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

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If in₀ denotes the permittivity of free space and ΦE is the flux of the electric field through the area bounded by the closed surface, then dimension of (in₀ dφEdt) are that of:

Solution & Explanation

Related Formula

According to the Maxwell-Ampere law, displacement current Id is defined as:

Id = ε₀ dΦEdt
Core Logic

Since Id represents a physical current, its dimensions must match those of standard conduction electric current ([A] or [I]).

Step 1: Dimensional Analysis

Let us verify using fundamental dimensions:

  • Permittivity of free space [ε₀] = [M⁻¹ L⁻³ T⁴ A²]
  • Electric flux [ΦE] = [M L³ T⁻³ A⁻¹]
  • Time [t] = [T]
[ ε₀ dΦEdt ] = [M⁻¹ L⁻³ T⁴ A²] × [M L³ T⁻³ A⁻¹][T] = [A]

This confirms the quantity is dimensionally equivalent to Electric current.

Pattern Recognition

Sees: ε₀ dΦEdt. Shortcut: Instantly identify this as Maxwell's expression for displacement current. Displacement current has the exact same dimensions as regular conduction current.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Reference Study Guides

Electromagnetic Waves

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_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%)