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

Q47 jee_main_2026_23_january_morning Energy Density and Intensity

The equation of the electric field of an electromagnetic wave propagating through free space is given by :

E = \sqrt(377) (6.27 \times 10³ t - 2.09 \times 10⁻⁵ x) N/C

The average power of the electromagnetic wave is

((1)/(α))W/m²

. The value of

α

is ____

(Take μ₀ε₀ = 377 in SI units)

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

I = (1)/(2)ε₀ E₀² c

c = 1 μ₀ε₀

Core Logic

The average power per unit area (Intensity) of an electromagnetic wave in vacuum is $I = (1)/(2)ε₀ E₀² c$ . We can reorganize the formula using the intrinsic impedance of free space $Z₀ = μ₀ε₀ = 377Ω$ .

Step 1: Rewrite Intensity Formula

I = (1)/(2) ε₀ E₀² 1 μ₀ε₀

I = (1)/(2) E₀² ε₀μ₀

We are given the intrinsic impedance $μ₀ε₀ = 377 ε₀μ₀ = (1)/(377)$ .

Step 2: Extract Amplitude and Calculate

From the wave equation, $E₀ = \sqrt(377) N/C$ $E_{0} = \sqrt{377} \text{ N/C}$. Substitute into the intensity formula:

I = (1)/(2) (\sqrt(377))² \times (1)/(377)

I = (1)/(2) \times 377 \times (1)/(377) = (1)/(2) W/m²

Step 3: Compare with given format

I = (1)/(α) α = 2

Pattern Recognition

Sees: "average power... W/m^2" → the problem is asking for Intensity. When given intrinsic impedance ( $377Ω$ $377\Omega$), use the form $I = Erms² / Z₀ = E₀² / (2Z₀)$ directly.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q43 jee_main_2026_23_january_evening Speed of Electromagnetic Wave

The ratio of speeds of electromagnetic waves in vacuum and a medium, having dielectric constant

k = 3

and permeability of

μ = 2μ₀

, is (

μ₀ = permeability of vacuum

)

A. $36 : 1$ $36 : 1$
B. $3 : 2$ $3 : 2$
C. $6 : 1$ $6 : 1$
D. $\sqrt(6) : 1$ $\sqrt{6} : 1$

Solution

Related Formula

v = 1\sqrt(μ ε)

c = 1\sqrt(μ₀ ε₀)

n = (c)/(v) = \sqrt(μᵣ εᵣ)

Core Logic

The speed of light in a medium is slower than in vacuum by a factor equal to the refractive index $n$ $n$. The relative permittivity (dielectric constant) $εᵣ = k = 3$ $\epsilon_r = k = 3$. The relative permeability $μᵣ = (μ)/(μ₀) = 2$ $\mu_r = \frac{\mu}{\mu_0} = 2$.

Step 1: Calculate Ratio

(c)/(v) = \sqrt(μᵣ εᵣ)

(c)/(v) = \sqrt(2 \times 3) = \sqrt(6)

Thus, the ratio is $\sqrt(6) : 1$ $\sqrt{6} : 1$.

Pattern Recognition

Refractive index is geometrically derived from electrical properties: $n = \sqrt(μᵣ εᵣ)$ $n = \sqrt{\mu_r \epsilon_r}$. Just plug in the constants relative to vacuum values.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q35 jee_main_2026_24_january_morning Electromagnetic Spectrum

Match the List-I with List-II

List-I	List-II
A. Radio-wave	I. is produced by Magnetron value
B. Micro-wave	II. Due to change in the vibrational modes of atoms
C. Infrared-wave	III. Due to inner shell electrons moving from higher energy level to lower energy level
D. X-ray	IV. Due to rapid acceleration of electrons

Choose the correct answer from the options given below:

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

Solution

Core Logic

Mapping production sources to electromagnetic waves: Radio waves: Produced by rapid acceleration and deceleration of electrons in aerials. (A -> IV) Microwaves: Produced by special vacuum tubes like klystrons, magnetrons, and Gunn diodes. (B -> I) Infrared waves: Produced by hot bodies and molecules. They originate due to the change in vibrational modes of atoms/molecules. (C -> II) X-rays: Produced when high-energy electrons are stopped by a metal target, or by transitions of inner shell electrons from a higher to a lower energy level. (D -> III)

Step 1: Final Match

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

Pattern Recognition

Memorize fundamental EM wave sources: Radio (accelerating charges), Microwave (magnetrons), IR (vibrational modes/heat), X-ray (inner shell transitions).

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q32 jee_main_2026_28_january_morning Properties of Electromagnetic Waves

The electric field of an electromagnetic wave travelling through a medium is given by

E(x,t) = 25 (2.0\times 10¹⁵t - 10⁷x) n

then the refractive index of the medium is ____. (All given measurement are in SI units)

A. $1.2$ $1.2$
B. $2$ $2$
C. $1.5$ $1.5$
D. $1.7$ $1.7$

Solution

Related Formula

v = (ω)/(k)

μ = (c)/(v)

Core Logic

Compare the given equation with the standard wave equation $E = E₀ (ω t - kx)$ $E = E_0 \sin(\omega t - kx)$ to extract angular frequency ( $ω$ $\omega$) and wave number ( $k$ $k$). Find wave velocity $v$ $v$ and then refractive index.

Step 1: Extracting Constants

From the given equation $E(x,t) = 25 (2.0\times 10¹⁵t - 10⁷x) n$ :

ω = 2 \times 10¹⁵ ~rad / s

k = 10⁷ ~m⁻¹

Step 2: Calculating Wave Velocity

v = (ω)/(k) = 2 \times 10¹⁵10⁷ = 2 \times 10⁸ ~m/s

Step 3: Calculating Refractive Index

v = (c)/(μ) \Rightarrow 2 \times 10⁸ = (3 \times 10⁸)/(μ)

μ = (3 \times 10⁸)/(2 \times 10⁸) = 1.5

Pattern Recognition

For standard wave format $ω t \pm kx$ $\omega t \pm kx$, the ratio $ω/k$ $\omega/k$ immediately yields the phase velocity in the medium.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q27 jee_main_2026_28_january_evening Relation Between E and B

A plane electromagnetic wave is moving in free space with velocity

c = 3 \times 10⁸ m/s

and its electric field is given as

E = 54 (kz - ω t) j V/m

, where

j

is the unit vector along y-axis. The magnetic field vector

B

of the wave is :

A. $-1.8 \times 10⁻⁷ (kz - ω t) i T$
B. $1.4 \times 10⁻⁷ (kz - ω t) k T$
C. $1.4 \times 10⁻⁷ (kz - ω t) i T$
D. $+1.8 \times 10⁻⁷ (kz - ω t) i T$

Solution

Related Formula

B₀ = (E₀)/(c)

c = E \times B

Where $c$ $\hat{c}$ is the direction of wave propagation.

Core Logic

From the argument of sine $(kz - ω t)$ $(kz - \omega t)$, the wave is propagating along the positive z-axis. Thus, $c = k$ $\hat{c} = \hat{k}$. The electric field oscillates along the y-axis, so $E = j$ $\hat{E} = \hat{j}$.

Step 1: Find Magnetic Field Direction

Using the cross-product relation for directions:

B = c \times E = k \times j = - i

Step 2: Find Magnetic Field Amplitude

B₀ = (E₀)/(c) = (54)/(3 \times 10⁸) = 18 \times 10⁻⁸ = 1.8 \times 10⁻⁷ T

Step 3: Assemble the Final Vector

B = 1.8 \times 10⁻⁷ (kz - ω t) (- i) = -1.8 \times 10⁻⁷ (kz - ω t) i T

Pattern Recognition

Always use $E \times B = c$ $\hat{E} \times \hat{B} = \hat{c}$. Given $c = k$ $\hat{c} = \hat{k}$ and $E = j$ $\hat{E} = \hat{j}$, $B$ $\hat{B}$ must be $- i$ $-\hat{i}$ to satisfy $j \times (- i) = k$ $\hat{j} \times (-\hat{i}) = \hat{k}$.

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 2

Solution

Related Formula

Core Logic

Step 1: Rewrite Intensity Formula

Step 2: Extract Amplitude and Calculate

Step 3: Compare with given format

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Ratio

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Extracting Constants

Step 2: Calculating Wave Velocity

Step 3: Calculating Refractive Index

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find Magnetic Field Direction

Step 2: Find Magnetic Field Amplitude

Step 3: Assemble the Final Vector

Pattern Recognition

Chapter Mix

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