Question

A plane electromagnetic wave of frequency 25 MHz travel in free space along the x- direction. At a particular point in space and time, $\overrightarrow{E} = 6.3\widehat{j}Vm^{- 1.}$At this point $\overrightarrow{B}$is equal to:

• A. $8.33 \times 10^{- 8}\widehat{k}T$
• B. $18.9 \times 10^{- 8}\widehat{k}T$
• C. $2.1 \times 10^{- 8}\widehat{k}T$
• D. $2.2 \times 10^{- 8}\widehat{k}T$

Answer 1

Answer: (C)

$2.1 \times 10^{- 8}\widehat{k}T$

Answer 2

: Here, $\overrightarrow{E} = 6.3\widehat{j}Vm^{- 1}$

The magnitude of $\overrightarrow{B}$is

$B = \frac{E}{c} = \frac{6.3Vm^{- 1}}{3 \times 10^{8}ms^{- 1}} = 2.1 \times 10^{8}T$

$\overrightarrow{E}$ is along y- direction and the wave propagates along x –axis. Therefore, $\overrightarrow{B}$should be in a direction perpendicular to both x and y – axis. Using vector algebra $\overrightarrow{E} \times \overrightarrow{B}$ should be along x – direction.

Since, $( + \widehat{j}) \times ( + \widehat{k}) = \widehat{i},\overrightarrow{B}$ is along z – direction.

Thus, $2.1 \times 10^{- 8}\widehat{k}T$