  • Lenz’s law: induced current creates field to oppose the decrease (1)
  • Current direction creates field in same direction as original field (1)

Answer: \(\varepsilon = 60\) V; current flows to create a magnetic field opposing the decrease (maintaining the original field direction)

32.6.3 Hard (5 marks)

A step-up transformer has a turns ratio of 1:20. The primary is connected to 12 V AC and draws 2.4 A. The secondary delivers 0.10 A to a load. Calculate the secondary voltage, output power, and transformer efficiency.

  • Secondary voltage: \(V_s = 12 \times 20 = 240\) V (1)
  • Input power: \(P_{in} = 12 \times 2.4 = 28.8\) W (1)
  • Output power: \(P_{out} = 240 \times 0.10 = 24\) W (1)
  • Efficiency: \(\eta = 24/28.8 = 0.833 = 83.3\%\) (1)
  • Power lost = 28.8 - 24 = 4.8 W (1)

Answer: \(V_s = 240\) V, \(P_{out} = 24\) W, efficiency = 83%

32.7 Multiple Choice Questions

Test your understanding with these interactive questions:

32.8 Quick Quiz: Electromagnetic Induction

Test your understanding of induction with this timed quiz:

32.9 Extended Response Practice

32.10 Summary

Key Takeaways
  • Magnetic flux: \(\Phi = BA\cos\theta\) (maximum when field ⊥ surface)
  • Faraday’s law: \(\varepsilon = -N\Delta\Phi/\Delta t\)
  • Lenz’s law: induced current opposes the change in flux
  • Motional EMF: \(\varepsilon = Blv\)
  • Transformer turns ratio: \(V_p/V_s = N_p/N_s\)
  • Ideal transformer: \(V_p I_p = V_s I_s\) (power conserved)
  • Real losses: resistive (I²R), eddy currents, hysteresis, flux leakage

32.11 Self-Assessment

Check your understanding:

After studying this section, you should be able to: