Nearly $10 \%$ of the power of a $110\,W$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1\, m$ from the bulb to a distance of $5\,m$ is $a \times 10^{-2}\,W / m ^{2}$. The value of ' $a$ ' will be.

The electric field in an electromagnetic wave is given by ${E}=\left(50\, {NC}^{-1}\right) \sin \omega({t}-{x} / {c})$

The energy contained in a cylinder of volume ${V}$ is $5.5 \times 10^{-12} \, {J}$. The value of ${V}$ is $......{cm}^{3}$ $\left(\right.$ given $\left.\in_{0}=8.8 \times 10^{-12} \,{C}^{2} {N}^{-1} {m}^{-2}\right)$

An electromagnetic wave with frequency $\omega $ and wavelength $\lambda $ travels in the $+y$ direction. Its magnetic field is along $+x$ axis. The vector equation for the associated electric field (of amplitude $E_0$) is

The electric field of a plane electromagnetic wave is given by

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$ At $\mathrm{t}=0,$ a positively charged particle is at the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .$ If its instantaneous velocity at $(t=0)$ is $v_{0} \hat{\mathrm{k}},$ the force acting on it due to the wave is

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.

Electromagnetic waves travel in a medium with speed of $1.5 \times 10^8 \mathrm{~ms}^{-1}$. The relative permeability of the medium is $2.0$ . The relative permittivity will be :