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.]

Answer :

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:


......................(ii)


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



Or,



From equation (i) and (ii), we get


UE = UB


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