Class 12 Physics | Unit IV
Chapter 6: Electromagnetic Induction
Faraday's Laws • Lenz's Law • Motional EMF • Self Inductance • Mutual Inductance • AC Generator
1. Magnetic Flux
Magnetic Flux (ϕ): Total number of magnetic field lines passing through a surface.
ϕ = B·A = BA cosθ, where θ is the angle between B and area vector (normal to surface).
Unit: Weber (Wb) = T·m². Scalar quantity.
ϕ = B A cosθ
• θ = 0° (B ∥ to area normal): ϕmax = BA
• θ = 90° (B ⊥ to area normal, i.e. B along plane): ϕ = 0
• ϕ can be positive, negative or zero depending on θ.
• θ = 0° (B ∥ to area normal): ϕmax = BA
• θ = 90° (B ⊥ to area normal, i.e. B along plane): ϕ = 0
• ϕ can be positive, negative or zero depending on θ.
- Area vector is always taken normal to the surface. Direction by right-hand rule (curl fingers along boundary, thumb = area vector).
- SI unit: Wb = V·s = kg·m²/(A·s²). CGS unit: Maxwell. 1 Wb = 108 Maxwell.
- Flux through a closed surface is zero (Gauss’s law for magnetism: magnetic monopoles don't exist).
2. Faraday's Laws of Electromagnetic Induction
2.1 Faraday's First Law
Whenever there is a change in magnetic flux linked with a circuit (coil), an EMF is
induced in the circuit. This EMF lasts as long as the flux is changing.
Flux can be changed by: (a) Moving the magnet relative to coil, (b) Changing B (by changing current in adjacent coil), (c) Changing area of loop, (d) Changing angle between B and coil.
2.2 Faraday's Second Law
The magnitude of the induced EMF is directly proportional to the rate of change of magnetic
flux linked with the circuit.
ε = −N dϕ/dt = −N Δϕ/Δt
N = number of turns, ϕ = flux per turn, dϕ/dt = rate of change of flux
The negative sign is due to Lenz's law.
N = number of turns, ϕ = flux per turn, dϕ/dt = rate of change of flux
The negative sign is due to Lenz's law.
2.3 Lenz's Law
The direction of induced EMF (and thus induced current) is always such that it opposes the
cause that produces it (change in flux). This is a consequence of the law of
conservation of energy.
- If flux through a coil increases, induced current creates its own field to oppose the increase (opposes the incoming flux).
- If flux decreases, induced current tries to maintain the flux (its field aids the original).
- Lenz's law gives the direction of induced current; Faraday's second law gives the magnitude.
- Mathematical: the negative sign (−) in ε = −dϕ/dt represents Lenz's law.
- Experiment analogy: Bringing N-pole of magnet toward a coil → flux increases → induced current in coil creates a N pole facing the approaching magnet → repulsion → opposes motion.
⚠️ NEET (2014, 2017, 2019, 2022): ε = −NΔϕ/Δt.
Lenz's law: opposes change in flux. Induced current direction by Lenz (use right-hand or screw rule). If
flux increases → induced current opposes: creates opposing B. Unit of EMF = Volt = Wb/s. Induced
charge q = Δϕ/R (independent of time, depends on flux change and resistance).
3. Motional EMF
When a straight conductor of length L moves with velocity v perpendicular to a magnetic field B, free
electrons experience a Lorentz force (F = qv×B), which causes charge separation, producing an EMF. This
is called motional EMF.
ε = BLv sinθ (θ = angle between v and B)
For v ⊥ B (most common case): ε = BLv
Alternative derivation: ε = −dϕ/dt = −d(BLx)/dt = −BL(dx/dt) = BLv
(as conductor moves, the area swept = Lx increases, flux increases, EMF induced)
Power dissipated (if circuit R): P = ε²/R = B²L²v²/R
Retarding force on conductor: F = BIL = B²L²v/R
For v ⊥ B (most common case): ε = BLv
Alternative derivation: ε = −dϕ/dt = −d(BLx)/dt = −BL(dx/dt) = BLv
(as conductor moves, the area swept = Lx increases, flux increases, EMF induced)
Power dissipated (if circuit R): P = ε²/R = B²L²v²/R
Retarding force on conductor: F = BIL = B²L²v/R
- The kinetic energy of moving conductor is converted to electrical energy (then to heat in R).
- If conductor moves along B (parallel), EMF = 0 (no flux change).
- For a rotating rod: one end fixed, other sweeps area. ε = ½BL²ω (where ω = angular velocity).
- Induced charge: q = Δϕ/(R) = BLΔx/R (independent of speed; depends on area swept).
4. Eddy Currents
When a bulk piece of conductor is placed in a changing magnetic field, induced currents circulate within it
in closed loops (like whirlpools). These are called eddy currents (or Foucault currents).
- Direction: Lenz's law (oppose change in flux).
- Effect: produce heat (energy loss); this is the disadvantage in transformers and motors.
- Minimised by: Using laminated cores (thin insulated sheets of iron) → reduces loops → reduces eddy currents → reduces heat loss.
- Applications (useful eddy currents):
- Induction furnaces: melt metals by eddy current heating.
- Electromagnetic braking: Lenz's law causes opposing force on moving metals (used in speedometers, magnetic braking of trains).
- Metal detectors: change in eddy current pattern detects metal objects.
- Induction cooktops: eddy currents heat the cooking vessel.
5. Self Inductance
Self Inductance (L): The property of a coil by which it opposes any change in the current
flowing through it, by inducing a back EMF in itself. Defined as: ϕ = LI (flux linkage proportional to
current). Unit: Henry (H).
Flux linkage: Nϕ = LI
Induced EMF: ε = −L dI/dt
Self inductance of solenoid: L = μ0N²A/l = μ0n²Al
(N = total turns, n = N/l = turns/metre, A = cross-sectional area, l = length)
With core (μr): L = μ0μrN²A/l
Energy stored: U = ½LI²
Induced EMF: ε = −L dI/dt
Self inductance of solenoid: L = μ0N²A/l = μ0n²Al
(N = total turns, n = N/l = turns/metre, A = cross-sectional area, l = length)
With core (μr): L = μ0μrN²A/l
Energy stored: U = ½LI²
- 1 Henry: a coil has L = 1 H if an EMF of 1 V is induced when current changes at 1 A/s.
- Self inductance is the electromagnetic analogue of mass (inertia) — it resists change in current like mass resists change in velocity.
- Energy stored in inductor: U = ½LI² (stored in magnetic field, not in the inductor material).
- Self inductance depends only on geometry (N, A, l) and material (μ), NOT on current or EMF.
⚠️ NEET: ε = −LdI/dt. L = μ0N²A/l for
solenoid. U = ½LI². L is like inertia; opposes change in I. 1H = 1 Wb/A = 1 V·s/A.
6. Mutual Inductance
Mutual Inductance (M): The property of a pair of coils where a change in current in primary
coil induces an EMF in the secondary coil. M = μ0N1N2A/l for two
coaxial solenoids.
Flux in secondary due to primary: N2ϕ2 = MI1
Induced EMF in secondary: ε2 = −M dI1/dt
Mutual inductance of coaxial solenoids: M = μ0N1N2A/l
Relation: M = k√(L1L2) where k = coupling coefficient (0 ≤ k ≤ 1)
Induced EMF in secondary: ε2 = −M dI1/dt
Mutual inductance of coaxial solenoids: M = μ0N1N2A/l
Relation: M = k√(L1L2) where k = coupling coefficient (0 ≤ k ≤ 1)
- M is a property of the pair of coils, not individual coils.
- Depends on geometry: relative positions, orientations, sizes, and medium.
- M is maximum (k = 1) when coils are perfectly coupled (same core, flux linkage = 100%).
- M = 0 when coils are oriented perpendicular (no flux linkage between them).
- Mutual inductance is the operating principle of transformers.
6.1 Combinations of Inductors
Series (no mutual coupling): Leff = L1 + L2
Parallel (no mutual coupling): 1/Leff = 1/L1 + 1/L2
(Same rules as resistors — unlike capacitors!)
Parallel (no mutual coupling): 1/Leff = 1/L1 + 1/L2
(Same rules as resistors — unlike capacitors!)
7. AC Generator
AC Generator: A device that converts mechanical energy into electrical energy using the
principle of electromagnetic induction (Faraday's law). When a coil rotates in a uniform magnetic field,
the flux changes continuously → alternating EMF is induced.
ϕ = NBA cosωt (if at t=0, coil is in plane of B, i.e., ϕ starts at 0)
ε = −dϕ/dt = NBAω sinωt = ε0 sinωt
Peak EMF: ε0 = NBAω = NBA(2πf)
N = turns, B = magnetic field, A = area, ω = angular velocity, f = frequency
ε = −dϕ/dt = NBAω sinωt = ε0 sinωt
Peak EMF: ε0 = NBAω = NBA(2πf)
N = turns, B = magnetic field, A = area, ω = angular velocity, f = frequency
- EMF is sinusoidal: varies as sinωt if coil starts in plane of B (perpendicular to B → ϕ=0, rate of change of ϕ is maximum → max EMF).
- EMF is maximum when coil is parallel to B (ϕ = 0, dϕ/dt is maximum).
- EMF is zero when coil is perpendicular to B (ϕ is maximum, dϕ/dt = 0).
- Parts: Armature (rotating coil), permanent magnets (provides B), slip rings + carbon brushes (transfer current from rotating coil to external circuit without twisting wires).
- Increasing ε0: increase N, B, A, or ω (any one).
⚠️ NEET: ε = NBAω sinωt. Peak EMF = NBAω. Max EMF when
coil || B (ϕ = 0). Zero EMF when coil ⊥ B (ϕ = max). Slip rings (AC generator) vs Commutator
(DC generator). Principle: rotating coil in B → sinusoidal EMF.
8. Displacement Current and Maxwell's Equations
Displacement Current: Introduced by Maxwell. Even in the absence of a physical conductor
(e.g., between capacitor plates), a changing electric field (∂E/∂t) acts as a source of magnetic
field. Id = ε0 dϕE/dt.
Displacement current: Id = ε0 dE/dt × A = ε0
dϕE/dt
Modified Ampere's law: ∮B·dl = μ0(I + Id)
Id is continuous across capacitor gap; Id = Iconduction (same value outside = inside between plates).
Modified Ampere's law: ∮B·dl = μ0(I + Id)
Id is continuous across capacitor gap; Id = Iconduction (same value outside = inside between plates).
- Displacement current ensures continuity of current in circuits with capacitors.
- Maxwell's equations unify electricity, magnetism, and optics → electromagnetic waves predicted.
- Key insight: Changing E field → B field (displacement current). Changing B field → E field (Faraday's law). Together: self-sustaining EM waves.
9. Key Derivations & Concepts Summary
| Concept | Formula | Key Condition |
|---|---|---|
| Faraday's law (N turns) | ε = −N dϕ/dt | Magnitude: |ε| = N|Δϕ/Δt| |
| Lenz's law direction | opposes cause | Conservation of energy |
| Motional EMF | ε = BLv | v ⊥ B ⊥ L |
| Motional force (retarding) | F = B²L²v/R | Opposes motion (Lenz) |
| Induced charge | q = Δϕ/R = NBΔA/R | Independent of time taken |
| Self inductance (solenoid) | L = μ0n²Al = μ0N²A/l | Geometry only |
| Energy in inductor | U = ½LI² | Stored in B field of inductor |
| Mutual inductance | ε2 = −M dI1/dt | M = μ0N1N2A/l |
| AC generator EMF | ε = NBAω sinωt | Max when coil || B |
| Displacement current | Id = ε0dϕE/dt | Between capacitor plates |
🎓 NEET Previous Year Questions
Q1. [NEET 2022] A square coil of side 10 cm has 20 turns and resistance 10 Ω. It
is placed perpendicular to a field of 0.5 T. The field drops to zero in 0.1 s. Find induced current.
Answer
Δϕ = BΔAΔcosθ = (0.5−0)×(0.1)² = 0.5×0.01 =
5×10−3 Wb
ε = NΔϕ/Δt = 20×5×10−3/0.1 = 20×0.05 = 1 V
I = ε/R = 1/10 = 0.1 A
ε = NΔϕ/Δt = 20×5×10−3/0.1 = 20×0.05 = 1 V
I = ε/R = 1/10 = 0.1 A
Q2. [NEET 2021] A conducting rod of length 1 m moves with 4 m/s perpendicular to field
of 0.5 T. Find EMF and current if resistance = 2 Ω.
Answer
ε = BLv = 0.5×1×4 = 2 V. I = ε/R = 2/2 = 1 A.
Q3. [NEET 2020] The self inductance of a solenoid (N=300 turns, length=25cm,
area=5cm²) is:
Answer
L = μ0N²A/l =
4π×10−7×(300)²×5×10−4/0.25
= 4π×10−7×90000×2×10−3 = 4π×10−7×180 = 720π×10−7
= 7.2π×10−5 ≈ 2.26×10−4 H ≈ 0.226 mH
= 4π×10−7×90000×2×10−3 = 4π×10−7×180 = 720π×10−7
= 7.2π×10−5 ≈ 2.26×10−4 H ≈ 0.226 mH
Q4. [NEET 2019] The flux linked with a coil changes from 2 Wb to 10 Wb in 4 s. Find
induced EMF.
Answer
ε = Δϕ/Δt = (10−2)/4 = 8/4 = 2 V. (N = 1 assumed; for N
turns: multiply by N.)
Q5. [NEET 2018] A coil of 100 turns and area 2×10−3 m² is
placed in B = 0.05 T. The plane of coil rotates from B to perpendicular in 0.05 s. Find EMF.
Answer
Initially coil plane along B → ϕi = 0. Finally coil ⊥ B →
ϕf = NBA = 100×0.05×2×10−3 = 0.01 Wb (for N turns:
N×BA)
Wait: NΔϕ = N(ϕf−ϕi) = 100×(0.05×2×10−3−0) = 100×10−4 = 0.01 Wb
ε = NΔϕ/Δt = 0.01/0.05 = 0.2 V
Wait: NΔϕ = N(ϕf−ϕi) = 100×(0.05×2×10−3−0) = 100×10−4 = 0.01 Wb
ε = NΔϕ/Δt = 0.01/0.05 = 0.2 V
Q6. [NEET 2016] An AC generator with 100 turns, area 1 m² rotates at 50 rps in B
= 0.1 T. Peak EMF = ?
Answer
ω = 2πf = 2π×50 = 100π rad/s
ε0 = NBAω = 100×0.1×1×100π = 1000π ≈ 3142 V ≈ 3.14 kV
ε0 = NBAω = 100×0.1×1×100π = 1000π ≈ 3142 V ≈ 3.14 kV
💡 Rapid Revision — Key Formulas
- ϕ = BA cosθ | ε = −N dϕ/dt | Charge q = NΔϕ/R (time-independent)
- Motional EMF: ε = BLv | Retarding F = B²L²v/R | Rotating rod: ε = ½BL²ω
- Eddy currents: minimised by laminated core
- L (solenoid) = μ0N²A/l = μ0n²Al | ε = −LdI/dt | U = ½LI²
- Mutual M: ε2 = −MdI1/dt | M = μ0N1N2A/l | M = k√(L1L2)
- AC generator: ε = NBAω sinωt | Max EMF when coil || B | Zero EMF when coil ⊥ B
- Displacement current: Id = ε0dϕE/dt
CLASS 12 PHYSICS | NCERT SOLUTIONS
Chapter 6 — Electromagnetic Induction
25 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions
Key Formulas: ϕ=BAcosθ | ε=−Ndϕ/dt | q=NΔϕ/R |
ε=BLv | F=B²L²v/R | L=μ0N²A/l | ε=−LdI/dt | U=½LI²
| ε2=−MdI1/dt | ε0=NBAω
📝 NCERT Exercise Questions (6.1 – 6.16)
3 MarksQ1 (Ex 6.1). A circular coil of radius 8 cm
and 20 turns lies in a magnetic field of 3×10−2 T. Find (a) flux through coil when
its plane is perpendicular to B, (b) flux when its plane makes 30° angle with B.
✓ Solution
A = πr² = π×(0.08)² = π×64×10−4 = 2.011×10−2 m², B = 3×10−2 T
(a) Plane ⊥ to B ⇒ normal to plane is parallel to B ⇒ θ = 0°:
A = πr² = π×(0.08)² = π×64×10−4 = 2.011×10−2 m², B = 3×10−2 T
(a) Plane ⊥ to B ⇒ normal to plane is parallel to B ⇒ θ = 0°:
ϕ = NBA cos0° =
20×3×10−2×2.011×10−2×1 =
1.21×10−2 Wb
(b) Plane makes 30° with B ⇒ normal makes 60° with B ⇒ θ =
60°:ϕ = NBAcos60° = 1.21×10−2×0.5 =
6.05×10−3 Wb
3 MarksQ2 (Ex 6.2). A rectangular loop of sides 8 cm
and 2 cm with resistance = 4 Ω is pulled out of uniform B = 0.3 T with constant velocity 1 cm/s.
Calculate power consumed by loop.
✓ Solution
L = 2 cm = 0.02 m (side perpendicular to motion), v = 0.01 m/s, B = 0.3 T, R = 4 Ω
Motional EMF:
Power:
L = 2 cm = 0.02 m (side perpendicular to motion), v = 0.01 m/s, B = 0.3 T, R = 4 Ω
Motional EMF:
ε = BLv = 0.3×0.02×0.01 = 6×10−5 V
Current: I = ε/R = 6×10−5/4 = 1.5×10−5 APower:
P = ε²/R = (6×10−5)²/4 =
36×10−10/4 = 9×10−10 W
3 MarksQ3 (Ex 6.3). A long coaxial solenoid has outer
coil: 500 turns, radius 10 cm; inner coil: 1000 turns, radius 4 cm. Find mutual inductance assuming all
flux of inner links outer.
✓ Solution
Inner coil (primary): N1 = 1000, r1 = 0.04 m
Outer coil (secondary): N2 = 500, r2 = 0.10 m
B due to inner coil: B = μ0N1I/(l) (if same length l). Flux through outer coil:
(for l = 1 m and given N1N2): M = 4π×10−7×1000×500×5.027×10−3/1
M = 4π×10−7×5×105×5.027×10−3 ≈ 3.15×10−3 H ≈ 3.15 mH
Inner coil (primary): N1 = 1000, r1 = 0.04 m
Outer coil (secondary): N2 = 500, r2 = 0.10 m
B due to inner coil: B = μ0N1I/(l) (if same length l). Flux through outer coil:
M = μ0N1N2Ainner/l
Area of inner coil: A = π×(0.04)² = 5.027×10−3 m²(for l = 1 m and given N1N2): M = 4π×10−7×1000×500×5.027×10−3/1
M = 4π×10−7×5×105×5.027×10−3 ≈ 3.15×10−3 H ≈ 3.15 mH
3 MarksQ4 (Ex 6.4). A solenoid with iron core (800
turns, 25 cm long, cross-section 4 cm²). Relative permeability = 200. Current changes from 0 to 5A
in 1 s. Find induced EMF.
✓ Solution
L = μ0μrN²A/l = 4π×10−7×200×(800)²×4×10−4/0.25
L = 4π×10−7×200×640000×1.6×10−3
L = 4π×10−7×200×1024 = 4π×10−7×2.048×105
L ≈ 4×3.14159×10−7×2.048×105 ≈ 2.57×10−1 H ≈ 0.257 H
L = μ0μrN²A/l = 4π×10−7×200×(800)²×4×10−4/0.25
L = 4π×10−7×200×640000×1.6×10−3
L = 4π×10−7×200×1024 = 4π×10−7×2.048×105
L ≈ 4×3.14159×10−7×2.048×105 ≈ 2.57×10−1 H ≈ 0.257 H
ε = L ΔI/Δt = 0.257×5/1 = 1.285 V ≈ 1.3
V
3 MarksQ5 (Ex 6.5). A jet plane with wing span 25 m
is flying horizontally at 1800 km/h. Earth's field BV = 3.0×10−5 T.
Find EMF induced between wing tips.
✓ Solution
L = 25 m, v = 1800 km/h = 1800×1000/3600 = 500 m/s
Horizontal motion → vertical BV cuts the wings:
L = 25 m, v = 1800 km/h = 1800×1000/3600 = 500 m/s
Horizontal motion → vertical BV cuts the wings:
ε = BVLv = 3.0×10−5×25×500 =
3.0×10−5×12500 = 0.375 V
3 MarksQ6 (Ex 6.6). Current in a solenoid changes at
5 A/s. Induced EMF = 20 mV. Find self inductance.
✓ Solution
|ε| = L |dI/dt|
|ε| = L |dI/dt|
L = |ε| / |dI/dt| = 20×10−3 / 5 =
4×10−3 H = 4 mH
3 MarksQ7 (Ex 6.7). A 1.0 m rod rotates about one end
in a plane perpendicular to uniform B = 0.4 T. If rod completes 1 revolution per second, find EMF.
✓ Solution
ω = 2πf = 2π×1 = 2π rad/s
ω = 2πf = 2π×1 = 2π rad/s
ε = ½BL²ω = ½×0.4×(1.0)²×2π
= ½×0.4×2π = 0.4π
ε = 0.4×3.14159 ≈ 1.257 V ≈ 1.26 V3 MarksQ8 (Ex 6.8). A rectangular wire loop of sides
0.1 m × 0.05 m with resistance 1 Ω is placed in a B field that changes at 3 T/s. Find induced
EMF and current.
✓ Solution
Area A = 0.1×0.05 = 5×10−3 m²
Area A = 0.1×0.05 = 5×10−3 m²
ε = A × dB/dt = 5×10−3×3 =
1.5×10−2 V = 15 mV
I = ε/R = 15×10−3/1 = 15 mA3 MarksQ9 (Ex 6.9). A 100-turn coil of area 200
cm² in a field of 0.1 T. In 0.1 s, coil is rotated from initial position (ϕ = max) to final
position (ϕ = 0). Find average EMF.
✓ Solution
Initial ϕi = NBA = 100×0.1×(200×10−4) = 100×0.1×0.02 = 0.2 Wb
Final ϕf = 0
Initial ϕi = NBA = 100×0.1×(200×10−4) = 100×0.1×0.02 = 0.2 Wb
Final ϕf = 0
εavg = NØϕ/Δt =
(ϕi−ϕf)/Δt = 0.2/0.1 = 2 V
(N is already included in 0.2 Wb = Nϕ linkage)3 MarksQ10 (Ex 6.10). A pair of adjacent coils has
mutual inductance 1.5 H. If current in one coil changes from 0 to 20 A in 0.5 s, find (a) change in
flux, (b) induced EMF in other coil.
✓ Solution
M = 1.5 H, ΔI = 20 A, Δt = 0.5 s
(a) Change in flux linkage in secondary:
M = 1.5 H, ΔI = 20 A, Δt = 0.5 s
(a) Change in flux linkage in secondary:
NΔϕ = MΔI = 1.5×20 = 30 Wb (flux linkage
change)
(b) EMF in other coil:|ε| = M |ΔI/Δt| = 1.5×20/0.5 = 1.5×40 = 60
V
3 MarksQ11 (Ex 6.11). A coil (100 turns, area 1
cm²) rotates at 50 rev/s in field of 0.12 T. Find peak and rms EMFs.
✓ Solution
ω = 2πf = 2π×50 = 100π rad/s, A = 1×10−4 m²
εrms = ε0/√2 = 0.377/1.414 = 0.267 V
ω = 2πf = 2π×50 = 100π rad/s, A = 1×10−4 m²
ε0 = NBAω =
100×0.12×10−4×100π =
100×0.12×10−4×314.16
ε0 = 100×0.12×0.031416 = 100×3.77×10−3 =
0.377 V ≈ 0.38 Vεrms = ε0/√2 = 0.377/1.414 = 0.267 V
3 MarksQ12 (Ex 6.12). A circular coil of radius 12 cm
and 250 turns is rotated at 60 rev/s in field of 0.3 T. Find peak EMF.
✓ Solution
A = π×(0.12)² = π×0.0144 = 4.524×10−2 m²
ω = 2π×60 = 120π rad/s
A = π×(0.12)² = π×0.0144 = 4.524×10−2 m²
ω = 2π×60 = 120π rad/s
ε0 = NBAω =
250×0.3×4.524×10−2×120π
= 250×0.3×4.524×10−2×376.99 ≈ 250×0.3×17.06 =
250×5.117 = 1279 V ≈ 1.28 kV3 MarksQ13 (Ex 6.13). Two coils are wound on the same
iron core. Coil 1 has 3000 turns, coil 2 has 1500 turns. A current of 4A in coil 1 produces flux of
4×10−4 Wb per turn in coil 1. Find M and L1.
✓ Solution
All flux of coil 1 passes through coil 2 (iron core, k = 1)
N1ϕ1 = L1I1 ⇒ L1 = N1ϕ1/I1 = 3000×4×10−4/4
Verify: M = k√(L1L2). If k=1: 0.15 = √(0.3×L2) ⇒ L2 = 0.0225/0.3 = 0.075 H ✓
All flux of coil 1 passes through coil 2 (iron core, k = 1)
N1ϕ1 = L1I1 ⇒ L1 = N1ϕ1/I1 = 3000×4×10−4/4
L1 = 3000×10−4 = 0.3 H
M = N2ϕ1/I1 = 1500×4×10−4/4 =
1500×10−4 = 0.15 HVerify: M = k√(L1L2). If k=1: 0.15 = √(0.3×L2) ⇒ L2 = 0.0225/0.3 = 0.075 H ✓
3 MarksQ14 (Ex 6.14). An inductor (L = 2H, resistance
= 200 Ω) is connected to 220V supply. Find energy stored when steady state is reached.
✓ Solution
Steady state: no change in I → no back EMF → inductor acts as pure resistor.
Isteady = V/R = 220/200 = 1.1 A
Steady state: no change in I → no back EMF → inductor acts as pure resistor.
Isteady = V/R = 220/200 = 1.1 A
U = ½LI² = ½×2×(1.1)² = ½×2×1.21
= 1.21 J
3 MarksQ15 (Ex 6.15). A solenoid of 500 turns, 50 cm
long carries current 2A. Find (a) magnetic field at centre, (b) energy stored in inductor. (area = 2
cm²)
✓ Solution
n = N/l = 500/0.5 = 1000 turns/m, I = 2 A, A = 2×10−4 m²
(a) B = μ0nI = 4π×10−7×1000×2 = 8π×10−4 = 2.51×10−3 T ≈ 2.51 mT
L = μ0n²Al = 4π×10−7×(1000)²×2×10−4×0.5
L = 4π×10−7×106×10−4 = 4π×10−5 = 1.257×10−4 H
n = N/l = 500/0.5 = 1000 turns/m, I = 2 A, A = 2×10−4 m²
(a) B = μ0nI = 4π×10−7×1000×2 = 8π×10−4 = 2.51×10−3 T ≈ 2.51 mT
L = μ0n²Al = 4π×10−7×(1000)²×2×10−4×0.5
L = 4π×10−7×106×10−4 = 4π×10−5 = 1.257×10−4 H
U = ½LI² = ½×1.257×10−4×4 =
2.51×10−4 J
3 MarksQ16 (Ex 6.16). Obtain the expression for
mutual inductance between a long straight wire and a rectangular loop of width b, length a, with nearest
side at distance d from the wire.
✓ Solution
Field due to long wire at distance x: B = μ0I/(2πx)
Flux through thin strip of width dx and length a: dϕ = B×a×dx = μ0Ia/(2πx) dx
Total flux: ϕ = ∫dd+b μ0Ia/(2πx) dx = μ0Ia/(2π) × ln[(d+b)/d]
M = ϕ/I:
Field due to long wire at distance x: B = μ0I/(2πx)
Flux through thin strip of width dx and length a: dϕ = B×a×dx = μ0Ia/(2πx) dx
Total flux: ϕ = ∫dd+b μ0Ia/(2πx) dx = μ0Ia/(2π) × ln[(d+b)/d]
M = ϕ/I:
M = μ0a/(2π) × ln[(d+b)/d] =
μ0a ln(1+b/d)/(2π)
This is the standard NCERT derivation. M increases with a, b, and decreases with d.🌟 Additional / Exemplar Questions (Q17 – Q25)
3 MarksQ17. What is the induced charge in a coil (100
turns, 40 mΩ resistance) when flux changes by 4 mWb?
✓ Solution
Induced charge: q = NΔϕ/R (independent of time of change!)
Induced charge: q = NΔϕ/R (independent of time of change!)
q = 100×4×10−3 / (40×10−3) =
0.4/0.04 = 10 C
This is a key result: induced charge depends only on flux change and resistance, NOT on how quickly the
flux changes.3 MarksQ18. A conducting rod of mass 0.1 kg, length
0.5 m, can slide without friction on rails 0.5 m apart in B = 2 T (perpendicular to rails). A force of 1
N is applied to rod. Find: (a) velocity after 10 s, (b) EMF and current at t=10s. (R = 5 Ω)
✓ Solution
Net force = Applied − Retarding force = F − B²L²v/R
At terminal velocity: net force = 0 ⇒ F = B²L²vt/R
vt = FR/(B²L²) = 1×5/(4×0.25) = 5/1 = 5 m/s (terminal velocity)
EMF at v=5m/s: ε = BLv = 2×0.5×5 = 5 V
Current: I = 5/5 = 1 A
At terminal: Fretard = BIL = 2×1×0.5 = 1 N = applied force ✓
Net force = Applied − Retarding force = F − B²L²v/R
At terminal velocity: net force = 0 ⇒ F = B²L²vt/R
vt = FR/(B²L²) = 1×5/(4×0.25) = 5/1 = 5 m/s (terminal velocity)
EMF at v=5m/s: ε = BLv = 2×0.5×5 = 5 V
Current: I = 5/5 = 1 A
At terminal: Fretard = BIL = 2×1×0.5 = 1 N = applied force ✓
3 MarksQ19. Find the direction of induced current in
a coil: (a) a bar magnet moved toward the face of a coil such that the face is NE-SW and N pole
approaches. (b) If the same magnet is moved away.
✓ Solution
(a) N pole approaching: Flux through coil increases (into coil). By Lenz's law, induced current must create N pole on the facing side to repel the magnet. By right-hand rule: current flows anticlockwise when viewed from the approaching magnet side (counterclockwise to make N pole facing incoming N).
(b) N pole moving away: Flux decreases. Induced current must try to maintain flux → creates S pole on facing side to attract the magnet (oppose decrease). Current flows clockwise when viewed from the magnet side.
(a) N pole approaching: Flux through coil increases (into coil). By Lenz's law, induced current must create N pole on the facing side to repel the magnet. By right-hand rule: current flows anticlockwise when viewed from the approaching magnet side (counterclockwise to make N pole facing incoming N).
(b) N pole moving away: Flux decreases. Induced current must try to maintain flux → creates S pole on facing side to attract the magnet (oppose decrease). Current flows clockwise when viewed from the magnet side.
3 MarksQ20. Derive an expression for self inductance
of a toroid of circular cross-section area A, mean radius R, having N turns.
✓ Solution
By Ampere's law, B inside toroid = μ0NI/(2πR)
Flux per turn = B×A = μ0NIA/(2πR)
Total flux linkage = N × flux per turn = μ0N²IA/(2πR)
L = Nϕ/I:
With core: L = μrμ0N²A/(2πR).
By Ampere's law, B inside toroid = μ0NI/(2πR)
Flux per turn = B×A = μ0NIA/(2πR)
Total flux linkage = N × flux per turn = μ0N²IA/(2πR)
L = Nϕ/I:
Ltoroid = μ0N²A/(2πR)
Compare with solenoid: Lsol = μ0N²A/l (l = length of solenoid = 2πR
for toroid → same formula!)With core: L = μrμ0N²A/(2πR).
3 MarksQ21. An inductance of 0.5 H is connected to a
200 V AC source. At the instant when I = 0.1 A and increasing at 100 A/s, find the back EMF and the net
voltage.
✓ Solution
Back EMF (self-induced): εback = L dI/dt = 0.5×100 = 50 V (opposes increase in I, so opposes source).
The inductor's εback always opposes the source. Net driving voltage would be related to resistive part.
Back EMF (self-induced): εback = L dI/dt = 0.5×100 = 50 V (opposes increase in I, so opposes source).
The inductor's εback always opposes the source. Net driving voltage would be related to resistive part.
3 MarksQ22. A circular loop of radius 10 cm is placed
in a uniform magnetic field of 0.2 T. The plane of loop is perpendicular to B. If loop is pulled out of
B region in 0.3 s, find induced EMF.
✓ Solution
Initial: ϕi = B×πr² = 0.2×π×(0.10)² = 0.2×π×0.01 = 2π×10−3 Wb
Final: ϕf = 0 (outside field)
Initial: ϕi = B×πr² = 0.2×π×(0.10)² = 0.2×π×0.01 = 2π×10−3 Wb
Final: ϕf = 0 (outside field)
ε = Δϕ/Δt = 2π×10−3/0.3 =
(2×3.14159×10−3)/0.3 ≈ 0.0209 V ≈ 20.9 mV
3 MarksQ23. Explain with example why Lenz's law
is a consequence of conservation of energy.
✓ Solution
Consider a bar magnet approaching a coil. If Lenz's law were reversed (induced current aided the flux increase), the coil would attract the approaching magnet → magnet accelerates → induces more current → more attraction → self-amplifying process → creates energy from nothing → violates conservation of energy.
Instead, Lenz's law ensures the induced current repels the approaching magnet → work must be done against this repulsion to move the magnet → this work is converted to electrical energy (then heat in resistance) → energy is conserved. Lenz's law ensures no free energy is created, consistent with the law of conservation of energy.
Consider a bar magnet approaching a coil. If Lenz's law were reversed (induced current aided the flux increase), the coil would attract the approaching magnet → magnet accelerates → induces more current → more attraction → self-amplifying process → creates energy from nothing → violates conservation of energy.
Instead, Lenz's law ensures the induced current repels the approaching magnet → work must be done against this repulsion to move the magnet → this work is converted to electrical energy (then heat in resistance) → energy is conserved. Lenz's law ensures no free energy is created, consistent with the law of conservation of energy.
3 MarksQ24. Two identical solenoids of inductance L
each are connected (a) in series, (b) in parallel. Find effective inductance in each case. (Ignore
mutual inductance.)
✓ Solution
Inductors combine like resistors (for zero mutual inductance):
(a) Series:
Inductors combine like resistors (for zero mutual inductance):
(a) Series:
Leff = L + L = 2L
(b) Parallel:1/Leff = 1/L + 1/L = 2/L ⇒ Leff =
L/2
Same as resistor rules. (Capacitors are opposite: series → 1/C adds; parallel → C adds.)3 MarksQ25. A bicycle generator (dynamo) has 100
turns of area 10 cm² rotating at 100 rpm in field of 0.05 T. Find peak EMF and frequency.
✓ Solution
N = 100, A = 10 cm² = 10×10−4 m² = 10−3 m²
f = 100 rpm = 100/60 = 5/3 Hz
ω = 2πf = 2π×5/3 = 10π/3 rad/s
Frequency = 5/3 ≈ 1.67 Hz
N = 100, A = 10 cm² = 10×10−4 m² = 10−3 m²
f = 100 rpm = 100/60 = 5/3 Hz
ω = 2πf = 2π×5/3 = 10π/3 rad/s
ε0 = NBAω =
100×0.05×10−3×10π/3 =
100×0.05×10π×10−3/3
= 100×0.05×0.01047 = 100×5.236×10−4 =
5.24×10−2 V ≈ 52.4 mVFrequency = 5/3 ≈ 1.67 Hz
✍ Score Guide — 25 Questions
NCERT Ex 6.1–6.16: 3 marks each — 48 marks | Additional Q17–Q25: 3 marks each — 27 marks | Grand Total: 75 marks
NCERT Ex 6.1–6.16: 3 marks each — 48 marks | Additional Q17–Q25: 3 marks each — 27 marks | Grand Total: 75 marks
CLASS 12 PHYSICS | FORMULA CAPSULE
Electromagnetic Induction
Chapter 6 — All Key Formulas & High-Yield Facts for NEET/JEE
📅 Section A — Faraday's Law & Motional EMF
| Quantity | Formula | Key Note |
|---|---|---|
| Magnetic flux | ϕ = BAcosθ | Max when B ∥ normal; Zero when B ∥ plane |
| Faraday's law (N turns) | ε = −N dϕ/dt | |ε| = N |Δϕ/Δt| |
| Induced charge | q = NΔϕ/R | Independent of time of change! |
| Motional EMF | ε = BLv | v ⊥ B ⊥ L all mutually perpendicular |
| Motional EMF (angle) | ε = BLv sinθ | θ between v and B |
| Power dissipated | P = B²L²v²/R | External agent supplies this power |
| Retarding force | F = B²L²v/R = BIL | Opposes motion (Lenz's law) |
| Rotating rod EMF | ε = ½BL²ω | One end fixed, other rotating |
| Lenz's law | Opposes cause (change in flux) | Consequence of energy conservation |
📅 Section B — Self & Mutual Inductance
| Quantity | Formula | Key Note |
|---|---|---|
| Self inductance (def) | Nϕ = LI | Unit: Henry = Wb/A = V·s/A |
| Induced back EMF | ε = −L dI/dt | Opposes change in current |
| L of solenoid | L = μ0N²A/l = μ0n²Al | Geometry only; n = N/l |
| L of toroid | L = μ0N²A/(2πR) | R = mean radius of toroid |
| Energy in inductor | U = ½LI² | Stored in magnetic field |
| Inductors series | Leff = L1+L2 | Same as resistors in series |
| Inductors parallel | 1/Leff = 1/L1+1/L2 | Same as resistors in parallel |
| Mutual inductance (def) | N2ϕ2 = MI1 | Unit: Henry |
| Induced EMF (secondary) | ε2 = −M dI1/dt | Mutual coupling |
| M of coaxial solenoids | M = μ0N1N2A/l | k=1 (perfectly coupled) |
| M in terms of L | M = k√(L1L2) | k = coupling coefficient (0 to 1) |
📅 Section C — AC Generator & Displacement Current
| Quantity | Formula | Key Note |
|---|---|---|
| Flux (rotating coil) | ϕ = NBAcosωt | t=0: coil ⊥ to B (flux max) |
| Induced EMF | ε = NBAω sinωt = ε0 sinωt | Alternating (sinusoidal) |
| Peak EMF | ε0 = NBAω = NBA×2πf | Increase N, B, A, or ω to increase ε0 |
| EMF max when | Coil plane ∥ to B (ϕ=0, dϕ/dt = max) | Not when ϕ is max! |
| EMF zero when | Coil plane ⊥ to B (ϕ = max, dϕ/dt = 0) | Not when coil is moving! |
| Displacement current | Id = ε0 dϕE/dt | Between capacitor plates |
| Ampere-Maxwell law | ∮B·dl = μ0(I + Id) | Generalised; Id=Iconduction outside |
🧠 Memory Tricks
“q = Δϕ/R — Time Independent!”
Induced charge q = NΔϕ/R depends ONLY on the change in flux and resistance. It does NOT depend
on HOW FAST the flux changes. Faster change → higher EMF but for shorter time → same charge.
This is a classic NEET trap question.
“L is like Mass, EMF is like Friction”
Self inductance L is the electromagnetic inertia. Just as mass resists change in velocity (F=ma), L
resists change in current (ε=LdI/dt). The back EMF is like friction — always opposing. Energy
stored = ½LI² (like KE = ½mv²).
“Max EMF when coil PARALLEL to B”
Generator: Max EMF when coil plane is parallel to B (NOT perpendicular!). Why? When plane || B →
ϕ = 0 but dϕ/dt is maximum → max EMF. When plane ⊥ B → ϕ = max but
dϕ/dt = 0 → zero EMF. Counterintuitive — know it well!
“Inductors = Resistors (not Capacitors!)”
Inductors in series: L adds up (like R in series). In parallel: reciprocals add (like R in parallel).
Capacitors are OPPOSITE. So for any given combination rule, LCR uses L and R similarly but C is the odd
one out.
“Eddy Current Losses → Laminations”
Eddy currents form in bulk conductors → heat loss. Laminated cores break the bulk into thin
insulated sheets → small loops → high resistance path → less eddy current → less
loss. Silicon steel laminations used in transformers and motors for this reason.
“Lenz = Opposing Force = Energy Conservation”
If Lenz's law were reversed, moving a magnet would accelerate itself → creating energy from
nothing → violates conservation of energy. So Lenz's law is not just a rule — it is
mandated by conservation of energy. The minus sign in ε=−dϕ/dt IS Lenz's law.
🔢 Critical Values & Unit Conversions
1 Wb = 1 T·m² = 1 V·s
1 H = 1 Wb/A = 1 V·s/A
q = NΔϕ/R (time-independent)
ε0(generator) = NBAω
Rotating rod: ε = ½BL²ω
Max EMF when coil || B
L (solenoid) = μ0n²Al
M = k√(L1L2); 0 ≤ k ≤ 1
❌ Common Mistakes to Avoid
- Flux vs rate of change: EMF = rate of change of flux (dϕ/dt), NOT flux itself. Large ϕ with zero rate of change → zero EMF.
- Direction of induced current: Always apply Lenz's law first (opposing flux change). Then determine current direction by right-hand rule for the coil.
- Induced charge formula: q = NΔϕ/R, NOT Nε/R. It is independent of time — slow or fast change, same total charge.
- Self inductance of solenoid: L = μ0N²A/l. It scales as N² (NOT N). Doubling turns quadruples L.
- AC Generator — EMF max position: EMF is MAX when coil is PARALLEL to B (not perpendicular). Flux is ZERO at this point, but rate of change is maximum.
- Displacement current continuity: Id = Ic (conduction current outside = displacement current between capacitor plates). Both produce the same magnetic field around the circuit.
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📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App