Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
Try each problem first, then open the solution to check your working.
The plates of a capacitor have radius 10 cm and are 2 mm apart. Find its capacitance.
Given: R = 0.1 m; d = 2 × 10⁻³ m
Use: C = ε₀ A / d
A = π (0.1)² = 0.0314 m²
C = 8.85 × 10⁻¹² × 0.0314 / 0.002 = 1.39 × 10⁻¹⁰ F
Answer: 139 pF
A 100 pF capacitor is charged so that its voltage rises at 10⁶ V/s. Find the displacement current between its plates.
Given: C = 10⁻¹⁰ F; dV/dt = 10⁶ V/s
Use: I_d = I = C dV/dt
I_d = 10⁻¹⁰ × 10⁶ = 10⁻⁴ A
Answer: 100 μA
The plates of a capacitor have area 0.01 m². The field between them changes at 10¹² V m⁻¹ s⁻¹. Find the displacement current.
Given: A = 0.01 m²; dE/dt = 10¹² V/(m s)
Use: I_d = ε₀ A dE/dt
I_d = 8.85 × 10⁻¹² × 0.01 × 10¹² = 0.0885 A
Answer: 0.089 A
A 1 μF capacitor is connected to a 230 V (rms), 50 Hz supply. Find the rms conduction current. What is the rms displacement current?
Given: C = 10⁻⁶ F; V = 230 V; f = 50 Hz
Use: I = V ω C
I = 230 × 2π × 50 × 10⁻⁶ = 0.0723 A
The displacement current in the gap has the same value.
Answer: 72 mA; equal
The displacement current between circular plates of radius 5 cm is 2 A. Find B at the edge of the plates.
Given: R = 0.05 m; I_d = 2 A
Use: B = μ₀ I_d / 2π R
B = 2 × 10⁻⁷ × 2 / 0.05 = 8 × 10⁻⁶ T
Answer: 8 × 10⁻⁶ T
Calculate 1/√(μ₀ε₀).
Given: μ₀ = 4π × 10⁻⁷; ε₀ = 8.85 × 10⁻¹²
Use: c = 1/√(μ₀ε₀)
μ₀ε₀ = 1.2566 × 10⁻⁶ × 8.85 × 10⁻¹² = 1.112 × 10⁻¹⁷
c = 1/√(1.112 × 10⁻¹⁷) = 3.0 × 10⁸ m/s
Answer: 3 × 10⁸ m/s
Find the wavelength of an FM broadcast at 100 MHz.
Given: ν = 10⁸ Hz
Use: λ = c / ν
λ = 3 × 10⁸ / 10⁸ = 3 m
Answer: 3 m
Find the frequency of light of wavelength 600 nm.
Given: λ = 6 × 10⁻⁷ m
Use: ν = c / λ
ν = 3 × 10⁸ / 6 × 10⁻⁷ = 5 × 10¹⁴ Hz
Answer: 5 × 10¹⁴ Hz
The electric field amplitude of an EM wave is 60 V/m. Find the magnetic field amplitude.
Given: E₀ = 60 V/m
Use: B₀ = E₀ / c
B₀ = 60 / 3 × 10⁸ = 2 × 10⁻⁷ T
Answer: 200 nT
A plane EM wave has E₀ = 100 V/m. Find its intensity and its average energy density.
Given: E₀ = 100 V/m
Use: u = ½ ε₀ E₀²; I = u c
u = 0.5 × 8.85 × 10⁻¹² × 10⁴ = 4.43 × 10⁻⁸ J/m³
I = u c = 4.43 × 10⁻⁸ × 3 × 10⁸ = 13.3 W/m²
Answer: u = 4.4 × 10⁻⁸ J/m³; I = 13.3 W/m²
For the wave in the previous problem find the radiation pressure on (a) a perfect absorber and (b) a perfect reflector.
Given: I = 13.3 W/m²
Use: p = I/c and 2I/c
(a) p = 13.3 / 3 × 10⁸ = 4.4 × 10⁻⁸ Pa
(b) p = 2 × 4.4 × 10⁻⁸ = 8.9 × 10⁻⁸ Pa
Answer: 4.4 × 10⁻⁸ Pa; 8.9 × 10⁻⁸ Pa
Light of wavelength 600 nm in air enters glass of refractive index 1.5. Find its frequency, speed and wavelength in glass.
Given: λ₀ = 600 nm; n = 1.5
Use: Frequency is unchanged; v = c/n; λ = λ₀/n
ν = 3 × 10⁸ / 6 × 10⁻⁷ = 5 × 10¹⁴ Hz (unchanged)
v = 3 × 10⁸ / 1.5 = 2 × 10⁸ m/s
λ = 600 / 1.5 = 400 nm
Answer: 5 × 10¹⁴ Hz; 2 × 10⁸ m/s; 400 nm
Find the photon energy of light of wavelength 500 nm in eV.
Given: λ = 500 nm
Use: E = hc/λ = 1240 eV nm / λ
E = 1240 / 500 = 2.48 eV
Answer: 2.5 eV
Find the photon energy of X-rays of wavelength 0.1 nm.
Given: λ = 0.1 nm
Use: E = 1240 / λ (eV, nm)
E = 1240 / 0.1 = 12 400 eV = 12.4 keV
Answer: 12.4 keV
Find the photon energy of a 100 MHz radio wave in eV.
Given: ν = 10⁸ Hz; h = 6.63 × 10⁻³⁴ J s
Use: E = h ν
E = 6.63 × 10⁻³⁴ × 10⁸ = 6.63 × 10⁻²⁶ J
= 6.63 × 10⁻²⁶ / 1.6 × 10⁻¹⁹ = 4.1 × 10⁻⁷ eV
Answer: 4.1 × 10⁻⁷ eV (a tiny energy)
Find the photon energy of ultraviolet light of wavelength 300 nm.
Given: λ = 300 nm
Use: E = 1240 / λ
E = 1240 / 300 = 4.13 eV
Answer: 4.1 eV
Find the wavelength of microwaves of frequency 10 GHz.
Given: ν = 10¹⁰ Hz
Use: λ = c / ν
λ = 3 × 10⁸ / 10¹⁰ = 0.03 m
Answer: 3 cm
Find the time light takes to travel from the Sun to the Earth, 1.5 × 10¹¹ m.
Given: d = 1.5 × 10¹¹ m
Use: t = d / c
t = 1.5 × 10¹¹ / 3 × 10⁸ = 500 s ≈ 8.3 min
Answer: 500 s (8.3 minutes)
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(a) C = ε₀A/d = 8.85 × 10⁻¹² × π(0.12)² / 0.05 = 8.0 pF. Since I = C dV/dt, dV/dt = I / C = 0.15 / 8.0 × 10⁻¹² = 1.87 × 10¹⁰ V s⁻¹.
(NCERT's answer key gives 80.1 pF and 1.87 × 10⁹ V s⁻¹; those values correspond to a plate separation of 5.0 mm. The method is the same: C = ε₀A/d and dV/dt = I/C.)
(b) The displacement current equals the conduction current in the wires: I_d = ε₀ dΦ_E/dt = C dV/dt = 0.15 A.
(c) Yes. At each plate the conduction current 0.15 A flowing in through the wire is equal to the displacement current 0.15 A leaving through the gap. Including the displacement current, the total current is continuous and the junction rule is satisfied.
(a) X_C = 1/ωC = 1 / (300 × 100 × 10⁻¹²) = 3.33 × 10⁷ Ω. I_rms = V_rms / X_C = 230 / 3.33 × 10⁷ = 6.9 μA.
(b) Yes. The conduction current in the wires equals the displacement current between the plates at every instant.
(c) The peak current is I₀ = √2 × 6.9 μA = 9.76 μA. For r < R, B = μ₀ r I₀ / 2π R² = 2 × 10⁻⁷ × 0.03 × 9.76 × 10⁻⁶ / (0.06)² = 1.63 × 10⁻¹¹ T.
The speed in vacuum, c = 3 × 10⁸ m s⁻¹. (All are electromagnetic waves, which travel at the same speed in vacuum whatever their wavelength or frequency.)
Both E and B are perpendicular to the direction of propagation (the z-axis), so they lie in the x–y plane, and they are perpendicular to each other (for example E along x and B along y, so that E × B is along z).
λ = c / ν = 3 × 10⁸ / (30 × 10⁶) = 10 m.
λ = c / ν. At 7.5 MHz: λ = 3 × 10⁸ / 7.5 × 10⁶ = 40 m. At 12 MHz: λ = 3 × 10⁸ / 12 × 10⁶ = 25 m. The band is 25 m to 40 m.
An oscillating charge radiates electromagnetic waves at the same frequency as its oscillation: 10⁹ Hz.
E₀ = c B₀ = 3 × 10⁸ × 510 × 10⁻⁹ = 153 N C⁻¹.
(a) B₀ = E₀ / c = 120 / 3 × 10⁸ = 4.0 × 10⁻⁷ T (400 nT). ω = 2πν = 2π × 50 × 10⁶ = 3.14 × 10⁸ rad s⁻¹. k = ω/c = 1.05 rad m⁻¹. λ = c/ν = 3 × 10⁸ / 50 × 10⁶ = 6.0 m.
(b) Taking the wave along +x with E along y: E = 120 sin (1.05 x − 3.14 × 10⁸ t) N/C along ŷ, and B = 4.0 × 10⁻⁷ sin (1.05 x − 3.14 × 10⁸ t) T along ẑ.
Use E = hc/λ = 1240 eV nm / λ. Typical values:
Relation to the sources: the photon energy reflects the energy changes in the source. Gamma rays come from changes within nuclei (MeV); X-rays from changes of inner-shell electrons of atoms (keV); ultraviolet and visible light from changes in the outer electrons of atoms (a few eV); infrared from the vibrations and rotations of molecules (about 10⁻² to 1 eV); and microwaves and radio waves from oscillating currents in electronic circuits (the energies are tiny, so these appear as a classical wave).
(a) λ = c / ν = 3 × 10⁸ / 2.0 × 10¹⁰ = 0.015 m (1.5 cm).
(b) B₀ = E₀ / c = 48 / 3 × 10⁸ = 1.6 × 10⁻⁷ T.
(c) Average energy density of the E field: u_E = ¼ ε₀ E₀². Of the B field: u_B = B₀² / 4μ₀. Since B₀ = E₀/c, u_B = E₀² / (4μ₀ c²) = ¼ ε₀ E₀² (using c² = 1/μ₀ε₀). So u_E = u_B. Numerically u_E = ¼ × 8.85 × 10⁻¹² × 48² = 5.1 × 10⁻⁹ J m⁻³.