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

Year 2026 2025 2024 Total
Questions 10 12 9 31

Match List I with List II:
List IList II
A. ∮ B · d l = μ₀ ic + μ₀ ε₀ (dφE)/(dt)I. Gauss' law for electricity
B. ∮ E · d l = -(dφB)/(dt)II. Gauss' law for magnetism
C. ∮ E · d A = (Q)/(ε₀)III. Faraday law
D. ∮ B · d A = 0IV. Ampere - Maxwell law
Choose the correct answer from the options given below:

Solution & Explanation

Core Logic

Let's review the fundamental Maxwell's equations:

  • Ampere - Maxwell Law relates the magnetic path integral to conduction current and displacement current:
∮ B · d l = μ₀ ic + μ₀ ε₀ (dφE)/(dt) A - IV
  • Faraday's Law of Induction states that changing magnetic flux induces an electromotive force (EMF):
∮ E · d l = -(dφB)/(dt) B - III
  • Gauss's Law for Electricity relates net electric flux to enclosed charge:
∮ E · d A = (Q)/(ε₀) C - I
  • Gauss's Law for Magnetism states that magnetic monopoles do not exist:
∮ B · d A = 0 D - II
Step 1: Match Evaluation

The match configurations are:

  • A arrow IV
  • B arrow III
  • C arrow I
  • D arrow II
  • This perfectly corresponds to Option (3).

Pattern Recognition

Understand the integral geometries: Path integrals (line integrals ∮ · d l) correspond to circulating fields (induction laws like Ampere/Faraday). Surface integrals (flux integrals ∮ · d A) correspond to bounded charge states (Gauss laws).

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatics Class 12 Physics: Magnetism and Matter

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_29_jan_morning

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%)