What is the amplitude of the electric field in a parallel beam of light of intensity $\left(\frac{15}{\pi}\right) \text{ W/m}^2$ (in $\text{ N/C}$)? $\left[\text{Assume} \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\right]$

$A$ plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10^8 \ m/s$ and its electric field is given as $\vec{E} = 54 \sin(kz - \omega t) \hat{j} \ V/m$, where $\hat{j}$ is the unit vector along the $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:

When light propagates through a material medium of relative permittivity $\varepsilon_{r}$ and relative permeability $\mu_{r}$,the velocity of light,$v$,is given by: ($c$ = velocity of light in vacuum)

The magnetic field of an electromagnetic wave is given by $B_y = 3 \times 10^{-7} \sin(10^3 x + 3 \times 10^{11} t)$,where $x$ is in meters and $t$ is in seconds. The wavelength of the electromagnetic wave is: (in $cm$)

An electric bulb is rated as $200 \, W$. What will be the peak magnetic field $(\times 10^{-8} \, T)$ at $4 \, m$ distance produced by the radiations coming from this bulb? Consider this bulb as a point source with $3.5 \%$ efficiency.

The electric field part of an electromagnetic wave in a medium is represented by $E_x = 0$; $E_y = 2.5 \frac{N}{C} \cos \left[ (2\pi \times 10^6 \frac{rad}{s})t - (\pi \times 10^{-2} \frac{rad}{m})x \right]$; $E_z = 0$. The wave is: