CLASS 12 PHYSICS · CHAPTER 6 · STUDY NOTES
Electromagnetic Induction
A comprehensive study of how changing magnetic fields generate electric currents, covering Faraday's laws, Lenz's law, motional emf, inductance, and the principles of AC generators.
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The Discovery of Electromagnetic Induction
Compare a moving magnet with two coils that stay still.Hold the magnet: the meter reads zero. Deflection is qualitative; it depends on speed, geometry and resistance.
Why this works and model limits
Relative motion matters because it changes linked flux, not because motion alone guarantees induction.
- Electromagnetic induction
- An emf appears when the magnetic flux linkage of a circuit changes. A closed conducting circuit can then carry current.
A magnet approaching a coil produces a meter deflection. Holding it nearby gives a stronger but steady flux, so the deflection stops. Withdrawing it reverses the change and the deflection.
Induction also works without mechanical motion: changing the current in a nearby stationary coil changes the receiver’s flux. Holding that primary current steady produces no receiver emf.
Relative motion is one way to change flux, rather than the universal requirement. In every case distinguish emf from current: an open circuit can have an induced emf without a circulating conduction current.
Take a moment. What do you think?
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Question
Two coils stay still. The source current first rises and then stays constant. When does the receiver have an induced emf?
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Understanding Magnetic Flux
Use the orientation and patch areas to find signed flux.For B = 0.50 T and A = 0.040 m², θ = 0° gives Φ = BA cos θ = 0.0200 Wb. θ is measured to the surface normal, not to its plane.
Why this works and model limits
The projected area can be zero although B and the actual area are nonzero. Reversing the chosen normal reverses signed flux.
- Magnetic flux
- Φ = ∫ B·dA. For a plane surface in uniform B, Φ = BA cosθ, where θ is measured to its chosen normal.
| Quantity | Symbol | SI Unit | Nature |
|---|---|---|---|
| Magnetic Flux | ΦB | Weber (Wb) | Scalar |
| Magnetic Field | B | Tesla (T) | Vector |
| Area | A | m² | Vector (Normal to surface) |
The area vector points perpendicular to the surface. A field parallel to that normal gives maximum positive flux; a field parallel to the surface gives zero flux. The opposite normal changes the sign.
One weber is one tesla square metre. Flux is signed, while area is a positive magnitude. A strong field can give zero flux through a particular surface.
For a non-uniform field or curved surface, add B·dA over small patches, or use the surface integral. Using the largest local field for the whole area generally overestimates the flux.
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Question
A uniform 0.30 T field makes 60° with the normal to a 0.020 m² loop. Find the flux.
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Faraday's Law of Induction
Connect the flux graph to Faraday’s signed emf.Each of N = 20 turns links the same flux. Φ changes by +0.006 Wb in 2 s: ε = −NΔΦ/Δt = -0.060 V.
Why this works and model limits
The orange slope and blue emf share the chosen positive normal. This constant-rate interval has a constant emf; endpoints mark the start and stop of that rate. Changing B, area or orientation can change flux. An open loop still has emf, but it carries no conduction current.
- Faraday’s law
- ε = −dλ/dt, where λ is flux linkage. For N turns with identical flux per turn, λ = NΦ and ε = −N dΦ/dt.
The steeper the flux-linkage graph, the larger the emf magnitude. The minus sign relates its direction to the chosen normal and circulation; Lenz’s law supplies the physical meaning.
An unchanged field can still induce emf when a loop’s area or orientation changes. A stationary fixed loop in a steady field has constant flux and no induced emf.
For a closed resistive circuit whose inductive response is negligible, I = ε/R. Its sign follows the chosen positive circulation. A stronger magnet alone does not guarantee a current: the linked flux must change.
An original flux-change calculation
ε_average = −NΔΦ/ΔtA 25-turn loop of area 0.012 m² has its normal along a uniform field. B rises from 0.10 T to 0.30 T in 0.50 s. ΔΦ = 0.0024 Wb per turn, so the average emf is −0.120 V for the chosen positive normal.
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Question
Thirty identical turns each gain 0.0040 Wb in 0.20 s. Find the average emf for that positive normal.
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Lenz's Law and Energy Conservation
Predict the induced field before naming the current direction.North approaches: flux into the page increases. The coil creates an outward field; viewed from the magnet, conventional current is anticlockwise and the near face acts as N.
Why this works and model limits
Repulsion opposes approach. Lenz’s law opposes the change, not always the original field. With a closed resistive loop, your mechanical work supplies electrical heating. Open the loop and current—and its magnetic drag—vanish in this ideal picture, even though a changing-flux emf remains.
- Lenz’s law
- The induced emf tends to drive a current whose magnetic effect opposes the change of flux that produced it.
Viewed from an approaching north end, the original field enters the coil. Its increasing flux induces an outward field: anticlockwise current gives a north face and repels the magnet.
When that north end withdraws, inward flux decreases. A clockwise current makes an inward field and a south face, attracting the receding magnet. The induced field now supports the original field while opposing its decrease.
In the resistive closed-loop experiment, the external agent supplies the work that becomes Joule heat. Opening the circuit prevents circulating current and its magnetic drag in this ideal model; changing-flux emf can remain.
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Question
Does Lenz’s law require the induced field always to point against the original field?
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Motional Electromotive Force
Use v×B to predict polarity, then calculate the voltage.The rod moves right at 1 m/s in B = 0.40 T into the page. Positive charges feel v×B upward: the top end is positive.
Why this works and model limits
For l = 0.50 m, ε = Blv = 0.200 V. With total resistance 3 Ω, I = 0.067 A anticlockwise. Magnetic drag is left. External power Fv = εI = 0.013 W equals I²R. This model keeps v perpendicular to both B and the rod; changing position alone at rest is not a sustained voltage source.
- Motional emf
- For a straight rod perpendicular to both v and uniform B, |ε| = Blv. More generally its contribution is ∫(v×B)·dl.
A rod moving right in a field into the page pushes positive charge upward by q(v×B), leaving the top positive. In an isolated rod an electric field builds up against further separation.
On closed rails the separated charge drives a current. The rod experiences magnetic drag. The force maintaining a steady speed supplies power εI, equal to I²R for the ideal resistive circuit.
A rod rotating about one end in a field normal to its rotation plane has speed ωr along its length. Integration gives |ε| = ½BωR². Polarity depends on rotation and field directions; the straight-rod Blv expression cannot use one end’s speed for the entire rotating rod.
Integrating a spoke’s changing speed
|ε| = ½BωR²With B = 0.30 T, ω = 8.0 rad/s and R = 0.25 m, the rim-to-centre emf magnitude is 0.075 V. Multiple spokes join the same rim and axle in parallel, so this voltage does not multiply by the number of spokes.
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Question
A 0.40 m rod moves at 1.5 m/s perpendicular to a 0.20 T field. What happens if the speed doubles?
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Mutual Inductance
Identify the common cross-section and the changing source current.For long nested air-core solenoids, n₁ = 500 m⁻¹, n₂ = 1000 m⁻¹, inner A = 0.0020 m² and l = 0.50 m: M = μ₀n₁n₂Al = 0.628 mH. Source current is steady: ε_receiver = −M dI_source/dt = 0.000 mV.
Why this works and model limits
Exchanging source and receiver gives the same M for this fixed linear geometry. The inner cross-section sets shared flux; using the larger outer area would overcount it. End effects are neglected.
- Mutual inductance
- For fixed linear geometry, λ_receiver = M I_source and ε_receiver = −M dI_source/dt. M is measured in henries.
Coils couple magnetically without sharing a wire connection. The receiver’s emf depends on the rate of source-current change; a nonzero steady current gives steady linked flux rather than an induced emf.
For very long nested coaxial air-core solenoids, M = μ₀n₁n₂A_inner l. The common flux is limited by the smaller cross-section. The approximation neglects end fields and assumes the coils have the same length.
Exchanging source and receiver preserves M in this reciprocal, fixed linear arrangement. Separation, orientation, geometry and permeability affect coupling. Do not treat one constant M as universal when the geometry or nonlinear material state changes.
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Question
For M = 0.020 H, a source current falls at 3.0 A/s. Find the receiver emf for matching positive winding conventions.
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Self-Inductance and Magnetic Energy
Compare growth, holding, decay and the energy stored.For L = 0.40 H, dI/dt = +2 A/s gives ε_self = −0.80 V: the back emf resists growth. It opposes change in current, not current itself.
Why this works and model limits
For a long linear solenoid, L = μn²Al; geometry and permeability set L, not the instantaneous current. The graph uses an idealised prescribed ramp, not an exponential RL transient.
- Self-inductance
- For a fixed linear coil, λ = LI and ε_self = −L dI/dt. Its stored magnetic energy is U = ½LI².
| Property | Mechanical Analogue | Electrical Analogue |
|---|---|---|
| Inertia | Mass (m) | Inductance (L) |
| Velocity/Flow | Velocity (v) | Current (I) |
| Stored Energy | (1/2)mv² | (1/2)LI² |
The induced emf opposes increasing current and helps sustain decreasing current. A steady current produces no self-induced emf, although it can maintain a magnetic field and stored energy.
A long linear solenoid has L = μn²Al. Its turn density, cross-section, length and permeability determine L. Establishing current requires work against the induced emf; in the ideal lossless limit that work becomes magnetic energy.
The area under λ versus I gives ½LI² for a constant L. Doubling I quadruples the stored energy. In air or vacuum the magnetic energy density is B²/(2μ₀); do not apply that vacuum expression indiscriminately inside a material.
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Question
A 0.30 H coil carries 2.0 A. Calculate its stored energy, then its self-induced emf if the current is held steady.
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The AC Generator
Connect the coil, its normal, slip rings and the paired waveforms.A 40-turn coil of area 0.010 m² rotates at f = 2 Hz in B = 0.20 T. At θ = 0°, flux per turn = 2.000 mWb and ε = NBAω sin θ = 0.000 V; peak ε₀ = 1.005 V.
Why this works and model limits
Flux is largest at 0°/180°, where emf is zero; emf peaks at 90°/270°, where flux changes fastest. Continuous slip rings keep each coil end connected to its own brush; they do not rectify the output. A resistive load would carry alternating current. The top picture is an oblique projection; the angle dial defines θ to the normal. Playback slows one revolution to four seconds for teaching; the numerical example still uses f = 2 Hz.
- AC generator
- Mechanical rotation changes a coil’s flux. For uniform rotation, Φ = BA cosωt per turn and ε = NBAω sinωt.
A coil’s axis of rotation is perpendicular to B. At θ = 0 its normal is along B: flux is largest but its instantaneous rate of change is zero. At 90° the flux is zero while its rate of change—and positive emf—is greatest.
Two continuous slip rings and fixed brushes connect the rotating coil ends to the external circuit. They retain each end’s connection while the induced polarity reverses every half-turn. A resistive closed load carries alternating current.
The peak emf is NBAω, with ω = 2πf for this simple generator. Electrical frequency equals coil-rotation frequency for the model shown. Real multi-pole machines need not have the same mechanical and electrical frequencies; mechanical input supplies the electrical output and losses.
An original small generator
ε₀ = NBA(2πf)For N = 60, B = 0.15 T, A = 0.020 m² and f = 3.0 Hz, the peak emf is 3.393 V. This is peak voltage, not an RMS value. Halving the rotation frequency halves the peak emf for the same field and coil.
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Question
The coil turns from θ = 90° to 180°. Which reaches zero: flux or emf?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 17, 18, 19.