JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 12)
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 :
$$\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
Its magnetic field will be given by :
$${{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)$$
$${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
Explanation
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 $$\widehat E = {{\widehat i + \widehat j} \over {\sqrt 2 }}$$
$$ \therefore $$ $$\widehat k = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right) \times \overrightarrow B $$
$$ \Rightarrow $$ $$\widehat B = {{ - \widehat i + \widehat j} \over {\sqrt 2 }}$$
$$ \therefore $$ $$\widehat B$$ = $${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
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 $$\widehat E = {{\widehat i + \widehat j} \over {\sqrt 2 }}$$
$$ \therefore $$ $$\widehat k = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right) \times \overrightarrow B $$
$$ \Rightarrow $$ $$\widehat B = {{ - \widehat i + \widehat j} \over {\sqrt 2 }}$$
$$ \therefore $$ $$\widehat B$$ = $${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
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