JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 8)
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is
$$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$.
The magnetic field $$\overrightarrow B $$ , at the moment t = 0 is :
$$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$.
The magnetic field $$\overrightarrow B $$ , at the moment t = 0 is :
$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat j$$
$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat k$$
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat k$$
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat j$$
Explanation
$$\overrightarrow E = {E_0}\,\cos (\omega t - kx)\widehat j$$
We know, $$E = BC$$
$$B_{0} = {E_{0} \over C} = {{{E_0}} \over {{1 \over {\sqrt {{\mu _0}{ \in _0}} }}}}$$
$$ \Rightarrow $$ $$B_{0} = {E_0}\sqrt {{\mu _0}{ \in _0}} $$_3rd_September_Evening_Slot_en_8_1.png)
You can see direction of $$\overrightarrow B$$ is along z axis.
$$ \therefore $$ $$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (\omega t - kx)\widehat k$$
at t = 0
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (kx)\widehat k$$
We know, $$E = BC$$
$$B_{0} = {E_{0} \over C} = {{{E_0}} \over {{1 \over {\sqrt {{\mu _0}{ \in _0}} }}}}$$
$$ \Rightarrow $$ $$B_{0} = {E_0}\sqrt {{\mu _0}{ \in _0}} $$
You can see direction of $$\overrightarrow B$$ is along z axis.
$$ \therefore $$ $$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (\omega t - kx)\widehat k$$
at t = 0
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (kx)\widehat k$$
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