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

Year 2026 2025 2024 Total
Questions 10 12 9 31

If μ₀ and ε₀ are the permeability and permittivity of free space, respectively, then the dimension of ((1)/(μ₀ε₀)) is:

Solution & Explanation

Related Formula

Maxwell's relation for the speed of light (c) in vacuum:

c = 1√(μ₀ ε₀)
Core Logic

Square both sides of the speed of light formula:

c² = (1)/(μ₀ ε₀)

Since c represents the speed of light (velocity), its dimensional formula is:

[c] = [L T⁻¹]

Therefore, the dimensions of c² are:

[c²] = [L T⁻¹]² = [L² T⁻²]
Step 1: Match options

Express the dimensions in terms of the given variable ratios:

[c²] = (L²)/(T²)

This perfectly matches Option (2).

Pattern Recognition

Sees: Dimensions of vacuum permittivity-permeability inverse product. Trap: Trying to find the separate dimensions of both μ₀ and ε₀, then conducting manual division. This is extremely slow and prone to algebraic error. Shortcut: Directly identify the term as c². The velocity squared has units of m²s⁻², giving dimensions of L²T⁻² immediately.

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 11 Physics: Units and Measurements

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_02_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%)