The electric field of a plane electromagnetic wave propagating along the $x$ direction in vacuum is $\overrightarrow{ E }= E _{0} \hat{ j } \cos (\omega t - kx )$. The magnetic field $\overrightarrow{ B },$ at the moment $t =0$ is :
$\overrightarrow{ B }= E _{0} \sqrt{\mu_{0} \epsilon_{0}} \cos ( kx ) \hat{ j }$
$\overrightarrow{ B }=\frac{ E _{0}}{\sqrt{\mu_{0} \epsilon_{0}}} \cos ( kx ) \hat{ k }$
$\overrightarrow{ B }= E _{0} \sqrt{\mu_{0} \epsilon_{0}} \cos ( kx ) \hat{ k }$
$\overrightarrow{ B }=\frac{ E _{0}}{\sqrt{\mu_{0} \in_{0}}} \cos ( kx ){\hat{j}}$
The magnetic field in a travelling electromagnetic wave has a peak value of $20\ n T$. The peak value of electric field strength is......$Vm^{-1}$
The electric field of a plane electromagnetic wave is given by $\vec E = {E_0}\hat i\,\cos \,\left( {kz} \right)\,\cos \,\left( {\omega t} \right)$ The corresponding magnetic field $\vec B$ is then given by
A plane electromagnetic wave, has frequency of $2.0 \times 10^{10}\, Hz$ and its energy density is $1.02 \times 10^{-8}\, J / m ^{3}$ in vacuum. The amplitude of the magnetic field of the wave is close to$....nT$
$\left(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{\circ} \frac{ Nm ^{2}}{ C ^{2}}\right.$ and speed of $1 ight$ $\left.=3 \times 10^{8}\, ms ^{-1}\right)$
The speed of electromagnetic radiation in vacuum is
Even though an electric field $E$ exerts a force $qE$ on a charged particle yet the electric field of an $EM$ wave does not contribute to the radiation pressure (but transfers energy). Explain.