Question

A plane electromagnetic wave of frequency $25GHz$ is propagating in vacuum along the $z$-direction. At a particular point in space and time, the magnetic field is given by $\begin{array}{l}\overrightarrow{B}=5\times {10}^{-8}\hat{j}T\end{array}$. The corresponding electric field is given by $\begin{array}{l}\overrightarrow{E}\end{array}$ is ( speed of light $𝑐=3\times {10}^8\raisebox{1ex}{$𝑚$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ )

• A. $1.66\times {10}^{-16i}\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$
• B. $-1.66\times {10}^{-16i}\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$
• C. $15i\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$
• D. $-15i\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Answer 1

The correct option is C

$15i\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Explanation for the correct answer is,

Step 1. Given Data:

Frequency of electromagnetic wave $=25GHz$

Magnetic field $\begin{array}{l}\overrightarrow{B}=5\times {10}^{-8}\hat{j}T\end{array}$

Speed of EM wave $𝐶=3\times {10}^8\raisebox{1ex}{$𝑚$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Electric field $\overrightarrow{E}$

Step 2. Calculating Electric field:

Direction of by $\overrightarrow{E}$ will be perpendicular to $\overrightarrow{B}$ and $\overrightarrow{E}\times \overrightarrow{B}$ gives direction of propagation of wave which is in $z$-direction.

Therefore, $\overrightarrow{E}$ will be along positive $x$-direction.

$$\begin{gathered} E=\overrightarrow{B}\times C \\ \Rightarrow E=(5\times {10}^{-8}{j}^{})\times (3\times {10}^{8}k) \\ \Rightarrow E=15(j^{}\times k) \end{gathered}$$

$\therefore \overrightarrow{E}=15i\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Hence, the correct option is (c).