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NCERT Class XII Chapter
Electromagnetic Waves
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Question : 10 of 15
Marks: +1, -0
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 101010^{10} Hz and amplitude 48 V m−1m^{-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 E⃗\vec{E} field equals the average energy density of the B⃗\vec{B} field. [c = 3 × 10810^8 m s−1s^{-1}.]
Solution:  
(a) Wavelength of electromagnetic wave is
λ = cv\frac{c}{v} = 3×1082×1010\frac{3\times10^8}{2\times10^{10}} = 1.5 × 10−210^{-2} m
(b) Amplitude of magnetic field, E0B0\frac{E_0}{B_0} = c
B0B_0 = E0c\frac{E_0}{c} = 483×108\frac{48}{3\times10^8} = 16 × 10−810^{-8} T
(c) Energy density as electric field, uEu_E = 12ϵ0E2\frac{1}{2} \epsilon_0 E^2
Here, EB\frac{E}{B} = c
So, uEu_E = 12ϵ0c2B2\frac{1}{2} \epsilon_0 c^2 B^2
where speed of electromagnetic wave, c = 1μ0ϵ0\frac{1}{\sqrt{\mu_0 \epsilon_0}}
so uEu_E = 12ϵ0μ0ϵ0B2\frac{1}{2} \frac{\epsilon_0}{\mu_0 \epsilon_0} B^2 = B22μ0\frac{B^2}{2 \mu_0} = μB\mu_B
[Energy density as magnetic field]
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