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State Board · Class 12 · Physics

Lesson 12: Electromagnetic Induction

7 concepts · 56 practice questions with answers · Free to read

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Faraday's laws, magnetic flux and induced emf

From daily life

In a cycle dynamo, turning the wheel changes the flux through a coil and lights the lamp.

Simple definition

Magnetic flux = B × area × cos θ; the induced emf equals the rate of change of flux.

Changing flux induces an emf××××××××××××Φ = B A cosθe = −N dΦ/dtθ = angle between A and B

In the figure : Φ = BA cosθ; a changing Φ ⇒ an emf.

Rule 1: Magnetic flux — Φ = B A cos θ, unit weber (Wb = T m²), dimensions [M L² T⁻² A⁻¹]. θ = angle with the area vector.

  • Easy : B = 0.4 T, A = 20 cm², θ = 60° ⇒ Φ = 4 × 10⁻⁴ Wb.
  • Medium : B parallel to the plane ⇒ θ = 90° ⇒ Φ = 0.
  • Exam (CET) : 1 Wb = ? (1) 1 T m² (2) 1 T/m² (3) 1 T m (4) 1 N/A → answer 1)

Rule 2: Faraday's laws — (1) A changing flux induces an emf. (2) e = −N dΦ/dt; N turns give N times the emf.

  • Easy : N = 100, Φ changes 0.1 → 0.5 Wb in 0.2 s ⇒ e = 100 × 0.4 ÷ 0.2 = 200 V.
  • Medium : Current I = e ÷ R; charge q = ΔΦ ÷ R (independent of time).
  • Exam (CET) : q = ? (1) NΔΦ/R (2) NΔΦ R (3) e R (4) ΔΦ → answer 1) Trick : q = (turns × ΔΦ) ÷ R.

Rule 3: Ways to change the flux — Change B, change the area A or change the angle θ — each gives an emf.

  • Easy : Rotating the loop changes θ (generator).
  • Medium : Squeezing the loop changes A.
  • Exam (CET) : No emf with? (1) constant Φ (2) changing B (3) changing A (4) changing θ → answer 1)

Derivation : e = −N dΦ/dt and charge q = NΔΦ/R

  • Step 1 : Total flux linkage = NΦ; e = −d(NΦ)/dt.
  • Step 2 : I = e ÷ R = −(N/R) dΦ/dt.
  • Step 3 : dq = I dt = −(N/R) dΦ.
  • Step 4 : q = NΔΦ ÷ R (sign by direction).
Remember

Φ = BA cosθ; e = −N dΦ/dt; q = NΔΦ/R.

Common mistake / CET trap

emf depends on the rate of change, not on the size of Φ; q is independent of time.

Shortcut

Memory clue : "rate matters, not amount".

Spot the difference

QuantityFormulaUnit / dimensions
ΦBA cosθWb; [ML²T⁻²A⁻¹]
e−N dΦ/dtV
qNΔΦ/RC

Practice questions

Easy
  1. 1Unit of Φ?
  2. 2Faraday's law?
  3. 3When is an emf induced?
Medium
  1. 4B = 0.2, A = 0.05 m², θ = 0° : Φ?
  2. 5N = 50, ΔΦ = 0.04 Wb, Δt = 0.1 s : e?
  3. 6R = 10 Ω, ΔΦ = 0.5 Wb, N = 20 : q?
Challenging
  1. 7Φ = 5t² − 3t (Wb), N = 1, t = 2 s : e?
  2. 8Turns doubled : emf and q?
Show answers (try them yourself first)
  1. 1Weber
  2. 2e = −N dΦ/dt
  3. 3When Φ changes
  4. 4Φ = 0.01 Wb
  5. 5e = 20 V
  6. 6q = 1 C
  7. 7e = −(10t − 3) = −17 V
  8. 8Both double

Lenz's law and conservation of energy

From daily life

When someone pushes you, you resist; nature resists a change in the same way.

Simple definition

Lenz's law gives the direction of the induced current.

Lenz's law : magnet toward a coilNSN facing N : repulsionthe minus sign : opposes the change

In the figure : N approaches ⇒ N face ⇒ repulsion.

Live figure

Lenz's law and conservation of energy

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Rule 1: Lenz's law — The induced current flows so as to oppose the cause (the flux change) that produces it — that is the minus sign in e = −N dΦ/dt.

  • Easy : North pole approaching ⇒ the near face becomes a north pole (repulsion).
  • Medium : North pole receding ⇒ the near face becomes a south pole (attraction).
  • Exam (CET) : As the N pole approaches, the near face of the coil is? (1) N (2) S (3) neutral (4) varying → answer 1)

Rule 2: Energy conservation — Lenz's law is energy conservation. Without opposition the magnet would speed up by itself and energy would appear for free.

  • Easy : Work done pushing the magnet = electrical energy (heat).
  • Medium : In an open coil there is emf but no current ⇒ no opposition.
  • Exam (CET) : Lenz's law follows from? (1) energy (2) charge (3) momentum (4) mass → answer 1)

Rule 3: Finding the direction — (1) direction of the original B, (2) is Φ increasing or decreasing, (3) induced B opposes (if rising) or supports (if falling), (4) right-hand rule for I.

  • Easy : B into the page increasing ⇒ induced B out of the page ⇒ I anticlockwise.
  • Medium : B into the page decreasing ⇒ I clockwise.
  • Exam (CET) : B (into the page) rises; I direction? (1) anticlockwise (2) clockwise (3) zero (4) both → answer 1)

Derivation : Lenz's law and energy

  • Step 1 : Suppose the induced force pulled the magnet further in.
  • Step 2 : The magnet would speed up and the flux change would grow.
  • Step 3 : This would give energy with no input — impossible.
  • Step 4 : So the induced current opposes (the minus in e).
Remember

Direction : Φ rising ⇒ opposing B; Φ falling ⇒ supporting B.

Common mistake / CET trap

It opposes the change of Φ, not Φ itself.

Shortcut

Memory clue : "rise ⇒ opposite, fall ⇒ same".

Spot the difference

SituationΦInduced B
Φ risesincreaseopposite
Φ fallsdecreasesame

Practice questions

Easy
  1. 1State Lenz's law.
  2. 2What does the minus sign show?
  3. 3Lenz expresses which conservation?
Medium
  1. 4South pole approaches : near face?
  2. 5Φ decreases : induced B?
  3. 6B out of the page rises : I direction?
Challenging
  1. 7Why does a magnet fall slowly through a copper tube?
  2. 8Why no Lenz opposition in an open coil?
Show answers (try them yourself first)
  1. 1The induced current opposes its cause
  2. 2Opposition
  3. 3Energy
  4. 4S
  5. 5Same direction
  6. 6Clockwise
  7. 7Eddy currents give an opposing force
  8. 8No current ⇒ no magnetic force

Motional emf

From daily life

Iron train axles moving in Earth's field develop a small emf.

Simple definition

Motional emf is the emf due to a conductor moving in a magnetic field.

Motional emf : a rod sliding on railsv××××××××××××le = B l v

In the figure : The rod slides at v; e = Blv.

Rule 1: Motional emf — A rod of length l moved at v perpendicular to B : e = B l v. With an angle, use the perpendicular component of v.

  • Easy : B = 0.5 T, l = 0.4 m, v = 6 m/s ⇒ e = 1.2 V.
  • Medium : v ∥ B ⇒ e = 0.
  • Exam (CET) : e depends on? (1) B, l, v (2) B only (3) R (4) m → answer 1)

Rule 2: Current and force — With R = 2 Ω : I = e ÷ R; opposing force F = B I l = B² l² v ÷ R; mechanical power = F v.

  • Easy : With e = 1.2 V, R = 2 Ω ⇒ I = 0.6 A; F = 0.12 N; P = 0.72 W.
  • Medium : P = I² R = e²/R = heat (the same).
  • Exam (CET) : v doubled ⇒ F? (1) doubles (2) halves (3) four times (4) same → answer 1)

Rule 3: Direction and applications — Direction : Fleming's right-hand rule or Lenz. Uses : emf across aircraft wings and train axles.

  • Easy : Wingspan 40 m, v = 250 m/s, B = 5 × 10⁻⁵ T ⇒ e = 0.5 V.
  • Medium : The emf is only across the wing tips; no lamp lights.
  • Exam (CET) : The right-hand rule gives? (1) current direction (2) force direction (3) B direction (4) v → answer 1)

Derivation : e = B l v

  • Step 1 : In time dt the rod moves dx = v dt.
  • Step 2 : Area change dA = l v dt.
  • Step 3 : dΦ = B l v dt.
  • Step 4 : |e| = dΦ/dt = B l v.
Remember

e = Blv; F = B²l²v/R; P = B²l²v²/R.

Common mistake / CET trap

e = Blv only if v ⟂ B; otherwise use the perpendicular component.

Shortcut

Memory clue : "e = B l v : three letters, three quantities".

Spot the difference

QuantityFormulaUnit
eBlvV
FB²l²v/RN
PB²l²v²/RW

Practice questions

Easy
  1. 1Formula for motional emf?
  2. 2e when v ∥ B?
  3. 3Which rule gives the direction?
Medium
  1. 4B = 0.2, l = 0.5, v = 4 : e?
  2. 5With that e and R = 4 Ω : I?
  3. 6l tripled : e?
Challenging
  1. 7Force to keep v steady (B = 0.5, l = 0.4, v = 6, R = 2)?
  2. 8v at 30° to B : e?
Show answers (try them yourself first)
  1. 1B l v
  2. 2Zero
  3. 3Fleming right hand / Lenz
  4. 4e = 0.4 V
  5. 5I = 0.1 A
  6. 6Triple
  7. 7F = 0.12 N
  8. 8e = B l v sin30° (perpendicular component)

Generator, alternating emf and back emf

From daily life

A turbine at a dam spins and makes the household supply — that is a generator.

Simple definition

A generator converts mechanical energy to electrical energy.

AC generator : e = NBAω sin ωtNSe = e₀ sin ωte₀ = NBAω

In the figure : A rotating coil ⇒ a sine wave.

Live figure

Generator, alternating emf and back emf

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Rule 1: AC generator — A coil of N turns rotated in B at ω : e = N B A ω sin ωt; peak e₀ = N B A ω.

  • Easy : N = 200, B = 0.3 T, A = 0.02 m², f = 50 Hz ⇒ e₀ = 377 V.
  • Medium : ω = 2πf = 314 rad/s; period 0.02 s.
  • Exam (CET) : f doubled ⇒ e₀? (1) doubles (2) halves (3) four times (4) same → answer 1)

Rule 2: Peak and zero positions — When the plane is parallel to B, Φ = 0 but e is maximum; when the plane is ⟂ B, Φ is maximum but e = 0.

  • Easy : At ωt = 90° e = e₀.
  • Medium : At ωt = 0° e = 0 and Φ = NBA.
  • Exam (CET) : When Φ is maximum, e? (1) 0 (2) maximum (3) half (4) negative max → answer 1)

Rule 3: Back emf — A running motor induces an emf opposing the supply : I = (V − E_b) ÷ R. At start E_b = 0 so the current is large.

  • Easy : V = 220, R = 10 Ω : starting I = 22 A.
  • Medium : With E_b = 200 V ⇒ I = 2 A.
  • Exam (CET) : Motor current is maximum when? (1) starting (2) full speed (3) medium (4) always same → answer 1)

Derivation : e = N B A ω sin ωt

  • Step 1 : θ = ωt; Φ = N B A cos ωt.
  • Step 2 : e = −dΦ/dt = N B A ω sin ωt.
  • Step 3 : sin ωt = 1 ⇒ e₀ = N B A ω.
  • Step 4 : e = e₀ sin ωt.
Remember

e = NBAω sinωt; e₀ = NBAω; I = (V − E_b)/R.

Common mistake / CET trap

e is zero when Φ is maximum — do not confuse them.

Shortcut

Memory clue : "Φ max ⇒ e zero (cos and sin)".

Spot the difference

PositionΦe
Plane ⟂ Bmaximum0
Plane ∥ B0maximum
GeneralNBA cosωtNBAω sinωt

Practice questions

Easy
  1. 1Formula for e₀?
  2. 2What is back emf?
  3. 3e when the plane ⟂ B?
Medium
  1. 4N = 100, B = 0.5, A = 0.01, ω = 100 : e₀?
  2. 5V = 120, R = 6 Ω, E_b = 100 : I?
  3. 6f = 60 Hz : ω?
Challenging
  1. 7e = 311 sin 100πt : f?
  2. 8Current if the motor stalls?
Show answers (try them yourself first)
  1. 1N B A ω
  2. 2The opposing emf in a motor
  3. 3Zero
  4. 4e₀ = 50 V
  5. 5I = 3.33 A
  6. 6ω = 377 rad/s
  7. 7f = 50 Hz
  8. 8E_b = 0 ⇒ large I; the motor may burn

Energy transfer in induction and eddy currents

From daily life

An induction cooker has no flame, yet the pan heats.

Simple definition

Eddy currents are currents set up in a solid conductor by a changing flux.

Eddy currents in a solid plateeddy currents give heatlamination ⇒ less loss

In the figure : Circular currents in a solid plate.

Rule 1: Mechanical to electrical — Pulling a rod at steady v needs F = B²l²v/R; the work F v equals I² R heating — energy is conserved.

  • Easy : B = 0.5, l = 0.4, v = 6, R = 2 ⇒ F = 0.12 N; F v = 0.72 W; e²/R = 0.72 W.
  • Medium : Both are equal ⇒ work becomes heat.
  • Exam (CET) : F v = ? (1) I²R (2) I R (3) I/R (4) e I² → answer 1)

Rule 2: Eddy currents — A changing flux sets up circular currents in a solid conductor; they produce heat and an opposing force.

  • Easy : Induction cooker : eddy currents heat the pan.
  • Medium : A magnet falls slowly in a copper tube.
  • Exam (CET) : To reduce eddy currents? (1) lamination (2) thicker core (3) higher f (4) higher B → answer 1)

Rule 3: Uses and losses — Uses : induction furnace, electromagnetic brakes, meter damping. Harm : heating in transformer and motor cores.

  • Easy : Thin sheets ⇒ lower eddy loss.
  • Medium : Varnish between sheets ⇒ high resistance.
  • Exam (CET) : An electromagnetic brake uses? (1) eddy currents (2) friction (3) air (4) a spring → answer 1)

Derivation : F v = I² R

  • Step 1 : e = B l v; I = B l v ÷ R.
  • Step 2 : Force F = B I l = B² l² v ÷ R.
  • Step 3 : Power F v = B² l² v² ÷ R.
  • Step 4 : I² R = (B l v/R)² R = B² l² v² ÷ R — identical.
Remember

F v = I²R = B²l²v²/R; eddy loss ⇒ lamination.

Common mistake / CET trap

Electrical energy is not created from nothing; it comes from mechanical work.

Shortcut

Memory clue : "pull in, heat out".

Spot the difference

ItemCauseResult
Eddy currentschanging Φheat
Laminationhigher resistanceless loss
Induction cookereddypan heats

Practice questions

Easy
  1. 1What are eddy currents?
  2. 2One use?
  3. 3How is the loss reduced?
Medium
  1. 4F = 0.4 N, v = 5 : power?
  2. 5I = 2 A, R = 3 Ω : heating power?
  3. 6v doubled : power?
Challenging
  1. 7Why F v = I²R at steady v?
  2. 8Why a laminated transformer core?
Show answers (try them yourself first)
  1. 1Circular currents in a solid conductor
  2. 2Induction cooker
  3. 3Lamination
  4. 42 W
  5. 512 W
  6. 64 times
  7. 7Energy conservation
  8. 8To raise resistance and cut loss

Self-inductance and energy stored in a magnetic field

From daily life

A spark at a choke when a circuit is switched off — the coil resists a change of current.

Simple definition

Self-inductance L = Φ ÷ I; the coil opposes changes of its own current.

Self-inductance : coil and graphL = μ₀N²A ÷ lI(t) = I₀(1 − e^(−t/τ))U = ½ L I²

In the figure : L = μ₀N²A/l; energy is stored in the coil.

Rule 1: Self-inductance — e = −L dI/dt, Φ = L I; unit henry (H = V s A⁻¹), dimensions [M L² T⁻² A⁻²].

  • Easy : L = 2 H, dI/dt = 5 A/s ⇒ |e| = 10 V.
  • Medium : L = 0.5 H, I = 4 A ⇒ Φ = 2 Wb.
  • Exam (CET) : 1 H = ? (1) 1 V s/A (2) 1 V/A (3) 1 A s/V (4) 1 Wb → answer 1)

Rule 2: L of a solenoid — L = μ₀ N² A ÷ l (with a core multiply by μ_r).

  • Easy : N = 500, A = 4 cm², l = 25 cm ⇒ L = 5.03 × 10⁻⁴ H.
  • Medium : N doubled ⇒ L four times.
  • Exam (CET) : L ∝ ? (1) N² (2) N (3) 1/N (4) N⁰ → answer 1)

Rule 3: Stored energy — U = ½ L I²; magnetic energy density u = B² ÷ 2μ₀.

  • Easy : L = 5.03 × 10⁻⁴ H, I = 2 A ⇒ U = 1.01 × 10⁻³ J.
  • Medium : I doubled ⇒ U four times.
  • Exam (CET) : U = ? (1) ½LI² (2) LI² (3) LI (4) ½LI → answer 1)

Derivation : U = ½ L I²

  • Step 1 : While raising the current dW = |e| I dt = L I dI.
  • Step 2 : W = ∫ from 0 to I of L I dI.
  • Step 3 : = ½ L I².
  • Step 4 : U = ½ L I².
Remember

e = −L dI/dt; L = μ₀N²A/l; U = ½LI².

Common mistake / CET trap

L opposes a change of current, not the current itself.

Shortcut

Memory clue : "L is electrical inertia" — m ↔ L, v ↔ I.

Spot the difference

QuantityFormulaUnit / dimensions
Lμ₀N²A/lH; [ML²T⁻²A⁻²]
e−L dI/dtV
U½LI²J
uB²/2μ₀J/m³

Practice questions

Easy
  1. 1Unit of L?
  2. 2What does L depend on?
  3. 3Formula for U?
Medium
  1. 4L = 3 H, dI/dt = 4 : e?
  2. 5L = 0.2 H, I = 5 A : U?
  3. 6N tripled : L?
Challenging
  1. 7B = 0.2 T : u?
  2. 8L = 0.1 H, I falls 2 → 0 A in 0.01 s : e?
Show answers (try them yourself first)
  1. 1Henry
  2. 2Geometry, turns, core
  3. 3½LI²
  4. 412 V
  5. 52.5 J
  6. 69 times
  7. 7u = 1.59 × 10⁴ J/m³
  8. 8e = 20 V

Mutual inductance, combinations and the transformer

From daily life

A phone charger uses a transformer to step 220 V down.

Simple definition

Mutual inductance = emf in one coil due to a change in another.

TransformerN_pN_sV_s/V_p = N_s/N_p = I_p/I_s

In the figure : N_s/N_p = V_s/V_p; a laminated iron core.

Rule 1: Mutual inductance — e₂ = −M dI₁/dt, Φ₂ = M I₁; M = μ₀ N₁N₂ A ÷ l (coaxial solenoids); M = k √(L₁L₂), k ≤ 1.

  • Easy : L₁ = 0.1 H, L₂ = 0.2 H, k = 0.4 ⇒ M = 0.057 H.
  • Medium : M = 0.5 H, dI₁/dt = 4 A/s ⇒ e₂ = 2 V.
  • Exam (CET) : k for perfect coupling? (1) 1 (2) 0 (3) 0.5 (4) 2 → answer 1)

Rule 2: Combining inductors — Series (no coupling) : L = L₁ + L₂; parallel : 1/L = 1/L₁ + 1/L₂ (like resistors).

  • Easy : 2 H and 3 H in series : 5 H.
  • Medium : In parallel : 1.2 H.
  • Exam (CET) : Series L? (1) sum (2) smaller (3) product (4) zero → answer 1)

Rule 3: Transformer — Ideal : V_s ÷ V_p = N_s ÷ N_p = I_p ÷ I_s; step-up N_s > N_p, step-down N_s < N_p. Efficiency η = P_s ÷ P_p.

  • Easy : N_p = 1000, N_s = 100, V_p = 220 ⇒ V_s = 22 V.
  • Medium : P_p = 110 W, η = 90% ⇒ P_s = 99 W.
  • Exam (CET) : In a step-up transformer I_s is? (1) lower (2) higher (3) same (4) zero → answer 1)

Derivation : Transformer ratio

  • Step 1 : Perfect coupling : same flux in both coils.
  • Step 2 : e_p = −N_p dΦ/dt, e_s = −N_s dΦ/dt.
  • Step 3 : Dividing, V_s/V_p = N_s/N_p.
  • Step 4 : Ideal : V_p I_p = V_s I_s ⇒ V_s/V_p = I_p/I_s.
Remember

M = k√(L₁L₂); V_s/V_p = N_s/N_p = I_p/I_s.

Common mistake / CET trap

A transformer does not work on DC; in step-up the current drops.

Shortcut

Memory clue : "volts up, amps down".

Spot the difference

TypeNVI
Step-upN_s > N_prisesfalls
Step-downN_s < N_pfallsrises

Practice questions

Easy
  1. 1Unit of M?
  2. 2What is step-up?
  3. 3Does a transformer work on AC or DC?
Medium
  1. 4N_p = 500, N_s = 50, V_p = 100 : V_s?
  2. 5L₁ = 4, L₂ = 12 H in series : L?
  3. 6V_s = 11, I_s = 10, ideal : I_p (V_p = 220)?
Challenging
  1. 7η = 80%, P_s = 400 W : P_p?
  2. 8Why transmit power at high voltage?
Show answers (try them yourself first)
  1. 1Henry
  2. 2N_s > N_p, voltage rises
  3. 3AC only
  4. 4V_s = 10 V
  5. 5L = 16 H
  6. 6I_p = 0.5 A
  7. 7P_p = 500 W
  8. 8Lower I ⇒ lower I²R loss

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
FluxΦ = BA cosθWb; [ML²T⁻²A⁻¹]
Faradaye = −N dΦ/dt; q = NΔΦ/RV; C
Lenzopposes the change—
Motionale = Blv; F = B²l²v/RV; N
Generatore = NBAω sinωtV
Le = −L dI/dt; L = μ₀N²A/lH; [ML²T⁻²A⁻²]
EnergyU = ½LI²; u = B²/2μ₀J; J/m³
TransformerV_s/V_p = N_s/N_p = I_p/I_s—

🎯 Most important points for the exam

  • e = −N dΦ/dt; q = NΔΦ/R (no time).
  • Lenz : Φ rising ⇒ opposing B; falling ⇒ supporting B.
  • Motional e = Blv; Fv = I²R.
  • Generator e₀ = NBAω; Φ max ⇒ e = 0.
  • Eddy currents : lamination.
  • L = μ₀N²A/l; U = ½LI².
  • Transformer : V ∝ N; I ∝ 1/N.
Study mantra

Find Φ → rate of change → e = −N dΦ/dt → direction by Lenz → units.