In an electromagnetic wave,the energy density associated with the magnetic field will be:

During the propagation of electromagnetic waves in a medium,

An electromagnetic wave in vacuum has the electric and magnetic field $\vec{E}$ and $\vec{B}$,which are always perpendicular to each other. The direction of polarization is given by $\vec{X}$ and that of wave propagation by $\vec{k}$. Then:

$A$ plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10^8 \ m/s$ and its electric field is given as $\vec{E} = 54 \sin(kz - \omega t) \hat{j} \ V/m$, where $\hat{j}$ is the unit vector along the $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:

In fog, photographs of the objects taken with infrared radiations are more clear than those obtained during visible light because

$A$ plane electromagnetic wave of frequency $100 \, MHz$ is travelling in vacuum along the $x$-direction. At a particular point in space and time,$\overrightarrow{B} = 2.0 \times 10^{-8} \hat{k} \, T$ (where $\hat{k}$ is the unit vector along the $z$-direction). What is $\overrightarrow{E}$ at this point?