🧩 Problem Types & Solved Examples
6 Problem Categories — Solved
Every type of EMI question that appears in CBSE, NEET, JEE Main and JEE Advanced — analyzed by what the examiner is actually testing.
🎯 How to Use This Section
Don't just read the solutions — after seeing each problem, pause and try to identify: (1) what concept is being tested, (2) what formula to use, (3) what common mistake would trap a student here. This mental pattern-recognition is what separates top rankers.
Type 1 — Direct Formula Application
These test whether you can correctly identify the right formula and plug in values without errors. Most CBSE numericals and 40% of NEET MCQs fall here.
Problem 1 — Induced EMF in a Coil
Given
A coil of 200 turns has a magnetic flux of 0.05 Wb through it. If the flux drops to 0.02 Wb in 0.3 seconds, find the average induced EMF.
What Examiner Tests
Correct use of ε = −NΔΦ/Δt. Whether student uses magnitude or includes direction. Whether N is correctly multiplied.
Concept Selection
Faraday's Second Law: ε = −NΔΦ/Δt. Use magnitude for EMF value.
Solution
N = 200, ΔΦ = 0.05 − 0.02 = 0.03 Wb, Δt = 0.3 s
|ε| = N × |ΔΦ/Δt| = 200 × (0.03/0.3)
|ε| = 200 × 0.1 = 20 V
|ε| = N × |ΔΦ/Δt| = 200 × (0.03/0.3)
|ε| = 200 × 0.1 = 20 V
Shortcut Insight
Always verify units: N(turns) × Wb/s = V ✓. If Δt is very small → EMF is large. This is consistent with Lenz's Law opposing rapid changes.
Problem 2 — Motional EMF in Rail Problem
Given
A rod of length 0.5 m moves with velocity 4 m/s perpendicular to a magnetic field of 0.8 T. The circuit resistance is 0.4 Ω. Find: (a) Induced EMF, (b) Current, (c) Opposing force on rod.
What Examiner Tests
Whether student can chain ε → I → F correctly. The direction of opposing force (always opposite to motion). Concept of terminal velocity.
Solution
(a) ε = BLv = 0.8 × 0.5 × 4 = 1.6 V
(b) I = ε/R = 1.6/0.4 = 4 A
(c) F = BIL = 0.8 × 4 × 0.5 = 1.6 N (opposing velocity)
Alternatively: F = B²L²v/R = (0.64 × 0.25 × 4)/0.4 = 0.64/0.4 = 1.6 N ✓
(b) I = ε/R = 1.6/0.4 = 4 A
(c) F = BIL = 0.8 × 4 × 0.5 = 1.6 N (opposing velocity)
Alternatively: F = B²L²v/R = (0.64 × 0.25 × 4)/0.4 = 0.64/0.4 = 1.6 N ✓
Shortcut
Power = Fv = 1.6 × 4 = 6.4 W. Cross-check: P = ε²/R = 2.56/0.4 = 6.4 W ✓ Energy conservation verified.
Problem 3 — Inductance & Back EMF
Given
An inductor of L = 50 mH has a current that increases uniformly from 0 to 5 A in 0.1 s. Find the induced back-EMF.
Solution
L = 50 × 10⁻³ H, dI/dt = (5−0)/0.1 = 50 A/s
ε = L × dI/dt = 50 × 10⁻³ × 50 = 2.5 V
This is the back-EMF that opposes the increase in current.
ε = L × dI/dt = 50 × 10⁻³ × 50 = 2.5 V
This is the back-EMF that opposes the increase in current.
Key Point
Back EMF always opposes the change. When current is increasing, back EMF opposes the source EMF. This is why real inductors need time to build up current.
Type 2 — Conceptual / Direction-Based
These test your understanding, not calculations. This is where 60% of NEET EMI questions live. Most students answer wrong because they skip the reasoning step.
Problem 4 — Direction of Induced Current
Given
A rectangular coil lies in the plane of the page. A magnetic field is directed into the page (shown as ×). The coil is being pulled to the right, partially out of the field. Find the direction of induced current in the coil.
What Examiner Tests
Application of Lenz's Law — not the formula. Whether student can reason about flux change direction. Whether they use the correct hand rule for current direction.
Solution — Step by Step
Step 1: B is into the page (×). Flux = B × (area inside field) — pointing into page.
Step 2: As coil moves right (out of field), effective area inside B decreases → Flux decreases.
Step 3: Lenz's Law: induced current must oppose the decrease → induced current creates B into the page.
Step 4: For B into page inside loop: using right-hand rule, current must flow clockwise when viewed from front.
Answer: Clockwise
Step 2: As coil moves right (out of field), effective area inside B decreases → Flux decreases.
Step 3: Lenz's Law: induced current must oppose the decrease → induced current creates B into the page.
Step 4: For B into page inside loop: using right-hand rule, current must flow clockwise when viewed from front.
Answer: Clockwise
Shortcut
Direction of induced current: always try to maintain the original field direction. B was INTO page, it's decreasing, so induced current creates B into page → clockwise.
Problem 5 — Static Magnet Inside Coil
Given
A bar magnet is placed at rest inside a closed conducting ring. What is the induced EMF?
❌ Trap Answer
Most students say "some EMF is induced because of the magnetic field." WRONG. This is the #1 Faraday's Law conceptual trap.
EMF = −NdΦ/dt. The flux through the ring is constant (B doesn't change, area doesn't change, angle doesn't change). Therefore dΦ/dt = 0.
Induced EMF = 0
The magnetic field exists, but since it's not CHANGING, no EMF is induced. Faraday's Law requires RATE OF CHANGE of flux, not just flux.
Induced EMF = 0
The magnetic field exists, but since it's not CHANGING, no EMF is induced. Faraday's Law requires RATE OF CHANGE of flux, not just flux.
Problem 6 — Two Coils (Mutual Induction)
Given
Two coils A and B are placed coaxially. When current in coil A changes from 2A to 8A in 0.05 s, the EMF induced in B is 0.12 V. What is the mutual inductance of the system?
Using: ε₂ = M × |dI₁/dt|
|dI₁/dt| = (8−2)/0.05 = 120 A/s
M = ε₂ / |dI₁/dt| = 0.12 / 120
M = 1 × 10⁻³ H = 1 mH
|dI₁/dt| = (8−2)/0.05 = 120 A/s
M = ε₂ / |dI₁/dt| = 0.12 / 120
M = 1 × 10⁻³ H = 1 mH
Type 3 — Multi-Step Problems
These require chaining multiple formulas. A mistake in step 1 propagates. JEE Main is full of these. Build systematic habits.
Problem 7 — Rod on Rails with Resistance
Given
Two parallel rails separated by L = 1 m are connected by a resistance R₁ = 2 Ω at one end. A conducting rod of resistance R₂ = 1 Ω slides on the rails with v = 3 m/s in B = 0.5 T (perpendicular to rail plane). Find: (a) EMF, (b) current, (c) force on rod, (d) power dissipated.
What Examiner Tests
Whether student includes rod's own resistance in total resistance calculation. (This is the most common error in rail problems.)
(a) ε = BLv = 0.5 × 1 × 3 = 1.5 V
(b) Total R = R₁ + R₂ = 2 + 1 = 3 Ω (rod's resistance in series with external)
I = ε / R_total = 1.5 / 3 = 0.5 A
(c) F = BIL = 0.5 × 0.5 × 1 = 0.25 N (opposing motion)
(d) P_total = εI = 1.5 × 0.5 = 0.75 W
P_R₁ = I²R₁ = 0.25 × 2 = 0.5 W
P_R₂ = I²R₂ = 0.25 × 1 = 0.25 W
Check: 0.5 + 0.25 = 0.75 W ✓
(b) Total R = R₁ + R₂ = 2 + 1 = 3 Ω (rod's resistance in series with external)
I = ε / R_total = 1.5 / 3 = 0.5 A
(c) F = BIL = 0.5 × 0.5 × 1 = 0.25 N (opposing motion)
(d) P_total = εI = 1.5 × 0.5 = 0.75 W
P_R₁ = I²R₁ = 0.25 × 2 = 0.5 W
P_R₂ = I²R₂ = 0.25 × 1 = 0.25 W
Check: 0.5 + 0.25 = 0.75 W ✓
Critical Mistake Alert
If you use only R₁ = 2 Ω as total resistance, you get I = 0.75 A — wrong answer. The rod itself has resistance R₂. Always check: "Does the rod have resistance?" Most JEE problems include it.
Problem 8 — Solenoid + Energy
Given
A solenoid of 1000 turns, length 0.5 m, cross-section area 5 cm² carries a current of 2 A. Find: (a) self-inductance, (b) energy stored, (c) magnetic energy density.
A = 5 × 10⁻⁴ m², l = 0.5 m, N = 1000, I = 2A
(a) L = μ₀N²A/l = (4π × 10⁻⁷ × 10⁶ × 5 × 10⁻⁴) / 0.5
L = (4π × 10⁻⁷ × 10⁶ × 10⁻³) = 4π × 10⁻⁴ ≈ 1.26 mH
(b) U = ½LI² = ½ × 1.26 × 10⁻³ × 4 = 2.51 × 10⁻³ J ≈ 2.51 mJ
(c) B = μ₀nI = μ₀(N/l)I = 4π × 10⁻⁷ × 2000 × 2 = 4π × 10⁻³ ≈ 1.26 × 10⁻² T
u_B = B²/2μ₀ = (1.26 × 10⁻²)² / (8π × 10⁻⁷) ≈ 63 J/m³
(a) L = μ₀N²A/l = (4π × 10⁻⁷ × 10⁶ × 5 × 10⁻⁴) / 0.5
L = (4π × 10⁻⁷ × 10⁶ × 10⁻³) = 4π × 10⁻⁴ ≈ 1.26 mH
(b) U = ½LI² = ½ × 1.26 × 10⁻³ × 4 = 2.51 × 10⁻³ J ≈ 2.51 mJ
(c) B = μ₀nI = μ₀(N/l)I = 4π × 10⁻⁷ × 2000 × 2 = 4π × 10⁻³ ≈ 1.26 × 10⁻² T
u_B = B²/2μ₀ = (1.26 × 10⁻²)² / (8π × 10⁻⁷) ≈ 63 J/m³
Problem 9 — Terminal Velocity in Rail System
Given
A conducting rod of mass m = 0.1 kg on rails (L = 0.5 m, R = 1 Ω, B = 2T) is pushed horizontally. An external force F = 1 N is applied. Neglect friction. Find terminal velocity.
At terminal velocity: Net force = 0
Applied force = Opposing magnetic force
F = B²L²v_T/R
v_T = FR/(B²L²) = (1 × 1)/(4 × 0.25) = 1/1 = 1 m/s
Before terminal velocity: ma = F − B²L²v/R → rod accelerates toward v_T asymptotically.
Applied force = Opposing magnetic force
F = B²L²v_T/R
v_T = FR/(B²L²) = (1 × 1)/(4 × 0.25) = 1/1 = 1 m/s
Before terminal velocity: ma = F − B²L²v/R → rod accelerates toward v_T asymptotically.
JEE Insight
At terminal velocity: v_T = FR/B²L². This is direct and important. The mass m doesn't appear in the terminal velocity formula — only the force and electromagnetic parameters matter. Mass affects only HOW FAST you reach terminal velocity.
Type 4 — Graph-Based Problems
Flux-time, EMF-time, and current-time graphs. This type appears every year in JEE Main and increasingly in NEET.
Problem 10 — Φ vs t Graph → Find EMF
Given
A coil (N=100) has flux varying as: Φ = 3t² − 5t + 2 (in Weber, t in seconds). Find the induced EMF at t = 1 s. Is the EMF constant or varying?
What Examiner Tests
Application of calculus: ε = −NdΦ/dt. Whether student can differentiate and evaluate at specific t. Whether they know if EMF is constant (linear Φ) or varying (non-linear Φ).
Φ = 3t² − 5t + 2
dΦ/dt = 6t − 5
At t = 1: dΦ/dt = 6(1) − 5 = 1 Wb/s
ε = −N × dΦ/dt = −100 × 1 = −100 V (magnitude = 100 V)
EMF is varying because Φ is a quadratic function of t (non-linear).
dΦ/dt = 6t − 5
At t = 1: dΦ/dt = 6(1) − 5 = 1 Wb/s
ε = −N × dΦ/dt = −100 × 1 = −100 V (magnitude = 100 V)
EMF is varying because Φ is a quadratic function of t (non-linear).
Graph Insight
Rule: If Φ-t graph is LINEAR → EMF is CONSTANT. If Φ-t graph is CURVED (quadratic, sinusoidal) → EMF varies with time. The slope of the Φ-t graph at any point = dΦ/dt = EMF (up to factor N).
🔬 Graph Reading — Master These Patterns
1. Φ vs t linear (constant slope): ε = constant → I = constant → Force on rod = constant
2. Φ vs t parabolic: ε = linear in t → I = linear in t
3. Φ vs t sinusoidal: ε = cosinusoidal (phase shifted by 90°) — AC generator behavior
4. EMF vs t rectangular: Flux changes in steps (loop entering/leaving field abruptly)
2. Φ vs t parabolic: ε = linear in t → I = linear in t
3. Φ vs t sinusoidal: ε = cosinusoidal (phase shifted by 90°) — AC generator behavior
4. EMF vs t rectangular: Flux changes in steps (loop entering/leaving field abruptly)
Problem 11 — AC Generator EMF Graph
Given
A coil of 100 turns, area 0.01 m² rotates in B = 0.5 T with ω = 100 rad/s. Write the equation for instantaneous EMF. What is its peak value? At what angle is EMF maximum?
ε₀ = NBAω = 100 × 0.5 × 0.01 × 100 = 50 V
ε = ε₀ sin(ωt) = 50 sin(100t) V
EMF is maximum when sin(ωt) = 1 → ωt = π/2 → plane of coil parallel to B (θ = 90° between plane and B, or θ = 0° between normal and B initially, so normal ⊥ B → plane || B)
EMF is MAXIMUM when plane of coil is PARALLEL to B
EMF is ZERO when plane of coil is PERPENDICULAR to B
ε = ε₀ sin(ωt) = 50 sin(100t) V
EMF is maximum when sin(ωt) = 1 → ωt = π/2 → plane of coil parallel to B (θ = 90° between plane and B, or θ = 0° between normal and B initially, so normal ⊥ B → plane || B)
EMF is MAXIMUM when plane of coil is PARALLEL to B
EMF is ZERO when plane of coil is PERPENDICULAR to B
Critical Point
This is the most-tested confusion point: Flux is maximum when plane ⊥ B (EMF = 0). EMF is maximum when plane || B (Flux = 0). They are exactly 90° out of phase. Never mix these up.
Type 5 — Assertion & Reason
Both NEET and CBSE include A&R questions. The trap: both A and R can be individually true, but R may not explain A.
🎯 4-Step A&R Strategy
Step 1: Is Assertion true or false? (independently)
Step 2: Is Reason true or false? (independently)
Step 3: IF both true → Does R correctly explain A? (conceptually sound?)
Step 4: Select the matching option (A-E type)
Most mistakes: Students assume if both are true, R explains A. Test the causal link explicitly.
Step 2: Is Reason true or false? (independently)
Step 3: IF both true → Does R correctly explain A? (conceptually sound?)
Step 4: Select the matching option (A-E type)
Most mistakes: Students assume if both are true, R explains A. Test the causal link explicitly.
A&R Problem 1
Assertion (A)
The induced current in a loop always flows in a direction to oppose the change in magnetic flux that caused it.
Reason (R)
Lenz's Law is a consequence of the law of conservation of energy.
A is TRUE — correct statement of Lenz's Law.
R is TRUE — Lenz's Law is indeed derived from energy conservation.
R correctly explains A — if induced current aided the flux change, energy would be created from nothing.
Answer: Both A and R are true, and R is the correct explanation of A.
R is TRUE — Lenz's Law is indeed derived from energy conservation.
R correctly explains A — if induced current aided the flux change, energy would be created from nothing.
Answer: Both A and R are true, and R is the correct explanation of A.
A&R Problem 2
Assertion (A)
A bar magnet held stationary near a coil does not produce any current in the coil.
Reason (R)
Electromagnetic induction requires a changing magnetic flux.
A is TRUE — static magnet → constant flux → no EMF → no current.
R is TRUE — EMI requires dΦ/dt ≠ 0.
R correctly explains A.
Answer: Both A and R are true, and R is the correct explanation of A.
R is TRUE — EMI requires dΦ/dt ≠ 0.
R correctly explains A.
Answer: Both A and R are true, and R is the correct explanation of A.
A&R Problem 3 — Tricky
Assertion (A)
The self-inductance of a solenoid increases when a ferromagnetic core is inserted.
Reason (R)
Insertion of a ferromagnetic core increases the magnetic permeability of the medium.
A is TRUE — L = μ₀μᵣN²A/l. With ferromagnetic core, μᵣ >> 1 → L increases significantly.
R is TRUE — Ferromagnetic materials have very high μᵣ (up to 10⁵ for some materials).
R correctly explains A — A follows directly from R via the formula L = μ₀μᵣN²A/l.
Answer: Both A and R are true, and R is the correct explanation of A.
R is TRUE — Ferromagnetic materials have very high μᵣ (up to 10⁵ for some materials).
R correctly explains A — A follows directly from R via the formula L = μ₀μᵣN²A/l.
Answer: Both A and R are true, and R is the correct explanation of A.
Type 6 — Case-Based / Paragraph Questions
CBSE Board 2020+ introduced case-based questions. NEET 2022+ followed. Multiple sub-questions from one setup. Read the setup carefully — all answers come from the same given data.
Case Study — Rail System Analysis
Setup
Two parallel conducting rails are separated by distance L = 2 m. A resistance R = 4 Ω connects one end. A conducting rod of negligible resistance slides on the rails in a uniform magnetic field B = 0.5 T (perpendicular to the plane of the rails) with constant velocity v = 10 m/s. The rod has been pushed by an external force so it moves at constant velocity.
Q1: What is the induced EMF?
ε = BLv = 0.5 × 2 × 10 = 10 V
Q2: What is the current through the resistance?
I = ε/R = 10/4 = 2.5 A
Q3: What is the magnitude and direction of force on the rod?
F = BIL = 0.5 × 2.5 × 2 = 2.5 N, directed opposite to the rod's velocity (by Lenz's Law)
Q4: What power must the external agent supply to maintain constant velocity?
Since velocity is constant, external force = opposing force = 2.5 N
P = Fv = 2.5 × 10 = 25 W
Verify: P = ε²/R = 100/4 = 25 W ✓ (Energy supplied = energy dissipated in R)
P = Fv = 2.5 × 10 = 25 W
Verify: P = ε²/R = 100/4 = 25 W ✓ (Energy supplied = energy dissipated in R)