Self Induction, Mutual Induction and AC Generator | EMI Physics Notes
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Self Induction, Mutual Induction And AC Generator

Learn inductance, mutual induction and working of AC generators with correct diagrams, derivations, formula sheets, exam numericals and detailed solutions.

Self InductionCoefficient of Self InductionInductorEnergy Stored in InductorMutual InductionCoefficient of Mutual InductionAC GeneratorConstructionWorking PrincipleNumerical ProblemsPYQs
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1. Introduction

Electromagnetic induction becomes deeper when students understand that a changing current can create its own induced emf, can induce emf in a neighbouring circuit, can store magnetic energy, and can generate alternating voltage in a rotating coil. This page connects self induction, mutual induction, inductors, rotating disc emf and AC generators in one exam-oriented flow.

Concept map for the complete topic
ChangingFlux Self Inductione = −L dI/dtEnergy in LU = ½LI²Mutual Inductione₂ = −M dI₁/dtLong SolenoidL = μN²A/lRotating Disce = ½BωR²AC Generatore = NBAω sinωt

Self Induction

Changing current in the same coil changes its own flux linkage and produces back emf.

Mutual Induction

Changing current in one coil changes flux linked with another nearby coil.

AC Generator

Mechanical rotation changes magnetic flux and produces alternating emf.

2. Self Induction

Self induction is the property of a coil by which it opposes any change in current flowing through itself. When current changes, magnetic flux linked with the same coil changes, and an induced emf appears in the coil. This emf is called back emf because it opposes the cause producing it.

Self induction: changing current in a coil induces back EMF in the same coil
increasing current I back EMF opposes change The induced EMF is not against current always; it is against change in current.
Flux linkage relationNΦ = LI
Self-induced EMFe = −L(dI/dt)
One HenryL = 1 H when 1 V is induced for current change of 1 A s⁻¹
1

A changing current produces a changing magnetic field.

2

Changing magnetic field changes flux linkage with the same coil.

3

According to Faraday's law, induced emf appears.

4

According to Lenz law, this induced emf opposes the change in current.

3. Self Inductance of a Long Solenoid

Long solenoid: self inductance and stored magnetic energy
Long solenoid: N turns, length l, area A B = μ₀nI, n = N/l NΦ = μ₀N²AI/l → L = μ₀N²A/l Energy density u = B²/(2μ)
1

For a long solenoid, B = μ₀nI, where n = N/l.

2

Flux through each turn is Φ = BA = μ₀nIA.

3

Total flux linkage is NΦ = Nμ₀nIA = μ₀N²AI/l.

4

Since NΦ = LI, L = μ₀N²A/l = μ₀n²Al.

Air-core solenoidL = μ₀N²A/l = μ₀n²Al
With magnetic coreL = μN²A/l = μ₀μᵣN²A/l
MeaningMore turns, larger area and higher permeability increase inductance.

4. Energy Stored in an Inductor

1

At current I, back emf magnitude is e = L(dI/dt).

2

Small work done in time dt is dW = eI dt.

3

Substitute e: dW = L I dI.

4

Integrate from 0 to I: W = ∫ LI dI = ½LI².

Energy StoredU = ½LI²
Magnetic Energy Densityu = B²/(2μ)
Physical MeaningThe energy is stored in the magnetic field produced by current.

5. Inductors in Series and Parallel

These simple addition rules are valid when mutual coupling between inductors is negligible. If coils are magnetically coupled, mutual inductance terms must be included.

Inductors without mutual coupling: 0.75 H in series with two 0.5 H inductors in parallel
0.75 H 0.5 H 0.5 H L_parallel = 0.25 H, therefore L_total = 0.75 + 0.25 = 1 H
SeriesL = L₁ + L₂ + L₃ + ...
Parallel1/L = 1/L₁ + 1/L₂ + 1/L₃ + ...
Exam TrapThese formulas fail when flux of one inductor strongly links the other.

6. Mutual Induction

Mutual induction is the phenomenon in which changing current in one coil induces emf in a nearby coil. The first coil is called the primary coil and the second coil is called the secondary coil.

Mutual induction: changing current in primary coil changes flux through secondary coil
Primary coil P Secondary coil S changing Iₚ changing flux linkage G
Mutual Flux LinkageN₂Φ₂ = MI₁
Induced EMF in Secondarye₂ = −M(dI₁/dt)
Coupling RelationM = k√(L₁L₂)
Coefficient of Mutual InductionM = N₂Φ₂/I₁
One HenryM = 1 H when 1 A s⁻¹ current change induces 1 V
Best Couplingk = 1 for ideal complete flux linkage

7. Mutual Inductance of Two Coaxial Solenoids

Mutual inductance of two coaxial solenoids: only common flux linkage is counted
Primary: N₁ turns, length l Secondary: N₂ turns over common area A M = μ₀N₁N₂A/l (air core), M = μ₀μᵣN₁N₂A/l (core)
1

Magnetic field due to primary solenoid is B₁ = μ₀n₁I₁ = μ₀(N₁/l)I₁.

2

Flux through each turn of secondary is Φ₂ = B₁A.

3

Flux linkage of secondary is N₂Φ₂ = μ₀N₁N₂AI₁/l.

4

Therefore M = N₂Φ₂/I₁ = μ₀N₁N₂A/l.

Air CoreM = μ₀N₁N₂A/l
Magnetic CoreM = μ₀μᵣN₁N₂A/l
Common AreaA is the common area through which flux links both coils.

8. Rotating Disc EMF

A conducting disc rotating in a magnetic field develops emf between its centre and rim. Every point of the disc has a different speed, so integration is necessary.

Rotating conducting disc in a perpendicular magnetic field
×××××××××××××××××××××××××××××××××××××××××××××××× R ω de = Bωr dr, e = ∫₀ᴿ Bωr dr = ½BωR²
1

At distance r from the centre, speed v = rω.

2

Small radial element dr has motional emf de = Bvdr = Bωrdr.

3

Integrate from centre to rim: e = ∫₀ᴿ Bωrdr.

4

Therefore e = ½BωR².

Rotating Disc EMFe = ½BωR²
Average Velocity Methode = B(ωR/2)R = ½BωR²

9. AC Generator

An AC generator converts mechanical energy into electrical energy using electromagnetic induction. A coil rotates in a magnetic field. Flux linked with the coil changes continuously, so alternating emf is induced.

AC generator: rotating coil between magnetic poles produces sinusoidal EMF
N S brush B₁ brush B₂ Load Uniform B from N to S; rotating coil changes flux and induces alternating EMF.
AC generator output: e = NBAω sinωt
et emax = NBAω The direction reverses every half rotation, so the output is alternating.
1

For a coil of N turns and area A rotating in field B, flux linkage is NΦ = NBA cosωt.

2

Induced emf is e = −d(NΦ)/dt.

3

Differentiate: e = NBAω sinωt.

4

Maximum emf is e₀ = NBAω.

Flux LinkedNΦ = NBA cosωt
Instantaneous EMFe = NBAω sinωt
Maximum EMFe₀ = NBAω
Frequencyf = ω/(2π)

10. Losses in Practical AC Generators

LossCauseReduction Method
Copper LossI²R heating in armature and external circuitUse low-resistance thick copper windings
Eddy Current LossInduced currents in iron coreUse laminated core
Hysteresis LossRepeated magnetisation and demagnetisation of coreUse soft iron or silicon steel
Mechanical LossFriction at bearings and air resistanceUse lubrication and good mechanical design

11. Board-Style Solved Numericals

These solved examples are rewritten clearly so students can follow the exact method, units and final result.

19

Question 19: Self induction

An average induced emf of 0.20 V appears in a coil when the current changes from 5.0 A in one direction to 5.0 A in the opposite direction in 0.20 s. Find the self inductance.

Given: e = 0.20 V, ΔI = 10 A, Δt = 0.20 s.

Solution: e = L(ΔI/Δt). Hence L = eΔt/ΔI = (0.20 × 0.20)/10 = 4.0 × 10^-3 H.

Answer: 4 × 10^-3 H
20

Question 20: Self induction

Magnetic flux of 5 micro weber is linked with a coil when a current of 1 mA flows through it. Find self inductance.

Given: NΦ = 5 × 10^-6 Wb, I = 1 × 10^-3 A.

Solution: L = NΦ/I = (5 × 10^-6)/(1 × 10^-3) = 5 × 10^-3 H.

Answer: 5 mH
21

Question 21: Self induction

A 10 H inductor carries a steady current of 2 A. How can a 100 V self-induced emf be produced?

Given: L = 10 H, ΔI = 2 A, e = 100 V.

Solution: e = L(ΔI/Δt). Therefore Δt = LΔI/e = 10 × 2/100 = 0.20 s. Reduce the current from 2 A to zero in 0.20 s.

Answer: Reduce current to zero in 0.20 s
22

Question 22: Flux linkage

A 100-turn coil has self inductance 20 mH. Find flux linkage and flux per turn when current is 4 mA.

Given: L = 0.020 H, I = 0.004 A, N = 100.

Solution: NΦ = LI = 0.020 × 0.004 = 8.0 × 10^-5 Wb-turn. Φ = (8.0 × 10^-5)/100 = 8.0 × 10^-7 Wb.

Answer: NΦ = 8 × 10^-5 Wb-turn; Φ = 8 × 10^-7 Wb
23

Question 23: Faraday law

The magnetic flux through a coil of 50 turns changes from 0.3 Wb per turn to zero in 1 s. Find induced emf.

Given: N = 50, ΔΦ = 0.3 Wb, Δt = 1 s.

Solution: Average emf magnitude e = NΔΦ/Δt = 50 × 0.3/1 = 15 V.

Answer: 15 V
24

Question 24: Sinusoidal current in inductor

A coil of self inductance 10 mH carries current I = 0.1 sin(200t) A. Find maximum induced emf.

Given: L = 0.010 H, I = 0.1 sin(200t).

Solution: e = L|dI/dt|. dI/dt = 20 cos(200t), so maximum |dI/dt| = 20 A s^-1. emax = 0.010 × 20 = 0.20 V.

Answer: 0.2 V
25

Question 25: Turns scaling

A coil has 600 turns and self inductance 108 mH. Find self inductance of a similar 500-turn coil.

Given: L₁ = 108 mH, N₁ = 600, N₂ = 500.

Solution: For similar coils, L ∝ N². L₂ = 108(500/600)² = 75 mH.

Answer: 75 mH
26

Question 26: Solenoid self inductance

A solenoid of radius 3 cm and length 1 m has 600 turns per metre. Find its self inductance. Will it change on an iron core?

Given: r = 0.03 m, l = 1 m, n = 600 m^-1.

Solution: L = μ₀n²Al = 4π×10^-7 × 600² × π(0.03)² × 1 = 1.28 × 10^-3 H. An iron core increases μ, so L increases.

Answer: 1.28 × 10^-3 H; yes
27

Question 27: Inductor combination

A 0.75 H inductor is in series with two 0.5 H inductors in parallel. Calculate equivalent inductance.

Given: 0.75 H in series with 0.5 H || 0.5 H.

Solution: Parallel part = (0.5 × 0.5)/(0.5 + 0.5) = 0.25 H. Total = 0.75 + 0.25 = 1.0 H.

Answer: 1 H
28

Question 28: Relative permeability

Self inductance of an air-core solenoid increases from 0.04 mH to 16 mH on introducing a core. Find relative permeability.

Given: Lair = 0.04 mH, Lcore = 16 mH.

Solution: For the same solenoid, Lcore/Lair = μr. Hence μr = 16/0.04 = 400.

Answer: 400
29

Question 29: Mutual induction

An emf of 0.5 V is developed in secondary when primary current changes from 5.0 A to 2.0 A in 300 ms. Find mutual inductance.

Given: e = 0.5 V, ΔI = 3 A, Δt = 0.300 s.

Solution: M = eΔt/ΔI = 0.5 × 0.300/3 = 0.05 H.

Answer: 0.05 H
30

Question 30: Mutual induction

An air-core solenoid of length 0.3 m, area 1.2 × 10^-3 m² and 2500 turns has a 350-turn coil wound over its central part. Current reverses from 3 A to -3 A in 0.25 s. Find secondary emf.

Given: N₁ = 2500, N₂ = 350, l = 0.3 m, A = 1.2×10^-3 m², ΔI/Δt = 6/0.25.

Solution: M = μ₀N₁N₂A/l. Then e = MΔI/Δt = 4π×10^-7 × 2500 × 350 × 1.2×10^-3 × 24 /0.3 = 0.1056 V.

Answer: 0.1056 V
31

Question 31: Spark coil

In a car spark coil, 4000 V is induced in secondary when primary current changes from 4 A to zero in 10 μs. Find mutual inductance.

Given: e = 4000 V, ΔI = 4 A, Δt = 10^-5 s.

Solution: M = eΔt/ΔI = 4000 × 10^-5/4 = 0.01 H.

Answer: 0.01 H
32

Question 32: Coaxial solenoids

A 100-turn coil of radius 2 cm is wound near the centre of a 100 cm long solenoid of radius 2 cm having 1000 turns. Find mutual inductance.

Given: N₁ = 1000, N₂ = 100, r = 0.02 m, l = 1 m.

Solution: M = μ₀N₁N₂A/l = 4π×10^-7 × 1000 × 100 × π(0.02)²/1 = 1.58 × 10^-4 H.

Answer: 1.58 × 10^-4 H
33

Question 33: Mutual induction

A solenoid of length 20 cm, area 4.0 cm² and 4000 turns is placed inside another solenoid of 2000 turns, length 10 cm and area 8.0 cm². Find mutual inductance using common area.

Given: N₁ = 4000, N₂ = 2000, A = 4 × 10^-4 m², l = 0.20 m.

Solution: M = μ₀N₁N₂A/l = 4π×10^-7 × 4000 × 2000 × 4×10^-4/0.20 ≈ 2.0 × 10^-2 H.

Answer: 2 × 10^-2 H
34

Question 34: Flux linkage

A current of 10 A in primary changes flux by 500 Wb per turn in a 200-turn secondary. Find M and induced emf if the change occurs in 0.5 s.

Given: N₂ = 200, Φ = 500 Wb/turn, I = 10 A, Δt = 0.5 s.

Solution: M = N₂Φ/I = 200×500/10 = 10^4 H. e = N₂ΔΦ/Δt = 200×500/0.5 = 2×10^5 V.

Answer: 10^4 H; 2 × 10^5 V
35

Question 35: Coupled solenoids

A 1 m long solenoid of diameter 5 cm has 700 turns. A 50-turn solenoid is tightly wound over it. Find M and emf if current changes from 0 to 5 A in 0.01 s.

Given: r = 0.025 m, N₁ = 700, N₂ = 50, l = 1 m, ΔI/Δt = 500 A s^-1.

Solution: M = μ₀N₁N₂πr²/l = 8.6×10^-5 H. e = MΔI/Δt = 8.6×10^-5 × 500 = 4.3×10^-2 V.

Answer: 8.6 × 10^-5 H; 4.3 × 10^-2 V
36

Question 36: Circular coaxial coils

Coil X: radius 0.02 m, 100 turns. Coil Y: radius 0.2 m, 1000 turns. X is coaxial at the centre of Y. Current in Y changes from 7 A to 5 A in 4×10^-2 s. Find induced emf in X.

Given: Nx=100, Ny=1000, rx=0.02 m, Ry=0.2 m, ΔI/Δt=50 A s^-1.

Solution: M = μ₀NxNyπrx²/(2Ry) = 3.95×10^-4 H. e = MΔI/Δt = 3.95×10^-4 × 50 = 1.975×10^-2 V.

Answer: M = 39.5 × 10^-5 H; e = 19.75 mV

12. NEET Practice Bank

75 NEET MCQs

NEET-1A solenoid has 208 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-2Current through a 0.3 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-3In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-4Two coils have mutual inductance 0.13 H. Primary current changes at 32 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-5A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-6A solenoid has 248 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-7Current through a 0.2 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-8In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-9Two coils have mutual inductance 0.13 H. Primary current changes at 47 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-10A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-11A solenoid has 288 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-12Current through a 0.1 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-13In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-14Two coils have mutual inductance 0.13 H. Primary current changes at 62 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-15A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-16A solenoid has 328 turns, length 0.25 m and area 2 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-17Current through a 0.6 H inductor changes uniformly by 4 A in 0.03 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-18In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-19Two coils have mutual inductance 0.13 H. Primary current changes at 77 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-20A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-21A solenoid has 368 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-22Current through a 0.5 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-23In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-24Two coils have mutual inductance 0.13 H. Primary current changes at 92 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-25A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-26A solenoid has 408 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-27Current through a 0.4 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-28In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-29Two coils have mutual inductance 0.13 H. Primary current changes at 107 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-30A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-31A solenoid has 448 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-32Current through a 0.3 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-33In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-34Two coils have mutual inductance 0.13 H. Primary current changes at 122 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-35A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-36A solenoid has 488 turns, length 0.25 m and area 2 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-37Current through a 0.2 H inductor changes uniformly by 4 A in 0.03 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-38In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-39Two coils have mutual inductance 0.13 H. Primary current changes at 137 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-40A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-41A solenoid has 528 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-42Current through a 0.1 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-43In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-44Two coils have mutual inductance 0.13 H. Primary current changes at 152 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-45A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-46A solenoid has 568 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-47Current through a 0.6 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-48In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-49Two coils have mutual inductance 0.13 H. Primary current changes at 167 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-50A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-51A solenoid has 608 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-52Current through a 0.5 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-53In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-54Two coils have mutual inductance 0.13 H. Primary current changes at 182 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-55A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-56A solenoid has 648 turns, length 0.25 m and area 2 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-57Current through a 0.4 H inductor changes uniformly by 4 A in 0.03 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-58In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-59Two coils have mutual inductance 0.13 H. Primary current changes at 197 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-60A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-61A solenoid has 688 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-62Current through a 0.3 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-63In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-64Two coils have mutual inductance 0.13 H. Primary current changes at 212 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-65A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-66A solenoid has 728 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-67Current through a 0.2 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-68In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-69Two coils have mutual inductance 0.13 H. Primary current changes at 227 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-70A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
NEET-71A solenoid has 768 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: NEET

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
NEET-72Current through a 0.1 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: NEET

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
NEET-73In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: NEET

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
NEET-74Two coils have mutual inductance 0.13 H. Primary current changes at 242 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: NEET

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
NEET-75A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: NEET

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.

20 Assertion-Reason Questions

AR-1Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-2Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-3Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-4Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-5Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-6Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-7Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-8Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-9Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-10Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-11Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-12Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-13Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-14Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-15Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-16Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-17Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-18Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-19Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.
AR-20Assertion: Back emf in an inductor opposes the change in current. Reason: Lenz law says induced effect opposes the cause producing it.
  1. A) A and R true; R explains A
  2. B) A and R true; R does not explain A
  3. C) A true; R false
  4. D) A false; R true

Difficulty: NEET Assertion-Reason

Concept Tested: Lenz law in self induction

Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Detailed Explanation:

When current in a coil changes, flux linkage changes and an induced emf appears. Its direction is such that it opposes the increase or decrease of current. Therefore the assertion directly follows from Lenz law.

Common Student Mistake: Saying back emf always opposes current, not change in current.

20 Case-Study Questions

CASE-1A school laboratory uses an inductor connected to a DC source through a switch. When the switch is opened, a spark appears because the inductor tries to maintain current. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-2A generator coil rotates in a magnetic field. Its flux changes sinusoidally and the output is taken through slip rings. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-3Two coaxial coils are used in a demonstration of mutual induction. The secondary galvanometer deflects only when primary current changes. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-4A conducting disc rotates in a uniform magnetic field perpendicular to the disc. The centre and rim are connected to a voltmeter. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-5An iron-core solenoid stores magnetic energy and its inductance becomes much larger than the same air-core solenoid. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-6A school laboratory uses an inductor connected to a DC source through a switch. When the switch is opened, a spark appears because the inductor tries to maintain current. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-7A generator coil rotates in a magnetic field. Its flux changes sinusoidally and the output is taken through slip rings. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-8Two coaxial coils are used in a demonstration of mutual induction. The secondary galvanometer deflects only when primary current changes. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-9A conducting disc rotates in a uniform magnetic field perpendicular to the disc. The centre and rim are connected to a voltmeter. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-10An iron-core solenoid stores magnetic energy and its inductance becomes much larger than the same air-core solenoid. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-11A school laboratory uses an inductor connected to a DC source through a switch. When the switch is opened, a spark appears because the inductor tries to maintain current. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-12A generator coil rotates in a magnetic field. Its flux changes sinusoidally and the output is taken through slip rings. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-13Two coaxial coils are used in a demonstration of mutual induction. The secondary galvanometer deflects only when primary current changes. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-14A conducting disc rotates in a uniform magnetic field perpendicular to the disc. The centre and rim are connected to a voltmeter. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-15An iron-core solenoid stores magnetic energy and its inductance becomes much larger than the same air-core solenoid. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-16A school laboratory uses an inductor connected to a DC source through a switch. When the switch is opened, a spark appears because the inductor tries to maintain current. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-17A generator coil rotates in a magnetic field. Its flux changes sinusoidally and the output is taken through slip rings. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-18Two coaxial coils are used in a demonstration of mutual induction. The secondary galvanometer deflects only when primary current changes. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-19A conducting disc rotates in a uniform magnetic field perpendicular to the disc. The centre and rim are connected to a voltmeter. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-20An iron-core solenoid stores magnetic energy and its inductance becomes much larger than the same air-core solenoid. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.

13. JEE Main Practice Bank

75 JEE Main MCQs

JEE-MAIN-1A solenoid has 208 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-2Current through a 0.3 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-3In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-4Two coils have mutual inductance 0.13 H. Primary current changes at 32 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-5A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-6A solenoid has 248 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-7Current through a 0.2 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-8In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-9Two coils have mutual inductance 0.13 H. Primary current changes at 47 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-10A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-11A solenoid has 288 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-12Current through a 0.1 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-13In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-14Two coils have mutual inductance 0.13 H. Primary current changes at 62 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-15A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-16A solenoid has 328 turns, length 0.25 m and area 2 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-17Current through a 0.6 H inductor changes uniformly by 4 A in 0.03 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-18In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-19Two coils have mutual inductance 0.13 H. Primary current changes at 77 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-20A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-21A solenoid has 368 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-22Current through a 0.5 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-23In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-24Two coils have mutual inductance 0.13 H. Primary current changes at 92 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-25A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-26A solenoid has 408 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-27Current through a 0.4 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-28In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-29Two coils have mutual inductance 0.13 H. Primary current changes at 107 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-30A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-31A solenoid has 448 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-32Current through a 0.3 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-33In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-34Two coils have mutual inductance 0.13 H. Primary current changes at 122 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-35A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-36A solenoid has 488 turns, length 0.25 m and area 2 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-37Current through a 0.2 H inductor changes uniformly by 4 A in 0.03 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-38In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-39Two coils have mutual inductance 0.13 H. Primary current changes at 137 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-40A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-41A solenoid has 528 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-42Current through a 0.1 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-43In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-44Two coils have mutual inductance 0.13 H. Primary current changes at 152 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-45A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-46A solenoid has 568 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-47Current through a 0.6 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-48In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-49Two coils have mutual inductance 0.13 H. Primary current changes at 167 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-50A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-51A solenoid has 608 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-52Current through a 0.5 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-53In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-54Two coils have mutual inductance 0.13 H. Primary current changes at 182 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-55A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-56A solenoid has 648 turns, length 0.25 m and area 2 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-57Current through a 0.4 H inductor changes uniformly by 4 A in 0.03 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-58In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-59Two coils have mutual inductance 0.13 H. Primary current changes at 197 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-60A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-61A solenoid has 688 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-62Current through a 0.3 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-63In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-64Two coils have mutual inductance 0.13 H. Primary current changes at 212 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-65A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-66A solenoid has 728 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-67Current through a 0.2 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-68In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-69Two coils have mutual inductance 0.13 H. Primary current changes at 227 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-70A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JEE-MAIN-71A solenoid has 768 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JEE-MAIN-72Current through a 0.1 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JEE-MAIN-73In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JEE-MAIN-74Two coils have mutual inductance 0.13 H. Primary current changes at 242 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JEE-MAIN-75A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.

30 Numerical Problems

JM-NUM-1An inductor of 0.12 H has current changed uniformly by 2 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 12.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 2/0.02 = 12.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-2An inductor of 0.16 H has current changed uniformly by 3 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 16.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.16 × 3/0.03 = 16.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-3An inductor of 0.20 H has current changed uniformly by 4 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 20.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.20 × 4/0.04 = 20.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-4An inductor of 0.24 H has current changed uniformly by 5 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 24.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.24 × 5/0.05 = 24.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-5An inductor of 0.28 H has current changed uniformly by 6 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 28.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.28 × 6/0.06 = 28.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-6An inductor of 0.32 H has current changed uniformly by 7 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 112.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.32 × 7/0.02 = 112.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-7An inductor of 0.36 H has current changed uniformly by 8 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 96.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.36 × 8/0.03 = 96.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-8An inductor of 0.40 H has current changed uniformly by 2 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 20.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.40 × 2/0.04 = 20.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-9An inductor of 0.44 H has current changed uniformly by 3 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 26.4 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.44 × 3/0.05 = 26.4 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-10An inductor of 0.12 H has current changed uniformly by 4 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 8.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 4/0.06 = 8.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-11An inductor of 0.16 H has current changed uniformly by 5 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 40.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.16 × 5/0.02 = 40.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-12An inductor of 0.20 H has current changed uniformly by 6 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 40.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.20 × 6/0.03 = 40.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-13An inductor of 0.24 H has current changed uniformly by 7 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 42.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.24 × 7/0.04 = 42.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-14An inductor of 0.28 H has current changed uniformly by 8 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 44.8 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.28 × 8/0.05 = 44.8 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-15An inductor of 0.32 H has current changed uniformly by 2 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 10.7 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.32 × 2/0.06 = 10.7 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-16An inductor of 0.36 H has current changed uniformly by 3 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 54.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.36 × 3/0.02 = 54.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-17An inductor of 0.40 H has current changed uniformly by 4 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 53.3 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.40 × 4/0.03 = 53.3 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-18An inductor of 0.44 H has current changed uniformly by 5 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 55.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.44 × 5/0.04 = 55.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-19An inductor of 0.12 H has current changed uniformly by 6 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 14.4 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 6/0.05 = 14.4 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-20An inductor of 0.16 H has current changed uniformly by 7 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 18.7 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.16 × 7/0.06 = 18.7 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-21An inductor of 0.20 H has current changed uniformly by 8 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 80.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.20 × 8/0.02 = 80.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-22An inductor of 0.24 H has current changed uniformly by 2 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 16.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.24 × 2/0.03 = 16.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-23An inductor of 0.28 H has current changed uniformly by 3 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 21.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.28 × 3/0.04 = 21.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-24An inductor of 0.32 H has current changed uniformly by 4 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 25.6 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.32 × 4/0.05 = 25.6 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-25An inductor of 0.36 H has current changed uniformly by 5 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 30.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.36 × 5/0.06 = 30.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-26An inductor of 0.40 H has current changed uniformly by 6 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 120.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.40 × 6/0.02 = 120.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-27An inductor of 0.44 H has current changed uniformly by 7 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 102.7 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.44 × 7/0.03 = 102.7 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-28An inductor of 0.12 H has current changed uniformly by 8 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 24.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 8/0.04 = 24.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-29An inductor of 0.16 H has current changed uniformly by 2 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 6.4 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.16 × 2/0.05 = 6.4 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
JM-NUM-30An inductor of 0.20 H has current changed uniformly by 3 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: JEE Main Numerical

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 10.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.20 × 3/0.06 = 10.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.

20 Conceptual Questions

JM-CONCEPT-1A solenoid has 208 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main Conceptual

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JM-CONCEPT-2Current through a 0.3 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main Conceptual

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JM-CONCEPT-3In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main Conceptual

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JM-CONCEPT-4Two coils have mutual inductance 0.13 H. Primary current changes at 32 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main Conceptual

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JM-CONCEPT-5A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main Conceptual

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JM-CONCEPT-6A solenoid has 248 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main Conceptual

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JM-CONCEPT-7Current through a 0.2 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main Conceptual

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JM-CONCEPT-8In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main Conceptual

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JM-CONCEPT-9Two coils have mutual inductance 0.13 H. Primary current changes at 47 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main Conceptual

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JM-CONCEPT-10A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main Conceptual

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JM-CONCEPT-11A solenoid has 288 turns, length 0.25 m and area 5 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main Conceptual

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JM-CONCEPT-12Current through a 0.1 H inductor changes uniformly by 4 A in 0.02 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main Conceptual

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JM-CONCEPT-13In an AC generator, N, B and A are unchanged but angular speed is made 3 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 3 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main Conceptual

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 3 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JM-CONCEPT-14Two coils have mutual inductance 0.13 H. Primary current changes at 62 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main Conceptual

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JM-CONCEPT-15A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main Conceptual

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
JM-CONCEPT-16A solenoid has 328 turns, length 0.25 m and area 2 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: JEE Main Conceptual

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
JM-CONCEPT-17Current through a 0.6 H inductor changes uniformly by 4 A in 0.03 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: JEE Main Conceptual

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
JM-CONCEPT-18In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: JEE Main Conceptual

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
JM-CONCEPT-19Two coils have mutual inductance 0.13 H. Primary current changes at 77 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: JEE Main Conceptual

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
JM-CONCEPT-20A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: JEE Main Conceptual

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.

14. JEE Advanced Practice Bank

30 Single-Correct Questions

ADV-SC-1Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-2Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-3Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-4Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-5Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-6Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-7Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-8Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-9Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-10Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-11Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-12Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-13Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-14Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-15Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-16Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-17Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-18Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-19Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-20Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-21Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-22Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-23Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-24Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-25Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-26Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-27Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-28Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-29Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-SC-30Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.

20 Multiple-Correct Questions

ADV-MC-1Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-2Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-3Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-4Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-5Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-6Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-7Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-8Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-9Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-10Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-11Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-12Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-13Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-14Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-15Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-16Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-17Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-18Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-19Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MC-20Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.

15 Integer-Type Questions

ADV-INT-1Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-2Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-3Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-4Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-5Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-6Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-7Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-8Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-9Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-10Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-11Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-12Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-13Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-14Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-INT-15Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.

10 Matrix-Match Questions

ADV-MATRIX-1Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-2Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-3Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-4Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-5Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-6Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-7Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-8Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-9Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-MATRIX-10Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.

10 Paragraph-Type Questions

ADV-PARA-1Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-2Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-3Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-4Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-5Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-6Advanced problem on coupled coils: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coupled coils

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coupled coils, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-7Advanced problem on generator with changing speed: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: generator with changing speed

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For generator with changing speed, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-8Advanced problem on rotating disc: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: rotating disc

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For rotating disc, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-9Advanced problem on inductor energy: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: inductor energy

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For inductor energy, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.
ADV-PARA-10Advanced problem on coaxial solenoids: identify the governing relation and the key physical constraint before calculation.

Difficulty: JEE Advanced

Concept Tested: coaxial solenoids

Correct Answer: Use flux linkage conservation idea with Faraday-Lenz law
Detailed Explanation:

For coaxial solenoids, first write flux linkage. Then differentiate with respect to time for emf, or integrate power for energy. In multi-concept questions the most common correct path is: geometry → flux → flux linkage → rate of change → sign from Lenz law.

Shortcut: Write NΦ before writing emf.
Common Student Mistake: Jumping directly to a formula without checking what is changing.

15. IB, IGCSE, ICSE and A-Level Practice

CASE-1A school laboratory uses an inductor connected to a DC source through a switch. When the switch is opened, a spark appears because the inductor tries to maintain current. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-2A generator coil rotates in a magnetic field. Its flux changes sinusoidally and the output is taken through slip rings. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-3Two coaxial coils are used in a demonstration of mutual induction. The secondary galvanometer deflects only when primary current changes. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-4A conducting disc rotates in a uniform magnetic field perpendicular to the disc. The centre and rim are connected to a voltmeter. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-5An iron-core solenoid stores magnetic energy and its inductance becomes much larger than the same air-core solenoid. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-6A school laboratory uses an inductor connected to a DC source through a switch. When the switch is opened, a spark appears because the inductor tries to maintain current. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-7A generator coil rotates in a magnetic field. Its flux changes sinusoidally and the output is taken through slip rings. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-8Two coaxial coils are used in a demonstration of mutual induction. The secondary galvanometer deflects only when primary current changes. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-9A conducting disc rotates in a uniform magnetic field perpendicular to the disc. The centre and rim are connected to a voltmeter. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
CASE-10An iron-core solenoid stores magnetic energy and its inductance becomes much larger than the same air-core solenoid. Which physical law is the central reason for the observed induced emf?
  1. A) Ohm law only
  2. B) Faraday's law with Lenz law direction
  3. C) Coulomb law
  4. D) Hooke law

Difficulty: Case Study

Concept Tested: Induction mechanism

Correct Answer: Faraday's law with Lenz law direction
Detailed Explanation:

Faraday's law gives the magnitude of induced emf due to changing flux linkage. Lenz law gives the direction, ensuring opposition to the change that produces the emf.

Common Student Mistake: Treating induction as a simple resistance problem.
IB-HL-1An inductor of 0.12 H has current changed uniformly by 2 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 12.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 2/0.02 = 12.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-2An inductor of 0.16 H has current changed uniformly by 3 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 16.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.16 × 3/0.03 = 16.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-3An inductor of 0.20 H has current changed uniformly by 4 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 20.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.20 × 4/0.04 = 20.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-4An inductor of 0.24 H has current changed uniformly by 5 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 24.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.24 × 5/0.05 = 24.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-5An inductor of 0.28 H has current changed uniformly by 6 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 28.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.28 × 6/0.06 = 28.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-6An inductor of 0.32 H has current changed uniformly by 7 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 112.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.32 × 7/0.02 = 112.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-7An inductor of 0.36 H has current changed uniformly by 8 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 96.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.36 × 8/0.03 = 96.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-8An inductor of 0.40 H has current changed uniformly by 2 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 20.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.40 × 2/0.04 = 20.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-9An inductor of 0.44 H has current changed uniformly by 3 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 26.4 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.44 × 3/0.05 = 26.4 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IB-HL-10An inductor of 0.12 H has current changed uniformly by 4 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: IB HL

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 8.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 4/0.06 = 8.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
IGCSE-1A solenoid has 208 turns, length 0.25 m and area 3 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: IGCSE

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
IGCSE-2Current through a 0.3 H inductor changes uniformly by 4 A in 0.04 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: IGCSE

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
IGCSE-3In an AC generator, N, B and A are unchanged but angular speed is made 2 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 2 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: IGCSE

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 2 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
IGCSE-4Two coils have mutual inductance 0.13 H. Primary current changes at 32 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: IGCSE

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
IGCSE-5A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: IGCSE

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
IGCSE-6A solenoid has 248 turns, length 0.25 m and area 4 × 10^-4 m². Which expression should be used first to find self inductance?
  1. A) L = μ₀N²A/l
  2. B) L = μ₀NA/l
  3. C) L = μ₀A/Nl
  4. D) L = IR

Difficulty: IGCSE

Concept Tested: Long solenoid self inductance

Correct Answer: L = μ₀N²A/l
Detailed Explanation:

For a long air-core solenoid, magnetic field inside is B = μ₀NI/l. Flux linkage is NΦ = NBA. Therefore L = NΦ/I = μ₀N²A/l.

Common Student Mistake: Do not forget the square of number of turns.
IGCSE-7Current through a 0.2 H inductor changes uniformly by 4 A in 0.05 s. What is the correct method for induced emf?
  1. A) Use e = IR
  2. B) Use e = LΔI/Δt
  3. C) Use e = NBAω
  4. D) Use e = Bvl

Difficulty: IGCSE

Concept Tested: Back EMF

Correct Answer: Use e = LΔI/Δt
Detailed Explanation:

The induced emf is found from the rate of change of current, not from current alone. Substitute L, ΔI and Δt into e = LΔI/Δt.

Shortcut: Rate of change first, sign later.
Common Student Mistake: Using steady-current formulas in induction problems.
IGCSE-8In an AC generator, N, B and A are unchanged but angular speed is made 4 times. What happens to peak emf?
  1. A) Unchanged
  2. B) Becomes 4 times
  3. C) Becomes square of that ratio
  4. D) Becomes zero

Difficulty: IGCSE

Concept Tested: AC generator peak EMF

Correct Answer: It becomes 4 times
Detailed Explanation:

Peak emf is e0 = NBAω. If only angular speed changes, peak emf changes in the same ratio.

Common Student Mistake: Confusing peak emf with flux amplitude NBA.
IGCSE-9Two coils have mutual inductance 0.13 H. Primary current changes at 47 A s^-1. Find the magnitude relation for secondary emf.
  1. A) e₂ = M/I
  2. B) e₂ = M(dI₁/dt)
  3. C) e₂ = LI²
  4. D) e₂ = IR

Difficulty: IGCSE

Concept Tested: Mutual induction

Correct Answer: e₂ = M(dI₁/dt)
Detailed Explanation:

Mutual induction produces emf in the secondary because current in primary changes. The magnitude is M times the rate of change of primary current.

Common Student Mistake: Using self inductance L instead of mutual inductance M.
IGCSE-10A rotating conducting disc has radius R in uniform B. Which integral gives emf between centre and rim?
  1. A) ∫₀ᴿ Bωr dr
  2. B) ∫₀ᴿ BR dr
  3. C) BωR
  4. D) NBAω

Difficulty: IGCSE

Concept Tested: Rotating disc EMF

Correct Answer: ∫₀ᴿ Bωr dr
Detailed Explanation:

A small radial element dr at distance r has speed v = ωr. The small emf is de = Bvdr = Bωrdr. Integrating from 0 to R gives e = ½BωR².

Common Student Mistake: Taking all points on the disc to have the same speed.
A-LEVEL-1An inductor of 0.12 H has current changed uniformly by 2 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 12.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 2/0.02 = 12.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-2An inductor of 0.16 H has current changed uniformly by 3 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 16.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.16 × 3/0.03 = 16.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-3An inductor of 0.20 H has current changed uniformly by 4 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 20.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.20 × 4/0.04 = 20.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-4An inductor of 0.24 H has current changed uniformly by 5 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 24.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.24 × 5/0.05 = 24.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-5An inductor of 0.28 H has current changed uniformly by 6 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 28.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.28 × 6/0.06 = 28.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-6An inductor of 0.32 H has current changed uniformly by 7 A in 0.02 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 112.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.32 × 7/0.02 = 112.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-7An inductor of 0.36 H has current changed uniformly by 8 A in 0.03 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 96.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.36 × 8/0.03 = 96.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-8An inductor of 0.40 H has current changed uniformly by 2 A in 0.04 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 20.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.40 × 2/0.04 = 20.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-9An inductor of 0.44 H has current changed uniformly by 3 A in 0.05 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 26.4 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.44 × 3/0.05 = 26.4 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.
A-LEVEL-10An inductor of 0.12 H has current changed uniformly by 4 A in 0.06 s. Find induced emf magnitude and explain the sign physically.

Difficulty: A-Level

Concept Tested: Numerical use of e = LΔI/Δt

Correct Answer: 8.0 V
Detailed Explanation:

Magnitude e = LΔI/Δt = 0.12 × 4/0.06 = 8.0 V. The negative sign in e = -L dI/dt tells that the induced emf opposes the change in current according to Lenz law.

Shortcut: Keep L in henry and time in seconds.
Common Student Mistake: Writing negative answer for magnitude.

16. Common Student Mistakes

Mistake 1

Confusing self-induced emf with applied battery emf.

Mistake 2

Forgetting the minus sign represents Lenz law.

Mistake 3

Using L = Φ/I when Φ is not flux linkage.

Mistake 4

Forgetting N in flux linkage NΦ.

Mistake 5

Not converting mH, μs, mA and cm² into SI units.

Mistake 6

Using N instead of N² in solenoid self inductance.

Mistake 7

Using total area instead of common area in mutual inductance.

Mistake 8

Forgetting steady current gives zero induced emf.

Mistake 9

Drawing AC generator without slip rings and brushes.

Mistake 10

Confusing an AC generator with a DC dynamo split-ring commutator.

Mistake 11

Thinking an inductor stores energy in an electric field.

Mistake 12

Forgetting core permeability affects inductance.

Mistake 13

Using resistance-combination rules blindly for mutually coupled inductors.

Mistake 14

Ignoring coupling coefficient in practical mutual induction.

Mistake 15

Taking rotating disc speed same at centre and rim.

Mistake 16

Using rms emf when peak emf is asked.

Mistake 17

Forgetting e0 = NBAω is peak value.

Mistake 18

Writing AC generator output as cosine without defining initial position.

Mistake 19

Not distinguishing primary and secondary coils.

Mistake 20

Using iron-core formula without relative permeability.

Mistake 21

Using parallel-inductor formula when coils are magnetically coupled.

Mistake 22

Mixing flux Φ and magnetic field B.

Mistake 23

Writing unit of inductance as weber instead of henry.

Mistake 24

Not explaining physical direction of induced emf.

Mistake 25

Forgetting that losses reduce practical output power.

17. Rapid Revision Formula Sheet

Self InductanceNΦ = LI
Back EMFe = −L dI/dt
Long SolenoidL = μN²A/l
EnergyU = ½LI²
Energy Densityu = B²/(2μ)
Series InductorsL = L₁ + L₂ + ...
Parallel Inductors1/L = 1/L₁ + 1/L₂ + ...
Mutual InductanceN₂Φ₂ = MI₁
Secondary EMFe₂ = −M dI₁/dt
Coaxial SolenoidsM = μN₁N₂A/l
Rotating Disce = ½BωR²
Generator FluxNΦ = NBA cosωt
Generator EMFe = NBAω sinωt
Peak EMFe₀ = NBAω
Frequencyf = ω/2π

Still Confused in Self Induction, Mutual Induction or AC Generator?

If self induction, coefficient of self inductance, energy stored in an inductor, mutual induction, coupled coils, AC generator construction, working principle or exam numericals are not clear, students can contact Kumar Sir for one-to-one Physics guidance.

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