# The electric field of a plane electromagnetic wave propagating

Question:

The electric field of a plane electromagnetic wave propagating along the $x$ direction in vacuum is $\vec{E}=E_{0} \hat{j} \cos (\omega t-k x)$. The magnetic field $\vec{B}$, at the moment $t=0$ is :

1. (1) $\vec{B}=\frac{E_{0}}{\sqrt{\mu_{0} \varepsilon_{0}}} \cos (k x) \hat{k}$

2. (2) $\vec{B}=E_{0} \sqrt{\mu_{0} \varepsilon_{0}} \cos (k x) \hat{j}$

3. (3) $\vec{B}=E_{0} \sqrt{\mu_{0} \varepsilon_{0}} \cos (k x) \hat{k}$

4. (4) $\vec{B}=\frac{E_{0}}{\sqrt{\mu_{0} \varepsilon_{0}}} \cos (k x) \hat{j}$

Correct Option: , 3

Solution:

(3) Relation between electric field and magnetic field for

an electromagnetic wave in vacuum is $B_{0}=\frac{E_{0}}{c}$.

In free space, its speed $c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$

Here, $\mu_{0}=$ absolute permeability,$\varepsilon_{0}=$ absolute permittivity

$\therefore B_{0}=\frac{E_{0}}{c}=\frac{E_{0}}{1 / \sqrt{\mu_{0} \varepsilon_{0}}}=E_{0} \sqrt{\mu_{0} \varepsilon_{0}}$

As the electromagnetic wave is propagating along $x$ direction and electric field is along $y$ direction.

$\therefore \hat{E} \times \hat{B} \| \hat{C}$ (Here, $\hat{C}=$ direction of propagation of wave)

$\therefore \vec{B}$ should be in $\hat{k}$ direction.

$\therefore \mathrm{B}=\mathrm{E}_{\mathrm{o}} \sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}} \cos (\mathrm{Wt}-\mathrm{Kx}) \hat{\mathrm{k}}$

At $\mathrm{t}=0$

$\mathrm{B}=\mathrm{E}_{\mathrm{o}} \sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}} \cos (\mathrm{Kx}) \hat{\mathrm{k}}$