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 :
Correct Option: 1,
(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
$\overrightarrow{\mathrm{B}}=\frac{\mathrm{B}_{0}}{2}[\cos (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{j}}-\cos (\mathrm{kz}+\omega \mathrm{t}) \hat{\mathrm{j}}]$
$\overrightarrow{\mathrm{B}}=\frac{\mathrm{B}_{0}}{2} \times 2 \sin \mathrm{kz} \sin \omega \mathrm{t}$
$\vec{B}=\left(\frac{E_{0}}{C} \sin k z \sin \omega t\right) \hat{j}$