📌 Snapshot
- Establishes that a time-varying magnetic flux through a circuit induces an emf — the foundational link between electricity and magnetism discovered by Faraday and Henry around 1830.
- Develops the quantitative laws: magnetic flux Φ_B = B·A = BA cos θ, Faraday's law ε = −dΦ_B/dt (or −N dΦ_B/dt for an N-turn coil), and Lenz's law for the sign (rooted in energy conservation).
- Introduces motional emf ε = Blv from a moving conductor and shows it can be derived independently from the Lorentz force on free charges.
- Defines inductance (self and mutual) as the geometric/material constant relating flux linkage to current, with worked formulas for long solenoids and coaxial solenoid pairs.
- The AC generator applies these ideas: ε = NBAω sin ωt.
📖 Detailed Notes
2.1 Core concepts
- Oersted, Ampere and others established that moving charges produce magnetic fields; the converse — that a changing magnetic field can drive currents in a closed coil — was demonstrated independently by Faraday (England) and Henry (USA) around 1830 (NCERT §6.1, p. 154).
- Experiment 6.1: a galvanometer connected to a coil deflects when a bar magnet is moved towards or away from the coil; deflection reverses if the pole is reversed, is larger for faster motion, and is zero when the magnet is stationary — it is the relative motion between magnet and coil that matters (NCERT §6.2, p. 155).
- Experiment 6.2: replacing the bar magnet with a second current-carrying coil C2 reproduces the effect; moving C2 relative to C1 induces a current in C1, again driven by relative motion (NCERT §6.2, p. 155-156).
- Experiment 6.3: two stationary coils — pressing the tapping key in coil C2 produces a momentary deflection in C1; holding the key down gives no deflection; releasing it gives a momentary deflection in the opposite direction. Inserting an iron rod dramatically increases the effect, showing that change of flux (not motion per se) is the cause (NCERT §6.2, p. 156).
- Magnetic flux through a plane of area A in a uniform field B is Φ_B = B·A = BA cos θ; for non-uniform fields and curved surfaces, Φ_B = Σ B_i · dA_i; SI unit is the weber (Wb) = T·m²; flux is a scalar (NCERT §6.3, p. 156-157).
- Faraday's law: the magnitude of induced emf equals the time rate of change of magnetic flux, ε = −dΦ_B/dt; for a closely wound coil of N turns, ε = −N dΦ_B/dt; the negative sign encodes direction (Lenz) (NCERT §6.4, p. 157-158).
- Flux can be changed by varying B, by changing area A (stretching/shrinking the coil), or by changing θ (rotating the coil) — all produce induced emf (NCERT §6.4, p. 158).
- Lenz's law (Heinrich Friedrich Lenz, 1834): the polarity of induced emf is such that the induced current opposes the change in flux that produced it. If the opposite were true, a small push on a magnet would create runaway acceleration — a perpetual motion machine — violating conservation of energy; therefore work done by the agent moving the magnet appears as Joule heat in the induced current (NCERT §6.5, p. 160).
- Motional emf: for a rod of length l moving with velocity v perpendicular to a uniform field B (and to its own length), Φ_B = Blx, so ε = −dΦ_B/dt = Blv. The same expression follows from the Lorentz force qvB on each free charge — work qvBl per charge gives emf Blv (NCERT §6.6, p. 162-163).
- For a stationary conductor in a time-varying B, the force on charges is purely qE (since v = 0), so a time-varying magnetic field must generate an electric field — this induced electric field is different in nature from the electrostatic field produced by static charges (NCERT §6.6, p. 163).
- For a rod of length R rotating with angular speed ω about one end in a field B parallel to the axis, the motional emf between centre and tip is ε = BωR²/2 (NCERT §6.6, Example 6.6, p. 163-164).
- Inductance: in any geometry-fixed coil, flux linkage NΦ_B ∝ I; the constant of proportionality is the inductance. It depends only on geometry and intrinsic material properties; SI unit is the henry (H), dimensions [M L² T⁻² A⁻²] (NCERT §6.7, p. 165).
- Mutual inductance M for two long coaxial solenoids (inner radius r1, turns/length n1; outer turns/length n2; common length l): M = μ0 n1 n2 π r1² l. In a medium of relative permeability μ_r, M = μ_r μ0 n1 n2 π r1² l. M12 = M21 = M is general, even when only one direction is easy to compute (NCERT §6.7.1, p. 165-167).
- Mutual-induction emf: ε1 = −M dI2/dt — a changing current in one coil induces emf in a neighbouring coil (NCERT §6.7.1, p. 167).
- Self-inductance: when current in a single coil changes, NΦ_B = LI, and ε = −L dI/dt; the self-induced emf is also called the back emf because it opposes any change (increase or decrease) of current (NCERT §6.7.2, p. 168).
- Self-inductance of a long solenoid (cross-section A, length l, n turns per unit length): L = μ0 n² A l (air core); L = μ_r μ0 n² A l (core of relative permeability μ_r) (NCERT §6.7.2, p. 168).
- Work done against the back emf in establishing a current I is stored as magnetic energy W = ½ L I²; L plays the role of inertia (mass-analogue) for current; magnetic energy density u_B = B²/(2μ0) (NCERT §6.7.2, p. 168-170).
- For two simultaneously-current-carrying nearby coils, flux in coil 1 is N1Φ1 = M11 I1 + M12 I2 = L1 I1 + M I2, so ε1 = −L1 dI1/dt − M12 dI2/dt (NCERT §6.7.2, p. 169).
- AC generator (credited to Nicola Tesla in its developed form): an armature coil of N turns and area A is rotated with constant angular speed ω in a uniform field B. With θ = ωt, Φ_B = BA cos ωt, so ε = NBAω sin ωt = ε0 sin ωt with ε0 = NBAω; commercial frequency is 50 Hz in India, 60 Hz in the USA (NCERT §6.8, p. 170-172).
- Energy stored in an inductor is analogous to kinetic energy of a moving mass: W = ½LI² mirrors K = ½mv², with L taking the role of inertia and I that of velocity. This analogy underpins the conservation arguments used for LC oscillations in the next chapter (NCERT §6.7.2, p. 168-170).
- Magnetic energy density: u_B = B²/(2μ₀) for an air-cored region; this is the magnetic counterpart of the electric energy density u_E = ½ε₀E² that students met in Chapter 2, and integrating it over the solenoid volume reproduces ½LI² (NCERT §6.7.2, p. 170).
- The induced electric field generated by a time-varying B is non-conservative — its line integral around a closed loop equals −dΦ_B/dt, not zero. Hence one cannot define a single-valued potential for an induced field; this is the conceptual fork from electrostatics (NCERT §6.6, p. 163).
- For a stationary loop and a time-varying field, the flux change drives the emf via the induced E-field; for a moving loop in a static field, the flux change is geometric and the emf is the motional Blv computed via the Lorentz force. Both pictures give identical numerical answers, illustrating the elegance of Faraday's law (NCERT §6.6, p. 162-163).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Electromagnetic induction | The phenomenon in which electric current (emf) is generated in a circuit by a changing magnetic flux through it. | 154 |
| Magnetic flux (Φ_B) | Φ_B = B·A = BA cos θ for a uniform field over a plane area A; SI unit weber (Wb) = T·m²; a scalar quantity. | 156-157 |
| Faraday's law of induction | The induced emf equals the (negative) time rate of change of magnetic flux: ε = −dΦ_B/dt; for N turns, ε = −N dΦ_B/dt. | 157-158 |
| Lenz's law | The polarity of the induced emf is such that the induced current opposes the change in magnetic flux that produced it. | 160 |
| Motional emf | The emf Blv induced across a rod of length l moving with velocity v perpendicular to a uniform magnetic field B. | 162 |
| Mutual inductance (M) | Constant of proportionality between flux linkage in one coil and the current in a neighbouring coil: N1Φ1 = M12 I2; ε1 = −M dI2/dt. | 165-167 |
| Self-inductance (L) | Constant of proportionality between flux linkage and current in the same coil: NΦ_B = LI; ε = −L dI/dt. | 168 |
| Henry (H) | SI unit of inductance; 1 H gives 1 V of emf for a current changing at 1 A s⁻¹. | 165 |
| Back emf | The self-induced emf in a coil, so called because it opposes the change in current that produces it. | 168 |
| AC generator | A device that converts mechanical energy into electrical energy by rotating a coil in a uniform magnetic field, producing ε = NBAω sin ωt. | 170-172 |
| Weber (Wb) | SI unit of magnetic flux; 1 Wb = 1 T·m². | 157 |
| Flux linkage | Product NΦ_B of number of turns and flux through one turn. | 157, 165 |
| Induced current | Current flowing in a closed circuit as a result of induced emf. | 154 |
| Induced electric field | Non-conservative electric field set up by a time-varying magnetic field; drives the induced current. | 163 |
| Magnetic energy density | u_B = B²/(2μ₀); energy per unit volume stored in a magnetic field. | 170 |
| Energy stored in inductor | W = ½ L I². | 168-170 |
| Coaxial solenoids | Two long solenoids on the same axis, used to compute mutual inductance. | 166 |
| Lenz–energy argument | Statement that the negative sign in ε = −dΦ/dt is enforced by energy conservation. | 160 |
| Frequency of AC generator | ν = ω/(2π); 50 Hz in India, 60 Hz in the USA. | 172 |
2.3 Diagrams / processes to remember
- Figure 6.1: bar magnet pushed towards a coil C1 connected to galvanometer G — galvanometer pointer deflects only while the magnet moves (NCERT p. 155).
- Figure 6.2: coil C2 carrying steady current, moved relative to coil C1 — galvanometer in C1 deflects (NCERT p. 156).
- Figure 6.3: two stationary coils C1 and C2; tapping key K in C2 — momentary deflection in C1 on press and again (opposite direction) on release (NCERT p. 156).
- Figure 6.4: a plane area A in a uniform magnetic field B, used to define Φ_B = B·A (NCERT p. 157).
- Figure 6.5: non-uniform field over a curved/divided surface, with area-element vectors dA_i — basis for Φ_B = Σ B_i · dA_i (NCERT p. 157).
- Figure 6.6 (a, b): induced-current direction illustrations for Lenz's law with bar magnet approaching/receding from a coil (NCERT p. 160).
- Figure 6.7 / Figure 6.8: planar loops of various shapes moving into/out of a field region — used to argue when induced emf is constant (rectangular) vs varying (circular) (NCERT p. 161-162).
- Figure 6.10: rectangular conductor PQRS with movable arm PQ — geometric setup for motional emf Blv (NCERT p. 162).
- Figure 6.11: rotating rod hinged at the centre of a metallic ring — setup for ε = BωR²/2 (NCERT p. 164).
- Figure 6.12: two long coaxial solenoids of common length l — geometry from which M = μ0 n1 n2 π r1² l is derived (NCERT p. 166).
- Figure 6.13 / 6.14: schematic of an AC generator and the resulting sinusoidal emf vs time (NCERT p. 170-172).
2.4 Common confusions / NTA trap points
- "Strong magnet near a stationary coil" — students often expect current. NCERT Example 6.5(a) is explicit: however strong the magnet, no flux change ⇒ no induced current (NCERT p. 162).
- Confusing changing electric flux with changing magnetic flux — Example 6.5(b) hammers home that a loop moving through a static electric field between capacitor plates does NOT have an induced current; only changing magnetic flux induces emf in this chapter (NCERT p. 162).
- "Number of spokes matters" trap — for a rotating wheel, all spokes are emf sources in parallel, so their number does not affect the emf between axle and rim (NCERT Example 6.7, p. 165).
- Misremembering the area-vector angle — flux is BA cos θ, where θ is the angle between B and the area-vector normal, not between B and the plane of the coil. NCERT Example 6.2 sets up a 45° case to test exactly this (NCERT p. 158-159).
- For the rotating rod, the right formula is ε = BωR²/2 (not BωR² and not BωR), traceable to integrating dε = Bω r dr from 0 to R (NCERT p. 164).
- For the long solenoid, L = μ0 n² A l (where n is turns per unit length and A is cross-section), often confused with formulas using N (total turns) — be careful which the question gives (NCERT p. 168).
- Lenz's law is not an independent postulate that overrides Faraday's law — it is the physical content of the negative sign and is enforced by conservation of energy (NCERT p. 160).
- Forgetting the factor of N for multi-turn coils: ε = −N dΦ/dt — neglecting N gives an emf low by exactly the number of turns. Likewise self-inductance L scales as n², not n.
- Sign of motional emf direction: students get the magnitude Blv right but confuse polarity. Use Lenz: the induced current must oppose the change in flux through the circuit; the resulting magnetic force on the rod is opposite to v (a drag force consistent with energy conservation).
- M = M₁₂ = M₂₁ even when the two coils have very different sizes; the reciprocity follows from energy conservation, even though direct calculation in one direction may be hard.
- Wrong direction of induced E-field: the induced E-field circles the changing flux, with sense given by the right-hand rule applied to −dB/dt (not dB/dt).
- Misapplying ε = Blv when v is parallel to l: the formula requires v perpendicular to both l and B. If the rod slides along its own length, no emf is induced.
2.5 Key formulas table
| Quantity | Symbol / Formula | NCERT reference |
|---|---|---|
| Magnetic flux | Φ_B = B·A = BA cos θ | §6.3, Eq. 6.1, p. 156 |
| Flux linkage | NΦ_B | §6.4, p. 157 |
| Faraday's law (single loop) | ε = −dΦ_B/dt | §6.4, Eq. 6.3, p. 158 |
| Faraday's law (N turns) | ε = −N dΦ_B/dt | §6.4, Eq. 6.4, p. 158 |
| Motional emf | ε = Blv | §6.6, Eq. 6.5, p. 162 |
| Rotating rod emf | ε = BωR²/2 | §6.6, Ex. 6.6, p. 164 |
| Mutual inductance (coaxial solenoids) | M = μ₀ n₁ n₂ π r₁² l | §6.7.1, Eq. 6.9, p. 166 |
| Mutual-induction emf | ε₁ = −M dI₂/dt | §6.7.1, p. 167 |
| Self-inductance (definition) | NΦ_B = LI | §6.7.2, p. 168 |
| Self-inductance of solenoid | L = μ₀ n² A l | §6.7.2, Eq. 6.15, p. 168 |
| Back emf | ε = −L dI/dt | §6.7.2, Eq. 6.14, p. 168 |
| Energy stored in inductor | W = ½ L I² | §6.7.2, p. 170 |
| Magnetic energy density | u_B = B²/(2μ₀) | §6.7.2, p. 170 |
| AC generator emf | ε = NBAω sin ωt = ε₀ sin ωt | §6.8, Eq. 6.19, p. 171 |
| Peak emf (AC gen.) | ε₀ = NBAω | §6.8, p. 171 |
| Coupled coils total emf (coil 1) | ε₁ = −L₁ dI₁/dt − M dI₂/dt | §6.7.2, p. 169 |
| Reciprocity | M₁₂ = M₂₁ = M | §6.7.1, p. 167 |
| SI unit of flux | weber Wb = T·m² = V·s | §6.3, p. 157 |
| SI unit of inductance | henry H = V·s·A⁻¹ = Wb·A⁻¹ | §6.7, p. 165 |
| Indian / USA mains frequency | 50 Hz / 60 Hz | §6.8, p. 172 |
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. The SI unit of magnetic flux is the weber, which is equivalent to:
▸ Show answer & explanation
Answer: B
By definition Φ_B = B·A, so the unit is tesla times area, i.e. T·m². Option (A) would correspond to a line integral of B, not flux.
Q2. Faraday's law of electromagnetic induction states that the induced emf in a closed circuit is equal to:
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Answer: B
NCERT states explicitly that the magnitude of induced emf equals the time rate of change of flux, with the negative sign indicating the direction (Lenz). (A) confuses flux with rate of change of flux.
Q3. According to Lenz's law, the polarity of the induced emf is such that it tends to produce a current which:
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Answer: B
Lenz's law is the verbal statement of the negative sign in Faraday's law and is required by conservation of energy. (A) would allow a perpetual-motion machine, as NCERT discusses.
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Q4. The motional emf induced across a straight conductor of length l moving with velocity v perpendicular to a uniform magnetic field B (the velocity also being perpendicular to the conductor) is:
▸ Show answer & explanation
Answer: B
From Φ_B = Blx and ε = −dΦ_B/dt = −Bl(dx/dt) = Blv with dx/dt = −v. The same result follows from the Lorentz-force derivation (work qvBl per charge) on the same page.
Q5. Consider the following statements about magnetic flux and choose the correct option: I. Magnetic flux is a vector quantity. II. Magnetic flux through a plane area A in a uniform field B equals BA cos θ, where θ is the angle between B and the area vector. III. Magnetic flux can be increased by stretching or rotating a coil placed in a magnetic field.
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Answer: B
NCERT explicitly states that magnetic flux is a scalar (so I is wrong). II is the definition (6.1) and III is from §6.4.
Q6. A square loop of side 10 cm and resistance 0.5 Ω is placed vertically. A uniform magnetic field of 0.10 T makes an angle of 45° with the area vector of the loop. The field is reduced steadily to zero in 0.70 s. The magnitude of the induced emf is approximately:
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Answer: B
Initial flux = BA cos 45° = 0.10 × 10⁻² × (1/√2) = 10⁻³/√2 Wb. ε = ΔΦ/Δt = (10⁻³/√2)/0.70 ≈ 10⁻³ V = 1.0 mV.
Q7. A 1.0 m long metallic rod is rotated about an axis through one end, perpendicular to its length, with angular speed ω in a uniform magnetic field B parallel to the rotation axis. The emf induced between the centre and the free end of the rod is:
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Answer: C
Each element dr at radius r moves with v = ωr, contributing dε = Bv dr = Bωr dr. Integrating from 0 to R gives BωR²/2. (A) and (B) drop dimensional factors.
Q8. The mutual inductance of two long coaxial solenoids of common length l, with the inner solenoid of radius r₁ and turns-per-unit-length n₁, and outer solenoid of turns-per-unit-length n₂, is (air core):
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Answer: A
The field due to the outer solenoid is μ₀ n₂ I₂ and it threads the inner solenoid over its cross-section π r₁²; total turns in S₁ are n₁l, giving M = μ₀ n₁ n₂ π r₁² l. The inner radius (not outer) enters because that limits where the inner flux is non-zero.
Q9. The self-inductance of a long air-core solenoid of cross-sectional area A, length l, and n turns per unit length is:
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Answer: B
Field inside the solenoid is B = μ₀ n I; flux linkage is NΦ = (nl)(μ₀nI)(A) = μ₀ n² A l I, so L = μ₀ n² A l. (C) inverts the length dependence.
Q10. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside it normal to the axis. If the current in the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, the magnitude of the induced emf in the loop is approximately:
▸ Show answer & explanation
Answer: A
n = 1500 turns/m and dI/dt = (4.0 − 2.0)/0.1 = 20 A/s, so dB/dt = μ₀ n (dI/dt) = (4π × 10⁻⁷)(1500)(20) ≈ 3.77 × 10⁻² T/s. Then |ε| = (dB/dt)·A = 3.77 × 10⁻² × 2.0 × 10⁻⁴ ≈ 7.5 × 10⁻⁶ V.
Q11. A current in a coil falls from 5.0 A to 0.0 A in 0.1 s, and an average emf of 200 V is induced. The self-inductance of the coil is:
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Answer: C
|ε| = L |dI/dt| ⇒ L = |ε|/|dI/dt| = 200 / (5.0/0.1) = 200/50 = 4.0 H.
Q12. Assertion (A): In an AC generator, the maximum emf occurs when the plane of the coil is parallel to the magnetic field. Reason (R): The instantaneous emf in an AC generator is ε = NBAω sin ωt, and this is maximum when sin ωt = ±1, i.e., when θ between the area vector and the field is 90°.
▸ Show answer & explanation
Answer: A
When the area vector is at 90° to B, the plane of the coil is parallel to B; flux is zero but is changing fastest, so ε is maximum. R correctly explains this via ε₀ = NBAω at sin ωt = ±1.
Q13. The energy stored in an inductor of self-inductance 2.0 H carrying a current of 3.0 A is
▸ Show answer & explanation
Answer: C
W = ½ × 2.0 × 3.0² = ½ × 2.0 × 9 = 9.0 J. The 18.0 J distractor drops the ½ factor.
Q14. A coil of N turns and area A is rotated with angular speed ω in a uniform magnetic field B. The peak emf produced is
▸ Show answer & explanation
Answer: B
Differentiating Φ = NBA cos ωt gives ε = NBAω sin ωt; the maximum is NBAω. NCERT explicitly identifies ε₀ = NBAω.
Q15. Assertion (A): The mutual inductance of two coils does not change if their roles (source coil and pickup coil) are reversed. Reason (R): The flux linkage in coil 2 due to unit current in coil 1 equals the flux linkage in coil 1 due to unit current in coil 2; this reciprocity relation M₁₂ = M₂₁ follows from energy conservation.
▸ Show answer & explanation
Answer: A
M₁₂ = M₂₁ ≡ M, independent of which coil is the source. The argument from energy conservation explains the equality. Hence both true and R explains A.
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