Question

The magnetic field of a plane electromagnetic wave is $\overrightarrow{B}=3\times {10}^{-8}\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{i}T$ where$c=3\times {10}^{8}m{s}^{-1}$ is the speed of light . The corresponding electric field is

• A. $\overrightarrow{E}=-9\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{k}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$
• B. $\overrightarrow{E}=9\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{k}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$
• C. $\overrightarrow{E}=-{10}^{-6}\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{k}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$
• D. $\overrightarrow{E}=3\times {10}^{-8}\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{k}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Answer 1

The correct option is A

$\overrightarrow{E}=-9\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{k}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Step 1: Given that:

The magnetic field of a plane electromagnetic wave is $\overrightarrow{B}=3\times {10}^{-8}\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{i}T$

The speed of light $c=3\times {10}^{8}m{s}^{-1}$

Step 2: Formula used and draw the required diagram:

Know that an electromagnetic wave propagates in direction of the vector $E\times B$and the phase gives the velocity of the wave.

The electric field vector of the $EM$wave is,

$E=cB$

Therefore

$$\begin{gathered} B=B_0\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{i} \\ B=3\times {10}^{-8}\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{i} \\ {E}_0=c{B}_0 \\ {E}_0=3\times {10}^8\times 3\times {10}^{-8} \end{gathered}$$

Therefore the value of$E_0=9\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Know that the phase $=y+ct$ is constant.

Then differentiate,

$\Rightarrow \frac{dy}{dt}=-c$

Step 3: Solve the electric field $E$.

Note that the wave propagates in the negative y-direction.

$\Rightarrow E\times B$is parallel to $-\hat{j}$.

$\Rightarrow \hat{E}\times \hat{i}=\lambda \left(-\hat{j}\right)$

Therefore,

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

Hence, the electric field vector is,

$$\begin{gathered} E=E_0\sin \left(ky+\omega t\right)\left(-\hat{k}\right) \\ E=-9\sin \left[200\mathrm{\pi }\left(\mathrm{y}+\mathrm{ct}\right)\right]\hat{k}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right. \end{gathered}$$

Hence, the correct option is B.