In a plane electromagnetic wave,

Solution:

Frequency of the electromagnetic wave, ν = 2.0 × 1010 Hz

Electric field amplitude, E0 = 48 V m−1

Speed of light, c = 3 × 108 m/s

(a) Wavelength of a wave is given as:

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

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

(b) Magnetic field strength is given as:

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

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

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

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

And, energy density of the magnetic field is given as:

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

Where,

∈0 = Permittivity of free space

μ0 = Permeability of free space

We have the relation connecting E and B as:

E = cB … (1)

Where,

$c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} \ldots$   (2)

Putting equation (2) in equation (1), we get

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

Squaring both sides, we get

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

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

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

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