33  Applications of the Motor Effect

33.1 Syllabus inquiry question

  • How has knowledge of the motor effect been applied to develop DC motors, generators, and other technologies?
Feynman Insight

From The Feynman Lectures on Physics, Vol II, Chapter 16:

The motor and the generator are essentially the same machine—run current through, and it spins; spin it, and current comes out. This beautiful reciprocity lies at the heart of our electrical civilization.

33.2 Learning Objectives

  • Calculate torque on a current loop: \(\tau = nIAB\sin\theta\)
  • Explain the role of the commutator in DC motors
  • Analyse back EMF and its effect on motor current
  • Compare DC motors, DC generators, and AC generators
  • Relate Lenz’s law to energy conservation in generators

33.3 Content

33.3.1 Torque on a Current Loop

A rectangular coil carrying current in a magnetic field experiences a torque:

\[\tau = nIAB\sin\theta\]

where: - \(\tau\) = torque (N·m) - \(n\) = number of turns - \(I\) = current (A) - \(A\) = area of coil (m²) - \(B\) = magnetic field strength (T) - \(\theta\) = angle between coil normal and field

Maximum vs Zero Torque
  • Maximum torque when \(\theta = 90°\) (coil plane parallel to field)
  • Zero torque when \(\theta = 0°\) (coil plane perpendicular to field)

33.3.2 Interactive: DC Motor Operation

Visualise torque and rotation in a DC motor:

33.3.3 The Commutator

The commutator is a split-ring device that:

  1. Reverses current direction every half rotation
  2. Maintains torque direction so the motor keeps spinning
  3. Converts DC input to the alternating current needed by the coil

Without a commutator, the motor would oscillate back and forth instead of rotating continuously.

33.3.4 Ways to Increase Motor Torque

From \(\tau = nIAB\sin\theta\), torque can be increased by:

Method Effect Practical Consideration
More turns (n) Linear increase Adds weight and resistance
Higher current (I) Linear increase More heating (I²R losses)
Larger coil area (A) Linear increase Larger motor size
Stronger magnet (B) Linear increase Stronger/heavier magnets
Radial field design Keeps \(\theta = 90°\) Standard in real motors

33.3.5 Back EMF in Motors

When a motor spins, the rotating coil acts as a generator, inducing an EMF that opposes the supply voltage:

\[V = IR + \varepsilon_{back}\]

or

\[I = \frac{V - \varepsilon_{back}}{R}\]

Starting Current

When a motor starts, \(\varepsilon_{back} = 0\) (no rotation), so starting current is high: \[I_{start} = \frac{V}{R}\] As the motor speeds up, back EMF increases and current decreases.

33.3.6 Interactive: Back EMF Effects

33.3.7 DC Generator

A DC generator is a motor run in reverse:

  1. Mechanical energy rotates the coil
  2. Changing flux through the coil induces an EMF
  3. The commutator converts AC output to pulsating DC
  4. Brushes transfer current to the external circuit

The induced EMF varies with angle: \[\varepsilon = nBAω\sin(ωt)\]

33.3.8 AC Generator (Alternator)

An AC generator uses slip rings instead of a commutator:

  • Output is sinusoidal AC: \(\varepsilon = \varepsilon_0\sin(ωt)\)
  • No sparking at brushes (smoother contact)
  • Simpler construction than DC generator

33.3.9 Interactive: Generator Comparison

33.3.10 Lenz’s Law and Energy Conservation

In a generator, Lenz’s law ensures energy conservation:

  1. Turning the generator coil requires mechanical work
  2. This induces a current that creates a magnetic force opposing rotation
  3. More current drawn → more force to overcome → more mechanical energy needed
  4. Electrical energy out = Mechanical energy in (minus losses)
No Free Energy

If the induced current aided rotation (instead of opposing it): - The generator would accelerate itself - Energy would be created from nothing - This violates conservation of energy

33.4 Worked Examples

33.4.1 Example 1: Motor torque

A 50-turn rectangular coil (0.08 m × 0.05 m) carries 0.40 A in a 0.30 T field. Find the maximum torque.

Solution:

  1. Use \(\tau = nIAB\sin\theta\) with \(\theta = 90°\) (maximum)

  2. \(A = 0.08 \times 0.05 = 0.004\ \text{m}^2\)

  3. \(\tau = 50 \times 0.40 \times 0.004 \times 0.30 \times 1 = 0.024\ \text{N·m}\)

33.4.2 Example 2: Back EMF and current

A motor with coil resistance 2.0 Ω is connected to 12 V. At operating speed, the back EMF is 8.0 V. Find the current at operating speed and at start-up.

Solution:

  1. At operating speed: \(I = \frac{V - \varepsilon_{back}}{R} = \frac{12 - 8.0}{2.0} = 2.0\ \text{A}\)

  2. At start-up (\(\varepsilon_{back} = 0\)): \(I_{start} = \frac{V}{R} = \frac{12}{2.0} = 6.0\ \text{A}\)

  3. Starting current is 3× the operating current!

33.4.3 Example 3: Generator EMF

A 100-turn coil of area 0.02 m² rotates at 50 Hz in a 0.25 T field. Find the peak EMF.

Solution:

  1. Angular velocity: \(ω = 2\pi f = 2\pi \times 50 = 314\ \text{rad/s}\)

  2. Peak EMF: \(\varepsilon_0 = nBAω\)

  3. \(\varepsilon_0 = 100 \times 0.25 \times 0.02 \times 314 = 157\ \text{V}\)

33.4.4 Example 4: Power in motor at different speeds

A 24 V motor has resistance 4.0 Ω. At full speed, it draws 2.0 A. Calculate the mechanical power output.

Solution:

  1. Back EMF: \(\varepsilon_{back} = V - IR = 24 - (2.0)(4.0) = 16\ \text{V}\)

  2. Total power input: \(P_{in} = VI = 24 \times 2.0 = 48\ \text{W}\)

  3. Power dissipated as heat: \(P_{heat} = I^2R = 4.0 \times 4.0 = 16\ \text{W}\)

  4. Mechanical power output: \(P_{mech} = P_{in} - P_{heat} = 48 - 16 = 32\ \text{W}\)

Alternatively: \(P_{mech} = I \times \varepsilon_{back} = 2.0 \times 16 = 32\ \text{W}\)

33.4.5 Example 5: Increasing generator output

A generator produces 10 V peak. State three ways to double this output to 20 V.

Solution:

From \(\varepsilon_0 = nBAω\):

  1. Double the number of turns (n × 2)
  2. Double the magnetic field strength (B × 2)
  3. Double the rotation frequency (ω × 2)

Other valid combinations: double the coil area, or any combination that doubles the product nBAω.

33.5 Common Misconceptions

Common Misconceptions
  • Misconception: Motors and generators are completely different devices. Correction: They are the same device run in opposite modes—motors convert electrical to mechanical energy; generators do the reverse.

  • Misconception: Back EMF is a design flaw that wastes energy. Correction: Back EMF is essential—it limits current and represents the conversion of electrical to mechanical energy.

  • Misconception: The commutator increases the output voltage. Correction: The commutator reverses current direction to maintain rotation (motor) or converts AC to pulsating DC (generator).

  • Misconception: A generator creates energy. Correction: A generator converts mechanical energy to electrical energy. The mechanical input must equal or exceed the electrical output.

  • Misconception: Higher motor speed means higher current. Correction: Higher speed means higher back EMF and lower current (with constant supply voltage).

33.6 Practice Questions

33.6.1 Easy (2 marks)

State the formula for torque on a current-carrying coil and identify when torque is maximum.

  • Formula: \(\tau = nIAB\sin\theta\) (1)
  • Maximum when θ = 90° (coil plane parallel to field) (1)

Answer: \(\tau = nIAB\sin\theta\); maximum when coil plane is parallel to the magnetic field (θ = 90°)

33.6.2 Medium (4 marks)

A DC motor connected to 9.0 V has coil resistance 1.5 Ω and back EMF of 6.0 V at operating speed. Calculate the operating current and explain why starting current is higher.

  • \(I = (V - \varepsilon_{back})/R = (9.0 - 6.0)/1.5 = 2.0\) A (1)
  • Starting current = V/R = 9.0/1.5 = 6.0 A (1)
  • At start, motor is stationary so no back EMF (1)
  • With no back EMF opposing supply, more current flows (1)

Answer: Operating current = 2.0 A; starting current is higher (6.0 A) because there is no back EMF when the motor is stationary.

33.6.3 Hard (5 marks)

A generator coil of 80 turns and area 0.015 m² rotates at 60 Hz in a 0.40 T field. Calculate the peak EMF and explain, using Lenz’s law, why turning the generator faster requires more mechanical power.

  • \(ω = 2\pi \times 60 = 377\) rad/s (1)
  • \(\varepsilon_0 = nBAω = 80 \times 0.40 \times 0.015 \times 377 = 181\) V (1)
  • Faster rotation → greater rate of flux change → larger induced EMF (1)
  • Larger EMF → larger induced current (1)
  • By Lenz’s law, this current creates forces opposing rotation, requiring more mechanical work to maintain speed (1)

Answer: \(\varepsilon_0 = 180\) V (or 181 V); faster rotation increases induced EMF and current, which by Lenz’s law creates greater opposing forces, requiring more mechanical power input.

33.7 Multiple Choice Questions

Test your understanding with these interactive questions:

33.8 Summary

Key Takeaways
  • Motor torque: \(\tau = nIAB\sin\theta\) (maximum when coil plane ∥ field)
  • Commutator reverses current to maintain rotation direction
  • Back EMF: \(V = IR + \varepsilon_{back}\); limits current as motor speeds up
  • DC generator uses commutator → pulsating DC output
  • AC generator uses slip rings → sinusoidal AC output
  • Peak EMF: \(\varepsilon_0 = nBAω\)
  • Lenz’s law ensures mechanical energy in = electrical energy out

33.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to:

33.10 Module 6 Complete

Congratulations on completing Module 6: Electromagnetism!

What you’ve learned
  • Charged particles in electric and magnetic fields
  • The motor effect and force on current-carrying conductors
  • Electromagnetic induction, Faraday’s law, and Lenz’s law
  • DC motors, generators, and transformers