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

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

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