JEE Main · 2019 · 08 Apr 2019 · Shift 2 · EM Waves

For an electromagnetic wave the field amplitudes satisfy $E=cB$ with $c=3\times 10^8\text{m/s}$. The amplitude of $\vec{B}$ here is $B_0=1.6\times 10^{-6}\sqrt{2^2+1^2}=1.6\times 10^{-6}\sqrt{5}$, so $E_0=c{B}_0=4.8\times 10^2\sqrt{5}\text{V/m}$, which fixes the $4.8\times 10^2$ prefactor. The electric field must be perpendicular to $\vec{B}$, so $\vec{E}\cdot \vec{B}=0$; of the directions offered, $(-\hat{i}+2\hat{j})$ is perpendicular to $(2\hat{i}+\hat{j})$. Checking the direction of travel, the phase $(2\times 10^{7}z+6\times 10^{15}t)$ describes a wave moving along $-\hat{k}$, and $(-\hat{i}+2\hat{j})\times (2\hat{i}+\hat{j})=-5\hat{k}$ confirms it. So the electric field oscillates along $(-\hat{i}+2\hat{j})$ with amplitude $4.8\times 10^2\text{V/m}$.