JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 7)

A plane electromagnetic wave of wavelength $$\lambda $$ has an intensity I. It is propagatting along the positive Y-direction. The allowed expressions for the electric and magnetic fields are givn by :
JEE Main 2018 (Online) 16th April Morning Slot Physics - Electromagnetic Waves Question 138 English Option 1
JEE Main 2018 (Online) 16th April Morning Slot Physics - Electromagnetic Waves Question 138 English Option 2
JEE Main 2018 (Online) 16th April Morning Slot Physics - Electromagnetic Waves Question 138 English Option 3
JEE Main 2018 (Online) 16th April Morning Slot Physics - Electromagnetic Waves Question 138 English Option 4

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.

JEE Main 2018 (Online) 16th April Morning Slot Physics - Electromagnetic Waves Question 138 English Explanation

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

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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