Question

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 1

Frequency of the electromagnetic wave, v = 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 = "c"/"v"`

= `(3 xx 10^8)/(2 xx 10^10)`

= 0.015 m

(b) Magnetic field strength is given as:

`"B"_0 = "E"_0/"c"`

= `48/(3 xx 10^8)`

= 1.6 × 10⁻⁷ T

(c) Energy density of the electric field is given as:

`"U"_"E" = 1/2in_0"E"^2`

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

`"U"_"B"= 1 /(2μ_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 = `1/(sqrt(in_0μ_0))` ….......(2)

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

`"E" = 1/sqrt(in_0μ_0)"B"`

Squaring both sides, we get

`"E"^2 = 1/(in_0μ_0)"B"^2`

`in_0"E"^2 = "B"^2/μ_0`

`1/2in_0"E"^2 = 1/2 "B"^2/μ_0`

U_E = U_B