The magnetic field of a plane electromagnetic wave is
$\vec{B}=3 \times 10^{-8} \sin [200 \pi(y+c t)] \hat{i} \mathrm{~T}$
Where $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ is the speed of light.
The corresponding electric field is :
Correct Option: , 2
$\overrightarrow{\mathrm{B}}=3 \times 10^{-8} \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{i}} \mathrm{T}$
$\mathrm{E}_{0}=\mathrm{CB}_{0} \Rightarrow \mathrm{E}_{0}=3 \times 10^{8} \times 3 \times 10^{-8}$
$=9 \mathrm{~V} / \mathrm{m}$
and direction of wave propagation is given as
$(\vec{E} \times \vec{B}) \| \vec{C}$
$\hat{\mathrm{B}}=\hat{\mathrm{i}} \quad \& \quad \hat{\mathrm{C}}=-\hat{\mathrm{j}}$
so $\hat{\mathrm{E}}=-\hat{\mathrm{k}}$
$\therefore \overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \sin [200 \pi(\mathrm{y}+\mathrm{ct})](-\hat{\mathrm{k}}) \mathrm{V} / \mathrm{m}$
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