JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 7)
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Explanation
$$\overrightarrow E $$ 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 $$\overrightarrow B = {E \over c}\widehat i \Rightarrow {1 \over c}E\widehat i$$
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