Question

A plane electromagnetic wave of frequency $50MHz$ travels in free space along the X-direction. At a particular point in space $E=7.2\hat{j}V/m$. At this point, $B$ is equal to

• A. $8.4\times {10}^{-8}\hat{k}T$
• B. $2.4\times {10}^{-8}\hat{k}T$
• C. $7.4\times {10}^{-6}\hat{i}T$
• D. $2.4\times {10}^{-8}\hat{j}T$

Answer 1

Answer: (B)

$2.4\times {10}^{-8}\hat{k}T$

Given, $E=7.2\hat{j}V/m$
The magnitude of $B$ is
$B=\frac{E}{C}=\frac{7.2V/m}{3\times {10}^{8}m/s}=2.4\times {10}^{-8}T$
$E$ is along Y-direction and the wave propagates along X-axis. Therefore, $B$ should be in a direction perpendicular to both $X$ and $Y-axes$. Using vector algebra $E\times B$ should be along X-direction.
Since, $(+\hat{j})\times (+\hat{k})=\hat{i},B$ is along Z-direction.
Thus, $B=2.4\times {10}^{-8}\hat{k}T$.