A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction.
Question:

A plane electromagnetic wave of frequency $50 \mathrm{MHz}$ travels in free space along the positive $x$-direction. At a particular point in space and time, $\overrightarrow{\mathrm{E}}=6.3 \hat{j} \mathrm{~V} / \mathrm{m}$. The

corresponding magnetic field $\overrightarrow{\mathrm{B}}$, at that point will be:

1. (1) $18.9 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}$

2. (2) $2.1 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}$

3. (3) $6.3 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}$

4. (4) $18.9 \times 10^{8} \hat{\mathrm{k}} \mathrm{T}$

Correct Option: , 2

Solution:

(2) As we know,

$|\overrightarrow{\mathrm{B}}|=\frac{|\overrightarrow{\mathrm{E}}|}{\mathrm{C}}=\frac{6.3}{3 \times 10^{8}}=2.1 \times 10^{-8} \mathrm{~T}$

and $\hat{\mathrm{E}} \times \hat{\mathrm{B}}=\hat{\mathrm{C}}$

$\hat{\mathrm{J}} \times \hat{\mathrm{B}}=\hat{\mathrm{i}}[\because$ EM wave travels along $+($ ve $) x$-direction. $]$

$\therefore \hat{\mathrm{B}}=\hat{\mathrm{k}}$ or $\overrightarrow{\mathrm{B}}=2.1 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}$