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Solutions

Overview

Step-by-step NCERT solutions for Electromagnetic Waves (Chapter 8, CBSE Class 12 Physics) — every question and answer worked out in full, not just the final result. You can also read the Electromagnetic Waves textbook chapter.

Solved

What these solutions cover

All 10 questions in Electromagnetic Waves are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

  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 0.15 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…
  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…
  3. What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?
  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?
  5. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
  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?
  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?
  8. Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz.
    • (a) Determine B₀, ω, k, and λ.
    • (b) Find expressions for E and B.
  9. The 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?
  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.]
Sample

Worked examples

3 questions from Electromagnetic Waves, solved step by step.

Exercises

What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?

X-rays (λ = 10⁻¹⁰ m), red light (λ = 6800 Å = 6.8 × 10⁻⁷ m), and radiowaves (λ = 500 m) are all electromagnetic waves — members of the electromagnetic spectrum.

Despite having vastly different wavelengths and frequencies, all electromagnetic waves travel through vacuum with the same speed:

c = 3 × 10⁸ m/s

Verification of frequencies:

X-rays: f = c/λ = 3 × 10⁸ / 10⁻¹⁰ = 3 × 10¹⁸ Hz

Red light: f = 3 × 10⁸ / 6.8 × 10⁻⁷ = 4.4 × 10¹⁴ Hz

Radiowaves: f = 3 × 10⁸ / 500 = 6 × 10⁵ Hz

The frequencies differ enormously, but the speed in vacuum is the same for all.

Answer: The speed in vacuum: c = 3 × 10⁸ m/s is the same for all three — X-rays, red light, and radiowaves are all electromagnetic waves traveling at the same speed in vacuum.

Exercises

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?

For an electromagnetic wave traveling along the z-direction:

The electric field vector E and magnetic field vector B must each be perpendicular to the direction of propagation (z-axis) and perpendicular to each other. Therefore, E and B both lie in the xy-plane and are mutually perpendicular.

For example: E along x-axis (î direction), B along y-axis (ĵ direction), so that E × B points along z (the propagation direction).

Wavelength calculation:

Given: frequency ν = 30 MHz = 30 × 10⁶ Hz, c = 3 × 10⁸ m/s

λ = c / ν = (3 × 10⁸) / (30 × 10⁶)

λ = 10 m

Answer: E and B lie in the xy-plane, perpendicular to each other and to z (the propagation direction). Wavelength λ = c/ν = 3 × 10⁸ / 30 × 10⁶ = 10 m.

Exercises

A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Given: frequency band f₁ = 7.5 MHz = 7.5 × 10⁶ Hz to f₂ = 12 MHz = 12 × 10⁶ Hz

c = 3 × 10⁸ m/s

Wavelength corresponding to f₁ (lower frequency, longer wavelength):

λ₁ = c / f₁ = (3 × 10⁸) / (7.5 × 10⁶) = 40 m

Wavelength corresponding to f₂ (higher frequency, shorter wavelength):

λ₂ = c / f₂ = (3 × 10⁸) / (12 × 10⁶) = 25 m

So the wavelength band is 25 m to 40 m.

Answer: Wavelength band: 25 m to 40 m (corresponding to frequencies 12 MHz and 7.5 MHz respectively).

All 10 questions in Electromagnetic Waves are solved step by step in the Solutions PDF above.

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