JEE Main 2018 — Electromagnetic Waves Question with Solution

$$\overrightarrow{E}$$ is electric field vector, $$\overrightarrow{B}$$ is magnetic field vector perpendicular to $$\overrightarrow{E}$$. The direction of propagation is $$(\overrightarrow{E}\times \overrightarrow{B})$$. The direction of propagation of wave is along + y axis, then $$\overrightarrow{E}$$ is along + z axis and $$\overrightarrow{B}$$ is along + x axis.

$$(E\hat{k}\times B\hat{i})=EB(\hat{k}\times \hat{i})=EB\hat{j}$$

As wave is travelling along + y axis with time, we will use (y $$-$$ ct) in wave equation.

Also, intensity is given by

$$I=\frac{1}{2}c{\epsilon }_0{E}_0^2$$

And $$E=cB$$

So, $$|E_0|=\sqrt{\frac{2I}{c{\epsilon }_0}}$$

Therefore, $$\overrightarrow{E}=\sqrt{\frac{2I}{c{\epsilon }_0}}\cos \left[\frac{2\pi }{\lambda }(y-ct)\right]\hat{k}$$

And $$\overrightarrow{B}=\frac{E}{c}\hat{i}\Rightarrow \frac{1}{c}E\hat{i}$$