For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is
Question:

For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is

$\overrightarrow{\mathrm{B}}(x, t)=\left[1.2 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{k}\right] \mathrm{T}$

is: (speed of light $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

1. (1) $\overrightarrow{\mathrm{E}}(x, t)=\left[-36 \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{j}\right] \frac{\mathrm{V}}{\mathrm{m}}$

2. (2) $\overrightarrow{\mathrm{E}}(x, t)=\left[36 \sin \left(1 \times 10^{3} x+0.5 \times 10^{11} t\right) \hat{j}\right] \frac{\mathrm{V}}{\mathrm{m}}$

3. (3) $\overrightarrow{\mathrm{E}}(x, t)=\left[36 \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{k}\right] \frac{\mathrm{V}}{\mathrm{m}}$

4. (4) $\overrightarrow{\mathrm{E}}(x, t)=\left[36 \sin \left(1 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{i}\right] \frac{\mathrm{V}}{\mathrm{m}}$

Correct Option: 1

Solution:

(1) Relation between electric field $E_{0}$ and magnetic field $B_{0}$ of an electromagnetic wave is given by

$c=\frac{E_{0}}{B_{0}}$

(Here, $c=$ Speed of light)

$\Rightarrow E_{0}=B_{0} \times c=1.2 \times 10^{-7} \times 3 \times 10^{8}=36$

As the wave is propagating along $x$-direction, magnetic field is along $z$-direction

and $(\hat{E} \times \hat{B}) \| \hat{C}$

$\therefore \vec{E}$ should be along $y$-direction.

So, electric field $\vec{E}=E_{0} \sin \vec{E} \cdot(x, t)$

$=\left[-36 \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{j}\right] \frac{V}{m}$