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 10⁻⁷Read more
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:
Energy of a photon is given as: E = hν = hc/λ Where, h = Planck’s constant = 6.6 x 10-34 Js c = Speed of light = 3 x 1o8 m/s λ = Wavelength of radiation Therefore, E = ( 6.6 x 10-34 x 3 x 1o8) /λ = (19.8 x 10⁻²⁶ )/λ J = (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹) = (12.375x10⁻7 )/λ eV The given table lRead more
Energy of a photon is given as:
E = hν = hc/λ
Where,
h = Planck’s constant = 6.6 x 10-34 Js
c = Speed of light = 3 x 1o8 m/s
λ = Wavelength of radiation
Therefore,
E = ( 6.6 x 10-34 x 3 x 1o8) /λ
= (19.8 x 10⁻²⁶ )/λ J
= (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹)
= (12.375×10⁻7 )/λ eV
The given table lists the photon energies for different parts of an electromagnetic spectrum for different λ are:
λ(m)
10³
1
10⁻³
10⁻⁶
10⁻⁸
10⁻¹⁰
10⁻¹²
E (eV)
12.375 x 10⁻¹⁰
12.375 x 10⁻⁷
12.375 x 10⁻⁴
12.375 x 10⁻¹
12.375 x 10¹
12.375 x 10³
12.375 x 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.
Electric field amplitude, Eo = 120 N/C Frequency of source, ν = 50.0 MHz = 50 x 106 Hz Speed of light, c = 3 x 108m/s Ans (a). Magnitude of magnetic field strength is given as: B0 =E0 /c = 120/(3 x 108) = 4 x 10⁻7 T= 400 nT Angular frequency of source is given as: ω =2πν = 2π x 50 x 106 = 3.14 x 108Read more
Electric field amplitude, Eo = 120 N/C
Frequency of source, ν = 50.0 MHz = 50 x 106 Hz
Speed of light, c = 3 x 108m/s
Ans (a).
Magnitude of magnetic field strength is given as:
B0 =E0 /c
= 120/(3 x 108)
= 4 x 10⁻7 T= 400 nT
Angular frequency of source is given as:
ω =2πν = 2π x 50 x 106 = 3.14 x 108 rad/s
Propagation constant is given as:
k = ω/c = (3.14 x 108 )/(3 x 108) =6.0 m
Ans (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.
Amplitude of magnetic field of an electromagnetic wave in a vacuum, Bo = 510 nT = 510 x 10⁻9T Speed of light in a vacuum, c = 3 x 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cBo = 3 x 108x 510 x 10⁻9= 153 N/C Therefore, the electric field part of theRead more
Amplitude of magnetic field of an electromagnetic wave in a vacuum, Bo = 510 nT = 510 x 10⁻9T Speed of light in a vacuum, c = 3 x 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cBo
= 3 x 108x 510 x 10⁻9= 153 N/C Therefore, the electric field part of the wave is 153 N/C.
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
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.]
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 10⁻⁷Read more
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
See lessThe 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?
Energy of a photon is given as: E = hν = hc/λ Where, h = Planck’s constant = 6.6 x 10-34 Js c = Speed of light = 3 x 1o8 m/s λ = Wavelength of radiation Therefore, E = ( 6.6 x 10-34 x 3 x 1o8) /λ = (19.8 x 10⁻²⁶ )/λ J = (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹) = (12.375x10⁻7 )/λ eV The given table lRead more
Energy of a photon is given as:
E = hν = hc/λ
Where,
h = Planck’s constant = 6.6 x 10-34 Js
c = Speed of light = 3 x 1o8 m/s
λ = Wavelength of radiation
Therefore,
E = ( 6.6 x 10-34 x 3 x 1o8) /λ
= (19.8 x 10⁻²⁶ )/λ J
= (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹)
= (12.375×10⁻7 )/λ eV
The given table lists the photon energies for different parts of an electromagnetic spectrum for different λ are:
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.
See lessSuppose that the electric field amplitude of an electromagnetic wave is Eo = 120 N/C and that its frequency is v = 50.0 MHz. (a) Determine, Bo, ω, k, and λ. (b) Find expressions for E and B.
Electric field amplitude, Eo = 120 N/C Frequency of source, ν = 50.0 MHz = 50 x 106 Hz Speed of light, c = 3 x 108m/s Ans (a). Magnitude of magnetic field strength is given as: B0 =E0 /c = 120/(3 x 108) = 4 x 10⁻7 T= 400 nT Angular frequency of source is given as: ω =2πν = 2π x 50 x 106 = 3.14 x 108Read more
Electric field amplitude, Eo = 120 N/C
Frequency of source, ν = 50.0 MHz = 50 x 106 Hz
Speed of light, c = 3 x 108m/s
Ans (a).
Magnitude of magnetic field strength is given as:
B0 =E0 /c
= 120/(3 x 108)
= 4 x 10⁻7 T= 400 nT
Angular frequency of source is given as:
ω =2πν = 2π x 50 x 106 = 3.14 x 108 rad/s
Propagation constant is given as:
k = ω/c = (3.14 x 108 )/(3 x 108) =6.0 m
Ans (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:
E = E0 sin (kx-ωt)j
= 120 sin [ 1.05 x -3.14 x 10⁸ t] j
And, Magnetic field vector is given as :
B = B0 sin (kx-ωt)k
= (4 x 10 sin⁻⁷ )[ 1.05 x -3.14 x 10⁸ t] k
See lessThe amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is Bo = 510 nT. What is the amplitude of the electric field part of the wave?
Amplitude of magnetic field of an electromagnetic wave in a vacuum, Bo = 510 nT = 510 x 10⁻9T Speed of light in a vacuum, c = 3 x 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cBo = 3 x 108x 510 x 10⁻9= 153 N/C Therefore, the electric field part of theRead more
Amplitude of magnetic field of an electromagnetic wave in a vacuum,
See lessBo = 510 nT = 510 x 10⁻9T
Speed of light in a vacuum, c = 3 x 108 m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cBo
= 3 x 108x 510 x 10⁻9= 153 N/C
Therefore, the electric field part of the wave is 153 N/C.
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
See less