Class 12 Physics
CBSE and UP Board
Electromagnetic Waves
Chapter-8 Exercise 8.10
NCERT Solutions for Class 12 Physics Chapter 8 Question 10
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 x 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 x 10⁸m s⁻¹.]
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Frequency of the electromagnetic wave,ν = 2.0 x 1010 Hz
Electric field amplitude, Eo = 48 V m⁻¹
Speed of light, c = 3 x 108 m/s
Ans (a).
Wavelength of a wave is given by :
λ = c/ν
= (3 x 108) /(2.0 x 1010) = 0.015 m
Ans (b).
Magnetic field strength is given by:
B0= E0/c = 48 /(3 x 108)= 1.6 x 10⁻⁷ T
Ans (c).
Energy density of the electric field is given by :
UE = 1/2 ε0E²
And, energy density of the magnetic field is given as:
UB = 1/2 μ0B²
Where,
ε0 = Permittivity of free space
μ0 = Permeability of free space We have the relation connecting E and B as:
E = cB———————–Eq-1
Where,
c = 1/√(ε0μ0)————-Eq-2
Putting equation (2) in equation (1), we get
E = [1/√(ε0μ0)]B
Squaring both sides, we get
E² = [1/(ε0μ0)]B²
=> E²ε0 = B²/μ0
=>1/2 x E²ε0 = 1/2 x B²/μ0
=> UE = UB