JEE Main 2020 Physics - Class 12 JEEM20Ph9d95 Solution: The electric field of a plane electromagnetic wave is given by…

QJEEM20Ph9d95: JEE Main 2020 Physics - Class 12 JEEM20Ph9d95 Solution: The electric field of a plane electromagnetic wave is given by…

The electric field of a plane electromagnetic wave is given by

\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)

Its magnetic field will be given by :

A ${{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$
B ${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)$
C ${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)$
D ${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$

Step-by-Step Verified Solution

Given,

\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)

We know, direction of propagation,

\overrightarrow C = \overrightarrow E \times \overrightarrow B

Here direction of propagation =

\widehat k

\therefore

\widehat k

\overrightarrow E \times \overrightarrow B

and

\widehat E = {{\widehat i + \widehat j} \over {\sqrt 2 }}

\therefore

\widehat k = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right) \times \overrightarrow B

\Rightarrow

\widehat B = {{ - \widehat i + \widehat j} \over {\sqrt 2 }}

\therefore

\widehat B

{{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)