JEE Main 2026 — Electromagnetic Waves Question with Solution

The direction of propagation of the electromagnetic wave is determined by the phase term $2\pi vt-\frac{2\pi x}{\lambda }$, which indicates that the wave is travelling along the $+x$ direction ($\hat{i}$).

The direction of the magnetic field $\vec{B}$ is given as $\hat{j}$.

In an electromagnetic wave, the direction of propagation is given by the direction of the cross product $\vec{E}\times \vec{B}$.

Let the direction of the electric field $\vec{E}$ be $\hat{n}$.

$\hat{n}\times \hat{j}=\hat{i}$

Since $(-\hat{k})\times \hat{j}=\hat{i}$, the direction of the electric field must be $-\hat{k}$.

The amplitude of the electric field is related to the magnetic field by $E_0=c{B}_0$, where $c$ is the speed of the wave.

The speed of the wave is $c=v\lambda$, where $v$ is the frequency.

$E_0=v\lambda B_0$

Therefore, the electric field vector is $\vec{E}=-v\lambda B_0\sin \left(2\pi vt-\frac{2\pi x}{\lambda }\right)\hat{k}$.

Answer: $\vec{E}=-v\lambda B_0\sin \left(2\pi vt-\frac{2\pi x}{\lambda }\right)\hat{k}$