Electromagnetic waves are a fascinating phenomenon where electric and magnetic fields oscillate perpendicular to each other and to the direction of wave propagation. These waves are responsible for a wide range of phenomena, from visible light to radio waves and X-rays.
Key Concepts:
Generation of Electromagnetic Waves:
Accelerating Charges: Electromagnetic waves are generated by accelerating charges. When a charged particle accelerates, it produces time-varying electric and magnetic fields that propagate outwards as electromagnetic waves.
Maxwell’s Equations: These equations, formulated by James Clerk Maxwell, unified the fields of electricity and magnetism, providing a theoretical foundation for electromagnetic waves.
Properties of Electromagnetic Waves:
Transverse Waves: Electromagnetic waves are transverse, meaning the electric and magnetic fields vibrate perpendicular to the wave’s direction of travel.
Speed of Light: Electromagnetic waves travel at the speed of light in vacuum, which is approximately 3 x 10^8 m/s.
Wavelength and Frequency: Electromagnetic waves are characterized by their wavelength (distance between two consecutive crests or troughs) and frequency (number of oscillations per second). The relationship between these two is expressed by the following equation:
c = λf
Energy: Electromagnetic waves carry energy, which is proportional to the square of the amplitude of the electric and magnetic fields.
Electromagnetic Spectrum:
The electromagnetic spectrum is a broad range of electromagnetic waves, classified based on their frequency or wavelength. It includes:
Radio Waves: Radio waves are utilized in communication systems, broadcasting, and radar technologies.
Microwaves : Microwaves find applications in communication systems, radar systems, and microwave ovens.
Infrared Radiation: Used for thermal imaging and remote sensing.
Visible Light: The range of electromagnetic waves that humans can see.
Ultraviolet Radiation: Used for sterilization and medical treatments.
X-rays: Used in material analysis & widely medical imaging .
Gamma Rays: Used in medical treatments and nuclear physics research.
Applications of Electromagnetic Waves:
Electromagnetic waves find numerous applications in our daily lives, such as:
Communication: Radio, television, mobile phones, and satellite communication.
Medicine: X-rays, CT scans, and MRI scans.
Remote Sensing: Weather forecasting, satellite imagery, and GPS.
Industry: Lasers, microwave ovens, and industrial heating.
Astronomy: Studying celestial objects and the universe.
Electromagnetic waves are a fundamental part of our modern world, enabling us to communicate, explore, and understand the universe. By understanding the properties and applications of electromagnetic waves, we can continue to innovate and develop new technologies that shape our future.
1. Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Ans :Analyzing the Capacitor and Charging Current
Given:
Radius of circular plates,
r = 12 cm
= 0.12 m
Separation distance,
d = 5.0 cm
= 0.05 m
Charging current,
I = 0.15 A
Part (a): Calculating Capacitance and Rate of Change of Potential Difference
Capacitance:
The capacitance C of a parallel-plate capacitor, a measure of its capability to store electrical charge, is:
C = ε₀ * (A/d)
where:
ε₀ = permittivity of free space
(8.85 × 10⁻¹² F/m)
A = area of the plates
(πr²)
d is the separation between the plates
Substituting the values:
C = (8.85 × 10⁻¹²) * (π * (0.12 m)² / 0.05 m) ≈ 8.01 × 10⁻¹² F
Rate of change of potential difference (dV/dt):
The current (I) flowing into a capacitor is related to the rate of change of charge (dQ/dt) on its plates:
I = dQ/dt
The charge (Q) on a capacitor is related to its capacitance (C) and the potential difference (V) across its plates:
Q = CV
Differentiating both sides with respect to time:
dQ/dt = C * dV/dt
Combining the two equations:
I = C * (dV/dt)
Solving for dV/dt:
dV/dt = I / C = 0.15 A / 8.01 × 10⁻¹² F ≈ 1.87 × 10¹⁰ V/s
Part (b): Displacement Current
As a capacitor charges, a displacement current (Id) flows between its plates. This current is equivalent to the rate of change of electric flux through the plates. As per Maxwell’s equations, this displacement current is identical to the conduction current (I).
Id = I = 0.15 A
Part (c): Kirchhoff’s First Rule
Kirchhoff’s first rule (junction rule) is valid at each plate of the capacitor. This is because the current flowing into the plate from the external source is equal to the displacement current flowing between the plates.
At the positive plate, the conduction current flows into the plate, and the displacement current flows out of the plate. At the negative plate, the conduction current flows out of the plate, and the displacement current flows into the plate. In both cases, the algebraic sum of currents at the junction is zero, satisfying Kirchhoff’s first rule.
2.A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1 .
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Ans :Given:
Radius of each circular plate,
R = 6.0 cm
= 0.06 m
Capacitance of the parallel plate capacitor,
C = 100 pF
= 100 x 10^-12 F
Supply voltage, V = 230 V
Angular frequency, ω = 300 rad/s
The RMS value of conduction current (I) is given by:
I = V / Xc
Where,
Xc = Capacitive reactance
= 1 / (ωC)
Therefore,
I = V x ωC
= 230 x 3 x 10^2 x 10^2 x 10^-12
= 6.9 x 10^-6 A
= 6.9 μA
hence, RMS value of conduction current is 6.9 microamperes.
(b) Yes, the conduction current is equal in magnitude to the displacement current.
(c) The magnetic field (B) at a distance r from the axis of a parallel plate capacitor can be calculated using the formula:
B = (μ₀r / 2πR²) * I₀
Where:
μ₀ = permeability of free space
(4π x 10^-7 Tm/A)
r = distance from the axis
(0.03 m)
R = radius of the plates
(0.06 m)
I₀ = maximum value of the current
(√2 * I)
Substituting the given values:
B = (4π x 10^-7 x 0.03 x √2 x 6.9 x 10^-6) / (2π x (0.06)^2)
B = 1.63 x 10^-11 T
Hence, the magnetic field at that point is 1.63 x 10^-11 Tesla.
3. What physical quantity is the same for X-rays of wavelength 10-10m, red light of wavelength 6800 Å and radiowaves of wavelength 500m?
Ans :The physical quantity that remains the same for all these electromagnetic waves is the speed of light in vacuum, which is approximately 3 × 10^8 m/s.
4. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Ans :A plane electromagnetic wave traveling in vacuum has its electric field (E) and magnetic field (B) vectors perpendicular to each other and also perpendicular to the direction of propagation (z-direction in this case). This indicates that the wave is transverse in nature.
we use following formula, to find the wavelength,
c = λf
Where:
* speed of light in vacuum (c)
(approximately 3 × 10^8 m/s)
* λ = wavelength
* f = frequency
(30 MHz = 30 × 10^6 Hz)
Rearranging the formula to solve for wavelength:
λ = c / f
Substituting the values:
λ = (3 × 10^8 m/s) / (30 × 10^6 Hz) = 10 m
Therefore, the wavelength of the electromagnetic wave is 10 meters.
5. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Ans :We can use the formula relating wavelength (λ), frequency (f), and the speed of light (c):
c = λf
where:
* speed of light in vacuum(c)= 3 × 10^8 m/s
* λ = wavelength (in meters)
* f = frequency (in Hertz)
For the lowest frequency
7.5 MHz
= 7.5 × 10^6 Hz
λ = c / f = (3 × 10^8 m/s) / (7.5 × 10^6 Hz) = 40 meters
For the highest frequency
12 MHz
= 12 × 10^6 Hz
λ = c / f = (3 × 10^8 m/s) / (12 × 10^6 Hz) = 25 meters
Therefore, the corresponding wavelength band is approximately 25 meters to 40 meters.
6. 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?
Ans :When a charged particle oscillates, it emits electromagnetic waves. The frequency of these electromagnetic waves is equal to the frequency of oscillation of the charged particle.
Therefore, the frequency of the electromagnetic waves produced by the oscillator is 10^9 Hz.
7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?
Ans :The amplitude of the electric field (E₀) and the magnetic field (B₀) in an electromagnetic wave in vacuum are related by the speed of light (c):
E₀ = cB₀
Where:
* speed of light in vacuum(c)
(approximately 3 × 10^8 m/s)
* Magnetic field amplitude (B₀)
(510 nT = 510 × 10^-9 T)
Substituting the values:
E₀ = (3 × 10^8 m/s) × (510 × 10^-9 T) = 153 V/m
Hence, the electric field amplitude of the wave is 153 volts per meter.
8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz.
(a) Determine, B0 ,w, k, and l.
(b) Find expressions for E and B.
Ans :(a)
The electric field amplitude (E₀) and the magnetic field amplitude (B₀) are related through the speed of light (c).
B₀ = E₀/c
= (120 N/C) / (3 × 10^8 m/s)
= 4 × 10^-7 T
*The angular frequency (ω) and the frequency (ν) are connected by the following relation:
ω = 2πν
= 2π × 50 × 10^6 Hz
≈ 3.14 × 10^8 rad/s
*k:The wave number (k) is related to the wavelength (λ) and angular frequency (ω):
k = 2π/λ = ω/c
= (3.14 × 10^8 rad/s)
——————–
(3 × 10^8 m/s)
≈ 1.05 rad/m
*λ: The wavelength (λ) can be calculated using the relation:
λ = c/ν
= (3 × 10^8 m/s) / (50 × 10^6 Hz)
= 6 m
(b)
Assuming the wave is propagating along the z-direction, the electric field (E) and magnetic field (B) can be expressed as:
E(z,t) = E₀ sin(kz – ωt) î
B(z,t) = B₀ sin(kz – ωt) ĵ
Substituting the values:
E(z,t) = 120 sin(1.05z – 3.14 × 10^8 t) î N/C
B(z,t) = 4 × 10^-7 sin(1.05z – 3.14 × 10^8 t) ĵ T
9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (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 :by using following formula, find energy (E) of a photon
E = hv = hc/λ
where:
h = Planck’s constant
(6.6 x 10^-34 Js)
c = the speed of light
(3 x 10^8 m/s)
λ = wavelength of radiation
Substituting the values:
E =
(6.6 x 10^-34 Js)*(3 x 10^8 m/s) / λ
= 19.8 x 10^-26 / λ J
Converting to electron volts (eV):
E =
(19.8 x 10^-26 / λ) J / (1.6 x 10^-19 J/eV)
= 12.375 x 10^-7 / λ eV
The provided table lists the photon energies for various parts of the electromagnetic spectrum.
Wavelength (λ) (m) Energy (E) (eV)
10^0 12.375 x 10^-7
1 12.375 x 10^-7
10^-3 12.375 x 10^-4
10^-6 12.375 x 10^-1
10^-8 12.375 x 10^1
10^-10 12.375 x 10^3
10^-12 12.375 x 10^5
The photon energies associated with different spectral regions of a source provide insights into the spacing of the source’s relevant energy levels.
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 m s-1.]
Ans :(a) Wavelength (λ) can be calculated using the formula:
λ = c/f
Where:
*
speed of light in vacuum(c)
(3 × 10^8 m/s)
frequency (f)
(2.0 × 10^10 Hz)
Substituting the values:
λ = (3 × 10^8 m/s) / (2.0 × 10^10 Hz) = 0.015 m = 1.5 cm
(b) The amplitude of the magnetic field (B₀) is related to the electric field amplitude (E₀) by the speed of light (c):
B₀ = E₀/c
Substituting the values:
B₀ = (48 V/m) / (3 × 10^8 m/s) = 1.6 × 10^-7 T
(c) The average energy density of the electric field (Ue) is given by:
Ue = (1/2)ε₀E₀²
The average energy density of the magnetic field (Ub) is given by:
Ub = B₀²/(2μ₀)
Where:
* ε₀ = permittivity of free space
(8.85 × 10^-12 C²/Nm²)
* μ₀ = permeability of free space
(4π × 10^-7 Tm/A)
To show that Ue = Ub, we can substitute the relation B₀ = E₀/c:
Ub = (E₀/c)² / (2μ₀) = ε₀E₀² / 2
Comparing the expressions for Ue and Ub, we see that they are equal:
Ue = Ub = (1/2)ε₀E₀² = B₀²/(2μ₀)
This equality demonstrates that the average energy density of the electric field is equal to the average energy density of the magnetic field in an electromagnetic wave.
