JEE Main 2025 Jan 28 Shift 2, Physics Q19: Electromagnetic Waves

For a plane EM wave travelling in direction ${\hat{k}}_{\text{wave}}$, $\vec{E}$, $\vec{B}$, ${\hat{k}}_{\text{wave}}$ form a right-handed orthogonal triad with $|\vec{E}|=c|\vec{B}|$, and $\vec{E}$ and $\vec{B}$ are in phase.

Here the propagation direction is $+\hat{k}$. The required relation is $\vec{E}=c\vec{B}\times \hat{k}$ (so that $\vec{E}\times \vec{B}$ points along $+\hat{k}$). Computing the cross product:

$\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)\times \hat{k}=\frac{\sqrt{3}}{2}(\hat{i}\times \hat{k})+\frac{1}{2}(\hat{j}\times \hat{k})=-\frac{\sqrt{3}}{2}\hat{j}+\frac{1}{2}\hat{i}$.

The amplitude becomes $30c$ and the same $\sin \left[\omega (t-z/c)\right]$ time-dependence is retained:

$\vec{E}=\left(\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j}\right)30c\sin \left[\omega \left(t-\frac{z}{c}\right)\right]$.