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.]
Solution
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:
`lambda=c/v`
`=(3xx10^8)/(2xx10^10)=0.015m`
(b) Magnetic field strength is given as:
`B_0=E_0/c`
`=48/(3xx10^8)=1.6xx10^-7 T`
(c) Energy density of the electric field is given as:
`U_E=1/2in_0E^2`
And, energy density of the magnetic field is given as:
`U_B=1/(2mu_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_0mu_0))` … (2)
Putting equation (2) in equation (1), we get
`E=1/sqrt(in_0mu_0)`
Squaring both sides, we get
`E=1/(in_0mu_0) B^2`
`in_0E^2=B^2/mu_0`
