JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 12)
The electric field component of a monochromatic radiation is given by
$$\overrightarrow E $$ = 2 E0 $$\widehat i$$ cos kz cos $$\omega $$t
Its magnetic field $$\overrightarrow B $$ is then given by :
$$\overrightarrow E $$ = 2 E0 $$\widehat i$$ cos kz cos $$\omega $$t
Its magnetic field $$\overrightarrow B $$ is then given by :
$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz cos $$\omega $$t
$$-$$ $${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega $$t
$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega $$t
$${{2{E_0}} \over c}$$ $$\widehat j$$ cos kz cos $$\omega $$t
Explanation
We have
$${{dE} \over {dz}} = {{ - dE} \over {dt}}$$
$${{dE} \over {dz}} = - 2{E_0}k\sin kz\cos \omega t = {{ - dB} \over {dt}}$$
Therefore, $$dB = + 2{E_0}k\sin kz\cos \omega t\,dt$$
That is, $$B = + 20{E_0}k\sin {k_z}\cos \omega t\,dt$$
$$B = + 2{E_0}k\sin {k_2}\int {\cos \omega t\,dt = + 2{E_0}{k \over \omega }\sin kz\sin \omega t} $$
Now, $${{{E_0}} \over {{B_0}}} = {\omega \over k} = c$$
Therefore, its magnetic field $$\overrightarrow B $$ is given by $$B = {{2{E_0}} \over c}\widehat j\sin kz\sin \omega t$$
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