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

Year 2026 2025 2024 Total
Questions 10 12 9 31

The equation of the electric field of an electromagnetic wave propagating through free space is given by : E = √(377) (6.27 × 10³ t - 2.09 × 10⁻⁵ x) N/C The average power of the electromagnetic wave is ((1)/(α))W/m². The value of α is ____ (Take μ₀ε₀ = 377 in SI units)

Numerical Answer Type:
Enter a numerical value Answer: 2 to 2 +4 marks

Solution & Explanation

Related Formula
I = (1)/(2)ε₀ E₀² c c = 1 μ₀ε₀
Core Logic

The average power per unit area (Intensity) of an electromagnetic wave in vacuum is I = (1)/(2)ε₀ E₀² c. We can reorganize the formula using the intrinsic impedance of free space Z₀ = μ₀ε₀ = 377Ω.

Step 1: Rewrite Intensity Formula
I = (1)/(2) ε₀ E₀² 1 μ₀ε₀ I = (1)/(2) E₀² ε₀μ₀

We are given the intrinsic impedance μ₀ε₀ = 377 ε₀μ₀ = (1)/(377).

Step 2: Extract Amplitude and Calculate

From the wave equation, E₀ = √(377) N/C. Substitute into the intensity formula:

I = (1)/(2) (√(377))² × (1)/(377) I = (1)/(2) × 377 × (1)/(377) = (1)/(2) W/m²
Step 3: Compare with given format
I = (1)/(α) α = 2
Pattern Recognition

Sees: "average power... W/m^2" → the problem is asking for Intensity. When given intrinsic impedance (377Ω), use the form I = Erms² / Z₀ = E₀² / (2Z₀) directly.

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

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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%)