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 a plane $EM$ wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} \mathrm{~Hz}$ and an amplitude of $50 \mathrm{Vm}^{-1}$. The total average energy density of the electromagnetic field of the wave is :

[Use $\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2$ ]

The energy of electromagnetic wave in vacuum is given by the relation

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 unit vector along $z-$direction) What is $\overrightarrow{ E }$ at this point ?

Light wave traveling in air along $x$-direction is given by $E _{ y }=540 \sin \pi \times 10^{4}( x - ct ) Vm ^{-1}$. Then, the peak value of magnetic field of wave will be $\dots \times 10^{-7}\,T$ (Given $c =3 \times 10^{8}\,ms ^{-1}$ )

A metal sample carrying a current along $X-$ axis with density $J_x$ is subjected to a magnetic field $B_z$ ( along $z-$ axis ). The electric field $E_y$ developed along $Y-$ axis is directly proportional io $J_x$ as well as $B_z$ . The constant of proportionality has $SI\, unit$.