Last Updated: May 2026
Magnetism and Electromagnetic Induction (EMI) are clubbed in JEE Main 2027 because they share the same fundamental field equation — Maxwell’s set. Together they contribute 3-5 questions per JEE Main shift, drawn from NCERT Class 12 Chapters 4, 5 and 6. The chapter rewards strong vector intuition (Biot-Savart, Lorentz force) and strong sign discipline (Lenz’s law, Faraday’s induced EMF). This guide condenses everything you need into one reference plus 35 JEE-pattern problems.
Chapter Snapshot
| Parameter | Detail |
|---|---|
| NCERT Chapters | Class 12: Ch 4 Moving Charges, Ch 5 Magnetism & Matter, Ch 6 EMI |
| JEE Main Questions / shift | 3-5 |
| JEE Advanced Difficulty | Calculus + vector heavy |
| Sub-topics typically tested | Solenoids, cyclotron, EMF in rotating rods, mutual inductance |
1. Magnetic Force — Lorentz Equation
F = q(E + v × B). Key consequences:
- Magnetic force is always perpendicular to velocity ⇒ does no work.
- Charged particle in uniform B (with v ⊥ B) moves in a circle: r = mv / qB.
- Time period: T = 2πm / qB. Independent of v and r — basis of cyclotron.
2. Biot-Savart Law (current-element field)
dB = (μ₀ / 4π) × (I dl × r̂) / r². Standard derived results:
- Long straight wire: B = μ₀I / (2πr).
- Circular loop axis: B = μ₀IR² / [2(R² + x²)^(3/2)]. At centre: B = μ₀I/(2R).
- Circular arc subtending angle θ: B = (μ₀I × θ) / (4πR).
3. Ampere’s Circuital Law
∮ B · dl = μ₀ I_enclosed. Useful for symmetric configurations.
- Solenoid: B = μ₀nI inside, ≈ 0 outside (for ideal long solenoid).
- Toroid: B = μ₀NI / (2πr) inside.
- Coaxial cable: field zero outside the outer conductor.
4. Force on a Current-Carrying Wire
F = I L × B (vector). For two parallel wires distance d apart carrying I₁ and I₂:
Force per unit length = μ₀I₁I₂ / (2πd). Attractive for parallel currents, repulsive for antiparallel.
5. Magnetic Properties of Matter
| Material | χ_m | μᵣ | Examples |
|---|---|---|---|
| Diamagnetic | Small −ve | < 1 | Cu, Au, Bi, water |
| Paramagnetic | Small +ve | ≥ 1 | Al, Pt, O₂ |
| Ferromagnetic | Large +ve | ≫ 1 | Fe, Ni, Co |
Curie’s Law: χ_m ∝ 1/T (paramagnetic). Above Curie temperature T_C, ferromagnetic → paramagnetic.
6. Faraday’s Law of EMI
EMF = −dΦ_B / dt. Lenz’s law gives the sign: induced current opposes the change in flux. Common configurations:
- Rod moving in B: EMF = Bvl.
- Rotating rod about one end: EMF = ½ Bωl².
- Coil rotating in B: EMF = NBAω sin(ωt) — basis of AC generator.
7. Self and Mutual Inductance
- Self-inductance: L = NΦ/I. EMF = −L dI/dt.
- Solenoid: L = μ₀n²Al = μ₀N²A/l.
- Mutual inductance M for two coils: M = k√(L₁L₂), 0 ≤ k ≤ 1.
- Energy stored: U = ½ LI².
8. AC Circuits — Essential Results
| Element | Phase between V and I |
|---|---|
| Resistor R | In phase |
| Inductor L | V leads I by 90° |
| Capacitor C | V lags I by 90° |
| Series LCR at resonance | X_L = X_C; circuit purely resistive |
Resonance frequency: ω₀ = 1/√(LC). Quality factor: Q = ω₀L/R.
35 Practice MCQs — Sample of 10
Q1. A proton enters a region of B = 0.1 T at 90°. Speed = 10⁷ m/s. Radius = (a) 1.04 m (b) 1.04 cm (c) 0.104 m (d) 10.4 m. Ans: (a) r = mv/qB = (1.67×10⁻²⁷ × 10⁷)/(1.6×10⁻¹⁹ × 0.1) ≈ 1.04 m.
Q2. Magnetic field at centre of a circular loop of radius 5 cm carrying 2 A is — (a) 4π × 10⁻⁶ T (b) 2π × 10⁻⁶ T (c) 8π × 10⁻⁶ T (d) 4π × 10⁻⁵ T. Ans: (c) B = μ₀I/(2R) = (4π×10⁻⁷ × 2)/(2×0.05) = 8π×10⁻⁶ T.
Q3. EMF induced in a rod of length 1 m moving at 5 m/s perpendicular to B = 0.5 T — (a) 0.5 V (b) 1.5 V (c) 2.5 V (d) 5 V. Ans: (c) ε = Bvl = 0.5×5×1 = 2.5 V.
Q4. A coil of 100 turns, area 10 cm² rotates at 50 rev/s in B = 0.5 T. Maximum EMF = (a) ~15.7 V (b) ~31.4 V (c) ~3.14 V (d) ~157 V. Ans: (a) NBAω = 100 × 0.5 × 10⁻³ × 100π ≈ 15.7 V.
Q5. Self-inductance of a solenoid of 1000 turns, length 50 cm, area 5 cm² — (a) 6.28 mH (b) 12.56 mH (c) 0.628 mH (d) 1.26 mH. Ans: (b) L = μ₀N²A/l = (4π×10⁻⁷ × 10⁶ × 5×10⁻⁴)/0.5 ≈ 12.56 mH.
Q6. Two parallel wires 10 cm apart carry currents 5 A and 10 A. Force per metre = (a) 10⁻⁵ N (b) 10⁻⁴ N (c) 10⁻⁶ N (d) 10⁻³ N. Ans: (b) μ₀I₁I₂/(2πd) = (4π×10⁻⁷ × 50)/(2π×0.1) = 10⁻⁴ N/m.
Q7. An LCR circuit has L = 0.1 H, C = 25 μF. Resonance frequency = (a) ~318 Hz (b) ~100 Hz (c) ~1000 Hz (d) ~50 Hz. Ans: (b) ω₀ = 1/√(LC) = 1/√(2.5×10⁻⁶) = 632 rad/s ⇒ f = 100 Hz.
Q8. A capacitor of 5 μF connected to AC of 50 Hz has reactance — (a) 637 Ω (b) 318 Ω (c) 1273 Ω (d) 159 Ω. Ans: (a) X_C = 1/(2πfC) ≈ 637 Ω.
Q9. Energy stored in inductor L = 0.5 H carrying I = 4 A — (a) 4 J (b) 8 J (c) 2 J (d) 1 J. Ans: (a) ½ LI² = ½ × 0.5 × 16 = 4 J.
Q10. A magnetic moment of bar magnet is 0.04 Am². Field at axial distance 0.2 m = (a) 1 × 10⁻⁶ T (b) 2 × 10⁻⁶ T (c) 4 × 10⁻⁶ T (d) 1 × 10⁻⁵ T. Ans: (a) B_axial = (μ₀/4π) × 2M/r³ = 10⁻⁷ × 2×0.04/0.008 = 10⁻⁶ T.
Common Pitfalls
- Forgetting the sign in Faraday’s law (Lenz’s contribution).
- Using R instead of l in solenoid inductance.
- Mixing X_L and X_C — remember X_L = ωL, X_C = 1/(ωC).
- Treating moving charge in pure B as gaining KE — magnetic force does no work.
Frequently Asked Questions
How many questions from EMI in JEE Main?
3-5 questions across magnetism, EMI and AC circuits per shift. EMI alone usually contributes 1-2 directly.
Is Maxwell’s set asked in JEE Main?
Maxwell’s equations as such are not derived in JEE Main. Their applied form (displacement current, EM waves) appears in the EM Waves chapter.
Best book for this chapter?
NCERT Class 12 plus HC Verma Volume 2 (Chapters on Magnetic Effect, Magnetism, EMI, AC). DC Pandey is a strong supplementary.
Can I skip the cyclotron section?
No. The cyclotron condition (qB = mω) and frequency formula appear in JEE Main almost yearly as direct one-mark questions.
How is mutual inductance tested?
Most often via two coaxial solenoids or two coplanar coils — compute M from M = k√(L₁L₂) or by integrating flux.