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

The electric field of a plane electromagnetic wave is given by $\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{i}} \cos (\mathrm{kz}) \cos (\omega \mathrm{t})$

The corresponding magnetic field $\vec{B}$ is then given by :

  1. (1) $\vec{B}=\frac{E_{0}}{C} \hat{j} \sin (k z) \sin (\omega t)$

  2. (2) $\vec{B}=\frac{E_{0}}{C} \hat{j} \sin (k z) \cos (\omega t)$

  3. (3) $\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_{0}}{\mathrm{C}} \hat{\mathrm{j}} \cos (\mathrm{kz}) \sin (\omega \mathrm{t})$

  4. (4) $\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_{0}}{\mathrm{C}} \hat{\mathrm{k}} \sin (\mathrm{kz}) \cos (\omega \mathrm{t})$


Correct Option: 1,

Solution:

(1)

$\frac{E_{0}}{B_{0}}=C$

$\Rightarrow \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}$

Given that $\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \cos (\mathrm{kz}) \cos (\omega \mathrm{t}) \hat{\mathrm{i}}$

$\overrightarrow{\mathrm{E}}=\frac{\mathrm{E}_{0}}{2}[\cos (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}}-\cos (\mathrm{kz}+\omega \mathrm{t}) \hat{\mathrm{i}}]$

Correspondingly

$\overrightarrow{\mathrm{B}}=\frac{\mathrm{B}_{0}}{2}[\cos (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{j}}-\cos (\mathrm{kz}+\omega \mathrm{t}) \hat{\mathrm{j}}]$

$\overrightarrow{\mathrm{B}}=\frac{\mathrm{B}_{0}}{2} \times 2 \sin \mathrm{kz} \sin \omega \mathrm{t}$

$\vec{B}=\left(\frac{E_{0}}{C} \sin k z \sin \omega t\right) \hat{j}$

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.