JEE Main 2018 — Electromagnetic Waves Question with Solution

Given : Frequency $$f=\frac{3}{2\pi }\times 10^{12}$$ Hz; direction $$=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$$.

The beam is polarised along $$\hat{k}$$ direction.

Direction of magnetic field $$\overrightarrow{B}=\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\times \hat{k}=\frac{-\hat{j}}{\sqrt{2}}+\frac{\hat{i}}{\sqrt{2}}\right)$$

Therefore, direction of $$\overrightarrow{B}=\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)$$

Now from wave equation, we have

$$B=B_0\cos (kr-\omega t)$$

$$\omega =2\pi f=2\pi \times \frac{3}{2\pi }\times 10^{12}Hz=3\times 10^{12}Hz$$

Now, $$k=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$$;

$$B_0=\frac{E_0}{c}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)$$

Therefore,

$$B=\frac{E_0}{c}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)\cos \left[{10}^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right).r-(3\times {10}^{12})t\right]$$