Question

The magnetic field of a plane electromagnetic wave is

$\overrightarrow{B}=3\times {10}^{-8}\sin \left[200\pi \right(y+ct\left)\right]\hat{i}\text{T}$
Where $c=3\times {10}^8{\text{ms}}^{-1}$ is the speed of light.
The corresponding electric field is-

• A. $\overrightarrow{E}=9\sin \left[200\pi \right(y+ct\left)\right]\hat{k}\text{V/m}$
• B. $\overrightarrow{E}=-{10}^{-6}\sin \left[200\pi \right(y+ct\left)\right]\hat{k}\text{V/m}$
• C. $\overrightarrow{E}=3\times {10}^{-8}\sin \left[200\pi \right(y+ct\left)\right]\hat{k}\text{V/m}$
• D. $\overrightarrow{E}=-9\sin \left[200\pi \right(y+ct\left)\right]\hat{k}\text{V/m}$

Answer 1

The correct option is D $\overrightarrow{E}=-9\sin \left[200\pi \right(y+ct\left)\right]\hat{k}\text{V/m}$

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

So, $B_0=3\times {10}^{-8}$

$E_0=c{B}_0$

$E_0=3\times {10}^8\times 3\times {10}^{-8}=9{\text{V m}}^{-1}$

Direction of wave propagation,

$(\overrightarrow{E}\times \overrightarrow{B})||c$

$B=\hat{i}\text{and}c=-\hat{j}$

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

$\therefore \overrightarrow{E}=E_0\sin \left[200\pi \right(y+ct\left)\right](-\hat{k}){\text{V m}}^{-1}$

$\overrightarrow{E}=-9\sin \left[200\pi \right(y+ct\left)\right]\hat{k}{\text{V m}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.