Question

For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is
$\overrightarrow{B}(x,t)=[1.2\times {10}^{-7}\sin (0.5\times {10}^{3}x+1.5\times {10}^{11}t\left)\hat{k}\right]\text{T}$
The instantaneous electric field $\overrightarrow{E}$ corresponding to $\overrightarrow{B}$ is (speed of light $c=3\times {10}^8{\text{ms}}^{-1})$

• A. $\overrightarrow{E}(x,t)=[-36\sin (0.5\times {10}^{3}x+1.5\times {10}^{11}t\left)\hat{j}\right]\frac{\text{V}}{\text{m}}$
• B. $\overrightarrow{E}(x,t)=[-36\sin (1\times {10}^{3}x+0.5\times {10}^{11}t\left)\hat{j}\right]\frac{\text{V}}{\text{m}}$
• C. $\overrightarrow{E}(x,t)=\left[36\sin \right(0.5\times {10}^{3}x+1.5\times {10}^{11}t\left)\hat{k}\right]\frac{\text{V}}{\text{m}}$
• D. $\overrightarrow{E}(x,t)=\left[36\sin \right(1\times {10}^{3}x+1.5\times {10}^{11}t\left)\hat{i}\right]\frac{\text{V}}{\text{m}}$

Answer 1

The correct option is A $\overrightarrow{E}(x,t)=[-36\sin (0.5\times {10}^{3}x+1.5\times {10}^{11}t\left)\hat{j}\right]\frac{\text{V}}{\text{m}}$

Relation between electric field $E_0$ and magnetic field $B_0$ of an electromagnetic wave is given by

$c=\frac{E_0}{B_0}\left(\text{Here, c= Speed of light}\right)$

$\Rightarrow E_0=B_0\times c=1.2\times {10}^{-7}\times 3\times {10}^8=36$

As the wave is propagating along $x-$ direction, magnetic field is along $z-$ direction and $(\hat{E}\times \hat{B})||\hat{C}$
$\therefore \overrightarrow{E}$ should be along $y-$ direction.

So, electric field will be,
$\overrightarrow{E}=[-36\sin (0.5\times {10}^{3}x+1.5\times {10}^{11}t\left)\hat{j}\right]\frac{\text{V}}{\text{m}}$

Hence, option $\left(\text{A}\right)$ is correct.