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

-1.8 × 10 -7 sin ( kz -ω t ) hat i T

1.4 × 10 -7 sin ( kz -ω t ) hat k T

1.4 × 10 -7 sin ( kz -ω t ) hat i T

+1.8 × 10 -7 sin ( kz -ω t ) hat i T

Physics · Electromagnetic Waves

JEE Main 2026 — 28 January, Evening Shift — Question 27

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

Option A:
$-1.8 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}} \mathrm{T}$
Correct
Option B:
$1.4 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{k}} \mathrm{T}$
Option C:
$1.4 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}} \mathrm{T}$
Option D:
$+1.8 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i} T}$

Answer: A

Step-by-step solution

$\hat{B}=\hat{C} \times \hat{E}=\hat{k} \times \hat{j}=-\hat{i}$ $\therefore \overrightarrow{\mathrm{B}}=\frac{54}{3 \times 10^{8}} \sin (\mathrm{kz}-\omega \mathrm{t})(-\hat{\mathrm{i}})$ $=-1.8 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}}$

Answer key and solution verified before publishing.

Practise Electromagnetic Waves

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Displacement Current and Equation of EM Waves