# In a plane electromagnetic wave,

Question:

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 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 field equals the average energy density of the field.

[= 3 × 108 m s−1.]

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

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

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

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

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