A point source of electromagnetic radiation has an average power output of $800\, W.$ The maximum value of electric field at a distance $4.0 \,m$ from the source is....$V/m$

The peak electric field produced by the radiation coming from the $8\, W$ bulb at a distance of $10\, m$ is $\frac{x}{10} \sqrt{\frac{\mu_{0} c }{\pi}} \,\frac{ V }{ m }$. The efficiency of the bulb is $10\, \%$ and it is a point source. The value of $x$ is ...... .

There exists a uniform magnetic and electric field of magnitude $1\, T$ and $1\, V/m$ respectively along positive $y-$ axis. A charged particle of mass $1\,kg$ and of charge $1\, C$ is having velocity $1\, m/sec$ along $x-$ axis and is at origin at $t = 0.$ Then the co-ordinates of particle at time $\pi$ seconds will be :-

The magnetic field in a plane electromagnetic wave is given by, $B=3.01 \times 10^{-7} \sin \left(6.28 \times 10^2 \times+2.2 \times 10^{10} t\right) \,T$. [where $x$ in $cm$ and $t$ in second]. The wavelength of the given wave is ....... $cm$

The electric field in an electromagnetic wave is given by $E =56.5 \sin \omega( t - x / c ) \;NC ^{-1}$. Find the intensity of the wave if it is propagating along $x-$axis in the free space. (Given $\left.\varepsilon_{0}=8.85 \times 10^{-12} \;C ^{2} N ^{-1} m ^{-2}\right)$

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10}\; Hz$ and amplitude $48\; Vm ^{-1}$

$(a)$ What is the wavelength of the wave?

$(b)$ What is the amplitude of the oscillating magnetic field?

$(c)$ Show that the average energy density of the $E$ field equals the average energy density of the $B$ field. $\left[c=3 \times 10^{8} \;m s ^{-1} .\right]$