Question

The electric field of a plane electromagnetic wave is given by $\overrightarrow{E}=E_0(\hat{x}+\hat{y})\sin (kz-\omega t)$.
Its magnetic field will be given by :

• A. $\frac{E_0}{c}(-\hat{x}+\hat{y})\sin (kz-\omega t)$
• B. $\frac{E_0}{c}(\hat{x}+\hat{y})\sin (kz-\omega t)$
• C. $\frac{E_0}{c}(\hat{x}-\hat{y})\sin (kz-\omega t)$
• D. $\frac{E_0}{c}(\hat{x}+\hat{y})\cos (kz-\omega t)$

Answer 1

The correct option is A $\frac{E_0}{c}(-\hat{x}+\hat{y})\sin (kz-\omega t)$

Given,

$\overrightarrow{E}=E_0(\hat{x}+\hat{y})\sin (kz-\omega t)$

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

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

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

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

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

$\therefore \overrightarrow{B}=\frac{E_0}{c}(-\hat{x}+\hat{y})\sin (kz-\omega t))$.

Hence, $\left(A\right)$ is the correct answer.