Question

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time $\underset{E}{\to }=6.3\hat{j}V/m$. The corresponding magnetic field $\underset{B}{\to }$ at that point will be :

• A. $18.9\times {10}^8\hat{K}T$
• B. $18.9\times {10}^{-8}\hat{K}T$
• C. $2.1\times {10}^{-8}\hat{K}T$
• D. $6.3\times {10}^{-8}\hat{K}T$

Answer 1

The correct option is A $18.9\times {10}^8\hat{K}T$

$\text{Given}$ : frequency 50 MHz

$\overrightarrow{E}$ = 6.3$\hat{j}$ V/m

$\text{To find}$: magnetic field

$\text{Solution}$ :

We know, $\left|B\right|$ = $\frac{\left|E\right|}{C}$

= $\frac{6.3}{3\times {10}^8}$ = 2.1$\times {10}^{-8}$ T

Now, $\hat{E}\times \hat{B}=\hat{C}$

Therefore, $\hat{j}\times \hat{B}=\hat{i}$

Therefore, $\hat{B}$ = $\hat{k}$

Hence, $\hat{B}$ = 2.1$\times {10}^{-8}$ $\hat{k}$ T

$\text{Hence C is the correct option}$