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प्रश्न
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1.
- What is the wavelength of the wave?
- What is the amplitude of the oscillating magnetic field?
- 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]
उत्तर
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`
UE = UB
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