8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120N/C and that its frequency is v = 50 MHz. (a) Determine B0, ω, k and λ (b) Find expressions for E and B.
Answer:
Electric field amplitude, E0 = 120N/C
Frequency of source, v = 50 MHz = 50 × 106 Hz
Speed of light, c = 3 × 108 m/s
(a) Magnitude of magnetic field strength is given as:
\(B_{0} = \frac{E_{0}}{c}\)
\(= \frac{120}{3\times 10^{8}}\)
= 40 × 10−8 = 400 × 10−9 T = 400 nT
Angular frequency of source is given by:
ω = 2πv = 2π × 50 × 106 = 3.14 × 108 rads−1
= 3.14 × 108 rad/s
Propagation constant is given as:
\(k = \frac{\omega }{c}\)
\(= \frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m\)
Wavelength of wave is given by:
\(\lambda = \frac{c}{v}\)
= \( \frac{3\times 10^{8}}{50\times 10^{6}}\) = 6.0 m
(b) 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:
\(\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j}\)
\(= 120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j}\)
And, magnetic field vector is given as:
\(\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k}\)
\(\overline{B} = (400 \times 10^{-9}) sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k}\)
9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for the 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?
Answer:
The energy of a photon is given as:
E = hv = \(\frac{hc}{\lambda}\)
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
If the wavelength λ is in metre and the energy is in Joule, then by dividing E by 1.6 × 10-19 will convert the energy into eV.
\(E=\frac{hc}{\lambda \times 1.6\times 10^{-19}}\,eV\)
a) For Gamma rays, the wavelength ranges from 10-10 to 10-14 m, therefore the photon energy can be calculated as follows:
\(E=\frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-10}\times 1.6\times 10^{-19}}=12.4\times 10^3\approx 10^4\,eV\)
Therefore,
λ = 10−10 m, energy = 104 eV
λ = 10−14 m, energy = 108 eV
The energy for Gamma rays ranges from 104 to 108 eV.
b) The wavelength for X-rays ranges between 10-8 m to 10-13 m
For λ = 10-8,
\(Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-8}\times 1.6\times 10^{-19}}=124\approx 10^2\,eV\)
For λ = 10-13 m, energy = 107 eV
c) For ultraviolet radiation, the wavelength ranges from 4 × 10-7 m to 6 × 10-7 m.
For 4 × 10-7 m,
\(Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{4\times 10^{-7}\times 1.6\times 10^{-19}}=3.1\approx 10^{10}\,eV\)
For 6 × 10-7 m, the energy is equal to 103 eV.
The energy of the ultraviolet radiation varies between 1010 to 103 eV.
d) For visible light, the wavelength ranges from 4 × 10-7 m to 7 × 10-7 m.
For 4 × 10-7, the energy is the same as above, that is 1010 eV
For 7 × 10-7 m, the energy is 100 eV
e) For infrared radiation, the wavelength ranges between 7 × 10-7 m to 7 × 10-14 m.
The energy for 7 × 10-7 m is 100 eV
The energy for 7 × 10-14 m is 10-3 eV
f) For microwaves, the wavelength ranges from 1 mm to 0.3 m.
For 1 mm, the energy is 10-3 eV.
For 0.3 m, the energy is 10-6 eV.
g) For radio waves, the wavelength ranges from 1 m to few km.
For 1 m, the energy is 10-6 eV.
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.
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.
(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 × 108 ms−1]
Answer:
Frequency of the electromagnetic wave, v = 2 × 1010 Hz
Electric field amplitude, E0 = 48 V m−1
Speed of light, c = 3 × 108 m/s
(a) Wavelength of a wave is given as:
\(\lambda = \frac{c}{v}\)
\(= \frac{3\times 10^{8}}{2\times 10^{10}} = 0.015\; m\)
(b) Magnetic field strength is given as:
\(B_{0} = \frac{E_{0}}{c}\)
\(= \frac{48}{3\times 10^{8}} = 1.6\times 10^{-7}\; T\)
(c) Energy density of the electric field is given as:
\(U_{E} = \frac{1}{2}\; \epsilon _{0} \;E^{2}\)
And, energy density of the magnetic field is given as:
\(U_{B} = \frac{1}{2\mu_{0}}B^{2}\)
Where,
ϵ0 = Permittivity of free space
μ0 = Permeability of free space
E = cB …(1)
Where,
\(c = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\) …(2)
Putting equation (2) in equation (1), we get
\(E = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\; B\)
Squaring on both sides, we get
\(E^{2} = \frac{1}{\epsilon_{0}\; \mu_{0}}\; B^{2}\)
\(\epsilon_{0}\; E^{2} = \frac{B^{2}}{\mu_{0}}\)
\(\frac{1}{2}\; \epsilon_{0}\; E^{2} = \frac{1}{2}\; \frac{B^{2}}{\mu_{0}}\)
⇒ UE = UB