Q. 103.6( 9 Votes )
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.]
Given: Frequency of electromagnetic wave f = 2.0×1010 Hz
Amplitude of the electromagnetic wave i.e. E0 = 48 V m–1
(a) Wavelength of the electromagnetic wave can be calculated as follows:
v = c / λ
also, λ = c/v
where c is the speed of light in vacuum
v is the frequency of electromagnetic wave
and λ is the wavelength of electromagnetic wave
Substituting the values
On calculating, we get
⇒ λ = 0.015 m
(b) Amplitude of the electromagnetic wave i.e. E0 = 48 V m–1
Amplitude of the oscillating magnetic field is given by the following expression:
B0 = E0/c
On calculating we get:
⇒ B0 = 1.6 × 10-7 Tesla
(c) The average energy density of the E field equals the average energy density of the B field can be proved as follows:
The energy density of electric field (i)
Where is the permittivity of free space
The energy density of magnetic field is given as follows:
Where is the permeability of free space
The relation between Electric field and magnetic field i.e. E and B is as follows:
On squaring both the sides, we get following relation
From equation (i) and (ii), we get
∴ UE = UB
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