NCERT Plus Two (Class 12) Physics • Chapter 6

Electromagnetic Induction

Q: State Faraday's laws of electromagnetic induction and Lenz's law. How is Lenz's law consistent with conservation of energy?

Faraday's Law: Induced EMF magnitude is directly proportional to the rate of change of magnetic flux: ε = −N (dΦ B /dt) . Lenz's Law: Induced current polarity always opposes the change in flux producing it. Energy Conservation: Mechanical work done against this opposing magnetic force turns into electrical energy I²Rt .

Q: Derive the expression for motional EMF ε = B l v induced across a straight conductor moving perpendicular to a uniform magnetic field.

Consider rod PQ of length l sliding on U-rails at speed v perpendicular to uniform B . Enclosed area at distance x is A = l x . Flux Φ B = B l x . By Faraday's law: |ε| = dΦ B /dt = B l (dx/dt) = B l v .

Q: Define self-inductance of a coil and state its SI unit. Derive the expression for self-inductance of a long solenoid.

Self-inductance (L): Flux linkage per unit current ( NΦ = L I ); SI unit: Henry (H) . For solenoid of length l , area A , n turns/m: B = μ₀nI , total turns N = nl . NΦ = (nl)(μ₀nI)A = (μ₀ n² A l) I ⇒ L = μ₀ n² A l (or μ₀ N² A / l ).

Q: Obtain an expression for the magnetic energy stored in an inductor of self-inductance L carrying current I.

Instantaneous rate of work done against back EMF ε = L(dI/dt) is dW/dt = ε I = L I (dI/dt) . Integrating from 0 to I : U = ∫₀ᴵ L I dI = ½ L I² .

Q: Explain the working principle of an AC generator with a neat expression for instantaneous EMF.

Principle: Electromagnetic induction — rotating an armature coil ( N turns, area A ) at angular speed ω in uniform field B varies flux. Φ(t) = N B A cos(ωt) . ε(t) = −dΦ/dt = N B A ω sin(ωt) = ε₀ sin(ωt) , delivered to external circuit via slip rings and brushes.

Q: A wheel with 10 metallic spokes each 0.5 m long rotates at 120 rev/min in a plane normal to Earth's horizontal field H_E = 0.4 G. Find induced EMF between axle and rim.

ν = 120/60 = 2 Hz ⇒ ω = 2πν = 4π rad/s ; B = 0.4 × 10⁻⁴ T ; R = 0.5 m . Spokes are in parallel ⇒ ε = ½ B ω R² = ½ × 0.4 × 10⁻⁴ × 4π × 0.25 = 6.28 × 10⁻⁵ V .

Q: Two concentric coplanar circular coils have radii r₁ and r₂ (r₁ ≪ r₂). Obtain an expression for their mutual inductance.

Current I₂ in outer coil produces centre field B₂ = μ₀ I₂ / (2 r₂) . Since r₁ ≪ r₂ , flux through inner coil is Φ₁ = π r₁² B₂ = μ₀ π r₁² I₂ / (2 r₂) . Therefore M = Φ₁ / I₂ = μ₀ π r₁² / (2 r₂) .

Q: Why are transformer and generator iron cores made of laminated thin sheets instead of a solid metal block?

Thin insulated laminations break up large swirling eddy current loops , sharply increasing resistance and reducing wasteful I²R Joule heating!

Faraday-Henry coil experiments, magnetic flux, Faraday's & Lenz's laws, motional EMF, self/mutual inductance & AC generator.

📄 Official NCERT Textbook PDF

Topic 6.2 – 6.3: Faraday-Henry Coil Experiments & Magnetic Flux (Φ = B·A)

Oersted proved electric current makes magnetism. Michael Faraday asked the brilliant reverse question: "Can a magnet create electric current?" He parked the strongest magnet right inside a copper coil and got… ZERO current! Only the instant he moved the magnet (or tapped a switch ON/OFF in a nearby coil) did the needle leap! Nature rewards CHANGE, not stillness!

FARADAY-HENRY RESULTRelative motion or changing current induces EMF — stationary magnet gives 0!

MAGNETIC FLUXΦ_B = B · A = B A cosθ (Scalar; θ = angle between B and normal!)

SI UNITWeber (Wb) = T · m² (1 Wb = 10⁸ Maxwell)

🧠 Memory Trick: Magnetic flux is like rain passing through a window — hold the frame straight to rain (θ = 0° with normal) for MAX flux; lay it flat (θ = 90°) for ZERO flux!

💡 Simple Kid-Check: A magnet sitting peacefully inside a loop creates lots of magnetic field lines, yet zero electric current! Current wakes up ONLY when the count of magnetic lines passing through the loop (called Magnetic Flux) changes with time!

1
Three Classic Faraday-Henry Experiments: (i) Push bar magnet IN or pull OUT of coil C₁ → galvanometer kicks opposite ways; hold still → zero deflection! (ii) Move a battery-powered coil C₂ toward/away from C₁ → same kick! (iii) Keep both coils still and tap switch ON/OFF in C₂ → momentary kick only at make & break!
2
Magnetic Flux Formula: For a uniform field B crossing flat area A whose outward perpendicular normal makes angle θ with B: Φ_B = B A cosθ. For non-uniform fields, sum patches: Φ_B = ∑ B_i · dA_i.
3
How Can Flux Change? Change field magnitude B, change loop area A, or rotate angle θ between B and normal n̂!

🧒 Super-Easy Example: A square coil of side 10 cm (A = 0.01 m²) sits in a 0.5 T magnetic field with its plane tilted at 30° to B. Angle with normal is θ = 90° − 30° = 60°, so Φ_B = 0.5 × 0.01 × cos 60° = 2.5 × 10⁻³ Wb!

Watch the #1 Flux Exam Trap: If a problem states "the plane of a coil makes 30° with magnetic field B", do NOT plug cos 30°! The angle θ in Φ = BA cosθ is ALWAYS between B and the normal perpendicular to the plane, so θ = 90° − 30° = 60°! Also remember: a huge stationary magnet induces ZERO current!

Topic 6.4: Faraday's Law of Electromagnetic Induction

Imagine magnetic flux like water pouring into a bucket, and induced EMF like a water-pressure gauge on the pipe! Pour slowly and barely a tiny voltage trickles out; slam a strong magnet fast through a 500-turn copper coil and the voltmeter shoots skyward — because voltage depends on the SPEED of flux change!

FARADAY'S LAWε = −N dΦ_Bdt (Magnitude: |ε| = N |ΔΦ_B|Δt)

INDUCED CURRENTI = |ε|R = NR |ΔΦ_B|Δt (Needs a closed circuit!)

INDUCED CHARGE SHORTCUTΔQ = I Δt = N |ΔΦ_B|R — INDEPENDENT of time Δt!

🧠 Memory Trick: Fast motion → HUGE EMF & Current (1/Δt); yet Total Charge ΔQ = N ΔΦ / R stays IDENTICAL whether fast or slow!

💡 Simple Kid-Check: Induced EMF cares about how quickly flux changes, NOT how big the flux happens to be right now! Even if magnetic flux is zero for a split second while flipping from +Φ to −Φ, the slope dΦ/dt is HUGE right there — so peak EMF strikes!

1
Faraday's First & Second Laws: Whenever magnetic flux linked with a circuit changes, an EMF is induced lasting only while the change lasts. Its magnitude equals the time rate of change of flux linkage: |ε| = N |dΦ_B/dt|.
2
Open vs Closed Circuit: An open wire loop STILL develops induced EMF ε across its ends, though induced current I = 0 until you close the loop!
3
Induced Charge Independence: Integrating I = (N/R)(dΦ/dt) gives ΔQ = N ΔΦR — pushing the magnet fast doubles current for half the time, keeping total charge ΔQ unchanged!

🧒 Super-Easy Example: A 200-turn coil of area 0.02 m² and resistance 4 Ω sits perpendicular to a 0.5 T field. Flip it 180° in 0.1 s:
• Φ₁ = +0.5 × 0.02 = +0.01 Wb; Φ₂ = −0.01 Wb ⇒ |ΔΦ| = 0.02 Wb (double, NOT zero!).
• |ε| = 200 × 0.02 / 0.1 = 40 V • I = 40/4 = 10 A • ΔQ = 10 × 0.1 = 1.0 C!

Watch two massive exam traps: (1) Rotating a coil by 180° changes flux from +BA to −BA, so |ΔΦ| = 2BA (students often forget the factor of 2!). (2) Induced Charge ΔQ = N ΔΦ / R is completely independent of how fast you move the magnet (Δt), whereas induced EMF & Current scale inversely with Δt!

Topic 6.5: Lenz's Law & Conservation of Energy

Push the North pole of a bar magnet toward a copper ring and something funny happens: the ring turns its facing side into a North pole too, pushing right back against your hand! Why is Nature so stubborn? Because if the ring pulled your magnet in instead, you'd get free speed AND free electricity forever — breaking the Law of Conservation of Energy!

LENZ'S LAWInduced current polarity ALWAYS opposes the change in flux that caused it (explains − sign!).

ENERGY CONSERVATIONMechanical work pushing against opposing force converts directly into Joule heat I² R t!

CLOCKWISE / CCW RULEMagnet N approaches → face becomes N (CCW); Magnet N leaves → face becomes S (CW)!

🧠 Memory Trick: Lenz's Law is the "Stubborn Gatekeeper" — Try to come IN? It pushes you OUT! Try to leave? It pulls you BACK! Never wants flux to change!

💡 Simple Kid-Check: Drop a small neodymium magnet down a plastic pipe vs a closed copper pipe. Through plastic it drops like a rock (a = g). Through copper, swirling induced currents make an opposing magnet behind and ahead — floating it slowly down as falling gravitational energy heats the copper!

1
Meaning of the Minus Sign: In ε = −N dΦ_B/dt, the minus sign represents Lenz's Law: induced current generates its own magnetic field B_ind opposing whatever change ΔΦ is happening!
2
Four-Case Direction Finder:
• N-pole moves IN: Flux into face rises → coil makes N-face (anticlockwise) to repel.
• N-pole moves OUT: Flux drops → coil makes S-face (clockwise) to attract.
• S-pole moves IN: Coil makes S-face (clockwise) to repel.
• S-pole moves OUT: Coil makes N-face (anticlockwise) to attract!
3
Why It Obeys Conservation of Energy: Pushing against repulsion (or pulling against attraction) requires mechanical work F·dx, which equals exact electrical energy ε I dt = I² R dt dissipated in the loop!

🧒 Super-Easy Example: A magnet falls vertically with its North pole pointing downward into a horizontal metal ring. Looking from above, approaching N induces an N-pole on top → current flows anticlockwise, slowing the fall (a < g). If the ring has a tiny cut/slit, no current flows → a = g!

Watch the #1 conceptual trap: Students say "Lenz's law opposes the magnetic flux"! NO — WRONG! It opposes the CHANGE in magnetic flux! When flux is increasing, induced field points OPPOSITE to B; when flux is decreasing, induced field points in the SAME direction as B to prop it up! Also: Lenz's law directly follows from Conservation of Energy (favourite 1-mark question!).

Topic 6.6: Motional EMF (ε = B l v) & Eddy Power Loss Intuition

Hold a straight copper rod horizontally and run fast across a perpendicular magnetic field — free electrons inside the copper get shoved sideways by Lorentz force until positive and negative charges pile at opposite tips, turning an ordinary sliding stick into a live battery! Even a passenger jet flying through Earth's field develops voltage across its wings!

MOTIONAL EMFε = B l v (for B ⊥ l ⊥ v; general: B l v sinθ)

ROTATING SPOKE / RODPivoted at one end: ε = ½ B ω l² = B π ν l² (all spokes in parallel!)

POWER & EDDY CURRENTSPull force F = B²l²vR • Power P = Fv = B²l²v²R = I²R • Laminate cores to cut eddy heat!

🧠 Memory Trick: Straight sliding rod → full B l v; Spinning spoke (centre speed 0, tip speed ωl, average ½ωl) → ½ B ω l²!

💡 Simple Kid-Check: Picture a rod sliding along U-shaped metal rails of width l in a perpendicular field B. Each second, it sweeps fresh rectangle area l · v, chopping B l v Webers of flux per second — so straight away, ε = B l v!

1
Two Equivalent Proofs of ε = Blv: (a) By Flux: Φ = B(lx) ⇒ |ε| = dΦ/dt = B l (dx/dt) = B l v. (b) By Lorentz Force: free charge q feels F_m = qvB pushing electrons to one end until electric force qE balances it: E = vB ⇒ ε = El = B l v!
2
Mechanical Work = Electrical Heat: Induced current I = Blv/R makes the field push back with braking force F = IlB = B²l²v/R. Pushing at steady v takes mechanical power P = Fv = B²l²v²/R = ε²/R!
3
Eddy Currents Intuition: When a solid bulk metal plate moves or sits in changing B, swirling loops of induced current (eddy currents) waste heat unless cut into insulated thin sheets (laminated cores) or slotted!

🧒 NCERT Example 6.7: A bicycle wheel with 10 metal spokes each 0.5 m long spins at 120 rpm (ν = 2 Hz ⇒ ω = 4π rad/s) normal to B = 0.4 G = 0.4 × 10⁻⁴ T.
All 10 spokes are connected in parallel between axle and rim, so ε = ½ B ω R² = ½ × (0.4 × 10⁻⁴) × 4π × (0.5)² = 6.28 × 10⁻⁵ V (NOT multiplied by 10!).

Two classic DHSE traps: (1) In a wheel with 10 or 20 spokes rotating in a magnetic field, NEVER multiply EMF by the number of spokes — all spokes share the same axle and rim in PARALLEL, so EMF stays ½ B ω R²! (2) If any two of vectors B, l, and v are parallel, motional EMF is strictly ZERO!

Topic 6.7.1 – 6.7.2: Self-Inductance, Mutual Inductance & Stored Energy (½ L I²)

Try spinning up a heavy iron flywheel — it fights back when you push it faster and fights back when you brake it! In electric circuits, an inductor coil acts as electrical inertia: it hates sudden current jumps inside itself (Self-Inductance L), and whispers every current ripple to a nearby coil (Mutual Inductance M)!

SELF-INDUCTANCE (L)NΦ = L I • ε = −L dIdt • Solenoid: L = μ₀ μ_r n² A l (Unit: Henry, H)

MUTUAL INDUCTANCE (M)N₁Φ₁ = M I₂ • ε₁ = −M dI₂dt • Coaxial solenoids: M = μ₀ n₁ n₂ π r₁² l (r₁ < r₂!)

MAGNETIC ENERGYU = ½ L I² (twin of ½mv²!) • Energy density: u_B = B²2μ₀

🧠 Memory Trick: L is the mass m of electricity! Kinetic energy ½ m v² ↔ Magnetic energy ½ L I²; Force m dv/dt ↔ Back EMF L dI/dt!

💡 Simple Kid-Check: Ever seen a tiny blue spark jump across a switch when unplugging a big motor or fan? Opening the switch tries to drop current I → 0 instantly (dI/dt huge!), so the motor coil's self-inductance fires a giant kickback voltage L dI/dt!

1
Self-Inductance of Long Solenoid: With n turns/m, length l, area A: field B = μ₀nI ⇒ total flux linkage NΦ = (nl)(μ₀nI)A = (μ₀n²Al)I ⇒ L = μ₀ μ_r n² A l (depends ONLY on geometry & core μ_r!).
2
Mutual Inductance & Reciprocity: Two long coaxial solenoids of inner radius r₁ and outer radius r₂ share identical mutual inductance M₁₂ = M₂₁ = M = μ₀ n₁ n₂ π r₁² l! Max possible coupling: M ≤ √(L₁ L₂).
3
Energy Stored in Inductor: Building current from 0 to I against back EMF does work W = ∫ L I dI = ½ L I², stored inside the magnetic field at density u_B = B² / (2μ₀)!

🧒 Super-Easy Example: Current in a 4 H coil drops steadily from 5 A to 1 A in 0.2 s.
• Back EMF magnitude: |ε| = L |ΔI/Δt| = 4 × (4 / 0.2) = 80 V.
• Initial stored magnetic energy: U = ½ L I² = ½ × 4 × 5² = 50 J!

Watch three classic traps: (1) For two coaxial solenoids of radii r₁ < r₂, mutual inductance M = μ₀ n₁ n₂ π r₁² l ALWAYS uses the SMALLER (inner) radius, because magnetic flux exists only inside the inner area! (2) Notice L ∝ N² (double turns in same length → 4× inductance!). (3) Steady DC current (dI/dt = 0) produces ZERO self-induced back EMF!

Topic 6.8: Principle, Flux Linkage & Output of an AC Generator

How does Idukki hydroelectric dam turn falling mountain water into the electricity running your fan in Kochi? Water turbines spin giant magnet rotors past stationary copper coils! As the coil angle θ = ωt spins round and round, magnetic flux oscillates like a cosine wave — giving birth to a smooth sine-wave AC voltage!

PRINCIPLE & FLUXElectromagnetic Induction: rotating coil changes flux Φ_B(t) = N B A cos(ωt)!

AC EMF EQUATIONε(t) = −dΦdt = N B A ω sin(ωt) = ε₀ sin(ωt)!

PEAK EMF & PHASEε₀ = N B A ω = 2π ν N B A • Max EMF when Φ = 0 (plane ∥ to B, θ = 90°)!

🧠 Memory Trick: Flux is cos(ωt), EMF is sin(ωt) — they are 90° out of step! Zero flux = Peak EMF; Peak flux = Zero EMF!

💡 Simple Kid-Check: Picture a rotating rectangular coil in a magnetic field like a revolving door in wind. When the door faces straight into the wind (θ = 0°), maximum wind passes through, yet at that split second the open area barely changes → ε = 0! When the door sweeps edge-on past the wind (θ = 90°), flux flips from + to − fastest → PEAK EMF ε₀!

1
Four Core Parts: Armature coil (N turns, area A), strong Field magnet (B), two full copper Slip Rings (give AC, unlike split-ring commutator for DC!), and Carbon Brushes leading to the load.
2
Step-by-Step Derivation: At time t, angle between B and area normal n̂ is θ = ωt.
Total flux linkage: Φ = N B A cos(ωt).
By Faraday's Law: ε = −dΦdt = −N B A (−ω sin ωt) = N B A ω sin(ωt)!
3
India Mains Frequency: Standard power station frequency in India is 50 Hz (meaning current reverses direction 100 times every second!).

🧒 NCERT Example 6.10: Kamla pedals a stationary bike hooked to a 100-turn coil of area 0.10 m² spinning at ν = 0.5 Hz inside B = 0.01 T.
Peak voltage generated: ε₀ = N B A (2πν) = 100 × 0.01 × 0.10 × 2 × 3.14 × 0.5 = 0.314 V!

Watch the #1 AC Generator conceptual trap: Examiners ask "At what coil position is induced EMF MAXIMUM?" Students instinctively shout "When flux is maximum!" DEAD WRONG! When coil face is perpendicular to B (θ = 0°), flux is maximum (NBA) and EMF is ZERO! Peak EMF ε₀ = NBAω happens when the coil plane is PARALLEL to B (θ = 90°, flux = 0)! Also note: AC generators use slip rings (full rings), NOT split rings!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): State Faraday's laws of electromagnetic induction and Lenz's law. How is Lenz's law consistent with conservation of energy?

Faraday's Law: Induced EMF magnitude is directly proportional to the rate of change of magnetic flux: ε = −N (dΦ_B/dt).
Lenz's Law: Induced current polarity always opposes the change in flux producing it.
Energy Conservation: Mechanical work done against this opposing magnetic force turns into electrical energy I²Rt.

Q2 (Kerala HSE Say • 2 Marks): Derive the expression for motional EMF ε = B l v induced across a straight conductor moving perpendicular to a uniform magnetic field.

Consider rod PQ of length l sliding on U-rails at speed v perpendicular to uniform B. Enclosed area at distance x is A = l x.
Flux Φ_B = B l x.
By Faraday's law: |ε| = dΦ_B/dt = B l (dx/dt) = B l v.

Q3 (Kerala HSE March • 3 Marks): Define self-inductance of a coil and state its SI unit. Derive the expression for self-inductance of a long solenoid.

Self-inductance (L): Flux linkage per unit current (NΦ = L I); SI unit: Henry (H).
For solenoid of length l, area A, n turns/m: B = μ₀nI, total turns N = nl.
NΦ = (nl)(μ₀nI)A = (μ₀ n² A l) I ⇒ L = μ₀ n² A l (or μ₀ N² A / l).

Q4 (Kerala HSE Model • 2 Marks): Obtain an expression for the magnetic energy stored in an inductor of self-inductance L carrying current I.

Instantaneous rate of work done against back EMF ε = L(dI/dt) is dW/dt = ε I = L I (dI/dt).
Integrating from 0 to I: U = ∫₀ᴵ L I dI = ½ L I².

Q5 (Kerala HSE March • 3 Marks): Explain the working principle of an AC generator with a neat expression for instantaneous EMF.

Principle: Electromagnetic induction — rotating an armature coil (N turns, area A) at angular speed ω in uniform field B varies flux.
Φ(t) = N B A cos(ωt).
ε(t) = −dΦ/dt = N B A ω sin(ωt) = ε₀ sin(ωt), delivered to external circuit via slip rings and brushes.

Q6 (NCERT / HSE Numerical • 2 Marks): A wheel with 10 metallic spokes each 0.5 m long rotates at 120 rev/min in a plane normal to Earth's horizontal field H_E = 0.4 G. Find induced EMF between axle and rim.

ν = 120/60 = 2 Hz ⇒ ω = 2πν = 4π rad/s; B = 0.4 × 10⁻⁴ T; R = 0.5 m.
Spokes are in parallel ⇒ ε = ½ B ω R² = ½ × 0.4 × 10⁻⁴ × 4π × 0.25 = 6.28 × 10⁻⁵ V.

Q7 (Kerala HSE Say • 2 Marks): Two concentric coplanar circular coils have radii r₁ and r₂ (r₁ ≪ r₂). Obtain an expression for their mutual inductance.

Current I₂ in outer coil produces centre field B₂ = μ₀ I₂ / (2 r₂).
Since r₁ ≪ r₂, flux through inner coil is Φ₁ = π r₁² B₂ = μ₀ π r₁² I₂ / (2 r₂).
Therefore M = Φ₁ / I₂ = μ₀ π r₁² / (2 r₂).

Q8 (Kerala HSE Model • 1 Marks): Why are transformer and generator iron cores made of laminated thin sheets instead of a solid metal block?

Thin insulated laminations break up large swirling eddy current loops, sharply increasing resistance and reducing wasteful I²R Joule heating!