QJEEM18Ph7d77: JEE Main 2018 Physics - Class 12 JEEM18Ph7d77 Solution: A plane electromagnetic wave of wavelength has an intensity I.…

$\overrightarrow E$ is electric field vector, $\overrightarrow B$ is magnetic field vector perpendicular to $\overrightarrow E$ . The direction of propagation is $(\overrightarrow E \times \overrightarrow B )$ . The direction of propagation of wave is along + y axis, then $\overrightarrow E$ is along + z axis and $\overrightarrow B$ is along + x axis.

$(E\widehat k \times B\widehat i) = EB(\widehat k \times \widehat i) = EB\widehat j$

As wave is travelling along + y axis with time, we will use (y $-$ ct) in wave equation.

Also, intensity is given by

$I = {1 \over 2}c{\varepsilon _0}E_0^2$

And $E = cB$

So, $|{E_0}| = \sqrt {{{2I} \over {c{\varepsilon _0}}}}$

Therefore, $\overrightarrow E = \sqrt {{{2I} \over {c{\varepsilon _0}}}} \cos \left[ {{{2\pi } \over \lambda }(y - ct)} \right]\widehat k$

And $\overrightarrow B = {E \over c}\widehat i \Rightarrow {1 \over c}E\widehat i$