Fields from time varying B

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:

(∇×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:

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

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.

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.