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Electromagnetic Induction Class 12 Physics Numericals 2026-27 | Chapter 6
Class 12 Physics • Chapter 6 • 2026–27
Electromagnetic Induction Class 12 Physics Numericals
40 solved numericals covering magnetic flux, Faraday’s law, Lenz’s law, induced charge, motional EMF, self-induction, magnetic energy and mutual induction.
Designed for CBSE 2026–27 practice with step-by-step formula selection, substitution, units and final answers.
How to use this page: First attempt every numerical without looking at the solution. Write the known quantities, identify the physical change, select the governing relation, substitute SI units, and then check the magnitude and unit of the answer.
2026–27 syllabus boundary: The official CBSE curriculum lists Chapter 6 as electromagnetic induction, Faraday’s laws, induced EMF/current, Lenz’s law, self-induction and mutual induction. AC generator and transformer are listed under Chapter 7: Alternating Current, so they are intentionally excluded from this Chapter 6 numerical bank.
Why these problem types? Current Chapter 6 practice resources repeatedly emphasise Faraday’s law, magnetic-flux change, Lenz’s law, motional EMF, self-induction and mutual induction; recent practice material also includes induced-charge, graph/slope and numerical applications. This page turns those recurring skills into original, step-by-step problems rather than copying third-party questions.
Formula & Method Map
| Problem type | Core relation | What to check |
|---|---|---|
| Average induced EMF | |εavg| = N|ΔΦ|/Δt | Flux change per turn, number of turns, time interval |
| Magnetic flux | Φ = BA cosθ | θ is between B and the area vector |
| Motional EMF | ε = Bℓv for the stated perpendicular geometry | Use the appropriate velocity component |
| Induced charge | q = N|ΔΦ|/R | For a simple closed circuit with constant R |
| Self-induced EMF | |ε| = L|dI/dt| | Use current-change rate, not current itself |
| Inductor energy | U = ½LI² | Energy depends on L and the square of current |
| Mutual induction | |ε₂| = M|dI₁/dt| | Only the rate of change of primary current matters for instantaneous induced EMF |
Section A — Magnetic Flux & Faraday’s Law
Start here. These problems build the habit of identifying exactly what changes: field, area, orientation, turns or time.
Numerical 1
A 200-turn coil has a flux per turn that changes from 0.020 Wb to 0.005 Wb in 0.30 s. Find the magnitude of the average induced EMF.
Given: Turns N = 200; Φ₁ = 0.020 Wb; Φ₂ = 0.005 Wb; Δt = 0.30 s.
Solution:
Use |ε_avg| = N|ΔΦ|/Δt. Here |ΔΦ| = |0.005 − 0.020| = 0.015 Wb. Therefore |ε_avg| = 200 × 0.015 / 0.30.
Use |ε_avg| = N|ΔΦ|/Δt. Here |ΔΦ| = |0.005 − 0.020| = 0.015 Wb. Therefore |ε_avg| = 200 × 0.015 / 0.30.
Final answer: 5.0 V
Numerical 2
A 500-turn coil experiences a change in flux per turn of 4.0 × 10⁻³ Wb in 0.20 s. Find the average induced EMF.
Given: N = 500; |ΔΦ| = 4.0 × 10⁻³ Wb; Δt = 0.20 s.
Solution:
|ε_avg| = N|ΔΦ|/Δt = 500 × 4.0×10⁻³ / 0.20.
|ε_avg| = N|ΔΦ|/Δt = 500 × 4.0×10⁻³ / 0.20.
Final answer: 10 V
Numerical 3
A 100-turn rectangular coil of area 0.020 m² is in a uniform 0.50 T field. The angle between B and the area vector changes from 0° to 60° in 0.40 s. Find the magnitude of average induced EMF.
Given: N = 100; A = 0.020 m²; B = 0.50 T; θ₁ = 0°; θ₂ = 60°; Δt = 0.40 s.
Solution:
Φ = BA cosθ. So |ΔΦ| = BA|cos60° − cos0°| = (0.50)(0.020)(0.5) = 0.005 Wb. Then |ε_avg| = 100×0.005/0.40.
Φ = BA cosθ. So |ΔΦ| = BA|cos60° − cos0°| = (0.50)(0.020)(0.5) = 0.005 Wb. Then |ε_avg| = 100×0.005/0.40.
Final answer: 1.25 V
Numerical 4
A 50-turn coil of area 0.010 m² is placed perpendicular to a 0.40 T field. The field falls uniformly to zero in 0.20 s. Find the average induced EMF.
Given: N = 50; A = 0.010 m²; B changes 0.40 T → 0; Δt = 0.20 s. Since the field is perpendicular to the plane, B is parallel to the area vector.
Solution:
Initial flux per turn = BA = 0.40×0.010 = 0.004 Wb. Final flux = 0. Hence |ε_avg| = 50×0.004/0.20.
Initial flux per turn = BA = 0.40×0.010 = 0.004 Wb. Final flux = 0. Hence |ε_avg| = 50×0.004/0.20.
Final answer: 1.0 V
Numerical 5
A 300-turn coil has resistance 6 Ω. The flux per turn changes by 0.012 Wb. Find the total induced charge that flows through the coil.
Given: N = 300; R = 6 Ω; |ΔΦ| = 0.012 Wb.
Solution:
From q = ∫I dt and Faraday's law, q = N|ΔΦ|/R. Therefore q = 300×0.012/6.
From q = ∫I dt and Faraday's law, q = N|ΔΦ|/R. Therefore q = 300×0.012/6.
Final answer: 0.60 C
Numerical 6
A 150-turn coil of resistance 5 Ω experiences the same flux change of 0.020 Wb once in 0.10 s and once in 1.0 s. Find the induced charge in each case.
Given: N = 150; R = 5 Ω; |ΔΦ| = 0.020 Wb.
Solution:
q = N|ΔΦ|/R, which contains no time interval. q = 150×0.020/5 in both cases.
q = N|ΔΦ|/R, which contains no time interval. q = 150×0.020/5 in both cases.
Final answer: 0.60 C in each case
Numerical 7
A 0.40 m rod moves at 5 m s⁻¹ perpendicular to a 0.30 T magnetic field. Find the motional EMF across the rod.
Given: ℓ = 0.40 m; v = 5 m s⁻¹; B = 0.30 T; v ⟂ B and the rod is oriented for maximum motional EMF.
Solution:
Use ε = Bℓv = 0.30×0.40×5.
Use ε = Bℓv = 0.30×0.40×5.
Final answer: 0.60 V
Numerical 8
A conducting rod of length 0.80 m moves at 3 m s⁻¹ in a 0.25 T field. Its motion is perpendicular to B and the rod. Find the induced current if the total circuit resistance is 2 Ω.
Given: ℓ = 0.80 m; v = 3 m s⁻¹; B = 0.25 T; R = 2 Ω.
Solution:
ε = Bℓv = 0.25×0.80×3 = 0.60 V. Then I = ε/R = 0.60/2.
ε = Bℓv = 0.25×0.80×3 = 0.60 V. Then I = ε/R = 0.60/2.
Final answer: 0.30 A
Numerical 9
A rod moves at 6 m s⁻¹ through a 0.40 T field. The effective rod length perpendicular to the motion is 0.50 m. Find the motional EMF.
Given: B = 0.40 T; ℓ = 0.50 m; v = 6 m s⁻¹.
Solution:
ε = Bℓv = 0.40×0.50×6.
ε = Bℓv = 0.40×0.50×6.
Final answer: 1.20 V
Numerical 10
A rod of length 0.50 m moves at 4 m s⁻¹ in a 0.20 T field. The velocity makes 30° with the field. Find the magnitude of q(v×B) per unit charge and hence the motional EMF for the stated geometry.
Given: ℓ = 0.50 m; v = 4 m s⁻¹; B = 0.20 T; angle between v and B = 30°.
Solution:
The magnetic force per unit charge is vB sin30° = 4×0.20×0.5 = 0.40 N C⁻¹. For a rod arranged so this component produces the separation along its length, ε = Bℓv sin30° = 0.20×0.50×4×0.5.
The magnetic force per unit charge is vB sin30° = 4×0.20×0.5 = 0.40 N C⁻¹. For a rod arranged so this component produces the separation along its length, ε = Bℓv sin30° = 0.20×0.50×4×0.5.
Final answer: 0.20 V
Numerical 11
A straight rod of length 0.60 m rotates about one end with angular speed 10 rad s⁻¹ in a uniform field of 0.50 T perpendicular to the plane of rotation. Find the EMF between the axis and the free end.
Given: ℓ = 0.60 m; ω = 10 rad s⁻¹; B = 0.50 T.
Solution:
For a rotating rod, ε = ½Bωℓ². Hence ε = 0.5×0.50×10×(0.60)².
For a rotating rod, ε = ½Bωℓ². Hence ε = 0.5×0.50×10×(0.60)².
Final answer: 0.90 V
Numerical 12
A 100-turn coil has flux per turn Φ = (0.020 + 0.005t) Wb, where t is in seconds. Find the induced EMF magnitude.
Given: N = 100; Φ(t) = 0.020 + 0.005t Wb.
Solution:
Faraday's law gives |ε| = N|dΦ/dt|. Since dΦ/dt = 0.005 Wb s⁻¹, |ε| = 100×0.005.
Faraday's law gives |ε| = N|dΦ/dt|. Since dΦ/dt = 0.005 Wb s⁻¹, |ε| = 100×0.005.
Final answer: 0.50 V
Numerical 13
A 200-turn coil has flux per turn Φ = 0.010t² Wb. Find the magnitude of induced EMF at t = 3 s.
Given: N = 200; Φ = 0.010t² Wb.
Solution:
dΦ/dt = 0.020t. At t=3 s, dΦ/dt=0.060 Wb s⁻¹. Therefore |ε| = N(dΦ/dt) = 200×0.060.
dΦ/dt = 0.020t. At t=3 s, dΦ/dt=0.060 Wb s⁻¹. Therefore |ε| = N(dΦ/dt) = 200×0.060.
Final answer: 12 V
Numerical 14
A 100-turn coil has flux per turn varying linearly from +0.020 Wb at t=0 to −0.010 Wb at t=0.50 s. Find the magnitude of average induced EMF.
Given: N = 100; Φ₁ = +0.020 Wb; Φ₂ = −0.010 Wb; Δt = 0.50 s.
Solution:
|ΔΦ| = |−0.010 − 0.020| = 0.030 Wb. Thus |ε_avg| = 100×0.030/0.50.
|ΔΦ| = |−0.010 − 0.020| = 0.030 Wb. Thus |ε_avg| = 100×0.030/0.50.
Final answer: 6.0 V
Section B — Motional EMF & Flux-Change Applications
These problems move from direct substitution to geometry and time-dependent flux reasoning.
Numerical 15
A 2 H inductor carries a current of 3 A. Find the energy stored in its magnetic field.
Given: L = 2 H; I = 3 A.
Solution:
Use U = ½LI² = ½×2×3².
Use U = ½LI² = ½×2×3².
Final answer: 9 J
Numerical 16
An inductor stores 18 J when carrying 3 A. Find its inductance.
Given: U = 18 J; I = 3 A.
Solution:
U = ½LI², so L = 2U/I² = 36/9.
U = ½LI², so L = 2U/I² = 36/9.
Final answer: 4 H
Section C — Self-Induction & Magnetic Energy
Focus on the distinction between current, rate of change of current, inductance and stored magnetic energy.
Numerical 17
A 5 H inductor experiences a current change from 2 A to 8 A in 0.30 s. Find the magnitude of average self-induced EMF.
Given: L = 5 H; ΔI = 8−2 = 6 A; Δt = 0.30 s.
Solution:
|ε_avg| = L|ΔI|/Δt = 5×6/0.30.
|ε_avg| = L|ΔI|/Δt = 5×6/0.30.
Final answer: 100 V
Numerical 18
A coil of inductance 0.80 H has current changing at 5 A s⁻¹. Find the magnitude of self-induced EMF.
Given: L = 0.80 H; |dI/dt| = 5 A s⁻¹.
Solution:
|ε| = L|dI/dt| = 0.80×5.
|ε| = L|dI/dt| = 0.80×5.
Final answer: 4.0 V
Numerical 19
The current in a 3 H inductor falls uniformly from 4 A to zero in 0.20 s. Find the average magnitude of induced EMF.
Given: L = 3 H; ΔI = 4 A; Δt = 0.20 s.
Solution:
|ε_avg| = LΔI/Δt = 3×4/0.20.
|ε_avg| = LΔI/Δt = 3×4/0.20.
Final answer: 60 V
Numerical 20
Two ideal inductors have L₁ = 2 H and L₂ = 8 H and carry the same current of 3 A. Compare their stored energies.
Given: L₁ = 2 H; L₂ = 8 H; same I = 3 A.
Solution:
Since U = ½LI², for the same current U₂/U₁ = L₂/L₁ = 8/2 = 4.
Since U = ½LI², for the same current U₂/U₁ = L₂/L₁ = 8/2 = 4.
Final answer: U₂ = 4U₁
Numerical 21
A coil has 500 turns and flux linkage of 0.25 Wb-turn when its current is 2 A. Find its self-inductance, assuming linear behaviour.
Given: N = 500; flux per turn = 0.0005 Wb, so flux linkage NΦ = 0.25 Wb-turn; I = 2 A.
Solution:
By definition, flux linkage = LI in a linear coil. Therefore L = 0.25/2.
By definition, flux linkage = LI in a linear coil. Therefore L = 0.25/2.
Final answer: 0.125 H
Numerical 22
The self-inductance of a long solenoid is proportional to N² for fixed length, area and core. If the number of turns is doubled, how does L change?
Given: Initial inductance L ∝ N²; N₂ = 2N₁.
Solution:
L₂/L₁ = (N₂/N₁)² = 2².
L₂/L₁ = (N₂/N₁)² = 2².
Final answer: L₂ = 4L₁
Numerical 23
A coil has self-inductance 0.50 H. If the number of turns is reduced to half while all other solenoid dimensions and core conditions remain unchanged, find the new inductance.
Given: L₁ = 0.50 H; N₂ = N₁/2.
Solution:
L ∝ N², so L₂ = L₁(1/2)² = 0.50/4.
L ∝ N², so L₂ = L₁(1/2)² = 0.50/4.
Final answer: 0.125 H
Section D — Mutual Induction
Use the changing current in one coil to calculate the induced EMF in the neighbouring coil.
Numerical 24
Two coils have mutual inductance 0.20 H. The current in the primary changes uniformly from 0 to 5 A in 0.10 s. Find the average induced EMF in the secondary.
Given: M = 0.20 H; ΔI₁ = 5 A; Δt = 0.10 s.
Solution:
|ε₂| = M|ΔI₁|/Δt = 0.20×5/0.10.
|ε₂| = M|ΔI₁|/Δt = 0.20×5/0.10.
Final answer: 10 V
Numerical 25
The current in one coil changes at 8 A s⁻¹ and induces an EMF of 2.4 V in a neighbouring coil. Find the mutual inductance.
Given: |dI₁/dt| = 8 A s⁻¹; |ε₂| = 2.4 V.
Solution:
|ε₂| = M|dI₁/dt|, so M = 2.4/8.
|ε₂| = M|dI₁/dt|, so M = 2.4/8.
Final answer: 0.30 H
Numerical 26
Two coils have mutual inductance 0.60 H. If the rate of change of current in the primary is doubled, what happens to the magnitude of induced EMF in the secondary?
Given: M = 0.60 H; dI/dt changes by factor 2.
Solution:
|ε| = M|dI/dt|, so induced EMF changes in the same ratio.
|ε| = M|dI/dt|, so induced EMF changes in the same ratio.
Final answer: It doubles
Numerical 27
A primary current changes from 2 A to 6 A in 0.40 s and induces an average EMF of 3 V in a secondary coil. Find M.
Given: ΔI = 4 A; Δt = 0.40 s; |ε₂| = 3 V.
Solution:
M = |ε₂|Δt/|ΔI| = 3×0.40/4.
M = |ε₂|Δt/|ΔI| = 3×0.40/4.
Final answer: 0.30 H
Section E — Mixed Board-Style Numericals
These combine two or more Chapter 6 ideas and are useful for checking whether you can choose the correct path without being told which formula to use.
Numerical 28
A 100-turn coil of area 0.050 m² is in a 0.20 T field. The field is reduced uniformly to zero in 0.10 s. The coil resistance is 5 Ω. Find (a) average induced EMF and (b) total induced charge.
Given: N = 100; A = 0.050 m²; B: 0.20 T → 0; Δt = 0.10 s; R = 5 Ω.
Solution:
Initial flux per turn = BA = 0.20×0.050 = 0.010 Wb. (a) |ε_avg| = 100×0.010/0.10 = 10 V. (b) q = N|ΔΦ|/R = 100×0.010/5 = 0.20 C.
Initial flux per turn = BA = 0.20×0.050 = 0.010 Wb. (a) |ε_avg| = 100×0.010/0.10 = 10 V. (b) q = N|ΔΦ|/R = 100×0.010/5 = 0.20 C.
Final answer: (a) 10 V; (b) 0.20 C
Numerical 29
A 400-turn coil of area 0.010 m² is rotated from a position where B is parallel to its area vector to one where B is perpendicular to the area vector. B = 0.50 T and the change occurs in 0.20 s. Find average induced EMF.
Given: N = 400; A = 0.010 m²; B = 0.50 T; θ: 0° → 90°; Δt = 0.20 s.
Solution:
Initial flux per turn = BA = 0.005 Wb; final flux = 0. Hence |ε_avg| = 400×0.005/0.20.
Initial flux per turn = BA = 0.005 Wb; final flux = 0. Hence |ε_avg| = 400×0.005/0.20.
Final answer: 10 V
Numerical 30
A loop has magnetic flux Φ(t) = 0.04 − 0.01t Wb for 0≤t≤4 s and resistance 2 Ω. Find the induced current magnitude and total charge that passes through the loop during this interval.
Given: Single turn: Φ(t)=0.04−0.01t Wb; R=2 Ω; dΦ/dt = −0.01 Wb s⁻¹.
Solution:
|ε| = |dΦ/dt| = 0.01 V. Hence I = ε/R = 0.01/2 = 0.005 A. Total flux change magnitude = 0.04 Wb, so q = |ΔΦ|/R = 0.04/2.
|ε| = |dΦ/dt| = 0.01 V. Hence I = ε/R = 0.01/2 = 0.005 A. Total flux change magnitude = 0.04 Wb, so q = |ΔΦ|/R = 0.04/2.
Final answer: I = 5 mA; q = 0.020 C
Numerical 31
A 250-turn coil has resistance 10 Ω. Its flux per turn changes from 0.008 Wb to 0.002 Wb. Find the average induced current.
Given: N = 250; R=10 Ω; ΔΦ = 0.006 Wb; Δt = 0.50 s.
Solution:
Average EMF = NΔΦ/Δt = 250×0.006/0.50 = 3 V. Then I_avg = ε_avg/R = 3/10.
Average EMF = NΔΦ/Δt = 250×0.006/0.50 = 3 V. Then I_avg = ε_avg/R = 3/10.
Final answer: 0.30 A
Numerical 32
A 100-turn coil has flux per turn Φ = 2×10⁻³ sin(20t) Wb. Find the maximum induced EMF.
Given: N=100; Φ₀=2×10⁻³ Wb; angular frequency ω=20 rad s⁻¹.
Solution:
ε = −N dΦ/dt = −NΦ₀ω cos(20t). Thus ε_max = NΦ₀ω = 100×2×10⁻³×20.
ε = −N dΦ/dt = −NΦ₀ω cos(20t). Thus ε_max = NΦ₀ω = 100×2×10⁻³×20.
Final answer: 4.0 V
Numerical 33
An inductor of 0.25 H carries a current of 4 A. If its current is reversed to −4 A in 0.20 s, find the magnitude of average self-induced EMF.
Given: L=0.25 H; I changes from +4 A to −4 A, so |ΔI|=8 A; Δt=0.20 s.
Solution:
|ε_avg| = L|ΔI|/Δt = 0.25×8/0.20.
|ε_avg| = L|ΔI|/Δt = 0.25×8/0.20.
Final answer: 10 V
Numerical 34
A coil has 300 turns and area 0.020 m². It is in a uniform 0.40 T field with B normal to the coil. If the coil is removed completely from the field in 0.10 s and has resistance 4 Ω, find the total induced charge.
Given: N=300; A=0.020 m²; B=0.40 T; R=4 Ω.
Solution:
Initial flux per turn = BA = 0.40×0.020=0.008 Wb; final flux=0. q=NΔΦ/R=300×0.008/4.
Initial flux per turn = BA = 0.40×0.020=0.008 Wb; final flux=0. q=NΔΦ/R=300×0.008/4.
Final answer: 0.60 C
Numerical 35
A single-turn loop of area 0.25 m² is in a uniform magnetic field perpendicular to its plane. The field changes from 0.80 T to 0.20 T in 0.15 s. Find the magnitude of average induced EMF.
Given: A = 0.25 m²; B changes 0.80 T → 0.20 T; Δt = 0.15 s; N = 1.
Solution:
Because B is perpendicular to the plane, θ = 0° and Φ = BA. Thus |ΔΦ| = A|ΔB| = 0.25×0.60 = 0.15 Wb. Then |εavg| = |ΔΦ|/Δt = 0.15/0.15.
Because B is perpendicular to the plane, θ = 0° and Φ = BA. Thus |ΔΦ| = A|ΔB| = 0.25×0.60 = 0.15 Wb. Then |εavg| = |ΔΦ|/Δt = 0.15/0.15.
Final answer: 1.0 V
Numerical 36
The magnetic flux through each turn of a 200-turn coil varies linearly with time with slope dΦ/dt = −0.020 Wb s⁻¹. The coil resistance is 5 Ω. Find the magnitude of induced EMF and induced current.
Given: N = 200; dΦ/dt = −0.020 Wb s⁻¹; R = 5 Ω.
Solution:
|ε| = N|dΦ/dt| = 200×0.020 = 4.0 V. Since the circuit is closed and resistive, I = ε/R = 4.0/5.
|ε| = N|dΦ/dt| = 200×0.020 = 4.0 V. Since the circuit is closed and resistive, I = ε/R = 4.0/5.
Final answer: |ε| = 4.0 V; I = 0.80 A
Numerical 37
A long air-core solenoid has 1000 turns, length 0.50 m and cross-sectional area 4.0 × 10⁻³ m². Calculate its self-inductance. Take μ0 = 4π × 10⁻⁷ H m⁻¹.
Given: N = 1000; ℓ = 0.50 m; A = 4.0×10⁻³ m²; μ0 = 4π×10⁻⁷ H m⁻¹.
Solution:
For a long air-core solenoid, L = μ0N²A/ℓ. Substitution gives L = (4π×10⁻⁷)(1000²)(4.0×10⁻³)/0.50 ≈ 1.01×10⁻² H.
For a long air-core solenoid, L = μ0N²A/ℓ. Substitution gives L = (4π×10⁻⁷)(1000²)(4.0×10⁻³)/0.50 ≈ 1.01×10⁻² H.
Final answer: L ≈ 10.1 mH
Numerical 38
Two long coaxial air-core solenoids have 800 and 1200 turns, common length 0.40 m and common cross-sectional area 2.0 × 10⁻³ m². Find their ideal mutual inductance. Take μ0 = 4π × 10⁻⁷ H m⁻¹.
Given: N1 = 800; N2 = 1200; A = 2.0×10⁻³ m²; ℓ = 0.40 m.
Solution:
For ideal long coaxial air-core solenoids, M = μ0N1N2A/ℓ. Hence M = (4π×10⁻⁷)(800)(1200)(2.0×10⁻³)/0.40 ≈ 6.03×10⁻³ H.
For ideal long coaxial air-core solenoids, M = μ0N1N2A/ℓ. Hence M = (4π×10⁻⁷)(800)(1200)(2.0×10⁻³)/0.40 ≈ 6.03×10⁻³ H.
Final answer: M ≈ 6.03 mH
Numerical 39
A 0.50 H inductor is connected to a circuit in which its current rises uniformly from 0 to 4 A in 0.20 s. Find the magnitude of the self-induced EMF. Also find the energy stored when the current reaches 4 A.
Given: L = 0.50 H; I changes 0 → 4 A in 0.20 s.
Solution:
|ε| = L|ΔI|/Δt = 0.50×4/0.20 = 10 V. At I = 4 A, U = ½LI² = ½×0.50×16.
|ε| = L|ΔI|/Δt = 0.50×4/0.20 = 10 V. At I = 4 A, U = ½LI² = ½×0.50×16.
Final answer: self-induced EMF = 10 V; stored energy = 4.0 J
Numerical 40
A secondary coil has mutual inductance 0.15 H with a primary coil. The primary current varies according to I = 2t² A. Find the magnitude of induced EMF in the secondary at t = 2 s.
Given: M = 0.15 H; I = 2t² A; t = 2 s.
Solution:
For mutual induction, |ε2| = M|dI1/dt|. Here dI/dt = 4t, so at t = 2 s the rate is 8 A s⁻¹. Therefore |ε2| = 0.15×8.
For mutual induction, |ε2| = M|dI1/dt|. Here dI/dt = 4t, so at t = 2 s the rate is 8 A s⁻¹. Therefore |ε2| = 0.15×8.
Final answer: 1.2 V
How to Approach Chapter 6 Numericals
1. Identify the changing quantity: Ask whether B, A, θ, current, or coil coupling is changing.
2. Write the governing equation before substituting: This prevents mixing up Faraday's law, motional EMF, self-induction and mutual induction.
3. Use SI units: Convert cm² to m², ms to s and mH to H before calculation.
4. Separate magnitude from direction: The magnitude gives the numerical answer; the negative sign in Faraday's or induction equations carries directional information through Lenz's law.
5. Check the limiting case: If ΔΦ = 0, induced EMF from electromagnetic induction is zero; if dI/dt = 0, self-induced EMF is zero.
2. Write the governing equation before substituting: This prevents mixing up Faraday's law, motional EMF, self-induction and mutual induction.
3. Use SI units: Convert cm² to m², ms to s and mH to H before calculation.
4. Separate magnitude from direction: The magnitude gives the numerical answer; the negative sign in Faraday's or induction equations carries directional information through Lenz's law.
5. Check the limiting case: If ΔΦ = 0, induced EMF from electromagnetic induction is zero; if dI/dt = 0, self-induced EMF is zero.
Frequently Asked Questions — Chapter 6 Numericals
What are the main numerical topics in Electromagnetic Induction Class 12?
Magnetic flux, Faraday's law, induced EMF/current, motional EMF, induced charge, self-induction, inductance, magnetic energy and mutual induction.
Which formula is used for average induced EMF?
For an N-turn coil, |εavg| = N|ΔΦ|/Δt.
Is induced charge dependent on the time taken?
For a simple closed circuit with constant resistance, q = N|ΔΦ|/R, so the time interval cancels when the same total flux change is considered.
Are AC generator and transformer included here?
No. For the current CBSE 2026–27 mapping, AC generator and transformer are listed under Chapter 7: Alternating Current.
Common Numerical Traps
Trap 1 — Area-vector angle: In Φ = BA cosθ, θ is the angle between B and the area vector, not the plane of the coil.
Trap 2 — Flux change, not field alone: A magnetic field can be present without induction. What matters is a change in linked magnetic flux.
Trap 3 — EMF vs current: Induced EMF can exist in an open circuit; induced current additionally requires a conducting closed path.
Trap 4 — Lenz’s law: The induced effect opposes the change in flux, not necessarily the original magnetic field itself.
Trap 5 — Self-induction: ε = −L(dI/dt), so a steady current in an ideal coil does not produce a continuing self-induced EMF.
Trap 6 — Induced charge: For a simple constant-resistance circuit, q = N|ΔΦ|/R is independent of how long the flux change takes.
Trap 7 — Units: Convert centimetres, milliseconds and other quantities to SI units before substitution.
Quick Self-Check
| Skill | Can you do it without looking at the solution? |
|---|---|
| Calculate magnetic flux using the correct angle | □ |
| Find average induced EMF from a flux change | □ |
| Handle a time-dependent flux function | □ |
| Calculate motional EMF and induced current | □ |
| Calculate induced charge | □ |
| Use self-inductance to find induced EMF | □ |
| Calculate energy stored in an inductor | □ |
| Use mutual inductance correctly | □ |
| Keep Chapter 6 separate from Chapter 7 AC-generator/transformer content | □ |
Continue Your Chapter 6 Preparation
Chapter 6 Notes: Electromagnetic Induction Class 12 Physics Notes
Chapter 6 Important Questions: Electromagnetic Induction Important Questions
Chapter 6 MCQs: Electromagnetic Induction MCQs
Previous chapter: Magnetism and Matter Class 12 Physics Notes
Official CBSE: Physics Curriculum 2026–27 · Class XII 2026–27 Sample Question Papers & Marking Schemes
Final Exam-Readiness Check
✓ I can identify what change in magnetic flux causes the induction.
✓ I can calculate average and instantaneous induced EMF.
✓ I can solve motional-EMF problems and convert EMF to current when resistance is given.
✓ I can calculate induced charge for a simple closed circuit.
✓ I can solve self-induction and inductor-energy numericals.
✓ I can solve mutual-induction numericals.
✓ I check signs, units, geometry and limiting behaviour before accepting an answer.
Source discipline: These are original practice problems. They are not presented as official CBSE questions or guaranteed predictions. The official CBSE 2026–27 curriculum is the authority for syllabus scope; current third-party resources were used only to identify recurring student search/practice needs.
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