The electric field component of a monochromatic radiation is given by \(\vec E = 2{E_0}\,\hat i\,\cos\, kz\,\cos\, \omega t\) Its magnetic field \(\vec B\) is then given by

\((\overrightarrow{\mathrm{E}})=2 \mathrm{E}_{0} \hat{\mathrm{i}} \cos \mathrm{kz} \cos \omega \mathrm{t}\)

We know that, \(\frac{d E}{d z}=-\frac{d B}{d t}\)

\(\frac{\mathrm{dE}}{\mathrm{dz}}=-2 \mathrm{E}_{0} \mathrm{k} \sin \mathrm{kz} \cos \omega \mathrm{t}=-\frac{\mathrm{dB}}{\mathrm{dt}}\)

\(\mathrm{dB}=+2 \mathrm{E}_{0} \mathrm{k} \,\sin \mathrm{kz} \cos\, \omega \mathrm{td}\, \mathrm{t}\) .... \((i)\)

Integrating \(eq^n\). \((i)\), we have

\(B=+2 E_{0} k \sin k z \int \cos \omega t d t\)

Magnetic field is given by,

\(=+2 \mathrm{E}_{0} \frac{\mathrm{k}}{\omega} \sin \mathrm{kz} \sin \,\omega \mathrm{t}\)

We also know that,

\(\frac{E_{0}}{B_{0}}=\frac{\omega}{k}=c\)

Magnetic field vector,

\(\overrightarrow{\mathrm{B}}=\frac{2 \mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{j}} \sin \mathrm{kz} \sin \omega \mathrm{t}\)