Question

The magnetic field of a plane electromagnetic wave is
$\vec{B}=3\times 10^{-8}\sin [200\pi (y+ct)]\hat{i}T$
Where $c=3\times 10^{8}m{s}^{-1}$ is the speed of light. The corresponding electric field is :

• A. $\vec{E}=-10^{-6}\sin [200\pi (y+ct)]\hat{k}V/m$
• B. $\vec{E}=-9\sin [200\pi (y+ct)]\hat{k}V/m$
• C. $\vec{E}=9\sin [200\pi (y+ct)]\hat{k}V/m$
• D. $\vec{E}=3\times 10^{-8}\sin [200\pi (y+ct)]\hat{k}V/m$

Answer 1

Answer: (B)

$\overrightarrow{B}=3\times 10^{-8}\sin [200\pi (y+ct)]\hat{i}T$
$E_0=C{B}_0\Rightarrow E_0=3\times 10^8\times 3\times 10^{-8}=9V/m$
and direction of wave propagation is given as
$(\overrightarrow{E}\times \overrightarrow{B})\Vert \overrightarrow{C}$
$\hat{B}=\hat{i},\&,\hat{C}=-\hat{j}$
so $\hat{E}=-\hat{k}$
$\therefore \overrightarrow{E}=E_0\sin [200\pi (y+ct)](-\hat{k})V/m$