Question

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

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

Answer 1

The correct option is D

$\frac{E_{\circ }}{c}\left(-\hat{x}+\hat{y}\right)\sin (kz-\omega t)$

Step 1: Given data

Electric field, $\overrightarrow{E}=E_{\circ }\left(\hat{x}+\hat{y}\right)\sin (kz-\omega t)$

The direction of propagation of the wave $=+\hat{k}$

Step 2: Compute unit vector

The unit vector in the direction of the electric field is,

$\hat{E}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Step 3: Determine the magnetic field

The direction of the electromagnetic wave is perpendicular to both the electric and magnetic fields. So,

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

So, the magnetic field would be given by $\hat{B}=\frac{E_{\circ }}{c}\left(-\hat{x}+\hat{y}\right)\sin \left(kz-\omega t\right)$

Hence, option (D) is the correct answer.