આ MCQ મોડ્યુલ આના પર આધારિત છે: NCERT Exercises and Solutions: Electromagnetic Waves
આ મૂલ્યાંકન આના પર આધારિત હશે: NCERT Exercises and Solutions: Electromagnetic Waves
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
| Quantity | Symbol | Formula | SI Unit |
|---|---|---|---|
| Displacement current | id | ε₀ dΦE/dt | A |
| Speed of EM wave (vacuum) | c | 1/√(μ₀ε₀) | m/s |
| Refractive index | n | c/v = √(μrεr) | — |
| Amplitude ratio | E₀/B₀ | c | m/s |
| Energy density | u | ε₀E² | J/m³ |
| Intensity | I | ½ c ε₀ E₀² | W/m² |
| Momentum | p | U/c | kg·m/s |
| Photon energy | E | hf = hc/λ | J |
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. 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) C = ε₀A/d = (8.854×10⁻¹²)(π×0.12²)/0.05 = 80.1 pF. dV/dt = I/C = 0.15/(80.1×10⁻¹²) = 1.87 × 10⁹ V/s.
(b) Displacement current id = ε₀ dΦE/dt = ic = 0.15 A.
(c) Yes, Kirchhoff's first rule holds if displacement current is included as part of the "current" leaving/entering the plate. Conduction current 0.15 A flows into plate; equal displacement current 0.15 A "leaves" through the gap to the other plate.
A parallel plate capacitor of plate area 90 cm² and separation 2.5 mm has a voltage \(V = V_0 \sin\omega t\) with V₀ = 78 V and ω = 300 rad/s. Find the amplitude of (a) the conduction current, (b) the displacement current, and (c) the magnetic field 3 cm from the central axis between the plates.
C = ε₀A/d = (8.854×10⁻¹²)(90×10⁻⁴)/(2.5×10⁻³) = 31.87 pF.
(a) Conduction current ic = C dV/dt = Cω V₀ cos ωt. Amplitude = CωV₀ = 31.87×10⁻¹² × 300 × 78 = 7.45 × 10⁻⁷ A ≈ 0.745 μA.
(b) id = ic = 0.745 μA.
(c) Plate radius R = √(A/π) = √(0.009/π) = 0.0535 m. At r = 0.03 m (inside): B(2πr) = μ₀ × (r/R)² × id. B = μ₀ id r / (2π R²) = (4π×10⁻⁷ × 7.45×10⁻⁷ × 0.03)/(2π × (0.0535)²) = 1.63 × 10⁻¹¹ T.
What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å and radio waves of wavelength 500 m?
All three are electromagnetic waves and travel through vacuum at the same speed c = 3 × 10⁸ m/s.
A plane EM wave travels in vacuum along z direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency is 30 MHz, what is its wavelength?
E and B both lie in the x-y plane, perpendicular to z and perpendicular to each other.
λ = c/f = 3×10⁸/(30×10⁶) = 10 m (a radio/short-wave wavelength).
A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
λmax = c/fmin = 3×10⁸/(7.5×10⁶) = 40 m.
λmin = c/fmax = 3×10⁸/(12×10⁶) = 25 m.
Band: 25 m to 40 m (short-wave radio).
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the EM waves produced by the oscillator?
Same as the oscillation frequency: 10⁹ Hz = 1 GHz. The accelerating charge radiates at its own oscillation frequency.
The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?
E₀ = c B₀ = (3 × 10⁸)(510 × 10⁻⁹) = 153 V/m.
Suppose that the electric field amplitude of an EM wave is E₀ = 120 N/C and frequency ν = 50.0 MHz. (a) Determine B₀, ω, k and λ. (b) Find expressions for E and B.
(a) B₀ = E₀/c = 120/(3×10⁸) = 4.0 × 10⁻⁷ T. ω = 2π ν = π × 10⁸ = 3.14 × 10⁸ rad/s. λ = c/ν = 3×10⁸/(5×10⁷) = 6.0 m. k = 2π/λ = 1.05 rad/m.
(b) If the wave travels in +x and E along y, B along z:
Ey = 120 sin(1.05 x − 3.14×10⁸ t) N/C
Bz = 4.0×10⁻⁷ sin(1.05 x − 3.14×10⁸ t) T.
The terminology for different parts of the EM spectrum is given in the text. Use the formula E = hf (for the energy of a photon in a quantum of EM radiation: photon) and obtain the photon energy in units of eV for different parts of the EM spectrum. In what way are the different scales of photon energies that you obtain related to the sources of EM radiation?
| Band | Typical f (Hz) | Photon energy E (eV) | Source mechanism |
|---|---|---|---|
| Radio | 10⁶ | 4 × 10⁻⁹ | oscillating LC circuit |
| Microwave | 10¹⁰ | 4 × 10⁻⁵ | magnetron, molecular rotation |
| Infrared | 10¹³ | 0.04 | molecular vibration |
| Visible | 5×10¹⁴ | 2 | outer-shell electron transitions |
| UV | 10¹⁶ | 40 | excited atomic states |
| X-ray | 10¹⁸ | 4000 | inner-shell transitions, bremsstrahlung |
| γ-ray | 10²⁰ | 4 × 10⁵ | nuclear transitions |
Higher-frequency bands require higher-energy sources - mechanical oscillators for radio, atomic transitions for visible-UV, nuclear processes for γ-rays.
In a plane EM wave, the electric field oscillates sinusoidally at f = 2.0 × 10¹⁰ Hz and amplitude 48 V/m. (a) What is the wavelength? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E-field equals that of the B-field. [c = 3 × 10⁸ m/s.]
(a) λ = c/f = 3×10⁸/(2×10¹⁰) = 1.5 × 10⁻² m = 1.5 cm.
(b) B₀ = E₀/c = 48/(3×10⁸) = 1.6 × 10⁻⁷ T.
(c) ⟨uE⟩ = ¼ ε₀ E₀² = ¼(8.854×10⁻¹²)(48)² = 5.1 × 10⁻⁹ J/m³. ⟨uB⟩ = B₀²/(4μ₀) = (1.6×10⁻⁷)²/(4 × 4π×10⁻⁷) = 5.1 × 10⁻⁹ J/m³. Equal, as expected.
What is the wavelength of EM waves of frequency (a) 3 × 10¹⁹ Hz, (b) 100 MHz, (c) 5 × 10¹⁴ Hz? Name the band each belongs to.
(a) λ = 10⁻¹¹ m = 0.01 nm → γ-ray.
(b) λ = 3 m → radio (FM band).
(c) λ = 600 nm → visible (orange).
Set the wavelength (logarithmic slider). The output shows frequency, photon energy, band and a typical use.
| Wavelength λ | 100 nm |
| Frequency f | 3.0 × 10¹⁵ Hz |
| Photon energy | 12.4 eV |
| Band | UV |
| Typical use | sterilisation, fluorescence |
1. The power output of the panel is closest to: L1
2. Describe how Maxwell's introduction of displacement current resolved a conceptual contradiction in Ampere's law. L2
3. Calculate the peak electric field of the sunlight reaching the panel. L3
4. The panel is tilted at 30° to the rays. How does the absorbed intensity change? L4
5. Design a satellite-borne solar sail that can be propelled by sunlight. State the key design choices and justify them. L6
Assertion: Light from a distant star reaches us through the vacuum of space.
Reason: EM waves do not require a material medium to propagate.
Assertion: An EM wave can transfer energy and momentum to a surface.
Reason: The Poynting vector S = (1/μ₀) E × B represents energy flux density and the wave also carries momentum p = U/c.
Assertion: The speed of EM waves in a dielectric medium of refractive index n > 1 is less than c.
Reason: In matter the effective permittivity ε > ε₀, so v = 1/√(με) < c.
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Section A · Section B · Section C · Section D · Section E