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Electromagnetic Waves
Physics XII - Chapter 08: Complete NCERT Exercise Solutions
Master Electromagnetic Waves with NCERT solutions for Maxwell's equations, wave propagation, electromagnetic spectrum, and applications.
Question 8.1
Figure 8.6 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 0.15 A.
(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.
(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.
Answer & Explanation:
(a) Capacitance: \( C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 \pi r^2}{d} \)
\( C \approx 8.0 \times 10^{-12} \, \text{F} \)
\( \frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8.0 \times 10^{-12}} \approx 1.875 \times 10^{10} \, \text{V/s} \)
(b) Displacement current: \( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = I = 0.15 \, \text{A} \)
(c) Yes, Kirchhoff’s rule is valid if displacement current is included; total current (conduction + displacement) is continuous.
Answer: (a) \( C \approx 8.0 \, \text{pF} \), \( dV/dt \approx 1.875 \times 10^{10} \, \text{V/s} \); (b) 0.15 A; (c) Yes, with displacement current.
\( C \approx 8.0 \times 10^{-12} \, \text{F} \)
\( \frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8.0 \times 10^{-12}} \approx 1.875 \times 10^{10} \, \text{V/s} \)
(b) Displacement current: \( I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = I = 0.15 \, \text{A} \)
(c) Yes, Kirchhoff’s rule is valid if displacement current is included; total current (conduction + displacement) is continuous.
Answer: (a) \( C \approx 8.0 \, \text{pF} \), \( dV/dt \approx 1.875 \times 10^{10} \, \text{V/s} \); (b) 0.15 A; (c) Yes, with displacement current.
Question 8.2
A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius \( R = 6.0 \, \text{cm} \) has a capacitance \( C = 100 \, \text{pF} \). 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.
(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.
Answer & Explanation:
(a) \( X_C = \frac{1}{\omega C} \approx 3.33 \times 10^7 \, \Omega \)
\( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} \approx 6.9 \times 10^{-6} \, \text{A} \)
(b) Yes, in a capacitor, conduction current in wires equals displacement current between plates.
(c) Using Ampere–Maxwell law: \( B = \frac{\mu_0 r}{2} \cdot \frac{I_d}{\pi R^2} \) (for r < R)
\( B_0 \approx 1.63 \times 10^{-11} \, \text{T} \)
Answer: (a) ~6.9 μA, (b) Yes, (c) ~1.63 × 10⁻¹¹ T.
\( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} \approx 6.9 \times 10^{-6} \, \text{A} \)
(b) Yes, in a capacitor, conduction current in wires equals displacement current between plates.
(c) Using Ampere–Maxwell law: \( B = \frac{\mu_0 r}{2} \cdot \frac{I_d}{\pi R^2} \) (for r < R)
\( B_0 \approx 1.63 \times 10^{-11} \, \text{T} \)
Answer: (a) ~6.9 μA, (b) Yes, (c) ~1.63 × 10⁻¹¹ T.
Question 8.3
What physical quantity is the same for X-rays of wavelength \( 10^{-10} \, \text{m} \), red light of wavelength 6800 Å and radiowaves of wavelength 500 m?
Answer & Explanation:
Speed in vacuum: All electromagnetic waves travel at the same speed \( c = 3 \times 10^8 \, \text{m/s} \) in vacuum.
Answer: Speed in vacuum (c).
Answer: Speed in vacuum (c).
Question 8.4
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?
Answer & Explanation:
Field directions: \( \vec{E} \) and \( \vec{B} \) are perpendicular to each other and to the direction of propagation (z-axis). They oscillate in phase.
Wavelength: \( \lambda = \frac{c}{f} = \frac{3 \times 10^8}{30 \times 10^6} = 10 \, \text{m} \)
Answer: E and B are perpendicular to z and to each other; λ = 10 m.
Wavelength: \( \lambda = \frac{c}{f} = \frac{3 \times 10^8}{30 \times 10^6} = 10 \, \text{m} \)
Answer: E and B are perpendicular to z and to each other; λ = 10 m.
Question 8.5
A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer & Explanation:
\( \lambda = \frac{c}{f} \)
For \( f_1 = 7.5 \, \text{MHz} \): \( \lambda_1 \approx 40 \, \text{m} \)
For \( f_2 = 12 \, \text{MHz} \): \( \lambda_2 \approx 25 \, \text{m} \)
Answer: 25 m to 40 m.
For \( f_1 = 7.5 \, \text{MHz} \): \( \lambda_1 \approx 40 \, \text{m} \)
For \( f_2 = 12 \, \text{MHz} \): \( \lambda_2 \approx 25 \, \text{m} \)
Answer: 25 m to 40 m.
Question 8.6
A charged particle oscillates about its mean equilibrium position with a frequency of \( 10^9 \, \text{Hz} \). What is the frequency of the electromagnetic waves produced by the oscillator?
Answer & Explanation:
The frequency of the emitted electromagnetic wave equals the oscillation frequency of the charged particle.
Answer: \( 10^9 \, \text{Hz} \)
Answer: \( 10^9 \, \text{Hz} \)
Question 8.7
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is \( B_0 = 510 \, \text{nT} \). What is the amplitude of the electric field part of the wave?
Answer & Explanation:
\( E_0 = c B_0 \)
\( E_0 = 3 \times 10^8 \times 510 \times 10^{-9} = 153 \, \text{N/C} \)
Answer: 153 N/C
\( E_0 = 3 \times 10^8 \times 510 \times 10^{-9} = 153 \, \text{N/C} \)
Answer: 153 N/C
Question 8.8
Suppose that the electric field amplitude of an electromagnetic wave is \( E_0 = 120 \, \text{N/C} \) and that its frequency is \( \nu = 50.0 \, \text{MHz} \).
(a) Determine \( B_0 \), \( k \), and \( \lambda \).
(b) Find expressions for \( \vec{E} \) and \( \vec{B} \).
(a) Determine \( B_0 \), \( k \), and \( \lambda \).
(b) Find expressions for \( \vec{E} \) and \( \vec{B} \).
Answer & Explanation:
(a) \( B_0 = \frac{E_0}{c} = 4 \times 10^{-7} \, \text{T} \)
\( \lambda = \frac{c}{\nu} = 6 \, \text{m} \)
\( k = \frac{2\pi}{\lambda} \approx 1.047 \, \text{rad/m} \)
(b) Assuming propagation along x‑axis:
\( E_y = E_0 \cos(kx - \omega t) \), \( B_z = B_0 \cos(kx - \omega t) \)
Answer: (a) \( B_0 = 4 \times 10^{-7} \, \text{T} \), \( \lambda = 6 \, \text{m} \), \( k \approx 1.047 \, \text{rad/m} \); (b) As above.
\( \lambda = \frac{c}{\nu} = 6 \, \text{m} \)
\( k = \frac{2\pi}{\lambda} \approx 1.047 \, \text{rad/m} \)
(b) Assuming propagation along x‑axis:
\( E_y = E_0 \cos(kx - \omega t) \), \( B_z = B_0 \cos(kx - \omega t) \)
Answer: (a) \( B_0 = 4 \times 10^{-7} \, \text{T} \), \( \lambda = 6 \, \text{m} \), \( k \approx 1.047 \, \text{rad/m} \); (b) As above.
Question 8.9
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula \( E = h \nu \) (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?
Answer & Explanation:
Photon energy \( E = \frac{hc}{\lambda} \) in eV:
• Gamma rays: ~ MeV
• X‑rays: ~ keV
• Visible: ~ 1.8–3.1 eV
• Radio waves: ~ μeV
Relation: Higher energy photons come from processes involving larger energy changes (nuclear, atomic inner shells); lower energy photons from molecular/electronic oscillations.
Answer: Energies range from μeV (radio) to MeV (gamma); related to source energy scales.
• Gamma rays: ~ MeV
• X‑rays: ~ keV
• Visible: ~ 1.8–3.1 eV
• Radio waves: ~ μeV
Relation: Higher energy photons come from processes involving larger energy changes (nuclear, atomic inner shells); lower energy photons from molecular/electronic oscillations.
Answer: Energies range from μeV (radio) to MeV (gamma); related to source energy scales.
Question 8.10
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of \( 2.0 \times 10^{10} \, \text{Hz} \) and amplitude 48 V/m.
(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 \( \mathbf{E} \) field equals the average energy density of the \( \mathbf{B} \) field. [\( c = 3 \times 10^8 \, \text{m/s} \)]
(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 \( \mathbf{E} \) field equals the average energy density of the \( \mathbf{B} \) field. [\( c = 3 \times 10^8 \, \text{m/s} \)]
Answer & Explanation:
(a) \( \lambda = \frac{c}{f} = \frac{3 \times 10^8}{2 \times 10^{10}} = 0.015 \, \text{m} \)
(b) \( B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7} \, \text{T} \)
(c) \( u_E = \frac{1}{2} \varepsilon_0 E_{\text{rms}}^2 \), \( u_B = \frac{1}{2\mu_0} B_{\text{rms}}^2 \)
Using \( E_{\text{rms}} = c B_{\text{rms}} \) and \( c^2 = \frac{1}{\mu_0 \varepsilon_0} \), we get \( u_E = u_B \).
Answer: (a) 0.015 m, (b) \( 1.6 \times 10^{-7} \, \text{T} \), (c) Proof shown.
(b) \( B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7} \, \text{T} \)
(c) \( u_E = \frac{1}{2} \varepsilon_0 E_{\text{rms}}^2 \), \( u_B = \frac{1}{2\mu_0} B_{\text{rms}}^2 \)
Using \( E_{\text{rms}} = c B_{\text{rms}} \) and \( c^2 = \frac{1}{\mu_0 \varepsilon_0} \), we get \( u_E = u_B \).
Answer: (a) 0.015 m, (b) \( 1.6 \times 10^{-7} \, \text{T} \), (c) Proof shown.
Question 8.11
Suppose that the electric field part of an electromagnetic wave in vacuum is \( \mathbf{E} = \{(3.1 \, \text{N/C}) \cos[(1.8 \, \text{rad/m}) \, y + (5.4 \times 10^6 \, \text{rad/s}) \, t]\} \, \hat{i} \).
(a) What is the direction of propagation?
(b) What is the wavelength \( \lambda \)?
(c) What is the frequency \( \nu \)?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
(a) What is the direction of propagation?
(b) What is the wavelength \( \lambda \)?
(c) What is the frequency \( \nu \)?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
Answer & Explanation:
(a) Propagation along –y direction (since ky and ωt have same sign).
(b) \( k = 1.8 \, \text{rad/m} \) → \( \lambda = \frac{2\pi}{k} \approx 3.49 \, \text{m} \)
(c) \( \omega = 5.4 \times 10^6 \, \text{rad/s} \) → \( \nu = \frac{\omega}{2\pi} \approx 859 \, \text{kHz} \)
(d) \( B_0 = \frac{E_0}{c} \approx 1.03 \times 10^{-8} \, \text{T} \)
(e) \( \vec{B} = B_0 \cos(ky + \omega t) \, \hat{k} \) (since E along i, propagation along –y → B along z).
Answer: (a) –y, (b) ~3.49 m, (c) ~859 kHz, (d) ~1.03 × 10⁻⁸ T, (e) \( \vec{B} = B_0 \cos(ky + \omega t) \hat{k} \).
(b) \( k = 1.8 \, \text{rad/m} \) → \( \lambda = \frac{2\pi}{k} \approx 3.49 \, \text{m} \)
(c) \( \omega = 5.4 \times 10^6 \, \text{rad/s} \) → \( \nu = \frac{\omega}{2\pi} \approx 859 \, \text{kHz} \)
(d) \( B_0 = \frac{E_0}{c} \approx 1.03 \times 10^{-8} \, \text{T} \)
(e) \( \vec{B} = B_0 \cos(ky + \omega t) \, \hat{k} \) (since E along i, propagation along –y → B along z).
Answer: (a) –y, (b) ~3.49 m, (c) ~859 kHz, (d) ~1.03 × 10⁻⁸ T, (e) \( \vec{B} = B_0 \cos(ky + \omega t) \hat{k} \).
Question 8.12
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Answer & Explanation:
Visible power \( P = 0.05 \times 100 = 5 \, \text{W} \)
Intensity \( I = \frac{P}{4\pi r^2} \) (isotropic)
(a) \( r = 1 \, \text{m} \): \( I \approx 0.398 \, \text{W/m}^2 \)
(b) \( r = 10 \, \text{m} \): \( I \approx 0.00398 \, \text{W/m}^2 \)
Answer: (a) ~0.398 W/m², (b) ~3.98 mW/m².
Intensity \( I = \frac{P}{4\pi r^2} \) (isotropic)
(a) \( r = 1 \, \text{m} \): \( I \approx 0.398 \, \text{W/m}^2 \)
(b) \( r = 10 \, \text{m} \): \( I \approx 0.00398 \, \text{W/m}^2 \)
Answer: (a) ~0.398 W/m², (b) ~3.98 mW/m².
Question 8.13
Use the formula \( \lambda_m T = 0.29 \, \text{cm·K} \) to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Answer & Explanation:
Wien’s displacement law: \( \lambda_m T = b \) (b = 0.29 cm·K)
• For γ‑rays (λ ~ 10⁻¹² m): T ~ 2.9 × 10⁹ K (nuclear processes)
• For visible (λ ~ 500 nm): T ~ 5800 K (sun’s surface)
• For radio (λ ~ 1 m): T ~ 0.0029 K (cosmic background)
Interpretation: Different parts of the EM spectrum correspond to black‑body temperatures of typical sources emitting dominantly in that range.
Answer: Temperature ranges from ~10⁹ K (gamma) to ~0.003 K (radio).
• For γ‑rays (λ ~ 10⁻¹² m): T ~ 2.9 × 10⁹ K (nuclear processes)
• For visible (λ ~ 500 nm): T ~ 5800 K (sun’s surface)
• For radio (λ ~ 1 m): T ~ 0.0029 K (cosmic background)
Interpretation: Different parts of the EM spectrum correspond to black‑body temperatures of typical sources emitting dominantly in that range.
Answer: Temperature ranges from ~10⁹ K (gamma) to ~0.003 K (radio).
Question 8.14
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space—thought to be a relic of the 'big‑bang' origin of the universe].
(d) 5890 Å – 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in \(^{57}\text{Fe}\) nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space—thought to be a relic of the 'big‑bang' origin of the universe].
(d) 5890 Å – 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in \(^{57}\text{Fe}\) nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].
Answer & Explanation:
(a) Radio waves (21 cm line)
(b) Radio waves (~28 cm wavelength)
(c) Microwave background (peak wavelength ~1 mm)
(d) Visible light (yellow region)
(e) X‑rays / gamma rays (nuclear transition)
Answer: (a) Radio, (b) Radio, (c) Microwave, (d) Visible, (e) X‑ray/Gamma.
(b) Radio waves (~28 cm wavelength)
(c) Microwave background (peak wavelength ~1 mm)
(d) Visible light (yellow region)
(e) X‑rays / gamma rays (nuclear transition)
Answer: (a) Radio, (b) Radio, (c) Microwave, (d) Visible, (e) X‑ray/Gamma.
Question 8.15
Answer the following questions:
(a) Long distance radio broadcasts use short‑wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X‑ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe 'nuclear winter' with a devastating effect on life on earth. What might be the basis of this prediction?
(a) Long distance radio broadcasts use short‑wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X‑ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe 'nuclear winter' with a devastating effect on life on earth. What might be the basis of this prediction?
Answer & Explanation:
(a) Short waves (HF) are reflected by ionosphere → long‑range transmission.
(b) TV signals are VHF/UHF, not reflected by ionosphere; satellites provide line‑of‑sight relay.
(c) X‑rays are absorbed by atmosphere; space‑based telescopes avoid this.
(d) Ozone absorbs harmful UV radiation, protecting life.
(e) Lower (no greenhouse effect).
(f) Soot and dust from fires/nuclear explosions block sunlight → global cooling.
Answer: Conceptual explanations above.
(b) TV signals are VHF/UHF, not reflected by ionosphere; satellites provide line‑of‑sight relay.
(c) X‑rays are absorbed by atmosphere; space‑based telescopes avoid this.
(d) Ozone absorbs harmful UV radiation, protecting life.
(e) Lower (no greenhouse effect).
(f) Soot and dust from fires/nuclear explosions block sunlight → global cooling.
Answer: Conceptual explanations above.
📘 Exam Preparation Tip:
These exercise questions will help you understand the nature and properties of electromagnetic waves. You'll learn about Maxwell's equations and their significance, understand electromagnetic wave characteristics (speed, wavelength, frequency), explore the complete EM spectrum from radio to gamma waves, and analyze practical applications in communication and technology. Fundamental for modern physics and communication systems.