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Physics · Electromagnetic Induction

Electromagnetic Induction, Inductance, and Alternating Current

A changing magnetic field creates a voltage. Investigate how the induced voltage depends on the rate of change of the field, the number of turns, and the area of a coil; explain self-inductance, the property by which a coil opposes changes in its own current; and see how a rotating coil makes the alternating voltage that powers the world. In each part you calculate the value yourself, then reveal the instrument and compare.

Theory — Induced Voltage, Inductance, and AC

One of the most useful discoveries in physics is that a changing magnetic field can create, or induce, a voltage. It is the change that matters, not the mere presence of a field. A steady magnetic field through a coil produces no voltage, but a field that grows or shrinks, or a coil that moves or turns, drives a current.

Magnetic flux and Faraday's law

The magnetic flux through a coil measures how much field passes through it, the field strength times the area it threads through. The induced voltage equals the number of turns times how fast that flux changes.

Flux and induced voltageMagnetic flux: Φ = B × A (field perpendicular to the coil)
Induced voltage: EMF = N × (change in Φ) / (change in time) = N × A × (ΔB / Δt)
More turns, a larger area, or a faster change in the field all increase the induced voltage

The minus sign in the full statement of the law, Lenz's law, says the induced voltage opposes the change that made it. For calculating sizes we use the magnitude.

Inductance

Inductance is the property by which a coil or circuit opposes changes in its own current by inducing a voltage that resists the change, much as inertia resists changes in motion. A coil with a large inductance reacts strongly to any attempt to change its current. For a long solenoid the inductance depends only on its geometry.

Self-inductance of a solenoidL = μ₀ × N² × A / ℓ
μ₀ = 4π × 10⁻⁷, N = turns, A = cross-section area, ℓ = length
The voltage a coil induces in itself is EMF = L × (ΔI / Δt)

Inductance is measured in henries. A changing current through an inductor induces a voltage across it proportional to how fast the current changes, which is why inductors smooth and oppose sudden current changes in circuits.

The AC generator

Turning a coil steadily inside a magnetic field changes the flux through it smoothly up and down, so the induced voltage rises and falls as a sine wave. This is alternating current, and the device is an electric generator. The faster the coil spins, the larger and more frequent the voltage swings.

Generator outputEMF(t) = N × B × A × ω × sin(ω t),  with ω = 2π f
Peak voltage: EMF₀ = N × B × A × ω
A transformer then changes the AC voltage in proportion to its turns: V_secondary / V_primary = N_secondary / N_primary

Change drives it

A steady field gives no voltage. Only a changing flux, from a changing field or a moving coil, induces a voltage.

Inductance opposes change

A coil induces a voltage in itself that resists any change in its current, the electrical version of inertia.

Generators make AC

A coil rotating in a field produces a sinusoidal voltage, the alternating current that runs the power grid.

QuantityRelationshipUnits
Magnetic fluxΦ = B Aweber (Wb)
Induced voltageEMF = N A (ΔB / Δt)volt (V)
InductanceL = μ₀ N² A / ℓhenry (H)
Inductor voltageEMF = L (ΔI / Δt)volt (V)
Generator peakEMF₀ = N B A ωvolt (V)

Apparatus

The equipment a real electromagnetic-induction experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

induction coil
Induction coil
The EMF is induced in this coil as the magnetic flux through it changes.
NSmagnetic field source
Bar magnet
Moved toward or away from the coil to change the flux.
0detects current
Galvanometer
Detects the small induced current and its direction.
coil with core
Solenoid
Carries a varying current to induce an EMF in a nearby coil.
displays waveforms
Oscilloscope
Displays the induced EMF as it varies with time.
sets frequency
Signal generator
Drives the primary coil with an alternating current.

Instructions — Running the Virtual Experiment

This is a record, calculate, and compare lab. In each part you set the apparatus, calculate the quantity by hand, enter your value, and only then does the simulation reveal its instrument reading so you can compare. Record every value in your worksheet.

Part A — Induced Voltage (Faraday's Law tab)
1
Open Simulation → Faraday's Law. Set the number of turns N, the coil area A, and the rate at which the field changes, ΔB/Δt.
2
Calculate the induced voltage from EMF = N A (ΔB/Δt), enter your value, and click Check. The voltmeter then reveals its reading so you can compare. Vary one quantity at a time and record how the induced voltage responds.
Part B — Inductance (Inductance tab)
1
Open Inductance. Set the turns N, area A, and length ℓ of a solenoid. With the target set to self-inductance, calculate L = μ₀ N² A / ℓ by hand, enter it in millihenries, and Check.
2
Switch the target to induced voltage and set a rate of current change ΔI/Δt. Calculate EMF = L (ΔI/Δt), enter it, and Check to compare.
Part C — AC Generator (Generator tab)
1
Open Generator. Set the turns N, field B, coil area A, and rotation frequency f. The coil spins and the output traces a sine wave.
2
Calculate the peak voltage from EMF₀ = N B A ω, where ω = 2π f, enter it, and click Check to compare. Note how raising the frequency raises both the peak voltage and how often it cycles.
For your reportInclude your data tables, your worked calculations for at least one case in each part, screenshots, and a short discussion comparing your calculated values with the simulation and explaining why a changing field, and not a steady one, induces a voltage.

Simulation — The Induction Bench

Electromagnetic Induction Virtual LabSet the apparatus, calculate, then reveal and compare
N (turns)A (cm²)ΔB/Δt (T/s)Your EMF (V)EMF actual (V)
No rows yet — set the coil, calculate EMF, enter it, and check.

Coil and changing field

Apparatus
Turns N100
Area A50 cm²
ΔB/Δt2.0 T/s
Voltmeter— hidden
TargetNA (cm²)ℓ (cm)Your valueActual
No rows yet — set the solenoid, calculate, enter, and check.

Solenoid

Apparatus
Turns N500
Area A5 cm²
Length ℓ20 cm
Inductance— hidden
NB (T)A (cm²)f (Hz)Your EMF₀ (V)EMF₀ actual (V)
No rows yet — set the generator, calculate the peak voltage, enter it, and check.

Rotating coil generator

Apparatus
Turns N50
Field B0.50 T
Area A100 cm²
Frequency f60 Hz
ω = 2πf
Peak voltage— hidden

Team Questions

Question 1. What must a magnetic field do to induce a voltage in a coil? (one word)
Question 2. A coil of 200 turns and area 0.01 m² sits in a field changing at 3 T/s. What is the induced voltage? (one number, in V)
Question 3. What is the property of a coil that opposes changes in its own current? (one word)
Question 4. In what unit is inductance measured? (one word)
Question 5. A 2 H inductor carries a current changing at 4 A/s. What voltage is induced across it? (one number, in V)
Question 6. What kind of voltage does a coil rotating steadily in a magnetic field produce? (two words)
Question 7 — Challenge. A transformer has 100 turns on the primary and 500 on the secondary. If the primary voltage is 120 V, what is the secondary voltage? (one number, in V)

Example Lab Report

A worked example showing the expected format and the record, calculate, and compare workflow.

Electromagnetic Induction, Inductance, and Alternating Current

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Objective

To describe and calculate induced voltages, to explain and calculate inductance, and to study the alternating voltage of a generator, comparing every calculated value with the simulation.

Part A — Induced Voltage (worked example)

Coil of N = 100 turns, area A = 50 cm² = 0.0050 m², field changing at ΔB/Δt = 2.0 T/s. EMF = N A (ΔB/Δt) = 100 × 0.0050 × 2.0 = 1.0 V. The voltmeter read 1.0 V.

Part B — Inductance (worked example)

Solenoid N = 500, A = 5 cm² = 5.0 × 10⁻⁴ m², ℓ = 0.20 m. L = μ₀ N² A / ℓ = (4π × 10⁻⁷)(500²)(5.0 × 10⁻⁴) / 0.20 = 7.85 × 10⁻⁴ H = 0.785 mH. With ΔI/Δt = 500 A/s, the induced voltage is L(ΔI/Δt) = 7.85 × 10⁻⁴ × 500 = 0.393 V. Both matched the simulation.

Part C — AC Generator (worked example)

N = 50, B = 0.50 T, A = 100 cm² = 0.010 m², f = 60 Hz, so ω = 2π(60) = 377 rad/s. Peak voltage EMF₀ = N B A ω = 50 × 0.50 × 0.010 × 377 = 94.2 V, matching the simulation.

Discussion and Conclusion

Every calculated value agreed with the simulation. A steady field produced no voltage; only a changing flux did. The induced voltage rose with turns, area, and the rate of field change; the solenoid inductance grew with the square of the turns; and the generator produced a sinusoidal voltage whose peak rose with frequency, confirming the principles of induction.

Practice Questions

Question 1
A 50-turn coil of area 0.020 m² is in a field that changes from 0.10 T to 0.40 T in 0.50 s. Find the induced voltage.
Hint: ΔB/Δt = 0.30/0.50 = 0.60 T/s; EMF = 50 × 0.020 × 0.60 = 0.60 V.
Question 2
A solenoid has 800 turns, area 4.0 × 10⁻⁴ m², and length 0.25 m. Find its inductance.
Hint: L = (4π × 10⁻⁷)(800²)(4.0 × 10⁻⁴)/0.25 = 1.29 × 10⁻³ H = 1.29 mH.
Question 3
The current through a 0.50 H inductor changes at 200 A/s. What voltage is induced across it?
Hint: EMF = L(ΔI/Δt) = 0.50 × 200 = 100 V.
Question 4
A generator coil of 80 turns and area 0.015 m² spins at 50 Hz in a 0.30 T field. Find the peak voltage.
Hint: ω = 2π(50) = 314 rad/s; EMF₀ = 80 × 0.30 × 0.015 × 314 = 113 V.
Question 5 — Challenge
Explain why holding a magnet still inside a coil produces no current, while pushing it in or pulling it out does.
Hint: only a changing flux induces a voltage; a stationary magnet gives a constant flux and so no induced voltage, while moving it changes the flux.