The electric field part of an electromagnetic wave in vacuum is
$E = 3.1 \, N C^{-1} \cos [ (1.8 \, rad \, m^{-1}) y + (5.4 \times 10^8 \, rad \, s^{-1}) t ] \hat{i}$
The wavelength of this part of the electromagnetic wave is ...... $m$.

The peak value of the magnetic field of an electromagnetic wave moving in a magnetic field is $20 \, nT$. The peak value of the electric field intensity is ..... $V/m$.

The electric field of a plane electromagnetic wave in a medium is given by $\vec{E}(x, y, z, t) = E_0 \hat{n} e^{i k_0[(x+y+z)-ct]}$, where $c$ is the speed of light in free space. The $\vec{E}$ field is polarized in the $x-z$ plane. If the speed of the wave in the medium is $v$, then:

$A$ laser beam has an intensity of $4.0 \times 10^{14} \ W/m^{2}$. The amplitude of the magnetic field associated with the beam is . . . . . . $T$. (Take $\epsilon_{0} = 8.85 \times 10^{-12} \ C^{2}/Nm^{2}$ and $c = 3 \times 10^{8} \ m/s$)

$A$ plane electromagnetic wave is given by $E_z = 100 \cos(6 \times 10^8 t + 4x) \text{ V/m}$. The refractive index of the medium in which the wave is traveling is:

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. $[c = 3 \times 10^{8} \; m s^{-1}]$.