Due to presence of an em-wave whose electric component is given by E = 100 (omega 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

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). 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 sqrt(4) = 2 to maintain balance. Class 12 Physics: Electromagnetic Waves

Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-1 , a cylinder of length 200~cm holds certain amount of em-energy inside it. If another cylinder of same len

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). 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 sqrt(4) = 2 to maintain balance. Class 1

Due to presence of an em-wave whose electric component is given by E = 100 (omega 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

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). 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 sqrt(4) = 2 to maintain balance. Class 12 Physics: Electromagnetic Waves

Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-1 , a cylinder of length 200~cm holds certain amount of em-energy inside it. If another cylinder of same len

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). 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 sqrt(4) = 2 to maintain balance. Class 1

Due to presence of an em-wave whose electric component is given by E = 100 (omega t

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 1: Final Equation Match

Pattern Recognition

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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