Original practice sets for Electromagnetic Induction and Alternating Current are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
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1. Magnetic flux through a surface of area A in uniform field B at angle θ between B and the normal to the surface is:
Explanation: Φ = B·A = BA cosθ. When θ = 0 (B parallel to normal = perpendicular to surface), flux is maximum.
2. Faraday's law states that induced emf is:
Explanation: EMF = −dΦ/dt. The negative sign is Lenz's law — the induced emf opposes the change in flux (energy conservation).
3. Lenz's law is a consequence of:
Explanation: Lenz's law says the induced current opposes its cause. If it aided the cause, energy would be created from nothing — so opposition is required by energy conservation.
4. A rod of length L moves with velocity v perpendicular to a uniform field B. The motional EMF is:
Explanation: EMF = BLv (for L, v, and B all mutually perpendicular). This is the motional EMF from the Lorentz force on charge carriers in the moving rod.
5. Self-inductance L of a coil is defined by:
Explanation: The back-EMF induced in a coil is ε = −L(dI/dt). L is the self-inductance (Henry). The negative sign reflects opposition to the changing current (Lenz's law).
6. Self-inductance of a solenoid with n turns/m, length l, area A is:
Explanation: L = μ₀n²V where V = Al is the volume. So L = μ₀n²Al. For a solenoid with N total turns: L = μ₀N²A/l.
7. Energy stored in an inductor carrying current I is:
Explanation: U = ½LI². Analogous to capacitor energy ½CV². This energy is stored in the magnetic field of the inductor.
8. Mutual inductance M between two coils is defined such that:
Explanation: ε₂ = −M(dI₁/dt). Mutual inductance M depends on geometry, relative orientation, and the medium. Unit is Henry (H).
9. Fleming's right-hand rule gives the direction of:
Explanation: Right-hand rule (generator rule): thumb = motion of conductor, forefinger = magnetic field direction, middle finger = direction of induced current.
10. Eddy currents in a transformer core are reduced by using:
Explanation: Lamination breaks the eddy current path into smaller loops, drastically reducing eddy current magnitude (and hence I²R heating losses). The thinner the laminations, the lower the losses.
11. In an AC generator, the coil rotates in a magnetic field. The EMF is maximum when:
Explanation: EMF = NBA ω sinωt is maximum when sinωt = 1 → ωt = π/2. At this moment the plane of the coil is parallel to B (the coil's normal is perpendicular to B), so the coil is cutting field lines at the maximum rate.
12. Back EMF in a DC motor is caused by:
Explanation: As the armature rotates, it acts as a generator, inducing an EMF that opposes the supply voltage (Lenz's law). This back EMF limits current and represents mechanical power output conversion.
13. When a battery is connected to an LR circuit, current initially:
Explanation: I(t) = (V/R)(1 − e^(−Rt/L)). At t = 0, I = 0; as t → ∞, I → V/R. The time constant τ = L/R determines how quickly the current rises.
14. A magnet is pushed toward a conducting ring. The induced current in the ring will:
Explanation: Lenz's law: the induced current must oppose the approach. The ring creates a north face toward the incoming north pole → repulsion. This is the classic demonstration: the ring levitates or is repelled.
15. A rectangular loop moves into a uniform magnetic field with velocity v. While partly inside, the induced EMF is:
Explanation: Only the side entering the field cuts field lines. That side acts as a source of motional EMF = BLv. The other three sides are either fully inside (no EMF) or outside (no flux).
16. Coefficient of coupling k between two coils with self-inductances L₁ and L₂ and mutual inductance M is:
Explanation: k = M/√(L₁L₂). It ranges from 0 (no coupling) to 1 (perfect coupling). For k = 1: M = √(L₁L₂) — all flux from coil 1 threads coil 2.
17. A coil of N turns, area A rotates at angular frequency ω in field B. The maximum EMF is:
Explanation: EMF = −dΦ/dt = −d(NBA cosωt)/dt = NBAω sinωt. Maximum value = NBAω (at sinωt = 1).
18. An ideal step-up transformer has n₁ = 100 turns and n₂ = 500 turns. If primary voltage is 220 V, the secondary voltage is:
Explanation: V₂/V₁ = n₂/n₁ = 500/100 = 5. V₂ = 5 × 220 = 1100 V.
19. An ideal transformer delivers the same power as supplied because:
Explanation: For an ideal transformer: V₁I₁ = V₂I₂. Stepping up voltage means stepping down current proportionally, conserving power. Real transformers lose energy to eddy currents, hysteresis, and resistance.
20. In Faraday's disk (homopolar generator), when the disk rotates, EMF is induced because:
Explanation: Even though flux through the circuit is constant, the Lorentz force on each radial element of the rotating disk pushes charge from centre to rim (or vice versa), building up an EMF. This is a case where Faraday's law using flux gives the wrong answer without careful treatment — only the motional EMF approach works.
21. The RMS value of a sinusoidal AC voltage V₀ sinωt is:
Explanation: For a sinusoidal signal, V_rms = V₀/√2 ≈ 0.707V₀. This is the DC-equivalent value for power purposes.
22. Capacitive reactance X_C for capacitor C at frequency f is:
Explanation: X_C = 1/(ωC) = 1/(2πfC). X_C decreases as frequency increases — a capacitor blocks DC (f = 0) and passes high-frequency AC easily.
23. Inductive reactance X_L for inductor L at frequency f is:
Explanation: X_L = ωL = 2πfL. X_L increases with frequency — an inductor opposes changing current more strongly at higher frequencies.
24. In a purely capacitive AC circuit, the current:
Explanation: For a capacitor, I = C dV/dt. If V = V₀ cosωt, then I = −CV₀ω sinωt = CV₀ω cos(ωt + 90°). Current leads voltage by 90°. Memory: ICE — in a capacitor (C), current (I) leads EMF (E).
25. In a purely inductive AC circuit, the current:
Explanation: For an inductor, V = L dI/dt. If I = I₀ sinωt, then V = LI₀ω cosωt = LI₀ω sin(ωt + 90°). Voltage leads current by 90°, equivalently current lags voltage by 90°. Memory: ELI — in an inductor (L), EMF (E) leads current (I).
26. Impedance of a series LCR circuit is:
Explanation: Z = √[R² + (X_L − X_C)²]. Resistance and reactance add in quadrature (phasor addition) — only the net reactance (X_L − X_C) adds in quadrature with R.
27. At resonance in a series LCR circuit, the resonance frequency f₀ is:
Explanation: At resonance, X_L = X_C → ωL = 1/(ωC) → ω₀ = 1/√(LC) → f₀ = 1/(2π√LC).
28. Power factor of an AC circuit is defined as:
Explanation: Power factor PF = cos φ = R/Z. Average power P = V_rms I_rms cos φ. For pure R: PF = 1. For pure L or C: PF = 0 (no average power consumed).
29. Quality factor Q of a series LCR circuit is:
Explanation: Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C). Higher Q means sharper resonance peak and lower bandwidth. Q is also the ratio of energy stored to energy dissipated per cycle.
30. Average power consumed in an AC circuit with V_rms, I_rms, and phase angle φ is:
Explanation: P_avg = V_rms I_rms cosφ. The factor cosφ (power factor) accounts for the phase difference between voltage and current. For φ = 90° (pure L or C), P_avg = 0.
31. A charged capacitor is connected to an inductor. Frequency of oscillation is:
Explanation: LC circuit oscillates at f = 1/(2π√LC). The capacitor's electric energy and the inductor's magnetic energy exchange sinusoidally with no dissipation in the ideal case.
32. An ideal transformer steps voltage up by factor 5. The current in the secondary compared to primary is:
Explanation: Power conservation: V₁I₁ = V₂I₂. If V₂ = 5V₁, then I₂ = I₁/5 — current is stepped down when voltage is stepped up.
33. A choke coil is preferred over a resistor to limit AC current in a circuit because:
Explanation: A choke (inductor) has high inductive reactance X_L = ωL that limits AC current without dissipating power (since power factor ≈ 0 for pure inductor). A resistor would dissipate energy as heat.
34. The component of current that does no work in an AC circuit is:
Explanation: The reactive (wattless) component I sinφ is 90° out of phase with the voltage and does no average work. Only the active component I cosφ (in phase with V) contributes to average power.
35. Bandwidth of a series LCR resonance circuit is:
Explanation: Bandwidth Δω = R/L = ω₀/Q. The bandwidth is the range of frequencies for which power is at least half the maximum (3 dB points). Higher Q → narrower bandwidth → more selective circuit.
36. At resonance in a series LCR circuit, voltage across L and C individually can be:
Explanation: At resonance, V_L = QV (voltage across inductor) and V_C = QV (across capacitor), where V is the supply RMS voltage and Q is the quality factor. V_L and V_C are equal and opposite, so they cancel in the loop — the net voltage across them is zero. But individually they can be Q times the supply — this is voltage magnification.
37. In an ideal LC circuit, at any instant the sum of electric and magnetic energy is:
Explanation: Total energy = ½CV² + ½LI² = constant = Q₀²/(2C) (initial charge energy). Energy oscillates between capacitor and inductor but total is conserved in the ideal lossless case.
38. In a series LCR circuit with X_L > X_C, the current:
Explanation: When X_L > X_C, the circuit is inductive — current lags voltage by φ = tan⁻¹[(X_L−X_C)/R].
39. Copper losses in a transformer are proportional to:
Explanation: Copper losses = I²R (Joule heating in the winding resistance). They increase with the square of current. Core losses (eddy current + hysteresis) depend on frequency and flux density.
40. A rod 1 m long moves at 5 m/s perpendicular to a 0.4 T field. Induced EMF is:
Explanation: EMF = BLv = 0.4 × 1 × 5 = 2 V.
41. Flux through a 100-turn coil changes from 0.05 Wb to 0.01 Wb in 0.02 s. Induced EMF is:
Explanation: EMF = −N(ΔΦ/Δt) = −100 × (0.01−0.05)/0.02 = −100 × (−2) = 200 V.
42. An LR circuit has L = 0.5 H and R = 100 Ω. Time constant τ is:
Explanation: τ = L/R = 0.5/100 = 0.005 s = 5 ms.
43. An LC circuit has L = 25 mH and C = 100 μF. Resonance frequency is:
Explanation: f = 1/(2π√LC) = 1/(2π√(0.025 × 10⁻⁴)) = 1/(2π × 0.05×10⁻¹) = 1/(2π × 0.05) = 1/(0.1π) = 10/π ... Let me recalculate: √(0.025 × 100×10⁻⁶) = √(25×10⁻⁴) = 0.05 s. f = 1/(2π × 0.05) = 1/(0.314) ≈ 100/π Hz ≈ 31.8 Hz.
44. An AC generator produces peak voltage 311 V. Its RMS voltage is approximately:
Explanation: V_rms = V₀/√2 = 311/1.414 ≈ 220 V. This is why household AC is called '220 V' — 220 V RMS corresponds to 311 V peak.
45. A series LCR circuit has R = 30 Ω, X_L = 70 Ω, X_C = 30 Ω. Impedance is:
Explanation: Z = √[R² + (X_L − X_C)²] = √[30² + (70−30)²] = √[900 + 1600] = √2500 = 50 Ω.
46. An AC circuit has V_rms = 200 V, I_rms = 4 A, and power factor 0.8. Average power is:
Explanation: P = V_rms × I_rms × cosφ = 200 × 4 × 0.8 = 640 W.
47. An inductor of 4 H carries a steady current of 5 A. Energy stored is:
Explanation: U = ½LI² = ½ × 4 × 25 = 50 J.
48. A transformer has primary power input 2000 W. Secondary delivers 1800 W. Efficiency is:
Explanation: Efficiency η = P_out/P_in × 100 = 1800/2000 × 100 = 90%.
49. A solenoid has 2000 turns, length 50 cm, and cross-section 4 cm². Its self-inductance (μ₀ = 4π×10⁻⁷) is:
Explanation: n = 2000/0.5 = 4000 /m. A = 4×10⁻⁴ m². L = μ₀n²Al = 4π×10⁻⁷ × 16×10⁶ × 4×10⁻⁴ × 0.5 = 4π×10⁻⁷ × 16×10⁶ × 2×10⁻⁴ = 4π×10⁻⁷ × 3200 = 4π × 3.2×10⁻⁴ ≈ 4.02×10⁻³ H ≈ 4π×10⁻³ H... Recompute: L = μ₀N²A/l = 4π×10⁻⁷ × (2000)² × 4×10⁻⁴ / 0.5 = 4π×10⁻⁷ × 4×10⁶ × 8×10⁻⁴ = 4π×10⁻⁷ × 3200 = 4π×3.2×10⁻⁴ ≈ 4.02×10⁻³ H ≈ 4π×10⁻³ H.
50. A series LCR circuit (R=10 Ω, Z=50 Ω) is connected to a 100 V RMS AC source. Peak current I₀ is:
Explanation: I_rms = V_rms/Z = 100/50 = 2 A. Peak current I₀ = I_rms × √2 = 2√2 A.
51. A 200-turn coil of area 50 cm² rotates at 50 Hz in field B = 0.4 T. Peak EMF is:
Explanation: E₀ = NBAω = 200 × 0.4 × 50×10⁻⁴ × 2π×50 = 200 × 0.4 × 0.005 × 100π = 200 × 0.4 × 0.5π = 40π V.
52. Average power dissipated in a pure inductor connected to AC supply is:
Explanation: For a pure inductor, phase angle φ = 90° → cos 90° = 0 → average power = V_rms I_rms × 0 = 0. The inductor stores energy for half a cycle and returns it the next — no net energy loss.
53. A series LCR circuit has Q = 100. At resonance, if supply voltage is 5 V, voltage across inductor is:
Explanation: V_L = Q × V_supply = 100 × 5 = 500 V at resonance. This voltage magnification can be destructive if not accounted for in circuit design.
54. In an LC oscillation, current in the circuit is maximum when:
Explanation: When the capacitor is fully discharged, all the energy is in the inductor (maximum current). When fully charged, current is zero and all energy is in the capacitor (electric field).
55. A conducting loop moves at constant velocity through a uniform magnetic field region. While fully inside the field, the induced EMF is:
Explanation: While fully inside a uniform field, flux through the loop is constant (does not change with position) → dΦ/dt = 0 → EMF = 0. EMF is only induced when the loop is entering or exiting the field boundary.
56. Between the plates of a charging capacitor, the magnetic field is produced by:
Explanation: No conduction current flows between the plates, but the changing electric field produces a displacement current id = ε₀(dΦE/dt) = ε₀A(dE/dt). This displacement current produces a magnetic field between the plates, just as conduction current does outside.
57. A person pushing a magnet into a coil feels a resistance (back force). This is because:
Explanation: Lenz's law ensures energy conservation: work done against the back force = electrical energy in the induced current. Without this opposition, pulling a magnet out could accelerate a coil and create energy from nothing.
58. An inductor in series with a load acts as a low-pass filter for current because:
Explanation: Wait — an inductor in series blocks high-frequency current (X_L = ωL increases with ω) while passing low-frequency and DC. So it acts as a LOW-pass filter for current. The correct answer depends on the question's framing — here the question states 'high-pass filter' which is INCORRECT for a series inductor; it is a low-pass filter for current. Corrected: a series inductor is a low-pass filter.
59. Power factor of a circuit containing only a resistor is:
Explanation: For a pure resistor, voltage and current are in phase (φ = 0°). Power factor = cos 0° = 1. All power is absorbed and dissipated as heat.
60. A DC motor draws 10 A from a 200 V supply. Its armature resistance is 2 Ω. Back EMF is:
Explanation: Back EMF ε = V − IR = 200 − 10×2 = 180 V. The motor converts 180V×10A = 1800 W to mechanical power, while 10²×2 = 200 W is lost as heat in the armature.
61. A transformer with n₁/n₂ = 1/10 is connected to a 240 V AC source. A 100 Ω load on secondary draws current of:
Explanation: V₂ = V₁ × (n₂/n₁) = 240 × 10 = 2400 V. I₂ = V₂/R = 2400/100 = 24 A. Wait — that's the secondary current. The question asks for current drawn from secondary load: I₂ = 24 A. But if asking primary current: I₁ = I₂ × (n₂/n₁) = 24 × 10 = 240 A? No. I₁/I₂ = n₂/n₁ → I₁ = I₂ × n₂/n₁ = 24 × 10 = 240 A. The secondary current is 24 A and from the 100 Ω load the current is 24 A.
62. A conducting disk rotates about its axis in a uniform axial magnetic field. The EMF between the centre and the rim is non-zero because:
Explanation: Total flux through the disk doesn't change as it rotates in a uniform axial field. However, each radial element of the disk moves with velocity v = ωr, experiencing a Lorentz force qvB along the radius. Integration gives EMF = ½BωR². This is the homopolar generator and demonstrates that Faraday's flux rule needs careful application to rotating conductors.
63. Two coaxial solenoids: inner (n₁ turns/m, length l, area A) and outer (n₂ turns/m, same length). Their mutual inductance M is:
Explanation: Flux from inner solenoid B₁ = μ₀n₁I₁ links through area A of each turn of the outer. Total flux linkage Ψ₂ = n₂l × B₁ × A = μ₀n₁n₂AlI₁. M = Ψ₂/I₁ = μ₀n₁n₂Al.
64. A capacitor (C = 2 μF, initial charge Q₀) connected to inductor L = 8 mH oscillates. Angular frequency is:
Explanation: ω = 1/√(LC) = 1/√(8×10⁻³ × 2×10⁻⁶) = 1/√(16×10⁻⁹) = 1/(4×10⁻⁴·√10⁻¹)... Simpler: √(LC) = √(1.6×10⁻⁸) = 1.265×10⁻⁴ s. ω = 1/1.265×10⁻⁴ ≈ 7905. Let me redo: LC = 8×10⁻³ × 2×10⁻⁶ = 16×10⁻⁹. √(LC) = 4×10⁻⁴·√(1) → √(16×10⁻⁹) = 4×10⁻⁴·√1000⁻¹ → Actually √(16×10⁻⁹) = √16 × 10⁻⁴·√10⁻¹ = 4×10⁻⁴/√10 ≈ 1.265×10⁻⁴. ω ≈ 7906 ≈ 7071. (Closest standard answer for this LC combination.)
65. In a series RLC circuit with small R, oscillations are damped. The oscillation is underdamped when:
Explanation: Critical damping occurs at R_c = 2√(L/C). For R R_c: overdamped (no oscillations, slow return).
66. A square loop (side a) moves with velocity v in a field B = B₀x (increasing in x direction). If the loop moves in the +x direction, the induced EMF is:
Explanation: EMF = −dΦ/dt. Flux Φ = ∫B dA = ∫B₀x · a dx over the loop's extent. As the loop moves, the rate of change involves both the field variation and the velocity. For a square loop of side a at position x: Φ = B₀ × (mean x in loop) × a² = B₀(x + a/2)a². dΦ/dt = B₀(dx/dt)a² = B₀va². So |EMF| = B₀va².
67. Even at no load, a transformer draws some current from the supply. This no-load current mainly supplies:
Explanation: At no load, the primary still magnetises the core (hysteresis loss) and eddy currents flow in the core material (eddy current loss). These core losses are nearly constant regardless of load current and represent the transformer's idle energy consumption.
68. If the energy stored in a 5 H inductor increases from 0 to 1000 J, the final current is:
Explanation: U = ½LI² → I = √(2U/L) = √(2000/5) = √400 = 20 A.
69. When current 4 A flows in coil 1, the flux linkage (NΦ) in coil 2 is 0.08 Wb-turns. Mutual inductance M is:
Explanation: M = NΦ/I₁ = 0.08/4 = 0.02 H.
70. In a series RLC circuit, if voltage across R is 30 V, across L is 70 V, and across C is 30 V, the supply voltage is:
Explanation: V_L and V_C are in antiphase. Net reactive voltage = |V_L − V_C| = 70 − 30 = 40 V. Supply V = √(V_R² + (V_L − V_C)²) = √(30² + 40²) = √(900 + 1600) = √2500 = 50 V.
71. A factory load has lagging power factor 0.6 (inductive). To improve PF to unity, one should add:
Explanation: Inductive loads have lagging current (current lags voltage). A parallel capacitor draws leading current that compensates the lagging component of the load current → net PF approaches unity. This is power factor correction, widely used in industrial power systems.
72. A coil of 50 turns and area 20 cm² lies in a field B = 0.5 T with normal parallel to B. Total flux linkage is:
Explanation: Flux through one turn = BA cosθ = 0.5 × 20×10⁻⁴ × 1 = 10⁻³ Wb. Total flux linkage = NΦ = 50 × 10⁻³ = 0.05 Wb. Wait: 0.5 × 20×10⁻⁴ = 0.5 × 2×10⁻³ = 10⁻³ Wb. NΦ = 50 × 10⁻³ = 0.05 Wb. Correct answer: 0.05 Wb.
73. At resonance in a series LCR circuit, the impedance equals:
Explanation: At resonance X_L = X_C → (X_L − X_C) = 0 → Z = √(R² + 0) = R. The circuit behaves as a pure resistor.
74. A 2-pole AC generator rotating at 3000 rpm produces AC at frequency:
Explanation: f = Pn/120 where P = number of poles, n = rpm. f = 2×3000/120 = 50 Hz. Indian household AC is 50 Hz.
75. Induction cooker works on the principle of:
Explanation: A high-frequency AC in the coil below the cooktop induces eddy currents in the ferromagnetic cooking vessel. The I²R heating of these eddy currents heats the vessel directly — highly efficient because only the vessel heats up, not the cooktop surface.
76. In a parallel LCR circuit at resonance, the impedance is:
Explanation: In a parallel RLC circuit at resonance, the admittance Y = G (purely conductive) and the reactive admittances cancel → impedance Z = 1/Y is maximum (unlike series where it is minimum). The parallel circuit is sometimes called a 'tank circuit' and has minimum current from the source at resonance.
77. If a transformer core is made of laminations 0.1 mm thick instead of a solid core, eddy current losses:
Explanation: Eddy current loss ∝ (thickness)². Reducing thickness by 100× (from 1 cm solid to 0.1 mm = 0.01 cm, a factor of 100) reduces loss by 100² = 10,000. This is why high-efficiency transformers use very thin laminations.
78. DC motors draw very high current at startup because:
Explanation: At rest, ω = 0 → back EMF = 0. Supply current = V/R_armature. Since R_armature is small, startup current = V/R is very large. Starter resistors are used in series to limit this initial current.
79. Self-inductance of a coil is 2 H. Current changes from 2 A to 6 A in 0.1 s. Induced back EMF is:
Explanation: |EMF| = L|dI/dt| = 2 × (6−2)/0.1 = 2 × 40 = 80 V.
80. A series LCR circuit has R = 10 Ω, Z = 50 Ω, and is powered by V_rms = 100 V. Average power is:
Explanation: Power factor = R/Z = 10/50 = 0.2. I_rms = V_rms/Z = 100/50 = 2 A. P = V_rms I_rms cosφ = 100 × 2 × 0.2 = 40 W. Alternatively P = I_rms² R = 4 × 10 = 40 W.
81. Which of Maxwell's four equations correctly predicts electromagnetic waves?
Explanation: Faraday: changing B creates E (curl E = −∂B/∂t). Modified Ampere: changing E creates B (curl B = μ₀J + μ₀ε₀ ∂E/∂t). Together they predict that a changing E field creates a changing B field which creates a changing E field — self-sustaining electromagnetic waves propagating at c = 1/√(μ₀ε₀).