JEE Main · Physics → Steady

Electromagnetic Waves appeared 31 times across 3 years — 3.6% of Physics. This question is from Impedance of Free Space.

Year 2026 2025 2024 Total
Questions 10 12 9 31

The dimension of μ₀ε₀ is equal to that of: (μ₀= Vacuum permeability and ε₀= Vacuum permittivity) [cite: 33, 34]

Solution & Explanation

Related Formula

L = (μ₀ N² A)/(l) μ₀ ∝ L [cite: 675]

C = (ε₀ A)/(d) ε₀ ∝ C [cite: 677]

Core Logic

From the basic formulas of inductance and capacitance, we can note the proportional parameters: [cite: 675, 677]

(μ₀)/(ε₀) ∝ (L)/(C) [cite: 678]

We know that the time constant for an LR circuit is τ = (L)/(R) and for a RC circuit is τ = RC[cite: 679]. Equating these time dimensions: [cite: 679]

(L)/(R) = RC (L)/(C) = R² [cite: 679]

Taking the square root or matching parameters from the text solution layout yields the characteristic dimension of resistance[cite: 679].

Pattern Recognition

The quantity √((μ₀)/(ε₀)) represents the intrinsic impedance of free space, which has the value ≈ 377 Ω[cite: 679]. Hence, its square matches the dimension of resistance squared, which maps to Resistance in the choice sets[cite: 38, 674].

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_07_april_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%)