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)

$A$ plane electromagnetic wave travels in a medium of relative permeability $\mu_{r} = 1.61$ and relative permittivity $\epsilon_{r} = 6.44$. If the magnitude of the magnetic intensity $H$ is $4.5 \times 10^{-2} \; A m^{-1}$ at a point,what will be the approximate magnitude of the electric field intensity $E$ at that point? (Given: $\mu_{0} = 4 \pi \times 10^{-7} \; N A^{-2}$,$c = 3 \times 10^{8} \; m s^{-1}$)

The electric field for an electromagnetic wave in free space is $E = i 30 \cos (k z - 5 \times 10^8 t)$,where the magnitude of $E$ is in $V/m$. The magnitude of the wave vector $k$ is (velocity of electromagnetic wave in free space $= 3 \times 10^8 \ m/s$).

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.

Statement-$I$ :- During propagation of electromagnetic wave,$\vec{E}$,$\vec{B}$ and direction of propagation are perpendicular to each other.
Statement-$II$ :- During propagation of electromagnetic wave,the energy density due to electric and magnetic fields are equal.

The relative permittivity of distilled water is $81$. The velocity of light in it will be $....\times 10^{7} \text{ m/s}$. (Given $\mu_{r} = 1$) (in $.33$)