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Physics

Electromagnetic waves

1. Characterize the nature and properties of electromagnetic waves.

Solution:

Every change in the magnetic field induces a varying electric field. Every change in the electric field induces a varying magnetic field. The magnetic and electric fields always exist simultaneously and form an electromagnetic field, which propagates through space at a speed of c = 3.108 m.s-1.

Phase velocity of electromagnetic wave is:

v=1εμv = \frac{1}{\sqrt{\varepsilon \mu}}

A progressive electromagnetic wave is characterized by the equation:

u=Umsin⁡(2π(tT−xλ)),λ=cT=cfu = U_m \sin \left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right), \quad \lambda = cT = \frac{c}{f} i=Imsin⁡(2π(tT−xλ)),x=distance from sourcei = I_m \sin \left(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\right), \quad x = \text{distance from source}

A dipole (radiator, antenna) is an open oscillating LC circuit that emits (receives) electromagnetic waves.

Half-wave radiator:

l=λ2=c2fl = \frac{\lambda}{2} = \frac{c}{2f}

Electromagnetic wave does not propagate along the conductors but arises between the conductors.

Two conductors can be connected:

  • Short-circuited – at the end of the line a conductor is connected; there is zero voltage and a current antinode.

  • Open-circuited – at the end of the line no conductor is connected; there is a voltage antinode and a current node.

The interference of the original and reflected wave arises between the conductors → “standing electromagnetic wave”

Amplitudes of standing electromagnetic wave:


Rλ Voltage Current
Short-circuited R→0R \to 0,     λ=4l\lambda    U0=2Umsin⁡2πxλU_0 = 2U_m \sin \frac{2\pi x}{\lambda}     I0=2Imcos⁡2πxλI_0 = 2I_m \cos \frac{2\pi x}{\lambda}
Open-circuited R→∞,     λ=2l\lambda    U0=2Umcos⁡2πxλU_0 = 2U_m \cos \frac{2\pi x}{\lambda}     I0=2Imsin⁡2πxλI_0 = 2I_m \sin \frac{2\pi x}{\lambda}

 

2.Characterize the individual types of electromagnetic radiation.

  1. Radio radiation
  2. Microwaves
  3. Infrared radiation
  4. Visible light
  5. Ultraviolet radiation
  6. X-ray radiation
  7. Gamma radiation
  8. Cosmic radiation

Solution:


  • Radio radiation: f=105 Hz–109 Hz, (λ=102 m–10−1 m)f = 10^{5}\,\text{Hz} – 10^{9}\,\text{Hz}, \; (\lambda = 10^{2}\,\text{m} – 10^{-1}\,\text{m})
    Used for sound and image transmission – radio, television, radar.

    Microwaves: f=109 Hz–1012 Hz, (λ=10−1 m–10−4 m)f = 10^{9}\,\text{Hz} – 10^{12}\,\text{Hz}, \; (\lambda = 10^{-1}\,\text{m} – 10^{-4}\,\text{m})
    Thermal vibrations of atoms and molecules.

    Infrared radiation: f=1012 Hz–1014 Hz, (λ=10−4 m–10−5 Hz)
    Sources are warm bodies – the Sun, heated metal, etc. Used as a heat source, in microwaves, remote controls, photography in the dark.

    Visible light: f=3.8⋅1014 Hz–7.8⋅1014 Hz, λ=7.9⋅10−7 m–3.86⋅10−7 m

    Ultraviolet radiation: f=1015 Hz–1016 Hz, (λ=10−7 m–10−8 m)
    The Sun, mercury vapor lamps, electric arcs. Causes tanning of the skin, kills bacteria, at high doses causes skin cancer. Used in so-called “mountain sunlamps.”

    X-ray radiation: f=1016 Hz–1020 Hz, (λ=10−8 m–10−12 m)
    Produced when cathode rays strike an anode target. Used in medicine for diagnosis.

    Gamma radiation: f=1020 Hz–1024 Hz, (λ=10−12 m–10−16 m)
    Very strong penetrating radiation; at small doses it treats cancer, at high doses it arises from nuclear explosions, decay of atomic nuclei, and atomic bombs.

    Cosmic radiation: f=1024 Hz ⇒ (λ=10−16 m)
    A stream of protons or photons falling to Earth from space.

    •  

    3.Calculate the speed of electromagnetic waves in a vacuum and in water, given that for water εr=81, μr = 1.

    Solution:

    Analysis:

    fyzika-elektromagneticke-vlnenie-3.gif 

    The speed of electromagnetic waves in a vacuum is c = 3.108 m.s-1, and in water v = 0,33.108 m.s-1.

    4.The VHF radio communication band has a frequency range of 66 MHz to 73 MHz. Determine the smallest and largest wavelength of electromagnetic waves in this band.

    Solution:

    Analysis:

    fyzika-elektromagneticke-vlnenie-4.gif 

    The wavelength range in the VHF radio communication band is from 4.1 m to 4.5 m.

    5.Determine the length of a half-wave dipole that has capacitance 10 pF and inductance 0.9 μH. (c = 3.108 m.s-1)

    Solution:

    Analysis:

    fyzika-elektromagneticke-vlnenie-5.gif 

    The half-wave dipole has length l = 2.826 m.

    6.The resonant circuit of a receiver is tuned to receive a transmission carried by electromagnetic waves with a wavelength of 5 m. Determine the inductance of the coil in the resonant circuit if its capacitance is 20 pF.

    Solution:

    Analysis:

    fyzika-elektromagneticke-vlnenie-6.gif 

    The inductance of the coil in the resonant circuit is 0.3522 μH.

    7.The resonant circuit of a transmitter consists of a coil with L = 50 μH and a capacitor whose capacitance can be varied from 60 pF to 240 pF. Determine the interval of wavelengths of the electromagnetic waves in which the transmitter operates.

    Solution:

    Analysis:

    fyzika-elektromagneticke-vlnenie-7.gif 

    The transmitter operates over a wavelength interval from 103.2 m to 206.3 m.

    8.A very long two-wire transmission line is connected to an AC voltage source with amplitude 1 V and frequency 75 MHz. Determine the instantaneous voltage between the wires at a distance of 5.5 m from the source at the moment when the source voltage is zero.

    Solution:

    Analysis:

    fyzika-elektromagneticke-vlnenie-8.gif 

    The instantaneous voltages are u1 = –0.71 V and u2 = 0.71 V.

    9.A standing electromagnetic wave is formed between two long conductors. The conductors are open-circuited. At what distance from the ends of the conductors will there be

    • a) a current antinode of the standing electromagnetic wave
    • b) a current node of the standing electromagnetic wave

    Solution:

    fyzika-elektromagneticke-vlnenie-9.gif

    10.An electromagnetic dipole for reception in air has length l = 1.8 m. Determine its length for receiving electromagnetic waves of the same frequency in water. (εr = 81, μr = 1)

    Solution:

    Analysis:

     fyzika-elektromagneticke-vlnenie-10.gif

    The length of the electromagnetic dipole in water is 20 cm.

    11. One resonant circuit is characterized by L1 = 3.10-3 H and C1 = 2.10-6 F, the other by L2 = 4.10-3 H and C2 = 1.10-6 F.

    • Are the circuits in resonance?
    • If not, make a correction.

    Solution:

    Analysis:

    L1 = 3.10-3 H and C1 = 2.10-6 F, L2 = 4.10-3 H and C2 = 1.10-6 F.

    Condition for resonance:
    L1·C1 = L2·C2
    3.10-3 H · 2.10-6 F = 4.10-3 H · 1.10-6 F
    6.10-6 H·F = 4.10-9 H·F — not valid

     fyzika-elektromagneticke-vlnenie-11.gif

    The circuits are not in resonance. There are several ways to fix this. For example, replace capacitor C2 = 1.10-6 F with another capacitor C2 = 1.5.10-6 F.

    12.Radio electromagnetic waves with wavelength λ1 = 375 m strike the interface of two media. The angle of incidence is α = 350, the angle of refraction β = 330. What is the wavelength of these waves λ2 in the second medium?

    Solution:

    Analysis:

    fyzika-elektromagneticke-vlnenie-12.gif 

    The wavelength in the second medium is λ2 = 357 m.