Physics · Electromagnetic waves · chapter 8 of 14

Class 12 Physics Chapter 8: Electromagnetic Waves case study question with answers

One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

In a laboratory, Kabir charges a capacitor whose circular plates, each of radius R = 5.0 cm, face each other across a small air gap, as shown. At one instant the current in the connecting wires is 0.40 A. A sensitive probe measures the magnetic field in the plane midway between the plates: inside the gap, and at the point P, 10 cm from the axis and outside the plates. Take μ₀ = 4π × 10⁻⁷ T m A⁻¹ and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²; ignore the spreading of the field at the edges.

Part 1 · 1 mark

The magnetic field at the rim of the plates, 5.0 cm from the axis, at that instant is

(a) 1.6 × 10⁻⁶ T(b) 8.0 × 10⁻⁷ T(c) 5.0 × 10⁻⁶ T(d) zero, as no charge flows across the gap

Show answer

Answer: (a) 1.6 × 10⁻⁶ T

The loop at the rim encloses the whole displacement current, equal to i: B = μ₀i/(2πR) = 2 × 10⁻⁷ × 0.40/0.050 = 1.6 × 10⁻⁶ T. 8.0 × 10⁻⁷ T is the field at P (10 cm); 5.0 × 10⁻⁶ T uses μ₀i/2R, the field at the centre of a loop; the displacement current produces a field although no charge crosses.

Part 2 · 1 mark

The magnetic field on the axis, midway between the plates, is

(a) zero(b) 1.6 × 10⁻⁶ T(c) 3.2 × 10⁻⁶ T(d) 8.0 × 10⁻⁷ T

Show answer

Answer: (a) zero

Inside, B = μ₀ i r/(2πR²), proportional to r: a loop of zero radius encloses no displacement current, so B = 0 on the axis.

Part 3 · 2 marks

Find the distance from the axis, inside the gap, at which the magnetic field equals the field at P.

Answer and marking scheme

At P, outside: B_P = μ₀i/(2π × 0.10) = 2 × 10⁻⁷ × 0.40/0.10 = 8.0 × 10⁻⁷ T. Inside: B = μ₀ i r/(2πR²). Setting them equal, r/R² = 1/0.10, so r = R²/0.10 = (0.050)²/0.10 = 0.025 m = 2.5 cm.

B inside = μ₀ir/(2πR²); B_P = μ₀i/(2π × 0.10) = 8.0 × 10⁻⁷ T (1)
r = R²/0.10 = 2.5 cm (1)

Find the rate at which the electric field between the plates is changing at that instant.

Answer and marking scheme

The displacement current equals the conduction current: ε₀ A dE/dt = i. dE/dt = i/(ε₀πR²) = 0.40/(8.85 × 10⁻¹² × π × 0.0025) ≈ 5.8 × 10¹² V m⁻¹ s⁻¹.

ε₀ A dE/dt = i, with A = πR² (1)
dE/dt ≈ 5.8 × 10¹² V m⁻¹ s⁻¹ (1)

← Chapter 7: Alternating Current Chapter 9: Ray Optics and Optical Instruments →