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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-I (Relation)List-II (Law)
A.∮ E· dl=-(d)/(dt)∮ B· daI.Ampere's circuital law.
B.∮ B· dl=μ₀(1+ε₀ dφEdt)II.Faraday's laws of electromagnetic induction.
(C)∮ E· da= 1ε₀∫ρ dvIII.Ampere-Maxwell law
(D)∮ B· dl=μ₀IIV.Gauss's law of electrostatics
Choose the correct answer from the options given below :

Solution & Explanation

Core Logic

Let's systematically identify each integral equation with its corresponding physical law:

(A) ∮ E· dl=- dφBdt corresponds to the line integral of the electric field around a closed loop being equal to the negative rate of change of magnetic flux, which is Faraday's Law of Electromagnetic Induction (II).

(B) ∮ B· dl=μ₀(I+ε₀ dφEdt) defines the Ampere-Maxwell Law (III), incorporating the displacement current.

(C) ∮ E· da= Qencε₀ relates the electric flux through a closed surface to the enclosed charge, mapping to Gauss's Law of Electrostatics (IV).

(D) ∮ B· dl=μ₀I is the original Ampere's Circuital Law (I) without Maxwell's correction.

Step 1: Final Conclusion

The correct matches are A-II, B-III, C-IV, D-I.

Pattern Recognition

Sees: Surface integral of E mapped to volume integral of charge → Gauss's Law. Circulation of E mapped to time derivative of B flux → Faraday's Law.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

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_2026_23_january_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%)