The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0 = 510 \;nT$.What is the amplitude of the electric field (in $N/C$) part of the wave?

The magnetic field vector of an electromagnetic wave is given by ${B}={B}_{o} \frac{\hat{{i}}+\hat{{j}}}{\sqrt{2}} \cos ({kz}-\omega {t})$; where $\hat{i}, \hat{j}$ represents unit vector along ${x}$ and ${y}$-axis respectively. At $t=0\, {s}$, two electric charges $q_{1}$ of $4\, \pi$ coulomb and ${q}_{2}$ of $2 \,\pi$ coulomb located at $\left(0,0, \frac{\pi}{{k}}\right)$ and $\left(0,0, \frac{3 \pi}{{k}}\right)$, respectively, have the same velocity of $0.5 \,{c} \hat{{i}}$, (where ${c}$ is the velocity of light). The ratio of the force acting on charge ${q}_{1}$ to ${q}_{2}$ is :-

The mean intensity of radiation on the surface of the Sun is about $10^{8}\,W/m^2.$ The $rms$ value of the corresponding magnetic field is closet to

The electric field of a plane electromagnetic wave is given by

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$ At $\mathrm{t}=0,$ a positively charged particle is at the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .$ If its instantaneous velocity at $(t=0)$ is $v_{0} \hat{\mathrm{k}},$ the force acting on it due to the wave is

Intensity of sunlight is observed as $0.092\, {Wm}^{-2}$ at a point in free space. What will be the peak value of magnetic field at that point? $\left(\sigma_{0}=8.85 \times 10^{-12}\, {C}^{2} \,{N}^{-1} \,{m}^{-2}\right.$ )

The vector $\vec E$ and $\vec B$ of an electromagnetic wave in vacuum are