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NEET-XII-Physics

08: Electromagnetic Waves

page 2

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  • #8-a
    Determine, B0, ω, k, and λ.
    Ans : Magnitude of magnetic field strength is given as:



    Angular frequency of source is given as:

    ω = 2``\pi``ν

    = 2``\pi`` × 50 × 106

    = 3.14 × 108 rad/s

    Propagation constant is given as:



    Wavelength of wave is given as:


  • #8-b
    Find expressions for E and B.
    Ans : Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.

    Equation of electric field vector is given as:



    And, magnetic field vector is given as:


  • Qstn #9
    The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
    Ans : Energy of a photon is given as:



    Where,

    h = Planck’s constant = 6.6 × 10-34 Js

    c = Speed of light = 3 × 108 m/s

    λ = Wavelength of radiation



    The given table lists the photon energies for different parts of an electromagnetic spectrum for differentλ.






    λ (m) 103 1 10-3 10-6 10-8 10-10 10-12
    E (eV) 12.375 × 10-10 12.375 × 10-7 12.375 × 10-4 12.375 × 10-1 12.375 × 101 12.375 × 103 12.375 × 105



    The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.
  • Qstn #10
    In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m-1.
    Ans : 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
  • #10-a
    What is the wavelength of the wave?
    Ans : Wavelength of a wave is given as:


  • #10-b
    What is the amplitude of the oscillating magnetic field?
    Ans : Magnetic field strength is given as:


  • #10-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.]
    Ans : Energy density of the electric field is given as:



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



    Where,

    ∈0 = Permittivity of free space

    μ0 = Permeability of free space

    We have the relation connecting E and B as:

    E = cB ... (1)

    Where,

    ... (2)

    Putting equation (2) in equation (1), we get



    Squaring both sides, we get


  • Qstn #11
    Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]} .
  • #11-a
    What is the direction of propagation?
    Ans : From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y direction i.e., .
  • #11-b
    What is the wavelength λ?
    Ans : It is given that,



    The general equation for the electric field vector in the positive x direction can be written as:



    On comparing equations (1) and (2), we get

    Electric field amplitude, E0 = 3.1 N/C

    Angular frequency, ω = 5.4 × 108 rad/s

    Wave number, k = 1.8 rad/m

    Wavelength, = 3.490 m
  • #11-c
    What is the frequency ν?
    Ans : Frequency of wave is given as:


  • #11-d
    What is the amplitude of the magnetic field part of the wave?
    Ans : Magnetic field strength is given as:



    Where,

    c = Speed of light = 3 × 108 m/s


  • #11-e
    Write an expression for the magnetic field part of the wave.
    Ans : On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:


  • Qstn #12
    About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
    Ans : Power rating of bulb, P = 100 W

    It is given that about 5% of its power is converted into visible radiation.

    Power of visible radiation,



    Hence, the power of visible radiation is 5W.
  • #12-a
    at a distance of 1 m from the bulb?
    Ans : Distance of a point from the bulb, d = 1 m

    Hence, intensity of radiation at that point is given as:


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