# 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: ......................(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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