A plane electromagnetic wave is moving in free space with velocity 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 :

(A) $-1.8 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}} \mathrm{T}$

(B) $1.4 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{k}} \mathrm{T}$

(C) $1.4 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}} \mathrm{T}$

(D) $+1.8 \times 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}} \mathrm{T}$

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · EM Waves

A plane electromagnetic wave is moving in free space with velocity c

= 3 \times 1 0^{8} m / s

and its electric field is given as

E = 54 sin (kz - ω t) \hat{j} V / m

, where

\hat{j}

is the unit vector along y-axis. The magnetic field vector

B

of the wave is :

Answer: (A)

The wave travels along

+ z

(the argument is

k z - ω t

) with the electric field along

\hat{j}

. For a plane wave the ordered triad is

\hat{E} \times \hat{B} = \overset{c}{^}

, the direction of propagation. Taking

\hat{B} = - \hat{i}

gives

\hat{j} \times (- \hat{i}) = \hat{k} = \overset{c}{^}

, which is correct. The amplitude follows from

B_{0} = \frac{E _{0}}{c} = \frac{54}{3 \times 1 0 ^{8}} = 1.8 \times 1 0^{- 7}

T. Therefore

B = - 1.8 \times 1 0^{- 7} sin (k z - ω t) \hat{i} T

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option