Question

A plane electromagnetic wave of frequency $500MHz$ is travelling in vacuum along $x,\hspace{0.17em}y,\hspace{0.17em}z$- direction. At a particular point in space and time, $\overrightarrow{B}=8.0\times {10}^{-8}\hat{z}T$ . The value of electric field at this point is: (speed of light$=3\times {10}^{8}m{s}^{–1}$) Assume $x,\hspace{0.17em}y,\hspace{0.17em}z$ are unit vectors along $\hat{x},\hat{y}\&\hat{z}$ directions.

• A. $\begin{array}{l}2.6\hat{x}\frac{V}{m}\end{array}$
• B. $\begin{array}{l}-2.6\hat{y}\frac{V}{m}\end{array}$
• C. $\begin{array}{l}24\hat{x}\frac{V}{m}\end{array}$
• D. $\begin{array}{l}-24\hat{x}\frac{V}{m}\end{array}$

Answer 1

The correct option is D

$\begin{array}{l}-24\hat{x}\frac{V}{m}\end{array}$

The explanation for the correct answer is,

Step 1. Given Data:

Frequency = $500MHz$

Magnetic field $\overrightarrow{B}=8.0\times {10}^{-8}\stackrel{\wedge }{z}T$

Since the EM wave is always perpendicular to the magnetic field hence the EM wave is moving in the direction of $\stackrel{\wedge }{y}$ and light is an Electromagnetic wave. Therefore, taking speed of light $\overrightarrow{C}=3\times {10}^8\stackrel{\wedge }{y}m{s}^{–1}$.

Step 2. Calculating electric field,

$$\begin{gathered} \overrightarrow{E}=\overrightarrow{B}\times \overrightarrow{C} \\ \Rightarrow \overrightarrow{E}=(8\times {10}^{–8})\stackrel{\wedge }{z}\times (3\times {10}^8)\stackrel{\wedge }{y} \\ \Rightarrow \overrightarrow{E}=-24\stackrel{\wedge }{z}\stackrel{\wedge }{y} \\ \Rightarrow \overrightarrow{E}=-24\stackrel{\wedge }{x}\frac{V}{m} \end{gathered}$$

Hence, the correct option is (D).