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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 6

Q40 jee_main_2024_29_jan_morning Maxwell Equations
Match List I with List II:
List IList II
A. ∮ B · d l = μ₀ ic + μ₀ ε₀ (dφE)/(dt)I. Gauss' law for electricity
B. ∮ E · d l = -(dφB)/(dt)II. Gauss' law for magnetism
C. ∮ E · d A = (Q)/(ε₀)III. Faraday law
D. ∮ B · d A = 0IV. Ampere - Maxwell law
Choose the correct answer from the options given below:
  • A. A-IV, B-I, C-III, D-II
  • B. A-II, B-III, C-I, D-IV
  • C. A-IV, B-III, C-I, D-II
  • D. A-I, B-II, C-III, D-IV

Solution

Core Logic

Let's review the fundamental Maxwell's equations:

  • Ampere - Maxwell Law relates the magnetic path integral to conduction current and displacement current:
∮ B · d l = μ₀ ic + μ₀ ε₀ (dφE)/(dt) A - IV
  • Faraday's Law of Induction states that changing magnetic flux induces an electromotive force (EMF):
∮ E · d l = -(dφB)/(dt) B - III
  • Gauss's Law for Electricity relates net electric flux to enclosed charge:
∮ E · d A = (Q)/(ε₀) C - I
  • Gauss's Law for Magnetism states that magnetic monopoles do not exist:
∮ B · d A = 0 D - II
Step 1: Match Evaluation

The match configurations are:

  • A arrow IV
  • B arrow III
  • C arrow I
  • D arrow II
  • This perfectly corresponds to Option (3).

Pattern Recognition

Understand the integral geometries: Path integrals (line integrals ∮ · d l) correspond to circulating fields (induction laws like Ampere/Faraday). Surface integrals (flux integrals ∮ · d A) correspond to bounded charge states (Gauss laws).

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatics Class 12 Physics: Magnetism and Matter

Q40 jee_main_2024_30_january_evening Momentum of EM Waves
If the total energy transferred to a surface in time t is 6.48 × 10⁵ ~J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:
  • A. 2.46 × 10⁻³ ~kg ~m / s
  • B. 2.16 × 10⁻³ ~kg ~m / s
  • C. 1.58 × 10⁻³ ~kg ~m / s
  • D. 4.32 × 10⁻³ ~kg ~m / s

Solution

Related Formula
p = (U)/(c)
Core Logic

For an electromagnetic wave incident on a surface that is completely absorbed, the total momentum transferred is equal to the total energy transferred divided by the speed of light in vacuum (c).

Step 1: Calculate Momentum

Given total energy E = 6.48 × 10⁵ ~J. Speed of light c = 3 × 10⁸ ~m/s.

p = (E)/(c) = (6.48 × 10⁵)/(3 × 10⁸) p = 2.16 × 10⁻³ ~kg~m/s
Pattern Recognition

Always check for "complete absorption" versus "perfect reflection". For complete absorption, p = E/c. For perfect reflection, p = 2E/c.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q49 jee_main_2024_30_january_evening Maxwell's Equations
Match List I with List II:
List-IList-II
A. Gauss's law of magnetostaticsI. ∮ E · d a = (1)/(ε₀) ∫ ρ dV
B. Faraday's law of electro magnetic inductionII. ∮ B · d a = 0
C. Ampere's lawIII. ∮ E · d l = - ddt∫ B · d a
D. Gauss's law of electrostaticsIV. ∮ B · d l = μ₀ I
Choose the correct answer from the options given below:
  • A. A-I, B-III, C-IV, D-II
  • B. A-III, B-IV, C-I, D-II
  • C. A-IV, B-II, C-III, D-I
  • D. A-II, B-III, C-IV, D-I

Solution

Core Logic

Match each law with its corresponding mathematical expression (Maxwell's equations). (A) Gauss's law of magnetostatics: The net magnetic flux through any closed surface is zero. ∮ B · d a = 0 (Matches II). (B) Faraday's law of electromagnetic induction: The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux. ∮ E · d l = - ddt ∫ B · d a (Matches III). (C) Ampere's law: The line integral of the magnetic field around a closed loop is proportional to the electric current passing through the loop. ∮ B · d l = μ₀ I (Matches IV). (D) Gauss's law of electrostatics: The electric flux through any closed surface is proportional to the enclosed electric charge. ∮ E · d a = (1)/(ε₀) ∫ ρ dV (Matches I).

Step 1: Final Match

A arrow II B arrow III C arrow IV D arrow I This matches option (4).

Pattern Recognition

These are the fundamental Maxwell equations in integral form. Memorizing their direct mappings guarantees quick marks.

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electromagnetic Induction

Q38 jee_main_2024_30_jan_morning Properties of EM Waves
The electric field of an electromagnetic wave in free space is represented as E = E₀ (ω t - kz) i The corresponding magnetic induction vector will be:
  • A. B = E₀C (ω t - kz) j
  • B. B = (E₀)/(C) (ω t - kz) j
  • C. B = E₀C (ω t + kz) j
  • D. B = (E₀)/(C) (ω t + kz) j

Solution

Related Formula
B₀ = (E₀)/(C) C = E × B
Core Logic

In an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by E₀ = c B₀. Thus, B₀ = E₀ / C.

The direction of wave propagation is given by the cross product of the electric field and magnetic field vectors: C = E × B.

Step 1: Determine Wave Direction and Magnetic Field Direction

Given the phase term (ω t - kz), the wave propagates in the +z direction, so C = k. The electric field oscillates in the +x direction, so E = i. We know:

k = i × B

Since i × j = k, the magnetic field must oscillate in the +y direction (j).

Step 2: Construct Final Vector

The full magnetic field vector shares the same phase and applies the above amplitude and direction:

B = (E₀)/(C) (ω t - kz) j
Pattern Recognition

Phase remains identical. Amplitude scales by 1/c. Direction satisfies the right-hand triad ( E, B, v) where v = E × B.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q40 jee_main_2024_31_jan_evening Properties of EM Waves
Given below are two statements: Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields. Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. Statement I is incorrect but Statement II is correct
  • B. Both Statement I and Statement II are correct
  • C. Both Statement I and Statement II are incorrect
  • D. Statement I is correct but Statement II is incorrect

Solution

Related Formula
ρE = (1)/(2)ε₀ Erms² ρB = Brms²2μ₀

pᵣ = (I)/(c) (radiation pressure for completely absorbing surface)

Core Logic

Statement I: In an EM wave, energy is stored in both the electric and magnetic fields. The energy densities are equal: UE = UB because E = cB and c² = (1)/(μ₀ ε₀). This is a correct fact.

Statement II: Electromagnetic waves carry momentum (p = U/c). When they strike a surface, they transfer this momentum, creating radiation pressure. This is also a correct fact.

Step 1: Final Evaluation

Since both facts are universally true theoretical properties of EM waves, both statements are correct.

Pattern Recognition

Standard NCERT theory. Energy density is strictly symmetric 50-50 between E and B. Momentum transfer → pressure is the definitive signature of the particle-like nature (photons) of EM waves.

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%)