A plane electromagnetic wave is moving in free space with velocity c = 3 x 108 m/s and its electric field is given as E = 54 (kz - omega t) j , V/m , where j is the unit vector along y-axis. The magnetic field vector B of the wave is :

Where c is the direction of wave propagation. From the argument of sine (kz - omega t), the wave is propagating along the positive z-axis. Thus, c = k. The electric field oscillates along the y-axis, so E = j. Using the cross-product relation for directions: Always use E x B = c. Given c = k and E = j, B must be -i to satisfy j x (-i) = k. Class 12 Physics: Electromagnetic Waves

A plane electromagnetic wave is moving in free space with velocity c = 3 x 108 m/s and its electric field is given as E = 54 (kz - omega t) j , V/m , where j is the unit vector along y-axis. The magnetic field vector B of the wave is :

Where c is the direction of wave propagation. From the argument of sine (kz - omega t), the wave is propagating along the positive z-axis. Thus, c = k. The electric field oscillates along the y-axis, so E = j. Using the cross-product relation for directions: Always use E x B = c. Given c = k and E = j, B must be -i to satisfy j x (-i) = k. Class 12 Physics: Electromagnetic Waves

A plane electromagnetic wave is moving in free space with velocity c = 3 x 108 m/s and its electric field is given as E…

Year	2026	2025	2024	Total
Questions	10	12	9	31

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

More Electromagnetic Waves Previous-Year Questions — Page 7

Q40 jee_main_2024_31_jan_morning Energy Density Of EM Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times 10¹⁰ ~Hz

and an amplitude of

50 ~Vm⁻¹

. The total average energy density of the electromagnetic field of the wave is : [Use

ε₀ = 8.85 \times 10⁻¹² C² / Nm²

]

A. $1.106 \times 10⁻⁸ Jm⁻³$
B. $4.425 \times 10⁻⁸ ~Jm⁻³$
C. $2.212 \times 10⁻⁸ ~Jm⁻³$
D. $2.212 \times 10⁻¹⁰ ~Jm⁻³$

Solution

Related Formula

Utotal average = (1)/(2)ε₀ E₀²

Core Logic

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:

Uavg = UE + UB = 2UE = 2 ( (1)/(4)ε₀ E₀² ) = (1)/(2)ε₀ E₀²

Where $E₀$ $E_0$ is the amplitude of the electric field.

Step 2: Substitution

Given: $E₀ = 50 V/m$ $E_0 = 50 \mathrm{\, V/m}$ $ε₀ = 8.85 \times 10⁻¹² C²/(N\cdot m²)$

Uavg = (1)/(2) \times (8.85 \times 10⁻¹²) \times (50)²

Uavg = (1)/(2) \times 8.85 \times 10⁻¹² \times 2500

Uavg = 1.10625 \times 10⁻⁸ J/m³

Chapter Mix

Class 12 Physics: Electromagnetic Waves

A plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10⁸ m/s$ $c = 3 \times 10^{8} \text{ m/s}$ and its electric field is given as $E = 54 (kz - ω t) j V/m$ , where $j$ $\hat{j}$ is the unit vector along y-axis. The magnetic field vector $B$ $\vec{B}$ of the wave is :

Solution & Explanation

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

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 2: Substitution

Chapter Mix

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Solution & Explanation

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

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 2: Substitution

Chapter Mix

More Electromagnetic Waves Questions — jee_main_2026_28_january_evening

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