A point source of electromagnetic radiation has an average power output of $800\,W$ . The maximum value of electric field at a distance $3.5\,m$ from the source will be…..$V/m$

A radiation is emitted by $1000\, W$ bulb and it generates an electric field and magnetic field at $P$, placed at a distance of $2\, m$. The efficiency of the bulb is $1.25 \%$. The value of peak electric field at $P$ is $x \times 10^{-1} \,V / m$. Value of $x$ is. (Rounded-off to the nearest integer)

[Take $\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}, c =3 \times 10^{8}$ $ms ^{-1}$ ]

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.

If electric field intensity of a uniform plane electro magnetic wave is given as

$E =-301.6 \sin ( kz -\omega t ) \hat{a}_{ x }+452.4 \sin ( kz -\omega t )$ $\hat{a}_{y} \frac{V}{m}$

Then, magnetic intensity $H$ of this wave in $Am ^{-1}$ will be

[Given: Speed of light in vacuum $c =3 \times 10^{8} ms ^{-1}$, permeability of vacuum $\mu_{0}=4 \pi \times 10^{-7} NA ^{-2}$ ]

An electron is constrained to move along the $y-$axis with a speed of $0.1\, c$ (c is the speed of light) in the presence of electromagnetic wave, whose electric field is $\overrightarrow{ E }=30 \hat{ j } \sin \left(1.5 \times 10^{7} t -5 \times 10^{-2} x \right)\, V / m$ The maximum magnetic force experienced by the electron will be: (given $c=3 \times 10^{8}\, ms ^{-1}$ and electron charge $\left.=1.6 \times 10^{-19} C \right)$