The energy of electromagnetic wave in vacuum is given by the relation
$\frac{{{E^2}}}{{2{\varepsilon _0}}} + \frac{{{B^2}}}{{2{\mu _0}}}$
$\frac{1}{2}{\varepsilon _0}{E^2} + \frac{1}{2}{\mu _0}{B^2}$
$\frac{{{E^2} + {B^2}}}{c}$
$\frac{1}{2}{\varepsilon _0}{E^2} + \frac{{{B^2}}}{{2{\mu _0}}}$
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 $c = \frac{1}{{\sqrt {{\mu _0}{ \in _0}} }}$
The velocity of certain ions that pass undeflected through crossed electric field $E = 7.7\,k\,V /m$ and magnetic field $B = 0.14\,T$ is.....$km/s$
Suppose that the electric field amplitude of an electromagnetic wave is $E_{0}=120\; N / C$ and that its frequency is $v=50.0\; MHz$.
$(a)$ Determine, $B_{0}, \omega, k,$ and $\lambda .$
$(b)$ Find expressions for $E$ and $B$
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
The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by
$E _{ x }= E _0 \sin ( kz -\omega t )$
$B _{ y }= B _0 \sin ( kz -\omega t )$
Then the correct relation between $E_0$ and $B_0$ is given by