The electric field of a plane electromagnetic wave is given

(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

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

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

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