For an electromagnetic wave,the electric field is given by $E_x = 36 \sin(1.20 \times 10^7 z - 3.6 \times 10^{15} t) \ V/m$. What is the intensity of the electric field in $W/m^2$?

The magnetic field of an $\text{E.M.}$ wave is given by $\overrightarrow{ B }=\left(\frac{\sqrt{3}}{2} \hat{ i }+\frac{1}{2} \hat{ j }\right) 30 \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right]$ ($\text{S.I.}$ units). The corresponding electric field in $\text{S.I.}$ units is:

The velocity factor of a transmission line is $x$. If the dielectric constant of the medium is $2.6$,the value of $x$ is:

An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant $\epsilon_r = 9$, has the electric field, $E_x = E_0 \sin (kz - 2\pi \times 10^6 t) \text{ Vm}^{-1}$ where $E_0$ is the amplitude and $k$ is the wave vector. Among the following options, the incorrect choice is :

If the amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $270 \ nT$,the amplitude of the electric field part of the wave is: (in $NC^{-1}$)

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.