The electric field of an electromagnetic wave in free space is represented as $\vec{E}=E_0 \cos (\omega t-k z) \hat{i}$.The corresponding magnetic induction vector will be :

A beam of light travelling along $X$-axis is described by the electric field $E _{ y }=900 \sin \omega( t - x / c )$. The ratio of electric force to magnetic force on a charge $q$ moving along $Y$-axis with a speed of $3 \times 10^{7}\,ms ^{-1}$ will be.

[Given speed of light $=3 \times 10^{8}\,ms ^{-1}$ ]

The electric field in an electromagnetic wave is given by $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right) N \mathrm{NC}^{-1}$. The magnetic field induction of this wave is (in SI unit):

If ${\varepsilon _0}$ and ${\mu _0}$ are respectively, the electric permittivity and the magnetic permeability of free space. $\varepsilon $ and $\mu $ the corresponding quantities in a medium, the refractive index of the medium is

The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }$, then the associated electric field will be :

The intensity of the light from a bulb incident on a surface is $0.22 \,W / m ^{2}$. The amplitude of the magnetic field in this light-wave is_______ $\times 10^{-9} \,T$. (Given : Permittivity of vacuum $\epsilon_{0}=8.85 \times 10^{-12} \,C ^{2} N ^{-1} m ^{-2}$, speed of light in vacuum $c =3 \times 10^{8} \,ms ^{-1}$ )