In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 x 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 x 10⁸m s⁻¹.]

Frequency of the electromagnetic wave,ν = 2.0 x 10¹⁰ Hz

Electric field amplitude, Eo = 48 V m⁻¹

Speed of light, c = 3 x 10⁸ m/s

Ans (a).

Wavelength of a wave is given by :

λ = c/ν 

=  (3 x 10⁸) /(2.0 x 10¹⁰) = 0.015 m

Ans (b).

Magnetic field strength is given by:

B₀= E₀/c = 48 /(3 x 10⁸)= 1.6 x 10⁻⁷ T

Ans (c).

Energy density of the electric field is given by :

U_E = 1/2 ε₀E²

And, energy density of the magnetic field is given as:

U_B = 1/2 μ₀B²

Where,

ε₀ = Permittivity of free space

μ₀ = Permeability of free space We have the relation connecting E and B as:

E = cB———————–Eq-1

Where,

c = 1/√(ε₀μ₀)————-Eq-2

Putting equation (2) in equation (1), we get

E = [1/√(ε₀μ₀)]B

Squaring both sides, we get

E² = [1/(ε₀μ₀)]B²

=>  E²ε₀ = B²/μ₀

=>1/2 x E²ε₀ = 1/2 x B²/μ₀

=> U_E = U_B