A radio transmitter transmits at $830\, kHz$. At a certain distance from the transmitter magnetic field has amplitude $4.82\times10^{-11}\,T$. The electric field and the wavelength are respectively

If $\overrightarrow E $ and $\overrightarrow B $ are the electric and magnetic field vectors of E.M. waves then the direction of propagation of E.M. wave is along the direction of

In the $EM$ wave the amplitude of magnetic field $H_0$ and the amplitude of electric field $E_o$ at any place are related as

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

$(i)$ Evaluate $\int {\vec E.\overrightarrow {dl} } $ over the rectangular loop $1234$ shown in figure.

$(ii)$ Evaluate $\int {\vec B} .\overrightarrow {ds} $ over the surface bounded by loop $1234$.

$(iii)$ $\int {\vec E.\overrightarrow {dl}  =  - \frac{{d{\phi _E}}}{{dt}}} $ to prove $\frac{{{E_0}}}{{{B_0}}} = c$

$(iv)$ By using similar process and the equation $\int {\vec B} .\overrightarrow {dl}  = {\mu _0}I + { \in _0}\frac{{d{\phi _E}}}{{dt}}$ , prove that  $c = \frac{1}{{\sqrt {{\mu _0}{ \in _0}} }}$ 

An electromagnetic wave is represented by the electric field $\vec E = {E_0}\hat n\,\sin \,\left[ {\omega t + \left( {6y - 8z} \right)} \right]$. Taking unit vectors in $x, y$ and $z$ directions to be $\hat i,\hat j,\hat k$ ,the direction of propogation $\hat s$, is

In propagation of electromagnetic waves the angle between the direction of propagation and plane of polarisation is