---
title: "Induced magnetic field, displacement current"
subject: "Engineering Physics"
chapter: 9
key: "semester_1/engineering_physics:displacement-current-induced-field-circular-plates"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/displacement-current-induced-field-circular-plates
site: BCT Engineering
---

# Induced magnetic field, displacement current

Engineering Physics · Chapter 9: Electromagnetic Waves

- Asked in: 2073 Chaitra, 5 marks
- Video: [Displacement current & Ampere Maxwell's law | Electromagnetic waves | Physics | Khan Academy](https://www.youtube.com/watch?v=t-W6_fn_1bI) — Khan Academy India - English
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/displacement-current-induced-field-circular-plates
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-9

## Question

A parallel plate capacitor with circular plates is being charged by a varying electric field of 1.5×10¹² V/m-s. Calculate the induced magnetic field if the diameter of the plate is 110 mm and the displacement current of this condition.

## Short answer

Displacement current I_d = ε₀ π R² dE/dt = 0.126 A. Induced magnetic field at the plate edge (r = 55 mm) is B = μ₀ ε₀ R (dE/dt)/2 = 4.6×10⁻⁷ T, increasing linearly with r inside.

## Answer

**Given:** dE/dt = 1.5×10¹² V/(m·s), plate diameter 110 mm, so R = 55 mm = 0.055 m. ε₀ = 8.85×10⁻¹² F/m, μ₀ = 4π×10⁻⁷ H/m.

The changing field E between the plates gives a displacement current I_d = ε₀ dΦ_E/dt = ε₀ A dE/dt, which induces a magnetic field.

#### Displacement current
```text
A   = π R² = π (0.055)² = 9.503×10⁻³ m²
I_d = ε₀ A dE/dt
    = 8.85×10⁻¹² × 9.503×10⁻³ × 1.5×10¹²
    = 0.126 A
```

#### Induced magnetic field
Ampere-Maxwell law on a circle of radius r ≤ R: B(2πr) = μ₀ε₀ (πr²) dE/dt, so B = μ₀ε₀ r/2 · dE/dt. At the edge of the plates (r = R, the maximum):

```text
B = μ₀ ε₀ R (dE/dt) / 2
  = (4π×10⁻⁷)(8.85×10⁻¹²)(0.055)(1.5×10¹²)/2
  = 4.59×10⁻⁷ T
(check: μ₀ I_d /(2πR) = 4π×10⁻⁷ × 0.126 /(2π × 0.055) = 4.59×10⁻⁷ T)
```

**Answer: I_d ≈ 0.126 A; B ≈ 4.6×10⁻⁷ T at the plate edge (r = 55 mm).** Inside, B grows linearly with r.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The displacement current between capacitor plates is given by:
   A. I_d = A dE/dt / ε₀
   B. I_d = μ₀ A dE/dt
   C. I_d = ε₀ dE/dt / A
   D. I_d = ε₀ A dE/dt
   Answer: D. I_d = ε₀ dΦ_E/dt and Φ_E = EA. (Key point: Displacement current formula)

2. The plate diameter is 110 mm. What radius and area are used?
   A. R = 0.055 m, A = πR² ≈ 9.5×10⁻³ m²
   B. R = 0.110 m, A ≈ 3.8×10⁻² m²
   C. R = 0.055 m, A = 2πR
   D. R = 0.11 m, A = πR
   Answer: A. Radius is half the diameter and area is πR². (Key point: Plate radius)

3. How does the induced B vary with r inside the plate region?
   A. B is constant
   B. B = μ₀ε₀ r/2 · dE/dt, growing linearly with r
   C. B falls as 1/r
   D. B is zero inside
   Answer: B. The enclosed displacement current grows as r², while the circumference grows as r. (Key point: Field inside plates)

## Related in Engineering Physics

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