January 3, 2024

Explore comprehensive solutions for Class 12 Science Physics Chapter 8 – Electromagnetic Waves in the NCERT textbook. These solutions offer step-by-step explanations, making them highly favored among students for quick completion of homework and effective exam preparation. The solutions cover all the questions from Chapter 8, providing valuable assistance to Class 12 Science Physics students. Access these free resources to enhance your understanding and excel in your studies. Page No 285: 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 charge 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: Radius of each circular plate, r = 12 cm = 0.12 m Distance between the plates, d = 5 cm = 0.05 m Charging current, I = 0.15 A Permittivity of free space, = 8.85 × 10−12 C2 N−1 m−2 (a) Capacitance between the two plates is given by the relation, C  Where, A = Area of each plate  Charge on each plate, q = CV Where, V = Potential difference across the plates Differentiation on both sides with respect to time (t) gives: Therefore, the change in potential difference between the plates is 1.87 ×109 V/s. (b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A. (c) Yes Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current. Page No 286: Question 8.2: 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 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: Radius of each circular plate, R = 6.0 cm = 0.06 m Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F Supply voltage, V = 230 V Angular frequency, ω = 300 rad s−1 (a) Rms value of conduction 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 to displacement current. (c) Magnetic field is given as: B  Where, μ0 = Free space permeability  I0 = Maximum value of current = 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. Page No 286: Question 8.3: 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? ANSWER: The speed of light (3 × 108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum. Page No 286: 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: The electromagnetic wave travels in 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. Frequency of the wave, ν = 30 MHz = 30 × 106 s−1 Speed of light in a vacuum, c = 3 × 108 m/s Wavelength of a wave is given as: Page No 286: 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: A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 × 106 Hz Maximum frequency, ν2 = 12 MHz = 12 × 106 Hz Speed of light, c = 3 × 108 m/s Corresponding wavelength for ν1 can be calculated as: Corresponding wavelength for ν2 can be calculated as: Thus, the wavelength band of the radio is 40 m to 25 m. Page No 286: Question 8.6: 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? ANSWER: The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz. Page No 286: Question 8.7: 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? ANSWER: Amplitude of magnetic field of an electromagnetic wave in a vacuum, B0 = 510 nT = 510 × 10−9 T Speed of light in a vacuum, c = 3 × 108 m/s Amplitude of 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. Page No 286: Question 8.8: 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. ANSWER: Electric field amplitude, E0 = 120 N/C Frequency of source, ν = 50.0 MHz = 50 × 106 Hz Speed of light, c = 3 × 108 m/s (a) Magnitude of magnetic field strength is given as: Angular frequency of source is given as: ω = 2πν = 2π × 50 × 106 = 3.14 × 108 rad/s Propagation constant is given as: Wavelength of wave is given as: (b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular. Equation of electric field vector is given as: And, magnetic field vector is given …

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