If $\mu_{0}$ is permeability of free space and $\varepsilon_{0}$ is permittivity of free space,the speed of light in vacuum is given by

An electromagnetic wave travelling in $x$-direction is described by field equation $E_y = 300 \sin \omega \left( t - \frac{x}{c} \right)$. If the electron is restricted to move in $y$-direction only with speed of $1.5 \times 10^6 \text{ m/s}$, then the ratio of maximum electric and magnetic forces acting on the electron is . . . . . . .

If the electric field intensity of a uniform plane electromagnetic wave is given as $E = -301.6 \sin (kz - \omega t) \hat{a}_{x} + 452.4 \sin (kz - \omega t) \hat{a}_{y} \text{ V/m}$. Then,the magnetic intensity $H$ of this wave in $\text{A/m}$ will be (Given: Speed of light in vacuum $c = 3 \times 10^{8} \text{ m/s}$,permeability of vacuum $\mu_{0} = 4\pi \times 10^{-7} \text{ N/A}^{2}$)

If the magnetic field of a plane electromagnetic wave is given by (The speed of light $c = 3 \times 10^8 \, m/s$):
$B = 100 \times 10^{-6} \sin \left[ 2\pi \times 2 \times 10^{15} \left( t - \frac{x}{c} \right) \right]$
Then the maximum electric field associated with it is:

$A$ lamp emits monochromatic green light uniformly in all directions. The lamp is $3\%$ efficient in converting electrical power to electromagnetic waves and consumes $100\,W$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $5\,m$ from the lamp will be nearly.......$V/m$

The quantities $x = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$,$y = \frac{E}{B}$,and $z = \frac{l}{CR}$ are defined,where $C$ is capacitance,$R$ is resistance,$l$ is length,$E$ is electric field,$B$ is magnetic field,and $\varepsilon_{0}, \mu_{0}$ are free space permittivity and permeability respectively. Then: