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 :
Correct Option: , 3
(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 $\mathrm{t}=0$
$\mathrm{B}=\mathrm{E}_{\mathrm{o}} \sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}} \cos (\mathrm{Kx}) \hat{\mathrm{k}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.