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Electromagnetic Waves appeared 31 times across 3 years — 3.6% of Physics. This question is from Energy Density of EM Waves.

Year 2026 2025 2024 Total
Questions 10 12 9 31

Due to presence of an em-wave whose electric component is given by E = 100 (ω t - kx)NC⁻¹ , a cylinder of length 200~cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds the same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

Solution & Explanation

Related Formula
Energy Density = (1)/(2) ε₀ E² Total Energy = Energy Density × Volume
Core Logic

Since both cylinders hold equal amounts of electromagnetic energy:

(Energy)₁ = (Energy)₂ (1)/(2) ε₀ E₁² · c π R₁² × L₁ = (1)/(2) ε₀ E₂² · c π R₂² × L₂

Since the lengths are identical (L₁ = L₂), this simplifies to:

E₁² R₁² = E₂² R₂² E₁ R₁ = E₂ R₂

Given the second cylinder has half the diameter (and radius) of the first (R₂ = R₁2):

100 × R₁ = E₂ × R₁2 E₂ = 200 N/C
Step 1: Final Equation Match

The wave equation adjusts its amplitude factor to 200 (ω t - kx)NC⁻¹, which matches option (2).

Pattern Recognition

When energy is constant and volume scales down inversely by a factor of 4 (due to R²), the electric field strength must increase by a factor of √(4) = 2 to maintain balance.

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_2025_28_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%)