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Electromagnetic Waves appeared 31 times across 3 years — 3.6% of Physics. This question is from Maxwell's Equations.

Year 2026 2025 2024 Total
Questions 10 12 9 31

Match List I with List II:
List-IList-II
A. Gauss's law of magnetostaticsI. ∮ E · d a = (1)/(ε₀) ∫ ρ dV
B. Faraday's law of electro magnetic inductionII. ∮ B · d a = 0
C. Ampere's lawIII. ∮ E · d l = - ddt∫ B · d a
D. Gauss's law of electrostaticsIV. ∮ B · d l = μ₀ I
Choose the correct answer from the options given below:

Solution & Explanation

Core Logic

Match each law with its corresponding mathematical expression (Maxwell's equations). (A) Gauss's law of magnetostatics: The net magnetic flux through any closed surface is zero. ∮ B · d a = 0 (Matches II). (B) Faraday's law of electromagnetic induction: The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux. ∮ E · d l = - ddt ∫ B · d a (Matches III). (C) Ampere's law: The line integral of the magnetic field around a closed loop is proportional to the electric current passing through the loop. ∮ B · d l = μ₀ I (Matches IV). (D) Gauss's law of electrostatics: The electric flux through any closed surface is proportional to the enclosed electric charge. ∮ E · d a = (1)/(ε₀) ∫ ρ dV (Matches I).

Step 1: Final Match

A arrow II B arrow III C arrow IV D arrow I This matches option (4).

Pattern Recognition

These are the fundamental Maxwell equations in integral form. Memorizing their direct mappings guarantees quick marks.

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electromagnetic Induction

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Q40 jee_main_2024_31_jan_morning Energy Density Of EM Waves
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ ~Hz and an amplitude of 50 ~Vm⁻¹. The total average energy density of the electromagnetic field of the wave is : [Use ε₀ = 8.85 × 10⁻¹² C² / Nm² ]
  • A. 1.106 × 10⁻⁸ Jm⁻³
  • B. 4.425 × 10⁻⁸ ~Jm⁻³
  • C. 2.212 × 10⁻⁸ ~Jm⁻³
  • D. 2.212 × 10⁻¹⁰ ~Jm⁻³

Solution

Related Formula
Utotal average = (1)/(2)ε₀ E₀²
Core Logic

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so:

Uavg = UE + UB = 2UE = 2 ( (1)/(4)ε₀ E₀² ) = (1)/(2)ε₀ E₀²

Where E₀ is the amplitude of the electric field.

Step 2: Substitution

Given: E₀ = 50 V/m ε₀ = 8.85 × 10⁻¹² C²/(N· m²)

Uavg = (1)/(2) × (8.85 × 10⁻¹²) × (50)² Uavg = (1)/(2) × 8.85 × 10⁻¹² × 2500 Uavg = 1.10625 × 10⁻⁸ J/m³
Chapter Mix

Class 12 Physics: Electromagnetic Waves

More Electromagnetic Waves Questions — jee_main_2024_30_january_evening

Practice all Electromagnetic Waves previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)