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Electromagnetic induction generators and motors

Learning objectives

  • Outcome

    Predict the direction of the induced current in a coil when a magnet approaches or moves away, using Lenz's law.

  • Outcome

    Calculate the average induced emf for a given change in magnetic flux and time interval using Faraday's law.

  • Outcome

    Compare a generator and an electric motor by naming the input and output energy forms and the role of the magnetic field and current.

  • Outcome

    Calculate the maximum emf of a rotating-coil generator using E0=NBAω\mathcal{E}_0 = N B A \omega and solve for the rotation frequency.

Understand the idea

Electromagnetic Induction: Generators and Motors

Text

When a magnetic field passes through a closed loop, the magnetic flux Φ=BAcos⁡θ\Phi = B A \cos\theta describes how many field lines pass through the loop. Faraday discovered that when this flux changes, an induced emf appears in the loop: E=−NΔΦΔt\mathcal{E} = -N \frac{\Delta\Phi}{\Delta t}. The minus sign expresses Lenz's law: the induced current is directed so that its magnetic field opposes the change in flux. In a generator, mechanical rotational energy is converted into electrical energy: a coil rotates in a magnetic field, the flux changes periodically, and an alternating emf is produced. In a motor, the process is reversed: current in conductors in a magnetic field experiences the Ampère force, creating a torque, and electrical energy is converted into mechanical energy. Thus, a generator and a motor are two aspects of the same electromechanical energy-conversion process: a motor is essentially a generator driven in reverse, and its rotating coil also exhibits electromagnetic induction, which produces a back-emf that opposes the supply current.

Worked examples

Direct

Problem

A coil has N=100N = 100 turns. The magnetic flux through each turn increases uniformly from 0.020.02 Wb to 0.080.08 Wb in Δt=0.40\Delta t = 0.40 s. Find the average induced emf.

Solution

The flux change is ΔΦ=0.08−0.02=0.06\Delta\Phi = 0.08 - 0.02 = 0.06 Wb. By Faraday's law: E=NΔΦΔt=100⋅0.060.40=15 V.\mathcal{E} = N \frac{\Delta\Phi}{\Delta t} = 100 \cdot \frac{0.06}{0.40} = 15\ \text{V}. The magnitude of the average emf is 1515 V; the sign depends on the chosen loop direction and Lenz's law.

Application

Problem

A resistor is connected to a coil. A magnet's north pole is moved toward the coil. Determine the direction of the induced current, which pole appears at the near end of the coil, and the direction of the force on the magnet. Justify with Lenz's law.

Solution

As the north pole approaches, the magnetic flux through the coil increases. By Lenz's law, the induced current must create a magnetic field that opposes this increase. Therefore the near end of the coil becomes a north pole, repelling the approaching north pole of the magnet. The current flows so that its magnetic field is directed opposite to the magnet's pole; the direction can be found by the right-hand rule. The force on the magnet is directed away from the coil and slows the motion. Work done by the external force against this force is converted into electrical energy in the resistor.

Transfer

Problem

In a generator, a rectangular coil of area A=0.020A = 0.020 m$^2$ has N=50N = 50 turns and rotates at frequency f=60f = 60 Hz in a uniform magnetic field B=0.10B = 0.10 T. 1) Find the maximum emf. 2) At what rotation frequency would the maximum emf be 1212 V? Take ω=2πf\omega = 2\pi f.

Solution

The maximum emf of the generator is: E0=NBAω=NBA⋅2πf.\mathcal{E}_0 = N B A \omega = N B A \cdot 2\pi f. Substituting: E0=50⋅0.10⋅0.020⋅2π⋅60≈37.7 V.\mathcal{E}_0 = 50 \cdot 0.10 \cdot 0.020 \cdot 2\pi \cdot 60 \approx 37.7\ \text{V}. For E0=12\mathcal{E}_0 = 12 V: f=E02πNBA=122π⋅50⋅0.10⋅0.020≈19.1 Hz.f = \frac{\mathcal{E}_0}{2\pi N B A} = \frac{12}{2\pi \cdot 50 \cdot 0.10 \cdot 0.020} \approx 19.1\ \text{Hz}. This is an example of calculating generator parameters; if the same coil were used as a motor, the current in it in field BB would create a torque.

Practice

Text

Safe self-study investigation with a simulation.

  1. Open an electromagnetic induction simulation (for example, PhET «Faraday's Electromagnetic Lab» or a similar one) in the «Pickup Coil» mode.
  2. Slowly move a magnet into and out of the coil; record the direction of the needle deflection.
  3. Repeat at a higher speed; compare the size of the deflection.
  4. Change the number of turns in the coil and check how the current changes.
  5. Switch to the «Generator» mode: rotate the source and observe the alternating current; then in the «Motor» mode, supply current and observe rotation.
    Do not connect real devices to the mains; for a real experiment use only a low-voltage source (for example, a battery) and a low-power LED.

Visual reference

Purpose

Show the link between changing magnetic flux, induced current, and Lenz's law, and compare a generator and a motor as mutually inverse devices.

Representation

Diagram: a coil with a galvanometer, a magnet, and arrows showing its motion; curved arrows show the direction of the induced current and the coil's magnetic field opposing the motion. Next to it, two panels: a generator (rotation of a coil in field BB → alternating current) and a motor (current in a coil in field BB → rotation), with labels for fluxes and forces.

Real-world examples

Bicycle dynamo (bike light)

Scenario

A cyclist rides at 18 km/h=5 m/s18\ \text{km/h} = 5\ \text{m/s}. The dynamo roller of radius r=1.5 cmr = 1.5\ \text{cm} is pressed against the tyre of radius R=0.34 mR = 0.34\ \text{m}, so the roller spins at f=v/(2πr)≈53 Hzf = v/(2\pi r) \approx 53\ \text{Hz}. A permanent magnet rotates inside, and the magnetic flux through the coil reverses sign 2f≈1062f \approx 106 times per second. The generator produces an alternating voltage and lights a 6 V6\ \text{V}, 3 W3\ \text{W} lamp (current I=0.5 AI = 0.5\ \text{A}).

Connection

This is a generator: rotational energy becomes electrical energy. Faraday's law ε=−NΔΦΔt\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t} applies; for steady rotation Φ(t)=BAcos⁡(ωt)\Phi(t) = BA\cos(\omega t) with ω=2πf\omega = 2\pi f, so the emf amplitude ε0=NBAω\varepsilon_0 = NBA\omega grows with rotation rate, field BB, coil area AA and number of turns NN.

Regenerative braking in an electric car

Scenario

A 1500 kg1500\ \text{kg} electric car brakes from 20 m/s20\ \text{m/s} (72 km/h72\ \text{km/h}); its kinetic energy is 12mv2=3.0×105 J\tfrac12 mv^2 = 3.0\times10^5\ \text{J}. The motor is switched to generator mode: the wheels turn the rotor and current flows into a 400 V400\ \text{V} battery. Returning about 60%60\% of the energy puts 1.8×105 J1.8\times10^5\ \text{J} into the battery, that is a charge Q=E/U=450 C≈0.13 A⋅hQ = E/U = 450\ \text{C} \approx 0.13\ \text{A·h}.

Connection

This shows machine reversibility: the same motor becomes a generator when the rotor is turned from outside. Lenz's law fixes the direction of the induced current, and the resulting braking torque M=BINAM = BINA (for NN turns, current II, coil area AA) opposes the rotation, while energy conservation links the stored charge to the loss of 12mv2\tfrac12 mv^2.

Induction hob in the kitchen

Scenario

An induction hob transfers 2000 W2000\ \text{W} into the base of a cast-iron pan of diameter 20 cm20\ \text{cm}. The coil under the glass is fed with alternating current at 25 kHz25\ \text{kHz}, so the magnetic flux through the steel base reverses 50 00050\,000 times per second. Eddy currents in the metal heat the base to 200 ∘C200\ ^\circ\text{C} and above, the glass stays near 60 ∘C60\ ^\circ\text{C}, and the transferred fraction of power reaches 90%90\%.

Connection

Faraday's law is applied to a solid conductor: ε=−ΔΦΔt\varepsilon = -\dfrac{\Delta\Phi}{\Delta t} drives eddy currents, and Joule heating P=I2RP = I^2R warms the base. Lenz's law explains why the eddy currents oppose the change of flux, while the high frequency and large field BB make ΔΦ/Δt\Delta\Phi/\Delta t large. Here induction works without any wire with terminals.