Ratio of electric field and magnetic field gives which physical quantity ? 

A plane electromagnetic wave of frequency $500\, MHz$ is travelling in vacuum along $y-$direction. At a particular point in space and

time, $\overrightarrow{ B }=8.0 \times 10^{-8} \hat{ z } \;T$. The value of electric field at this point is

(speed of light $\left.=3 \times 10^{8}\, ms ^{-1}\right)$

$\hat{ x }, \hat{ y }, \hat{ z }$ are unit vectors along $x , y$ and $z$ direction.

An electromagnetic wave of frequency $3\, GHz$ enters a dielectric medium of relative electric permittivity $2.25$ from vacuum. The wavelength of this wave in that medium will be $.......\,\times 10^{-2} \, cm$

The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to

The average value of electric energy density in an electromagnetic wave is :

A plane $EM$ wave travelling along $z-$ direction is described$\vec E = {E_0}\,\sin \,(kz - \omega t)\hat i$ and $\vec B = {B_0}\,\sin \,(kz - \omega t)\hat j$. Show that

$(i)$ The average energy density of the wave is given by $U_{av} = \frac{1}{4}{ \in _0}E_0^2 + \frac{1}{4}.\frac{{B_0^2}}{{{\mu _0}}}$

$(ii)$ The time averaged intensity of the wave is given by  $ I_{av}= \frac{1}{2}c{ \in _0}E_0^2$ વડે આપવામાં આવે છે.