Assume a bulb of efficiency $2.5\%$ as a point source. The peak values of electric field produced by the radiation coming from a $100\, W$ bulb at a distance of $3\, m$ is respectively…..$V\,{m^{ – 1}}$

If an electromagnetic wave propagating through vacuum is described by $E_y=E_0 \sin (k x-\omega t)$; $B_z=B_0 \sin (k x-\omega t)$, then

An infinitely long thin wire carrying a uniform linear static charge density $\lambda $ is placed along the $z-$ axis (figure). The wire is set into motion along its length with a uniform velocity $V = v{\hat k_z}$. Calculate the pointing vector $S = \frac{1}{{{\mu _0}}}(\vec E \times \vec B)$ .

In an electromagnetic wave, the amplitude of electric field is $1 V/m.$  the frequency of wave is $5 \times {10^{14}}\,Hz$. The wave is propagating along $z-$ axis. The average energy density of electric field, in $Joule/m^3$, will be

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 $\mu_{0}=4 \pi \times 10^{-7} NA ^{-2}$ ]