In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10}\; Hz$ and amplitude $48\; Vm ^{-1}$

$(a)$ 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. $\left[c=3 \times 10^{8} \;m s ^{-1} .\right]$

Frequency of the electromagnetic wave, $v=2.0 \times 10^{10} Hz$

Electric field amplitude, $E _{0}=48 V m ^{-1}$

Speed of light, $c=3 \times 10^{8} m / s$

$(a)$ Wavelength of a wave is given as:

$\lambda=\frac{c}{v}$

$=\frac{3 \times 10^{8}}{2 \times 10^{10}}=0.015 m$

$(b)$ Magnetic field strength is given as:

$B_{0}=\frac{E_{0}}{c}$

$=\frac{48}{3 \times 10^{8}}=1.6 \times 10^{-7} T$

$(c)$ Energy density of the electric field is given as:

$U_{E}=\frac{1}{2} \epsilon_{0} E^{2}$

And, energy density of the magnetic field is given as

$U_{B}=\frac{1}{2 \mu_{0}} B^{2}$

Where,

$\varepsilon_{0}=$ Permittivity of free space $\mu_{0}=$ Permeability of free space

We have the relation connecting $E$ and $B$ as:

$E = cB \ldots(i)$

Where,

$c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} \dots(ii)$

Putting equation $(ii)$ in equation $(i),$ we get

$E=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} B$

Squaring both sides, we get

$E^{2}=\frac{1}{\epsilon_{0} \mu_{0}} B^{2}$

$\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$

$\frac{1}{2} \epsilon_{0} E^{2}=\frac{1}{2} \frac{B^{2}}{\mu_{0}}$

$\Rightarrow U_{E}=U_{B}$