Question:
The magnetic field of a plane electromagnetic wave is
$\vec{B}=3 \times 10^{-8} \sin [200 \pi(y+c t)] \hat{i} T$
where $c=3 \times 10^{8} \mathrm{~ms}^{-1}$ is the speed of light.
The corresponding electric field is :
Correct Option: , 4
Solution:
(4) Given: $\bar{B}=3 \times 10^{-8} \sin [200 \pi(y+c t)] \hat{i} T$
$\therefore B_{0}=3 \times 10^{-8}$
$E_{0}=C B_{0} \Rightarrow E_{0}=3 \times 10^{8} \times 3 \times 10^{-8}=9 \mathrm{~V} / \mathrm{m}$ Directiono f wave propagation
$(\bar{E} \times \bar{B}) \| \bar{C}$
$\hat{B}=\hat{i}$ and $\hat{C}=-\hat{j}$ $\therefore \hat{E}=-\hat{k}$
$\therefore \bar{E}=E_{0} \sin [200 \pi(y+c t)](-\hat{k}) \mathrm{V} / \mathrm{m}$
or, $\bar{E}=-9 \sin [200 \pi(y+c t)] \hat{k} \mathrm{~V} / \mathrm{m}$