The ratio of the amplitude of the magnetic field to the amplitude of the electric field for an electromagnetic wave propagating in vacuum is equal to:

$A$ radio can tune in to any station in the $7.5\; MHz$ to $12\; MHz$ band. What is the corresponding wavelength band?

In an electromagnetic wave,if $\overrightarrow{E}$ is the electric field and $\overrightarrow{B}$ is the magnetic field,then the direction of propagation of the wave is given by:

The electric field in an electromagnetic wave is given by $E = (50 \, NC^{-1}) \sin \omega(t - x/c)$. The energy contained in a cylinder of volume $V$ is $5.5 \times 10^{-12} \, J$. The value of $V$ is $...... \, cm^3$ (given $\epsilon_0 = 8.8 \times 10^{-12} \, C^2 N^{-1} m^{-2}$).

The electric field component of an electromagnetic wave in vacuum is given as $\vec E = 3\cos(1.8y + 5.4 \times 10^8 t)\hat i$. Its direction of propagation and wavelength are:

In a plane electromagnetic wave,the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \; Hz$ and amplitude $48 \; V m^{-1}$.
$(a)$ What is the wavelength of the wave?
$(b)$ What is the amplitude of the oscillating magnetic field?
$(c)$ Show that the average energy density of the $E$ field equals the average energy density of the $B$ field. $[c = 3 \times 10^{8} \; m s^{-1}]$.