Chapter 8 Electromagnetic Waves Solutions
Question - 1 : - Figure 8.6 shows a capacitor madeof two circular plates each of radius 12 cm, and separated by 5.0 cm. Thecapacitor is being charged by an external source (not shown in the figure). Thecharging current is constant and equal to 0.15 A.
(a) Calculate the capacitance and therate of charge of potential difference between the plates.
(b) Obtain the displacement currentacross the plates.
(c) Is Kirchhoff’s first rule (junctionrule) valid at each plate of the capacitor? Explain.
Answer - 1 : -
Radius of each circular plate, r = 12 cm = 0.12 m
Distancebetween the plates, d = 5 cm =0.05 m
Chargingcurrent, I = 0.15 A
Permittivityof free space, 
= 8.85 × 10−12 C2 N−1 m−2
(a) Capacitance between the two plates isgiven by the relation,
C 
Where,
A =Area of each plate 
Charge on each plate, q = CV
Where,
V =Potential difference across the plates
Differentiationon both sides with respect to time (t)gives:
Therefore, the change inpotential difference between the plates is 1.87 ×109 V/s.
(b) The displacement current across theplates is the same as the conduction current. Hence, the displacement current, id is0.15 A.
(c) Yes
Kirchhoff’sfirst rule is valid at each plate of the capacitor provided that we take thesum of conduction and displacement for current.
A parallel plate capacitor (Fig.8.7) 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 theconduction current?
(b) Is the conduction current equal to thedisplacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axisbetween the plates.
Answer - 2 : -
Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitanceof a parallel plate capacitor, C =100 pF = 100 × 10−12 F
Supplyvoltage, V = 230 V
Angularfrequency, ω = 300 rad s−1
(a) Rms value ofconduction current, I 
Where,
XC =Capacitive reactance
∴ I = V × ωC
= 230 ×300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence,the rms value of conduction current is 6.9 μA.
(b) Yes, conduction current is equal todisplacement current.
(c) Magnetic field is given as:
B 
Where,
μ0 = Free spacepermeability 
I0 = Maximum value ofcurrent =
r =Distance between the plates from the axis = 3.0 cm = 0.03 m
∴B 
= 1.63 × 10−11 T
Hence,the magnetic field at that point is 1.63 × 10−11 T.
What physical quantity is thesame for X-rays of wavelength 10−10 m,red light of wavelength 6800 Å and radiowaves of wavelength 500 m?
Answer - 3 : -
The speed of light (3 × 108 m/s) in a vacuum is the same forall wavelengths. It is independent of the wavelength in the vacuum.
Question - 4 : - A plane electromagnetic wavetravels in vacuum along z-direction. What can you say about the directions ofits electric and magnetic field vectors? If the frequency of the wave is 30MHz, what is its wavelength?
Answer - 4 : -
The electromagnetic wave travelsin a vacuum along the z-direction. The electric field (E)and the magnetic field (H) are in the x-y plane.They are mutually perpendicular.
Frequencyof the wave, ν = 30 MHz = 30 × 106 s−1
Speed oflight in a vacuum, c = 3 × 108 m/s
Wavelengthof a wave is given as:
Question - 5 : - A radio can tune in to anystation in the 7.5 MHz to 12 MHz band. What is the corresponding wavelengthband?
Answer - 5 : -
A radio can tune to minimumfrequency, ν1 = 7.5 MHz= 7.5 × 106 Hz
Maximumfrequency, ν2 = 12 MHz = 12 × 106 Hz
Speed oflight, c = 3 × 108 m/s
Correspondingwavelength for ν1 can be calculatedas:Corresponding wavelength for ν2 can be calculated as:
Thus, the wavelength band of theradio is 40 m to 25 m.
A charged particle oscillatesabout its mean equilibrium position with a frequency of 109 Hz. What is the frequency of theelectromagnetic waves produced by the oscillator?
Answer - 6 : -
The frequency of anelectromagnetic wave produced by the oscillator is the same as that of acharged particle oscillating about its mean position i.e., 109 Hz.
Question - 7 : - The amplitude of the magneticfield part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electricfield part of the wave?
Answer - 7 : -
Amplitude of magnetic field of anelectromagnetic wave in a vacuum,
B0 =510 nT = 510 × 10−9 T
Speed oflight in a vacuum, c = 3 × 108 m/s
Amplitudeof electric field of the electromagnetic wave is given by the relation,
E = cB0
= 3 × 108 × 510 × 10−9 =153 N/C
Therefore,the electric field part of the wave is 153 N/C.
Question - 8 : - Suppose that the electric fieldamplitude of an electromagnetic wave is E0 = 120 N/C and thatits frequency is ν = 50.0 MHz.(a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B.
Answer - 8 : -
Electric field amplitude, E0 = 120 N/C
Frequencyof source, ν = 50.0 MHz = 50 ×106 Hz
Speed oflight, c = 3 × 108 m/s
(a) Magnitude of magneticfield strength is given as:Angular frequency of source isgiven as:
ω = 2πν
= 2π × 50× 106
= 3.14 ×108 rad/s
Propagationconstant is given as:
Wavelength of wave is given as:
(b) Supposethe wave is propagating in the positive x direction.Then, the electric field vector will be in the positive y direction and the magnetic fieldvector will be in the positive z direction.This is because all three vectors are mutually perpendicular.
Equation of electric field vectoris given as:
And, magnetic field vector isgiven as:
The terminology of differentparts of the electromagnetic spectrum is given in the text. Use the formula E = hν (forenergy of a quantum of radiation: photon) and obtain the photon energy in unitsof eV for different parts of the electromagnetic spectrum. In what way are thedifferent scales of photon energies that you obtain related to the sources ofelectromagnetic radiation?
Answer - 9 : -
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 photonenergies 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 thedifferent parts of the spectrum of a source indicate the spacing of therelevant energy levels of the source.
Question - 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 theoscillating magnetic field?
(c) Show that the average energydensity of the E fieldequals the average energy density of the B field. [c = 3 ×108 m s−1.]
Answer - 10 : -
Frequency of the electromagneticwave, ν = 2.0 × 1010 Hz
Electricfield amplitude, E0 = 48 V m−1
Speed oflight, c = 3 × 108 m/s
(a) Wavelength of a wave is given as:
(b) Magneticfield strength is given as:
(c) Energydensity of the electric field is given as:
And, energy density of themagnetic field is given as:
Where,
∈0 = Permittivity of free space
μ0 = Permeability offree space
We havethe relation connecting E and B as:
E = cB … (1)
Where,
… (2)
Puttingequation (2) in equation (1), we get
Squaring both sides, we get
