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

Q35 jee_main_2026_21_jan_morning Electric and Magnetic Fields

The electric field a plane electromagnetic wave is given by :

Ey = 69 [ 0.6 \times 10³ x - 1.8 \times 10¹¹ t ] V/m.

The expression for magnetic field associated with this electromagnetic wave is ____ T.

A. $Bz = 2.3\times 10⁻⁷ [0.6\times 10³x - 1.8\times 10¹¹t]$
B. $Bz = 2.3\times 10⁻⁷ [0.6\times 10³x + 1.8\times 10¹¹t]$
C. $By = 69 [0.6\times 10³x + 1.8\times 10¹¹t]$
D. $By = 2.3\times 10⁻⁷ [0.6\times 10³x - 1.8\times 10¹¹t]$

Solution

Related Formula

B₀ = (E₀)/(c)

c = E \times B

Core Logic

The phase of the wave is $(0.6 \times 10³ x - 1.8 \times 10¹¹ t)$ . This indicates the wave propagates in the $+x$ $+x$ direction, so $c = i$ $\hat{c} = \hat{i}$. The electric field oscillates along the $y$ $y$-axis, so $E = j$ $\hat{E} = \hat{j}$. From $B = c \times E$ $\hat{B} = \hat{c} \times \hat{E}$, we have $B = i \times j = k$ . So, the magnetic field is along the $z$ $z$-axis ( $Bz$ $B_z$).

Step 1: Calculate Amplitude of B

Wave speed $v = c = (ω)/(k) = 1.8 \times 10¹¹0.6 \times 10³ = 3 \times 10⁸ m/s$ . The amplitude of the magnetic field is:

B₀ = (E₀)/(c) = (69)/(3 \times 10⁸) = 23 \times 10⁻⁸ = 2.3 \times 10⁻⁷ T

The phase remains exactly the same as the electric field:

Bz = 2.3 \times 10⁻⁷ (0.6 \times 10³ x - 1.8 \times 10¹¹ t)

Pattern Recognition

$B₀ = E₀/c$ $B_0 = E_0/c$ gives the magnitude. The vector identity $B = v \times E$ $\hat{B} = \hat{v} \times \hat{E}$ gives the direction. Phase part never changes sign or terms between E and B equations.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q46 jee_main_2026_21_jan_evening Displacement Current

An electromagnetic wave of frequency

100 MHz

propagates through a medium of conductivity,

σ = 10 mho/m

. The ratio of maximum conducting current density to maximum displacement current density is ________.

[Take 14πε₀ = 9 \times 10⁹ N ²/C²]

Numerical Answer. Answer: 1800 to 1800

Solution

Related Formula

$Jc = σ E$ $J_c = \sigma E$

Jd = ε₀ (\partial E)/(\partial t)

Core Logic

Let the electric field of the wave be $E = E₀ (ω t - kx)$ $E = E_0 \sin(\omega t - kx)$.

The conduction current density is:

Jc = σ E₀ (ω t - kx)

The maximum conduction current density is:

(Jc)max = σ E₀ --- (i)

The displacement current density is:

Jd = (1)/(A) ( ε₀ (\partial (EA))/(\partial t) ) = ε₀ (\partial E)/(\partial t)

Jd = ε₀ E₀ ω (ω t - kx)

The maximum displacement current density is:

(Jd)max = ε₀ E₀ ω --- (ii)

Step 1: Taking the Ratio

Dividing (i) by (ii):

Ratio = (Jc)max(Jd)max = (σ E₀)/(ε₀ ω E₀) = (σ)/(ε₀ ω)

Step 2: Substitution and Calculation

We know $f = 100 MHz = 10⁸ Hz$ $f = 100 \text{ MHz} = 10^8 \text{ Hz}$, so $ω = 2π f = 2π \times 10⁸ rad/s$ . $σ = 10 mho/m$ $\sigma = 10 \text{ mho/m}$. Also, $(1)/(4πε₀) = 9 \times 10⁹ (1)/(ε₀) = 4π \times 9 \times 10⁹$ .

Ratio = (10 \times 4π \times 9 \times 10⁹)/(2π \times 10⁸)

Ratio = (360π \times 10⁹)/(2π \times 10⁸) = (3600)/(2) = 1800

Step 3: Final Conclusion

The required ratio is 1800.

Pattern Recognition

The ratio of conduction to displacement current density in any medium is universally $σ / (ω ε₀)$ $\sigma / (\omega \epsilon_0)$. This dictates whether a medium behaves as a good conductor or a dielectric at a given frequency.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q47 jee_main_2026_22_january_morning Dielectric Constant of Medium

The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by,

Ey=20 (3\times 10⁶x-4.5\times 10¹⁴t)

V/m (where x, t and other values have S.I. units). The dielectric constant of the medium is \_\_\_\_. (speed of light in free space is

3 \times 10⁸ ~m/s

)

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

v = (ω)/(k), n = (c)/(v) = \sqrt(μᵣ εᵣ)

Core Logic

Wave velocity in medium:

v = (ω)/(k) = 4.5 \times 10¹⁴3 \times 10⁶ = 1.5 \times 10⁸ m/s

Refractive index:

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

For non-magnetic medium ( $μᵣ = 1$ $\mu_r = 1$):

n = \sqrt(εᵣ) 2 = \sqrt(εᵣ) εᵣ = 4

Pattern Recognition

Sees: EM wave equation in dielectric medium. Shortcut: Extract phase velocity from wave equation coefficients, find refractive index and dielectric constant. Check: Numerical answer is 4. ✓

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q43 jee_main_2026_22_january_evening Intensity and Field Amplitudes of EM Waves

A laser beam has intensity of

4.0 \times 10¹⁴

W/m

²

. The amplitude of magnetic field associated with beam is ____ T. (Take

ε₀ = 8.85 \times 10⁻¹²

C

²

/Nm

²

and c = 3 × 10

⁸

m/s)

A. $2.0$ $2.0$
B. $18.3$ $18.3$
C. $5.5$ $5.5$
D. $1.83$ $1.83$

Solution

Related Formula

I = (1)/(2) ε₀ E₀² c

B₀ = (E₀)/(c) = (1)/(c) \sqrt((2I)/(ε₀ c))

Core Logic

Expressing electric field amplitude $E₀$ $E_0$ in terms of intensity $I$ $I$:

E₀ = \sqrt((2I)/(ε₀ c))

Relating magnetic field amplitude $B₀$ $B_0$ to $E₀$ $E_0$ via $B₀ = (E₀)/(c)$ $B_0 = \frac{E_0}{c}$:

B₀ = (1)/(c)\sqrt((2I)/(ε₀ c)) = (1)/(3 \times 10⁸) 2 \times 4.0 \times 10¹⁴8.85 \times 10⁻¹² \times 3 \times 10⁸

Simplifying terms under the radical:

B₀ = (1)/(3 \times 10⁸) 8 \times 10¹⁴2.655 \times 10⁻³ = (1)/(3 \times 10⁸) 3.013 \times 10¹⁷ = (10)/(3) \sqrt((8)/(8.85 \times 3)) \approx 1.83 ~T

Step 1: Final Conclusion

The amplitude of the magnetic field is $1.83 ~T$ $1.83 \mathrm{~T}$.

Pattern Recognition

EM Wave Intensity: $I = (1)/(2) c (B₀²)/(μ₀) = (1)/(2) ε₀ E₀² c$ . Direct sub: $B₀ = \sqrt((2 μ₀ I)/(c))$ $B_0 = \sqrt{\frac{2 \mu_0 I}{c}}$ or $B₀ = (1)/(c) \sqrt((2I)/(ε₀ c)) \approx 1.83~T$ .

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q28 jee_main_2026_23_january_morning Maxwell's Equations

Match List-I with List-II.

List-I (Relation)		List-II (Law)
A.	$\oint E\cdot dl=-(d)/(dt)\oint B\cdot da$	I.	Ampere's circuital law.
B.	$\oint B\cdot dl=μ₀(1+ε₀ dφEdt)$	II.	Faraday's laws of electromagnetic induction.
(C)	$\oint E\cdot da= 1ε₀\intρ dv$	III.	Ampere-Maxwell law
(D)	$\oint B\cdot dl=μ₀I$ $\oint\vec{B}\cdot\vec{dl}=\mu_{0}I$	IV.	Gauss's law of electrostatics

Choose the correct answer from the options given below :

A. A-II, B-III, C-I, D-IV
B. A-II, B-III, C-IV, D-I
C. A-I, B-IV, C-III, D-II
D. A-IV, B-I, C-II, D-III

Solution

Core Logic

Let's systematically identify each integral equation with its corresponding physical law:

(A) $\oint E\cdot dl=- dφBdt$ corresponds to the line integral of the electric field around a closed loop being equal to the negative rate of change of magnetic flux, which is Faraday's Law of Electromagnetic Induction (II).

(B) $\oint B\cdot dl=μ₀(I+ε₀ dφEdt)$ defines the Ampere-Maxwell Law (III), incorporating the displacement current.

(C) $\oint E\cdot da= Qencε₀$ relates the electric flux through a closed surface to the enclosed charge, mapping to Gauss's Law of Electrostatics (IV).

(D) $\oint B\cdot dl=μ₀I$ $\oint\vec{B}\cdot\vec{dl}=\mu_{0}I$ is the original Ampere's Circuital Law (I) without Maxwell's correction.

Step 1: Final Conclusion

The correct matches are A-II, B-III, C-IV, D-I.

Pattern Recognition

Sees: Surface integral of E mapped to volume integral of charge → Gauss's Law. Circulation of E mapped to time derivative of B flux → Faraday's Law.

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

Solution

Related Formula

Core Logic

Step 1: Calculate Amplitude of B

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Taking the Ratio

Step 2: Substitution and Calculation

Step 3: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

More Electromagnetic Waves Questions — jee_main_2025_28_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back