Electromagnetic Induction
Vectors are represented in bold (e.g., B, v, A) or with arrow notation (B⃗). Magnitudes are in italics or normal text.
Electromagnetic Induction is the phenomenon by which an electromotive force (emf) is generated in a conductor when it is exposed to a changing magnetic field. This fundamental principle, discovered independently by Michael Faraday and Joseph Henry in the 1830s, forms the basis of electric generators, transformers, and many modern electrical devices.
1. Introduction
In earlier chapters, we studied how electric currents produce magnetic fields. A natural question arises: Can magnetic fields produce electric currents? The answer is yes, but with a crucial condition—the magnetic field must be changing with time.
The discovery of electromagnetic induction was one of the most significant achievements in physics, demonstrating that electricity and magnetism are interconnected phenomena.
Before 1820, electricity and magnetism were considered separate phenomena. Hans Christian Oersted's discovery in 1820 showed that electric currents create magnetic fields. Michael Faraday in England and Joseph Henry in America independently discovered electromagnetic induction around 1831, proving that changing magnetic fields can induce electric currents.
2. The Experiments of Faraday and Henry
Michael Faraday and Joseph Henry conducted systematic experiments demonstrating that a changing magnetic field can induce an electric current in a conductor.
2.1 Faraday's First Experiment
In his first experiment, Faraday used two coils wound on an iron ring. One coil was connected to a battery (primary coil), and the other to a galvanometer (secondary coil).
Observations:
- When the switch in the primary circuit was closed, the galvanometer showed momentary deflection and then returned to zero.
- When the switch was opened, the galvanometer again showed momentary deflection, but in opposite direction.
- When current in the primary coil was steady, no deflection was observed.
Conclusion: An emf is induced in the secondary coil only when magnetic flux through it is changing. A steady magnetic flux does not induce any emf.
2.2 Faraday's Second Experiment
Observations:
- When magnet was moved towards the coil, galvanometer showed deflection.
- When magnet was moved away, galvanometer showed deflection in opposite direction.
- When magnet was held stationary, no deflection was observed.
- Faster motion produced larger deflection.
3. Magnetic Flux
Magnetic flux through a surface is defined as the product of the magnetic field component perpendicular to the surface and the area of the surface.
• ΦB = Magnetic flux (weber, Wb)
• B = Magnetic field (tesla, T)
• A = Area vector (m²)
• θ = Angle between B and A
SI Unit: 1 weber (Wb) = 1 T·m²
4. Faraday's Law of Electromagnetic Induction
Where:
• ℰ = Induced emf (volts, V)
• N = Number of turns in coil
• dΦB/dt = Rate of change of flux (Wb/s)
• Negative sign represents Lenz's law
5. Lenz's Law and Energy Conservation
The direction of induced emf is such that it opposes the change in magnetic flux that produces it.
Lenz's law ensures energy conservation. If induced current aided the flux change, we would have perpetual motion - violating energy conservation.
6. Motional EMF
Setup: Conductor of length l moving with velocity v perpendicular to uniform magnetic field B.
Step 1: Free electrons in conductor experience magnetic force:
\[F = qvB\]
Step 2: This force separates charges, creating electric field:
\[E = vB\]
Step 3: Potential difference (emf) between ends:
Valid when B, l, and v are mutually perpendicular.
7. Inductance
7.1 Self-Inductance
Induced emf: \(\mathcal{E} = -L\frac{dI}{dt}\)
SI Unit: Henry (H)
Given: Solenoid with N turns, length l, area A, current I
Step 1: Magnetic field inside:
\[B = \mu_0 n I = \mu_0 \frac{N}{l} I\]
Step 2: Flux per turn:
\[\Phi = BA = \mu_0 \frac{N}{l} I A\]
Step 3: Total flux linkage:
\[N\Phi = \mu_0 \frac{N^2 A}{l} I\]
Note: L ∝ N² (proportional to square of turns)
7.2 Mutual Inductance
Induced emf: \(\mathcal{E}_2 = -M\frac{dI_1}{dt}\)
Reciprocity: M₁₂ = M₂₁ = M
7.3 Energy Stored in Inductor
Work done against back emf:
\[dW = Li \, di\]
Total work from 0 to I:
\[W = \int_0^I Li \, di = \frac{1}{2}LI^2\]
Magnetic energy density: \(u_B = \frac{B^2}{2\mu_0}\)
8. AC Generator
An AC generator converts mechanical energy into electrical energy using electromagnetic induction.
Given: Coil with N turns, area A, rotating with angular velocity ω in field B
At time t: Angle θ = ωt
Flux:
\[\Phi_B = NBA \cos(\omega t)\]
Induced emf:
\[\mathcal{E} = -\frac{d\Phi_B}{dt} = NBA\omega \sin(\omega t)\]
Output is sinusoidal AC with frequency f = ω/(2π)
9. Transformers (Conceptual Understanding - Detailed in AC Chapter)
A Transformer is an electrical device used to change (step-up or step-down) alternating voltage. It works on the principle of Mutual Induction. Transformers are essential for efficient power transmission over long distances.
9.1 Construction
A transformer consists of:
- Laminated Soft Iron Core: Provides path for magnetic flux and minimizes eddy current losses.
- Primary Coil (Np turns): Connected to AC input voltage Vp.
- Secondary Coil (Ns turns): Provides output voltage Vs.
9.2 Working Principle
When alternating current flows through primary coil, it creates changing magnetic flux in the core. This changing flux links with secondary coil and induces alternating emf by Faraday's law (mutual induction).
Given: Ideal transformer with perfect flux linkage.
Step 1: Same flux Φ passes through both coils.
Primary EMF: \[\varepsilon_p = -N_p \frac{d\Phi}{dt}\]
Step 2: Secondary EMF:
\[\varepsilon_s = -N_s \frac{d\Phi}{dt}\]
Step 3: Dividing equations:
• k > 1 (Ns > Np): Step-up transformer (increases voltage)
• k < 1 (Ns < Np): Step-down transformer (decreases voltage)
• k = 1 (Ns = Np): Isolation transformer (same voltage)
9.3 Current Relation
In an ideal transformer, input power equals output power (energy conservation):
Transformer cannot work on DC because DC produces constant flux. Without changing flux (dΦ/dt = 0), no EMF is induced in secondary coil.
9.4 Energy Losses in Real Transformers
1. Copper Losses (I²R losses):
- Due to resistance of windings
- Minimized by: Using thick copper wires
2. Eddy Current Losses:
- Circulating currents in iron core produce heat
- Minimized by: Laminated core (thin insulated sheets)
3. Hysteresis Losses:
- Energy lost in repeated magnetization cycles
- Minimized by: Soft iron core (low retentivity)
4. Flux Leakage:
- All flux doesn't link both coils
- Minimized by: Proper core design and tight coupling
9.5 Efficiency
Applications of Transformers
- Power Transmission: Step-up at generation (reduce I²R losses), step-down for distribution
- Mobile Chargers: Step-down 230V AC to 5V DC
- Welding: Step-down to high current
- Radio/TV: Impedance matching
10. Applications of Electromagnetic Induction
1. Electric Generators
All power plants use generators to convert mechanical energy (from turbines) to electrical energy using EM induction.
2. Transformers
Use mutual induction to step voltage up or down for efficient power transmission and distribution.
3. Induction Cooktops
Rapidly changing magnetic fields induce eddy currents in cookware, heating it directly.
4. Electromagnetic Braking
Trains use strong magnets to induce eddy currents in wheels, creating opposition force (Lenz's law) for smooth braking.
5. Metal Detectors
Metal passing through coil disturbs magnetic field, inducing detectable currents.
Important Constants and Units
| Quantity | Symbol | SI Unit | Dimension |
|---|---|---|---|
| Magnetic Flux | ΦB | Weber (Wb) = T·m² | [ML²T⁻²A⁻¹] |
| Magnetic Field | B | Tesla (T) = Wb/m² | [MT⁻²A⁻¹] |
| Inductance (Self/Mutual) | L, M | Henry (H) = Wb/A | [ML²T⁻²A⁻²] |
| Induced EMF | ε | Volt (V) | [ML²T⁻³A⁻¹] |
| Permeability (Free Space) | μ₀ | 4π × 10⁻⁷ H/m | [MLT⁻²A⁻²] |
| Energy | U | Joule (J) | [ML²T⁻²] |
| Energy Density | u | J/m³ | [ML⁻¹T⁻²] |
Key Formulas Quick Reference:
- Faraday's Law: ε = -N(dΦB/dt)
- Motional EMF: ε = Blv
- Self-Inductance (Solenoid): L = μ₀N²A/l
- Energy in Inductor: U = ½LI²
- Energy Density: u = B²/(2μ₀)
- Transformer: Vs/Vp = Ns/Np
11. Summary
1. Magnetic Flux: ΦB = BA cos θ
2. Faraday's Law: ε = -N(dΦB/dt)
3. Motional EMF: ε = Blv
4. Self-Inductance: L = ΦB/I = μ₀N²A/l
5. Energy in Inductor: U = ½LI²
6. AC Generator: ε₀ = NBAω
7. Energy Density: uB = B²/(2μ₀)
- Faraday's experiments (3 marks)
- Lenz's law and energy conservation (5 marks)
- Solenoid inductance derivation (5 marks)
- AC generator working and derivation (5 marks)
- Numerical problems on flux, emf, inductance
Practice Questions
Section A: Multiple Choice Questions (50 MCQs - 1 mark each)
Instructions: Choose the correct option.
Q1. The phenomenon of electromagnetic induction was discovered by:
(a) Oersted
(b) Ampère
(c) Faraday and Henry
(d) Maxwell
Michael Faraday (England) and Joseph Henry (USA) independently discovered electromagnetic induction around 1831.
Q2. The SI unit of magnetic flux is:
(a) Tesla
(b) Weber
(c) Henry
(d) Ampere
1 Weber (Wb) = 1 T·m² = 1 kg·m²·s⁻²·A⁻¹
Q3. According to Lenz's law, induced current:
(a) Aids the flux change
(b) Opposes the flux change
(c) Is independent of flux
(d) Doubles the flux
Lenz's law states that induced current opposes the change that produces it.
Q4. The negative sign in Faraday's law represents:
(a) Direction of current
(b) Lenz's law
(c) Magnitude of emf
(d) Type of flux
The negative sign embodies Lenz's law.
Q5. SI unit of self-inductance is:
(a) Weber
(b) Tesla
(c) Henry
(d) Farad
1 Henry (H) = 1 Wb/A
Q6. When magnet moves towards coil, induced current creates field that:
(a) Attracts magnet
(b) Repels magnet
(c) Has no effect
(d) Stops magnet
By Lenz's law, induced field opposes approach.
Q7. Self-inductance of solenoid is proportional to:
(a) N
(b) N²
(c) 1/N
(d) 1/N²
L = μ₀N²A/l, so L ∝ N²
Q8. Energy stored in inductor is:
(a) LI
(b) L²I
(c) ½LI²
(d) LI²
U = ½LI²
Q9. In AC generator, emf is maximum when coil plane is:
(a) Perpendicular to field
(b) Parallel to field
(c) At 45°
(d) At any angle
When parallel, flux rate of change is maximum.
Q10. Motional emf formula is:
(a) Bl/v
(b) Bv/l
(c) Blv
(d) B/lv
ε = Blv when B, l, v are perpendicular.
Q11. Mutual inductance depends on:
(a) Current only
(b) Geometry only
(c) Both
(d) Neither
M depends on geometry and medium, not current.
Q12. Self-inductance is proportional to:
(a) r
(b) r²
(c) N
(d) N²
L ∝ N² for any coil.
Q13. Dimensional formula of inductance is:
(a) [ML²T⁻²A⁻²]
(b) [MLT⁻²A⁻¹]
(c) [ML²T⁻¹A⁻²]
(d) [M²LT⁻²A⁻²]
L = Φ/I gives [ML²T⁻²A⁻²]
Q14. If turns double, self-inductance becomes:
(a) Double
(b) Four times
(c) Half
(d) One-fourth
L ∝ N², so doubling N makes L four times.
Q15. Transformer works on principle of:
(a) Self-inductance
(b) Mutual inductance
(c) EM induction only
(d) Magnetic field
Transformer uses mutual inductance.
Q16. AC frequency in India is:
(a) 50 Hz
(b) 60 Hz
(c) 100 Hz
(d) 120 Hz
50 Hz (50 cycles per second)
Q17. Eddy currents are minimized by:
(a) Thick sheets
(b) Laminated cores
(c) Solid cores
(d) Copper cores
Lamination breaks eddy current paths.
Q18. Current changes 2A to 6A in 0.01s, emf 20V. Inductance is:
(a) 0.05 H
(b) 0.5 H
(c) 5 H
(d) 50 H
L = ε/(dI/dt) = 20/(4/0.01) = 0.05 H
Q19. Flux through 0.1 m² area perpendicular to 0.2 T field:
(a) 0.02 Wb
(b) 0.2 Wb
(c) 2 Wb
(d) 20 Wb
Φ = BA = 0.2 × 0.1 = 0.02 Wb
Q20. Back emf in motor:
(a) Aids applied voltage
(b) Opposes applied voltage
(c) Has no relation
(d) Doubles voltage
Back emf opposes applied voltage (Lenz's law).
Q21. Fleming's right-hand rule finds:
(a) Force direction
(b) Induced current direction
(c) Field direction
(d) EMF magnitude
Right-hand rule determines induced current direction.
Q22. Magnetic energy density is:
(a) B²/(2μ₀)
(b) μ₀B²/2
(c) B²/μ₀
(d) μ₀/B²
u = B²/(2μ₀) J/m³
Q23. Ideal transformer has:
(a) 100% efficiency
(b) No energy losses
(c) Perfect flux linkage
(d) All of these
Ideal transformer has all these properties.
Q24. In step-up transformer:
(a) Vs>Vp and Is>Ip
(b) Vs>Vp and Is
(d) Both equal
Step-up increases voltage, decreases current.
Q25. Motional emf depends on:
(a) Length only
(b) Velocity only
(c) Both l and v
(d) Neither
ε = Blv depends on both.
Q26. Slip rings in AC generator:
(a) Convert AC to DC
(b) Provide contact
(c) Increase voltage
(d) Decrease current
Slip rings maintain continuous contact.
Q27. Coil 100 turns, flux 0.01 Wb each. Total linkage:
(a) 0.01 Wb
(b) 0.1 Wb
(c) 1 Wb
(d) 10 Wb
NΦ = 100 × 0.01 = 1 Wb
Q28. Lenz's law follows from:
(a) Charge conservation
(b) Energy conservation
(c) Momentum conservation
(d) Newton's 3rd law
Lenz's law ensures energy conservation.
Q29. Peak AC 300V. RMS value:
(a) 150 V
(b) 212 V
(c) 300 V
(d) 424 V
Vrms = V₀/√2 = 300/1.414 ≈ 212 V
Q30. Induced emf independent of:
(a) Flux rate
(b) Turns
(c) Resistance
(d) Field strength
EMF independent of circuit resistance.
Q31. Solenoid 500 turns, 0.5m, area 10⁻³m². Inductance:
(a) 6.28×10⁻⁴ H
(b) 6.28×10⁻³ H
(c) 6.28×10⁻² H
(d) 6.28×10⁻¹ H
L = μ₀N²A/l = 6.28×10⁻⁴ H
Q32. EM induction NOT used in:
(a) Generator
(b) Transformer
(c) Battery
(d) Induction cooktop
Batteries use chemical reactions.
Q33. N-pole recedes, induced current creates:
(a) North facing
(b) South facing
(c) No field
(d) Alternating poles
South pole attracts receding north (Lenz's law).
Q34. Mutual inductance if coil reversed:
(a) 2M
(b) M
(c) -M
(d) Zero
Reversing coil reverses flux direction.
Q35. ΔΦ/Δt applies for:
(a) Uniform change
(b) Non-uniform change
(c) Both
(d) Neither
Discrete form for uniform change; dΦ/dt for general.
Q36. In DC motor, back emf is:
(a) Greater than applied
(b) Less than applied
(c) Equal to applied
(d) Zero
Back emf always less; difference drives current.
Q37. Eddy currents produce:
(a) Heating only
(b) Damping only
(c) Both
(d) Neither
Both heat (I²R) and damping (opposition).
Q38. Inductor opposes:
(a) DC current
(b) Change in current
(c) Constant current
(d) Voltage
Inductor opposes current change.
Q39. Loop emf zero when plane is:
(a) Perpendicular to field
(b) Parallel to field
(c) At 45°
(d) At 30°
When ⊥, flux max but dΦ/dt = 0.
Q40. Self-induction analogous to:
(a) Capacitance
(b) Resistance
(c) Inertia
(d) Friction
Like inertia opposes velocity change.
Q41. Transformer core should have:
(a) High R, high μ
(b) Low R, low μ
(c) High R, low μ
(d) Low R, high μ
High permeability for flux; high resistivity reduces eddy currents.
Q42. Max power when load resistance:
(a) Equals internal R
(b) Very high
(c) Very low
(d) Independent
Max power when Rload = Rinternal
Q43. Rod 1m, 5 m/s, 0.5T field. EMF:
(a) 0.5 V
(b) 1.0 V
(c) 2.5 V
(d) 5.0 V
ε = Blv = 0.5×1×5 = 2.5 V
Q44. Series inductors add like:
(a) Resistors series
(b) Capacitors series
(c) Resistors parallel
(d) Capacitors parallel
Ltotal = L₁ + L₂ + ...
Q45. Transformer conserves:
(a) Voltage
(b) Current
(c) Power
(d) Impedance
Pin = Pout in ideal transformer.
Q46. EM braking uses:
(a) Magnets
(b) Eddy currents
(c) Self-induction
(d) Capacitors
Eddy currents oppose motion.
Q47. LR circuit time constant:
(a) L/R
(b) R/L
(c) LR
(d) √(L/R)
τ = L/R seconds
Q48. Coil energy 2J, current 2A. Inductance:
(a) 0.5 H
(b) 1 H
(c) 2 H
(d) 4 H
U = ½LI², 2 = ½L(4), L = 1 H
Q49. Flux 5×10⁻³ Wb, current 2A. Mutual inductance:
(a) 2.5×10⁻³ H
(b) 2.5×10⁻² H
(c) 10×10⁻³ H
(d) 10 H
M = Φ/I = 5×10⁻³/2 = 2.5×10⁻³ H
Q50. AC vs DC generator differs in:
(a) Coils
(b) Commutator
(c) Field strength
(d) Speed
DC has commutator; AC has slip rings.
Section B: Very Short Answer Questions (50 Questions - 2 marks each)
Q1. Define magnetic flux.
Answer: Magnetic flux through a surface is the product of the magnetic field component perpendicular to the surface and the area: ΦB = BA cos θ.
Q2. State Faraday's law of electromagnetic induction.
Answer: The magnitude of induced emf in a circuit equals the time rate of change of magnetic flux: ε = -N(dΦB/dt).
Q3. State Lenz's law.
Answer: The direction of induced emf opposes the change in magnetic flux that produces it.
Q4. What is self-inductance?
Answer: Property of a coil by which it opposes change in its own current by inducing back emf.
Q5. SI unit of inductance?
Answer: Henry (H) = Wb/A = V·s/A.
Q6. Define mutual inductance.
Answer: Property where current change in one coil induces emf in another coil.
Q7. What is motional emf?
Answer: EMF induced in conductor moving through magnetic field: ε = Blv.
Q8. Name two EM induction devices.
Answer: Electric generator and transformer.
Q9. Why Lenz's law conserves energy?
Answer: It ensures induced effects oppose cause, requiring work input which converts to electrical energy.
Q10. AC generator principle?
Answer: Rotating coil in magnetic field induces alternating emf by Faraday's law.
Q11. Energy in inductor formula?
Answer: U = ½LI² joules.
Q12. Magnetic energy density?
Answer: uB = B²/(2μ₀) J/m³.
Q13. What are eddy currents?
Answer: Circulating currents induced in bulk conductors in changing magnetic flux.
Q14. How minimize eddy currents?
Answer: Use laminated cores to break current paths.
Q15. What is back emf?
Answer: Self-induced emf in motor opposing applied voltage during rotation.
Q16. Fleming's right-hand rule?
Answer: Thumb=motion, forefinger=field, middle finger=induced current direction.
Q17. What are slip rings?
Answer: Continuous metal rings maintaining electrical contact via brushes in AC generator.
Q18. Why laminated transformer core?
Answer: To reduce eddy current losses.
Q19. AC frequency in India?
Answer: 50 Hz.
Q20. Coefficient of mutual inductance?
Answer: M = Φ₂/I₁, flux in coil 2 per unit current in coil 1.
Q21. Dimensional formula of flux?
Answer: [ML²T⁻²A⁻¹]
Q22. If dΦ/dt doubles, emf?
Answer: EMF also doubles (ε ∝ dΦ/dt).
Q23. Law for induced current direction?
Answer: Lenz's law.
Q24. Relation M₁₂ and M₂₁?
Answer: M₁₂ = M₂₁ = M (reciprocity).
Q25. Self-inductance depends on?
Answer: Number of turns N, geometry (A, l), medium permeability μ.
Q26. Physical meaning of inductance?
Answer: Electrical inertia - opposes current change like mass opposes velocity change.
Q27. Why soft iron in transformer?
Answer: High permeability for flux linkage, low retentivity for low hysteresis losses.
Q28. Peak value of 220V AC?
Answer: V₀ = 220√2 ≈ 311 V.
Q29. Coefficient of coupling?
Answer: k = M/√(L₁L₂), fraction of flux linkage.
Q30. Ideal transformer?
Answer: 100% efficiency, no losses, perfect flux linkage.
Q31. Max power transfer condition?
Answer: Load resistance = internal resistance.
Q32. Electromagnetic damping?
Answer: Retarding force due to eddy currents opposing motion.
Q33. Magnet in copper tube?
Answer: Falls slowly due to eddy current damping.
Q34. LR time constant?
Answer: τ = L/R seconds.
Q35. Inductance vs turns?
Answer: L ∝ N².
Q36. Energy density unit?
Answer: J/m³.
Q37. Transformer on DC?
Answer: No - needs changing flux for induction.
Q38. Step-up transformer?
Answer: Ns > Np, increases voltage.
Q39. Transformer energy conservation?
Answer: VpIp = VsIs.
Q40. Conductor parallel to field?
Answer: No flux change, no induced emf.
Q41. Turn ratio?
Answer: n = Ns/Np.
Q42. Induction cooktop?
Answer: Eddy currents in cookware produce I²R heating.
Q43. EMF direction vs flux?
Answer: Opposes flux change (Lenz's law).
Q44. Commutator function?
Answer: Converts AC to DC in generator.
Q45. Inductors in series/parallel?
Answer: Series: Ltotal = ΣL; Parallel: 1/Ltotal = Σ(1/L).
Q46. Hysteresis loss?
Answer: Energy lost in repeated magnetization cycles.
Q47. Thick copper wire why?
Answer: Reduces I²R copper losses.
Q48. EM braking?
Answer: Eddy currents create retarding force for friction-free braking.
Q49. Coil area effect?
Answer: ε ∝ A (larger area → more emf).
Q50. Zero emf condition?
Answer: Coil plane ⊥ field (flux max, dΦ/dt = 0).
Section C: Short Answer Questions (High-Quality Selection)
Q1. Explain Faraday's experiment with bar magnet and coil. What did he conclude?
Answer: Faraday connected a coil to galvanometer. Moving magnet towards coil caused deflection (induced current). Moving away caused opposite deflection. Stationary magnet gave no deflection. Faster motion produced larger deflection. Conclusion: Changing magnetic flux induces emf; magnitude depends on rate of change (dΦ/dt).
Q2. State and explain Lenz's law with example.
Answer: Lenz's Law: Induced current direction opposes flux change. Example: N-pole approaching coil → induced current creates N-pole facing it (repulsion). N-pole receding → induced S-pole attracts it. This ensures energy conservation by requiring external work.
Q3. Derive motional emf formula ε = Blv.
Answer: Conductor length l moves with velocity v perpendicular to field B. Electrons experience force F = qvB, creating electric field E = vB. EMF: ε = El = vBl = Blv. Verification by Faraday's law: Area swept in dt: dA = lvdt, flux change dΦ = Blvdt, so ε = dΦ/dt = Blv.
Q4. What is self-inductance? Write unit and dimensional formula.
Answer: Self-inductance: Property opposing change in own current. Φ = LI, ε = -L(dI/dt). SI Unit: Henry (H) = Wb/A = V·s/A. Dimensional formula: L = (BA)/I = [MT⁻²A⁻¹][L²]/[A] = [ML²T⁻²A⁻²].
Q5. Derive energy stored in inductor: U = ½LI².
Answer: Back emf ε = -L(di/dt) opposes current change. Work done: dW = εi·dt = Li·di. Total work: W = ∫₀ᴵ Li·di = L[i²/2]₀ᴵ = ½LI². This work is stored as magnetic energy.
Q6. Distinguish between self and mutual inductance.
Answer: Self-inductance: Single coil opposes its own current change (ε = -LdI/dt). Mutual inductance: Current in one coil induces emf in another (ε₂ = -MdI₁/dt). Self depends on one geometry; mutual depends on both geometries and coupling.
Q7. Why does Lenz's law not violate energy conservation?
Answer: If induced current aided flux change, magnet would be attracted (gaining KE without work → perpetual motion). Lenz ensures opposition, requiring external work against magnetic force. This work = electrical energy generated, conserving total energy.
Q8. Derive self-inductance of solenoid: L = μ₀N²A/l.
Answer: Solenoid: N turns, length l, area A, current I. Field: B = μ₀(N/l)I. Flux per turn: Φ = BA = μ₀(N/l)IA. Total linkage: NΦ = μ₀(N²A/l)I. Self-inductance: L = NΦ/I = μ₀N²A/l.
Q9. What are eddy currents? Give two uses and two disadvantages.
Answer: Eddy currents: Circulating currents in bulk conductors in changing flux. Uses: (1) Induction furnaces, (2) EM braking. Disadvantages: (1) Heat loss in transformers, (2) Energy wastage in motors. Minimized by lamination.
Q10. Explain working of AC generator with diagram.
Answer: Coil (N turns, area A) rotates in magnetic field B with angular velocity ω. Flux: Φ = NBA cos(ωt). By Faraday's law: ε = -dΦ/dt = NBAω sin(ωt). Slip rings maintain continuous contact, delivering sinusoidal AC output with peak emf ε₀ = NBAω.
Note: Additional practice problems available in NCERT textbook and exemplar.
Section D: Long Answer Questions (High-Quality Selection)
Q1. State Faraday's laws. Describe experiments. Explain Lenz's law and energy conservation.
Answer: Faraday's Law: ε = -N(dΦ/dt). Experiments: (1) Two coils on iron ring - switching caused momentary deflection. (2) Bar magnet movement caused deflection; faster motion gave larger deflection. Lenz's Law: Induced current opposes flux change. Energy Conservation: If aided change, perpetual motion would occur. Opposition requires work = electrical energy generated, conserving energy.
Q2. Derive self-inductance of solenoid. Find energy stored and magnetic energy density.
Answer: Derivation: B = μ₀(N/l)I, Φ = BA = μ₀(N/l)IA, NΦ = μ₀(N²A/l)I, L = μ₀N²A/l. Energy: U = ½LI². Energy density: u = B²/(2μ₀) obtained from U/(volume) where volume = Al.
Q3. Explain AC generator: construction, working, diagram, EMF derivation.
Answer: Construction: Field magnet, armature (N turns, area A), slip rings, brushes. Working: Coil rotates with ω. Derivation: At time t, θ = ωt. Flux: Φ = NBA cos(ωt). EMF: ε = -dΦ/dt = NBAω sin(ωt). Peak: ε₀ = NBAω. Output is sinusoidal AC.
Q4. Explain transformer: principle, construction, voltage/current relations, energy losses.
Answer: Principle: Mutual induction. Construction: Laminated core, primary (Nₚ), secondary (Nₛ). Relations: Vₛ/Vₚ = Nₛ/Nₚ, Iₛ/Iₚ = Nₚ/Nₛ. Losses: (1) Copper (I²R), (2) Eddy currents (lamination reduces), (3) Hysteresis (soft iron reduces), (4) Flux leakage.
Q5. Derive motional EMF. Verify using Faraday's law. Give applications.
Answer: Force method: F = qvB on electrons → E = vB → ε = El = Blv. Faraday verification: dA = lvdt, dΦ = Blvdt, ε = dΦ/dt = Blv. Applications: AC/DC generators, electromagnetic flow meters, metal detectors.
Note: For detailed numerical problems, refer to NCERT solved examples.
Full-Length Test Paper
Section A: Multiple Choice (20 × 1 = 20 marks)
Q1-Q20 selected from 50 MCQs above covering all topics.
Section B: Very Short Answer (5 × 2 = 10 marks)
Q21-Q25 from VSA section above.
Section C: Short Answer (5 × 3 = 15 marks)
Q26-Q30 from SA section above.
Section D: Long Answer (5 × 5 = 25 marks)
Q31-Q35 from LA section above.
Complete solutions available in respective sections above.
No comments:
Post a Comment