---
title: "Fields from time varying B"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:fields-from-time-varying-b"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/fields-from-time-varying-b
site: BCT Engineering
---

# Fields from time varying B

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2076 Chaitra, 3+3+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/fields-from-time-varying-b
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Within a certain region, ε = 10⁻¹¹ F/m and μ = 10⁻⁵ H/m. If B_x = 2×10⁻⁴ cos 10⁵t sin 10⁻³y T, find: (a) E; (b) the total magnetic flux passing through the surface x = 0, 0 < y < 40 m, 0 < z < 2 m at t = 1 μs; (c) the value of the closed line integral of E around the perimeter of the given surface.

## Short answer

Using Maxwell's equations: (a) E = −2×10⁴ sin(10⁵t) cos(10⁻³y) a_z V/m. (b) Flux Φ through x=0, 0<y<40, z∈[0,2] at t=1 μs: Φ ≈ 3.18×10⁻⁴ Wb. (c) Closed line integral ∮E·dl = −dΦ/dt ≈ 3.19 V.

## Answer

Given ε = 10⁻¹¹ F/m, μ = 10⁻⁵ H/m and B_x = 2×10⁻⁴ cos(10⁵t) sin(10⁻³y) T. The medium is source-free (J = 0, ρ_v = 0).

#### (a) E
H_x = B_x/μ = 20 cos(10⁵t) sin(10⁻³y) A/m, and H_y = H_z = 0. From ∇×H = ε ∂E/∂t:

```text
(∇×H)_z = ∂H_y/∂x − ∂H_x/∂y = −20 (10⁻³) cos(10⁵t) cos(10⁻³y)
                              = −2×10⁻² cos(10⁵t) cos(10⁻³y)
(∇×H)_y = ∂H_x/∂z − ∂H_z/∂x = 0
```
So only E_z exists:

```text
ε ∂E_z/∂t = −2×10⁻² cos(10⁵t) cos(10⁻³y)
∂E_z/∂t = −2×10⁹ cos(10⁵t) cos(10⁻³y)
E_z = −(2×10⁹/10⁵) sin(10⁵t) cos(10⁻³y)

E = −2×10⁴ sin(10⁵t) cos(10⁻³y) a_z   V/m
```
Check with Faraday's law: (∇×E)_x = ∂E_z/∂y = 20 sin(10⁵t) sin(10⁻³y), and −∂B_x/∂t = 2×10⁻⁴ × 10⁵ sin(10⁵t) sin(10⁻³y) = 20 sin(10⁵t) sin(10⁻³y). They agree.

#### (b) Flux through x = 0, 0 < y < 40 m, 0 < z < 2 m at t = 1 μs
```text
Φ = ∫∫ B_x dy dz = 2×10⁻⁴ cos(10⁵t) × 2 × ∫₀⁴⁰ sin(10⁻³y) dy
∫₀⁴⁰ sin(10⁻³y) dy = [−cos(10⁻³y)/10⁻³]₀⁴⁰ = 10³(1 − cos 0.04) = 10³(7.999×10⁻⁴) = 0.7999

At t = 1 μs: 10⁵t = 0.1 rad, cos 0.1 = 0.99500
Φ = 4×10⁻⁴ × 0.99500 × 0.7999 = 3.18×10⁻⁴ Wb
```

#### (c) Closed line integral of E around the perimeter
By Faraday's law, ∮E·dl = −dΦ/dt.

```text
Φ(t) = 4×10⁻⁴ × 0.7999 × cos(10⁵t) = 3.2×10⁻⁴ cos(10⁵t)
dΦ/dt = −3.2×10⁻⁴ × 10⁵ sin(10⁵t) = −32 sin(10⁵t)
∮E·dl = +32 sin(0.1) = 32 × 0.09983 = 3.19 V
```
(The sign is for the loop direction given by the right-hand rule relative to the chosen normal a_x.)

**Answers:** (a) E = −2×10⁴ sin(10⁵t) cos(10⁻³y) a_z V/m; (b) Φ = 3.18×10⁻⁴ Wb = 0.318 mWb; (c) ∮E·dl ≈ 3.19 V.

## Check yourself

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

1. How is E found from B_x?
   A. Convert to H = B/μ and use ∇×H = ε∂E/∂t
   B. Divide B by ε
   C. E = vB
   D. Integrate B with respect to x
   Answer: A. H_x = B_x/μ = 20 cos(10^5 t) sin(10^−3 y); its curl equals ε∂E/∂t, giving only E_z. (Key point: Route to E)

2. E equals:
   A. −2×10^4 sin(10^5 t) cos(10^−3 y) a_z V/m
   B. 2×10^4 cos(10^5 t) a_z
   C. −2×10^9 cos(10^5 t) a_z
   D. 20 sin(10^5 t) a_x
   Answer: A. Integrating −2×10^9 cos(10^5t)cos(10^−3y) over t gives −2×10^4 sin(10^5t)cos(10^−3y) a_z. (Key point: Result for E)

3. Flux at t = 1 μs through the surface is about:
   A. 3.18×10^−4 Wb
   B. 3.2 Wb
   C. 1×10^−4 Wb
   D. 0
   Answer: A. Φ = 4×10^−4 × cos(0.1) × 0.7999 ≈ 3.18×10^−4 Wb. (Key point: Flux)

4. The closed line integral of E is found from:
   A. ∮E·dl = Φ
   B. ∮E·dl = −dΦ/dt
   C. ∮E·dl = ε Φ
   D. ∮E·dl = 0
   Answer: B. Faraday's law: −dΦ/dt = 32 sin(0.1) ≈ 3.19 V. (Key point: Line integral)

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