In a certain region uniform electric field $E$ and magnetic field $B$ are present in the opposite direction. At the instant $t = 0,$ a particle of mass $m$ carrying a charge $q$ is given velocity $v_o$ at an angle $\theta ,$ with the $y$ axis, in the $yz$ plane. The time after which the speed of the particle would be minimum is equal to

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

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 $10m$  from the lamp will be……..$V/m$

For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric $\left( U _{ e }\right)$ and magnetic $\left( U _{ m }\right)$ fields is

A plane $EM$ wave travelling along $z-$ direction is described$\vec E = {E_0}\,\sin \,(kz – \omega t)\hat i$ and $\vec B = {B_0}\,\sin \,(kz – \omega t)\hat j$. Show that

$(i)$ The average energy density of the wave is given by $U_{av} = \frac{1}{4}{ \in _0}E_0^2 + \frac{1}{4}.\frac{{B_0^2}}{{{\mu _0}}}$

$(ii)$ The time averaged intensity of the wave is given by  $ I_{av}= \frac{1}{2}c{ \in _0}E_0^2$ વડે આપવામાં આવે છે.