Question

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

• A. $2.6\hat{x}\frac{\text{V}}{\text{m}}$
• B. $-2.6\hat{y}\frac{\text{V}}{\text{m}}$
• C. $24\hat{x}\frac{\text{V}}{\text{m}}$
• D. $-24\hat{x}\frac{\text{V}}{\text{m}}$

Answer 1

The correct option is D $-24\hat{x}\frac{\text{V}}{\text{m}}$

Given, $\overrightarrow{B}=B\hat{z}=8.0\times {10}^{-8}\hat{z}\text{T}$

We know that, at any point
$E=B.c$
$\Rightarrow E=(8\times {10}^{-8})\times (3\times {10}^8)$
$\Rightarrow E=24\frac{\text{V}}{\text{m}}$

The wave travels in the direction of $\overrightarrow{E}\times \overrightarrow{B}$
As $(-\hat{x})\times \hat{z}=+\hat{y}$
$\therefore \overrightarrow{E}=-24\hat{x}\frac{\text{V}}{\text{m}}$

Hence, option (d) is correct.