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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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Class 12 Physics
CBSE and UP Board
Electromagnetic Waves
Chapter-8 Exercise 8.10
NCERT Solutions for Class 12 Physics Chapter 8 Question 10

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1 Answer

  1. 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 ε0

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

    UB = 1/2 μ0

    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

     

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