Question 1

Q1: 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 externa...

Accepted Answer

Given: Radius of circular plates, r = 12 cm = 0.12 m Separation between plates, d = 5.0 cm = 0.05 m Charging current, I = 0.15 A (a) Capacitance and rate of change of potential difference 1. Capacitance (C): The capacitance of a parallel plate capacitor is given by: C = ε₀A / d where A is the area o...

Question 2

Q10: 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 wa...

Accepted Answer

Given: Frequency, ν = 2.0 × 10¹⁰ Hz Electric field amplitude, E₀ = 48 V/m Speed of light, c = 3 × 10⁸ m/s (a) Wavelength of the wave (λ) The relationship between wavelength, frequency, and speed is λ = c/ν. λ = (3 × 10⁸ m/s) / (2.0 × 10¹⁰ Hz) λ = 1.5 × 10⁻² m = 1.5 cm (b) Amplitude of the oscillatin...

Question 3

Q2: 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...

Accepted Answer

Given: Radius of plates, R = 6.0 cm = 0.06 m Capacitance, C = 100 pF = 100 × 10⁻¹² F RMS voltage of AC supply, V_rms = 230 V Angular frequency, ω = 300 rad s⁻¹ Distance from the axis, r = 3.0 cm = 0.03 m (a) RMS value of the conduction current First, we calculate the capacitive reactance (X_C): X_C...

Question 4

Q3: 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 ?

Accepted Answer

The physical quantity that is the same for all these electromagnetic waves (X-rays, red light, and radio waves) is their speed in a vacuum. All electromagnetic waves travel through a vacuum at the same speed, which is the speed of light, c. c = 3 × 10⁸ m/s.

Question 5

Q4: 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...

Accepted Answer

Directions of Electric and Magnetic Field Vectors: As stated in the chapter, for an electromagnetic wave, the electric field vector (E), the magnetic field vector (B), and the direction of propagation are mutually perpendicular to each other. Since the wave travels along the z-direction, the electri...

Question 6

Q5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Accepted Answer

The relationship between frequency (ν) and wavelength (λ) for electromagnetic waves is λ = c/ν, where c = 3 × 10⁸ m/s. We need to find the wavelengths corresponding to the lower and upper frequency limits of the band. For the lower frequency limit (ν₁ = 7.5 MHz): ν₁ = 7.5 MHz = 7.5 × 10⁶ Hz λ₁ = c /...

Question 7

Q6: A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produ...

Accepted Answer

According to Maxwell's theory, an accelerating charged particle radiates electromagnetic waves. An oscillating charged particle is an example of an accelerating charge. The frequency of the electromagnetic waves produced by an oscillating charge is equal to the frequency of oscillation of the charge...

Question 8

Q7: 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...

Accepted Answer

Given: Amplitude of the magnetic field, B₀ = 510 nT = 510 × 10⁻⁹ T The amplitudes of the electric field (E₀) and magnetic field (B₀) in an electromagnetic wave in a vacuum are related by the speed of light (c): E₀ = c × B₀ where c = 3 × 10⁸ m/s. E₀ = (3 × 10⁸ m/s) × (510 × 10⁻⁹ T) E₀ = 1530 × 10⁻¹ V...

Question 9

Q8: Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is v = 50.0 MHz. (a) Determine, B₀, ω, k,...

Accepted Answer

Given: Electric field amplitude, E₀ = 120 N/C Frequency, ν = 50.0 MHz = 50 × 10⁶ Hz (a) Determination of B₀, ω, k, and λ    Magnetic field amplitude (B₀):     B₀ = E₀ / c = (120 N/C) / (3 × 10⁸ m/s) = 40 × 10⁻⁸ T = 400 nT    Angular frequency (ω):     ω = 2πν = 2π × (50 × 10⁶ Hz) = 100π × 10⁶ rad/s...

Question 10

Q9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation:...

Accepted Answer

To calculate the photon energy in eV, we use the formula E = hν, where h (Planck's constant) = 6.63 × 10⁻³⁴ J·s. To convert from Joules to eV, we divide by the charge of an electron, e = 1.602 × 10⁻¹⁹ J/eV. Here are the approximate photon energies for different parts of the spectrum:    Radio waves...

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