The magnetic field of a plane electromagnetic wave is

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 :

  1. (1) $\vec{E}=9 \sin [200 \pi(y+c t)] \hat{k} \mathrm{~V} / \mathrm{m}$

  2. (2) $\vec{E}=-10^{-6} \sin [200 \pi(y+c t)] \hat{k} \mathrm{~V} / \mathrm{m}$

  3. (3) $\vec{E}=3 \times 10^{-8} \sin [200 \pi(y+c t)] \hat{k} \mathrm{~V} / \mathrm{m}$

  4. (4) $\vec{E}=-9 \sin [200 \pi(y+c t)] \hat{k} \mathrm{~V} / \mathrm{m}$


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, $\bar{E}=-9 \sin [200 \pi(y+c t)] \hat{k} \mathrm{~V} / \mathrm{m}$

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