# The magnetic field of a plane electromagnetic wave is

Question:

The magnetic field of a plane electromagnetic wave is

$\vec{B}=3 \times 10^{-8} \sin [200 \pi(y+c t)] \hat{i} \mathrm{~T}$

Where $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ is the speed of light.

The corresponding electric field is :

1. $\overrightarrow{\mathrm{E}}=-10^{-6} \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$

2. $\overrightarrow{\mathrm{E}}=-9 \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$

3. $\overrightarrow{\mathrm{E}}=9 \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$

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

Correct Option: , 2

Solution:

$\overrightarrow{\mathrm{B}}=3 \times 10^{-8} \sin [200 \pi(\mathrm{y}+\mathrm{ct})] \hat{\mathrm{i}} \mathrm{T}$

$\mathrm{E}_{0}=\mathrm{CB}_{0} \Rightarrow \mathrm{E}_{0}=3 \times 10^{8} \times 3 \times 10^{-8}$

$=9 \mathrm{~V} / \mathrm{m}$

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 $\hat{\mathrm{E}}=-\hat{\mathrm{k}}$

$\therefore \overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \sin [200 \pi(\mathrm{y}+\mathrm{ct})](-\hat{\mathrm{k}}) \mathrm{V} / \mathrm{m}$