Exercises
(8.1) 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 Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain.
Answer
Given: r = 12 cm = 0.12 m, d = 5.0 cm = 0.05 m, I = 0.15 A, ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
(a) A = πr² = 3.14 × (0.12)² ≈ 4.52 × 10⁻² m².
C = ε₀A/d
⇒ C = 8.85 × 10⁻¹² × 4.52 × 10⁻² / 0.05
⇒ C ≈ 8.0 × 10⁻¹² F = 80 pF
Since Q = CV, I = dQ/dt = C dV/dt
⇒ dV/dt = I/C
⇒ dV/dt = 0.15 / (8.0 × 10⁻¹²)
⇒ dV/dt ≈ 1.87 × 10¹⁰ V s⁻¹
(b) Displacement current i_d = ε₀ dΦ_E/dt. Here Φ_E = EA = (Q/ε₀A) × A = Q/ε₀, so i_d = dQ/dt = I
⇒ i_d = 0.15 A.
(c) Yes. Kirchhoff's junction rule holds at each plate provided the current is taken as the sum of conduction and displacement currents. The conduction current of 0.15 A flowing into a plate equals the displacement current of 0.15 A leaving it across the gap, so the total current is continuous.
(8.2) 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 between the plates.
Answer
Given: R = 0.06 m, C = 100 × 10⁻¹² F, V = 230 V (rms), ω = 300 rad s⁻¹.
(a) X_C = 1/(ωC)
⇒ X_C = 1/(300 × 10⁻¹⁰)
⇒ X_C ≈ 3.33 × 10⁷ Ω
I = V/X_C = VωC
⇒ I = 230 × 300 × 10⁻¹⁰
⇒ I = 6.9 × 10⁻⁶ A = 6.9 μA
(b) Yes. The displacement current between the plates equals the conduction current in the wires at every instant (as shown in Q1(b)), so both are 6.9 μA (rms).
(c) Peak current i₀ = √2 I = 1.414 × 6.9 × 10⁻⁶ ≈ 9.76 × 10⁻⁶ A.
The displacement current is spread uniformly over the plate area, so the part enclosed by a circle of radius r = 0.03 m is i₀(r²/R²). Applying the Ampere–Maxwell law around this circle: B × 2πr = μ₀ i₀ (r²/R²)
⇒ B₀ = μ₀ r i₀ / (2πR²)
B₀ = (4π × 10⁻⁷ × 0.03 × 9.76 × 10⁻⁶) / (2π × (0.06)²)
⇒ B₀ = (2 × 10⁻⁷ × 0.03 × 9.76 × 10⁻⁶)/0.0036
⇒ B₀ ≈ 1.63 × 10⁻¹¹ T
(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
All three are electromagnetic waves, so their speed in vacuum is the same: c = 3 × 10⁸ m s⁻¹. (Their wavelengths, frequencies and photon energies differ.)
(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
Electromagnetic waves are transverse. The electric field E and magnetic field B both lie in the x-y plane, perpendicular to each other and to the direction of propagation (z), with E × B pointing along +z.
λ = c/ν
⇒ λ = 3 × 10⁸ / (30 × 10⁶)
⇒ λ = 10 m
(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
λ = c/ν.
For ν = 7.5 MHz: λ = 3 × 10⁸ / (7.5 × 10⁶)
⇒ λ = 40 m
For ν = 12 MHz: λ = 3 × 10⁸ / (12 × 10⁶)
⇒ λ = 25 m
The wavelength band is 25 m to 40 m.
(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
An oscillating charge produces electromagnetic waves of the same frequency as its oscillation. Hence the frequency of the waves is 10⁹ Hz.
(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
E₀ = cB₀
⇒ E₀ = 3 × 10⁸ × 510 × 10⁻⁹
⇒ E₀ = 153 N C⁻¹
(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
Given: E₀ = 120 N C⁻¹, ν = 50.0 × 10⁶ Hz.
(a) B₀ = E₀/c
⇒ B₀ = 120 / (3 × 10⁸)
⇒ B₀ = 4 × 10⁻⁷ T = 400 nT
ω = 2πν
⇒ ω = 2 × 3.14 × 50 × 10⁶
⇒ ω ≈ 3.14 × 10⁸ rad s⁻¹
k = ω/c
⇒ k = 3.14 × 10⁸ / (3 × 10⁸)
⇒ k ≈ 1.05 rad m⁻¹
λ = c/ν
⇒ λ = 3 × 10⁸ / (50 × 10⁶)
⇒ λ = 6.0 m
(b) Taking the wave to travel along the x-axis with E along y and B along z:
E = 120 sin(1.05 x − 3.14 × 10⁸ t) ĵ N C⁻¹
B = 4 × 10⁻⁷ sin(1.05 x − 3.14 × 10⁸ t) k̂ T
(8.9) 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?
Answer
E = hν = hc/λ. With h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹ and 1 eV = 1.6 × 10⁻¹⁹ J: hc ≈ 1.24 × 10⁻⁶ eV m, so E (in eV) ≈ 1.24 × 10⁻⁶ / λ (in m). For typical wavelengths of each region:
Region | Typical λ | Photon energy (approx.) |
Radio waves | 1 m | 1.24 × 10⁻⁶ eV |
Microwaves | 1 mm (10⁻³ m) | 1.24 × 10⁻³ eV |
Infrared | 10 μm (10⁻⁵ m) | 0.12 eV |
Visible light | 500 nm (5 × 10⁻⁷ m) | 2.5 eV |
Ultraviolet | 100 nm (10⁻⁷ m) | 12.4 eV |
X-rays | 1 nm (10⁻⁹ m) | 1.24 × 10³ eV (1.24 keV) |
Gamma rays | 10⁻¹² m | 1.24 × 10⁶ eV (1.24 MeV) |
The photon energy reflects the spacing of energy levels in the source that emits it. Radio waves and microwaves (tiny energies) come from oscillating currents in circuits and rotations of molecules; infrared from vibrations of atoms in molecules; visible and ultraviolet light (a few eV) from transitions of outer electrons in atoms; X-rays (keV) from transitions of inner electrons in heavy atoms or decelerated fast electrons; gamma rays (MeV) from energy changes inside atomic nuclei.
(8.10) 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.]
Answer
Given: ν = 2.0 × 10¹⁰ Hz, E₀ = 48 V m⁻¹.
(a) λ = c/ν
⇒ λ = 3 × 10⁸ / (2.0 × 10¹⁰)
⇒ λ = 1.5 × 10⁻² m = 1.5 cm
(b) B₀ = E₀/c
⇒ B₀ = 48 / (3 × 10⁸)
⇒ B₀ = 1.6 × 10⁻⁷ T
(c) Energy density of the electric field u_E = ½ε₀E²; of the magnetic field u_B = B²/(2μ₀).
In the wave E = cB at every instant, and c² = 1/(μ₀ε₀). So u_E = ½ε₀(cB)² = ½ε₀ × B²/(μ₀ε₀) = B²/(2μ₀) = u_B.
Since the two are equal at every instant, their averages over a cycle are also equal: average u_E = ¼ε₀E₀² = B₀²/(4μ₀) = average u_B.
Numerically: ¼ε₀E₀² = ¼ × 8.85 × 10⁻¹² × (48)² ≈ 5.1 × 10⁻⁹ J m⁻³, and B₀²/(4μ₀) = (1.6 × 10⁻⁷)² / (4 × 4π × 10⁻⁷) ≈ 5.1 × 10⁻⁹ J m⁻³.
Important Extra Questions
What is displacement current? Why did Maxwell introduce it?
Answer
Displacement current is i_d = ε₀ dΦ_E/dt, the current associated with a changing electric flux. Maxwell introduced it because Ampere's circuital law gave inconsistent results for a charging capacitor (zero current through a surface passing between the plates). Adding i_d makes the total current continuous and shows that a changing electric field produces a magnetic field.
What is the source of electromagnetic waves?
Answer
Accelerated (for example, oscillating) charges. An oscillating charge produces an oscillating electric field, which produces an oscillating magnetic field, and so on; the fields regenerate each other and travel through space as a wave with the frequency of the oscillating charge.
Write the expression for the speed of electromagnetic waves in vacuum and in a medium.
Answer
In vacuum c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m s⁻¹. In a medium of permeability μ and permittivity ε, v = 1/√(με).
Name the electromagnetic waves used (a) to kill germs in water purifiers, (b) in radar systems, (c) for imaging bone fractures.
Answer
(a) Ultraviolet rays, (b) microwaves, (c) X-rays.
Why are infrared waves called heat waves?
Answer
Infrared waves are readily absorbed by most substances, including water molecules; they set whole atoms and molecules vibrating, which increases the internal energy and hence the temperature of the material. That is why they are called heat waves.