In a plane electromagnetic wave,
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 E field equals the average energy density of the B field.
[c = 3 × 108 m s−1.]
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
$E^{2}=\frac{1}{\epsilon_{0} \mu_{0}} B^{2}$
$\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$
$\frac{1}{2} \in_{0} E^{2}=\frac{1}{2} \frac{B^{2}}{\mu_{0}}$
$\Rightarrow U_{E}=U_{B}$