The velocity of an electromagnetic wave is in the direction of

The intensity of light from a source is $\left( \frac{500}{\pi} \right) \, W/m^2$. Find the amplitude of the electric field in this wave.

An electromagnetic wave of frequency $5\, GHz$ is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are $2$. Its velocity in this medium is $\times 10^{7}\, m/s$.

$A$ plane electromagnetic wave of frequency $25\; GHz$ is propagating in vacuum along the $z$-direction. At a particular point in space and time,the magnetic field is given by $\overrightarrow{B} = 5 \times 10^{-8} \hat{j}\; T$. The corresponding electric field $\overrightarrow{E}$ is (speed of light $c = 3 \times 10^{8}\; m/s$):

Intensity of sunlight is observed as $0.092 \, W/m^2$ at a point in free space. What will be the peak value of the magnetic field at that point? (Given: $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$,$c = 3 \times 10^8 \, m/s$)

Light is an electromagnetic wave. Its speed in vacuum is given by the expression