In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is : [Use 0 = 8.85 x 10-12 , C2 / Nm2 ]

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: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is

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: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is : [Use 0 = 8.85 x 10-12 , C2 / Nm2 ]

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: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50 ~Vm-1. The total average energy density of the electromagnetic field of the wave is

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: Where E0 is the amplitude of the electric field. Given: E0 = 50 , V/m epsilon0 = 8.85 x 10-12 , C2/(Ncdot m2) Class 12 Physics: Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 x 1010 ~Hz and an amplitude of 50…

Year	2026	2025	2024	Total
Questions	10	12	9	31

Q40 jee_main_2024_29_jan_morning Maxwell Equations

Match List I with List II:

List I	List II
A. $\oint B \cdot d l = μ₀ ic + μ₀ ε₀ (dφE)/(dt)$	I. Gauss' law for electricity
B. $\oint E \cdot d l = -(dφB)/(dt)$	II. Gauss' law for magnetism
C. $\oint E \cdot d A = (Q)/(ε₀)$	III. Faraday law
D. $\oint B \cdot d A = 0$ $\oint \vec{B} \cdot d\vec{A} = 0$	IV. 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:

\oint B \cdot 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):

\oint E \cdot d l = -(dφB)/(dt) B - III

Gauss's Law for Electricity relates net electric flux to enclosed charge:

\oint E \cdot d A = (Q)/(ε₀) C - I

Gauss's Law for Magnetism states that magnetic monopoles do not exist:

\oint B \cdot d A = 0 D - II

Step 1: Match Evaluation

The match configurations are:

A $arrow$ $\rightarrow$ IV
B $arrow$ $\rightarrow$ III
C $arrow$ $\rightarrow$ I
D $arrow$ $\rightarrow$ II

This perfectly corresponds to Option (3).

Pattern Recognition

Understand the integral geometries: Path integrals (line integrals $\oint \cdot d l$ $\oint \cdot d\vec{l}$) correspond to circulating fields (induction laws like Ampere/Faraday). Surface integrals (flux integrals $\oint \cdot d A$ $\oint \cdot d\vec{A}$) 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 \times 10⁵ ~J

, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

A. $2.46 \times 10⁻³ ~kg ~m / s$
B. $2.16 \times 10⁻³ ~kg ~m / s$
C. $1.58 \times 10⁻³ ~kg ~m / s$
D. $4.32 \times 10⁻³ ~kg ~m / s$

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 ( $c$ $c$).

Step 1: Calculate Momentum

Given total energy $E = 6.48 \times 10⁵ ~J$ $E = 6.48 \times 10^5 \mathrm{~J}$. Speed of light $c = 3 \times 10⁸ ~m/s$ $c = 3 \times 10^8 \mathrm{~m/s}$.

p = (E)/(c) = (6.48 \times 10⁵)/(3 \times 10⁸)

p = 2.16 \times 10⁻³ ~kg~m/s

Pattern Recognition

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

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q49 jee_main_2024_30_january_evening Maxwell's Equations

Match List I with List II:

List-I	List-II
A. Gauss's law of magnetostatics	I. $\oint E \cdot d a = (1)/(ε₀) \int ρ dV$
B. Faraday's law of electro magnetic induction	II. $\oint B \cdot d a = 0$
C. Ampere's law	III. $\oint E \cdot d l = - ddt\int B \cdot d a$
D. Gauss's law of electrostatics	IV. $\oint B \cdot d l = μ₀ I$

Choose the correct answer from the options given below:

A. $A-I, B-III, C-IV, D-II$ $\text{A-I, B-III, C-IV, D-II}$
B. $A-III, B-IV, C-I, D-II$ $\text{A-III, B-IV, C-I, D-II}$
C. $A-IV, B-II, C-III, D-I$ $\text{A-IV, B-II, C-III, D-I}$
D. $A-II, B-III, C-IV, D-I$ $\text{A-II, B-III, C-IV, D-I}$

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. $\oint B \cdot d a = 0$ (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. $\oint E \cdot d l = - ddt \int B \cdot d a$ (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. $\oint B \cdot d l = μ₀ I$ (Matches IV). (D) Gauss's law of electrostatics: The electric flux through any closed surface is proportional to the enclosed electric charge. $\oint E \cdot d a = (1)/(ε₀) \int ρ dV$ (Matches I).

Step 1: Final Match

A $arrow$ $\rightarrow$ II B $arrow$ $\rightarrow$ III C $arrow$ $\rightarrow$ IV D $arrow$ $\rightarrow$ 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 \times B

Core Logic

In an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by $E₀ = c B₀$ $E_0 = c B_0$. Thus, $B₀ = E₀ / C$ $B_0 = E_0 / C$.

The direction of wave propagation is given by the cross product of the electric field and magnetic field vectors: $C = E \times B$ $\hat{C} = \hat{E} \times \hat{B}$.

Step 1: Determine Wave Direction and Magnetic Field Direction

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

k = i \times B

Since $i \times j = k$ $\hat{i} \times \hat{j} = \hat{k}$, the magnetic field must oscillate in the $+y$ $+y$ direction ( $j$ $\hat{j}$).

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 $1/c$ $1/c$. Direction satisfies the right-hand triad $( E, B, v)$ $(\vec{E}, \vec{B}, \vec{v})$ where $v = E \times B$ $\vec{v} = \vec{E} \times \vec{B}$.

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

$pᵣ = (I)/(c)$ $$p_r = \frac{I}{c}$$ (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: $UE = UB$ $U_E = U_B$ because $E = cB$ $E = cB$ and $c² = (1)/(μ₀ ε₀)$ $c^2 = \frac{1}{\mu_0 \epsilon_0}$. This is a correct fact.

Statement II: Electromagnetic waves carry momentum ( $p = U/c$ $p = U/c$). 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 $\to$ $\to$ pressure is the definitive signature of the particle-like nature (photons) of EM waves.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Solution & Explanation

Related Formula

Core Logic

Step 2: Substitution

Chapter Mix

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 6

Solution

Core Logic

Step 1: Match Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Momentum

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Wave Direction and Magnetic Field Direction

Step 2: Construct Final Vector

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Evaluation

Pattern Recognition

Chapter Mix

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