The ratio of the magnitude of the magnetic field $(B_0)$ and electric field intensity $(E_0)$ of a plane electromagnetic wave in free space with permeability $\mu_0$ and permittivity $\varepsilon_0$ is (Given that $c$ is the velocity of light in free space):

On a particular day,the sun delivers an average power of $\left(\frac{6}{\pi} \times 10^3\right) \frac{W}{m^2}$ to the top of Earth's atmosphere. Find the amplitude of the magnetic field for the electromagnetic waves above the atmosphere. (Take $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

For a plane electromagnetic wave,the magnetic field at a point $x$ and time $t$ is $\overrightarrow{ B }( x , t ) = [1.2 \times 10^{-7} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{ k }] \text{ T}$. The instantaneous electric field $\overrightarrow{ E }$ corresponding to $\overrightarrow{ B }$ is: (speed of light $c = 3 \times 10^{8} \text{ m/s}$)

$A$ plane electromagnetic wave has a 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 $....nT$. (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ Nm^2/C^2$ and speed of light $c = 3 \times 10^8 \ m/s$)

When $\mu_r$ and $\epsilon_r$ are the relative permeability and relative permittivity (dielectric constant) respectively,the refractive index is given by:

Even though an electric field $E$ exerts a force $qE$ on a charged particle, the electric field of an $EM$ wave does not contribute to the radiation pressure (but transfers energy). Explain.