JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 11)
The electric field of a plane electromagnetic
wave is given by
$$\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)$$
The corresponding magnetic field $$\overrightarrow B $$ is then given by
$$\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)$$
The corresponding magnetic field $$\overrightarrow B $$ is then given by
$$\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\sin (\omega t)$$
$$\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\cos (\omega t)$$
$$\overrightarrow B = {{{E_0}} \over C}\widehat j\cos (kz)\sin (\omega t)$$
$$\overrightarrow B = {{{E_0}} \over C}\widehat k\sin (kz)\cos (\omega t)$$
Explanation
$$ \therefore \overrightarrow E \times \overrightarrow B \parallel \overrightarrow v $$
Given that wave is propagating along positive z-axis and $$\overrightarrow E $$ along positive x-axis. Hence $$\overrightarrow B $$ along y-axis.
From Maxwell equation
$$\overrightarrow V \times \overrightarrow E = - {{\partial B} \over {\partial t}}$$
i.e. $${{\partial E} \over {\partial Z}} = - {{\partial B} \over {dt}}$$ and $${B_0} = {{{E_0}} \over C}$$
Given that wave is propagating along positive z-axis and $$\overrightarrow E $$ along positive x-axis. Hence $$\overrightarrow B $$ along y-axis.
From Maxwell equation
$$\overrightarrow V \times \overrightarrow E = - {{\partial B} \over {\partial t}}$$
i.e. $${{\partial E} \over {\partial Z}} = - {{\partial B} \over {dt}}$$ and $${B_0} = {{{E_0}} \over C}$$
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