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, ν = 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:

(b) Magnetic field strength is given as:

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

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

Where,

∈₀ = Permittivity of free space

μ₀ = Permeability of free space

We have the relation connecting E and B as:

E = cB … (1)

Where,

… (2)

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

Squaring both sides, we get