The electric field of a plane electromagnetic wave is given
Question:

The electric field of a plane electromagnetic wave is given by $\vec{E}=E_{0}(\hat{x}+\hat{y}) \sin (k z-\omega t)$ Its magnetic field will be given by:

1. (1) $\frac{E_{0}}{c}(-\hat{x}+\hat{y}) \sin (k z-\omega t)$

2. (2) $\frac{E_{0}}{c}(\hat{x}+\hat{y}) \sin (k z-\omega t)$

3. (3) $\frac{E_{0}}{c}(\hat{x}-\hat{y}) \sin (k z-\omega t)$

4. (4) $\frac{E_{0}}{c}(\hat{x}-\hat{y}) \cos (k z-\omega t)$

Correct Option: 1

Solution:

(1) $\vec{E}=E_{0}(\hat{x}+\hat{y}) \sin (k z-\omega t)$

Direction of propagation of em wave $=+\hat{k}$

Unit vector in the direction of electric field, $\hat{E}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

The direction of electromagnetic wave is perpendicular to both electric and magnetic field.

$\therefore \hat{k}=\hat{E} \times \hat{B}$

$\Rightarrow \hat{k}=\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \times \hat{B} \Rightarrow \hat{B}=\frac{-\hat{i}+\hat{j}}{\sqrt{2}}$

$\therefore \vec{B}=\frac{E_{0}}{c}(-\hat{x}+\hat{y}) \sin (k z-\omega t)$