An electromagnetic wave of frequency $2 \text{ MHz}$ propagates from vacuum to a non-magnetic medium of relative permittivity $9$. Then its wavelength

The efficiency of a bulb of power $60 \ W$ is $16 \%$. The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of $2 \ m$ from the bulb is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}\right)$ (in $Vm^{-1}$)

$A$ radio transmitter transmits at $830 \, kHz$. At a certain distance from the transmitter,the magnetic field has an amplitude of $4.82 \times 10^{-11} \, T$. The electric field and the wavelength are respectively:

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.

Show that the average value of radiant flux density $S$ over a single period $T$ is given by $S = \frac{1}{2c\mu_0}E_0^2$.

An electromagnetic wave is represented by the electric field $\vec{E} = E_0 \hat{n} \sin [\omega t + (6y - 8z)]$. Taking unit vectors in $x, y$ and $z$ directions to be $\hat{i}, \hat{j}$ and $\hat{k}$,the direction of propagation $\hat{s}$ is: