QJEEM17Phe657: JEE Main 2017 Physics - Class 12 JEEM17Phe657 Solution: The electric field component of a monochromatic radiation is…

We have

${{dE} \over {dz}} = {{ - dE} \over {dt}}$

${{dE} \over {dz}} = - 2{E_0}k\sin kz\cos \omega t = {{ - dB} \over {dt}}$

Therefore, $dB = + 2{E_0}k\sin kz\cos \omega t\,dt$

That is, $B = + 20{E_0}k\sin {k_z}\cos \omega t\,dt$

$B = + 2{E_0}k\sin {k_2}\int {\cos \omega t\,dt = + 2{E_0}{k \over \omega }\sin kz\sin \omega t}$

Now, ${{{E_0}} \over {{B_0}}} = {\omega \over k} = c$

Therefore, its magnetic field $\overrightarrow B$ is given by $B = {{2{E_0}} \over c}\widehat j\sin kz\sin \omega t$