Electromagnetic Fields ELEC-E4130
Homework answers
Fall 2022
Homework 3
Problem 3.1:
Let’s derive governing partial differential equation for E(t) from Maxwell’s equations using Faraday’s and Ampère’s laws
∂H
∇ × E = − ∂B
∂t = −µ ∂t ,
∇×H=J+
∂D
∂t
= σE + ϵ ∂E
∂t .
(1)
Let’s use vector identity
∇ × (∇ × E) = ∇(∇ · E) − ∇2 E = −∇2 E,
(2)
where ∇ · E = 0 since ∇ · D = 0. As ∇ is spatial and not a time-dependent operator, it can be moved inside time derivate by rearranging terms and substituting
Faraday’s law
∂
∇2 E = −∇ × (∇ × E) = −∇ × (−µ ∂H
∂t ) = µ ∂t (∇ × H)),
∂
(σE + ϵ ∂E
= µ ∂t
∂t )),
(3)
2
∂ E
⇒ ∇2 E − µσ ∂E
∂t − µϵ ∂t2 = 0.
The partial differential equation for H(t) is obtained similarly from Ampère’s
law as
∇2 H = −∇ × (∇ × H) = −∇ × (σE + ϵ ∂E
∂t )
∂
∂
∂H
= −σ∇ × E − ϵ ∂t
∇ × E = −σ(−µ ∂H
∂t ) − ϵ ∂t (−µ ∂t )
(4)
2
∂ H
⇒ ∇2 H − µσ ∂H
∂t − µϵ ∂t2 = 0.
Problem 3.2:
A single vector potential π e , known as electric Hertz potential, can be define
such that
H = jωϵ0 ∇ × π e .
(5)
1
Let’s substitute Eq. (5) to Faraday’s law as
∇ × E = −jωµ0 H = ω 2 µ0 ϵ0 ∇ × π e ,
(6)
⇒ ∇ × (E − k02 π e ) = 0,
where ω 2 µ0 ϵ0 = k02 . Let’s define
E − k02 π e = ∇Ve .
(7)
Next, let’s substitute polarized medium relation to Ampères’s law as
∇ × H = jωD = jω(ϵ0 E + P) = jωϵ0 (E +
P
ϵ0 ).
(8)
Now we can substitute Eq. (5) and Eq. (7) into Eq. (8) and use vector identity
as
∇ × H = jωϵ0 ∇ × (∇ × π e ) = jωϵ0 (k02 π e + ∇Ve + ϵP0 ),
(9)
= jωϵ0 (∇∇ · π e − ∇2 π e ).
(b) Show that π e satisfies the non-homogeneous Helmholtz’s equation. Let’s
choose ∇ · π e = Ve and substitute to Eq. (9), we obtain
∇2 π e + k02 π e = − ϵP0 ,
k0 =
√
µ0 ϵ0 .
(10)
(a) The electric field intensity E in terms of π e and P. Let’s use relation
∇ · π e = Ve as
E − k02 π e = ∇Ve ⇒
(11)
E = k02 π e + ∇(∇ · π e ) = k02 π e + ∇2 π e + ∇ × (∇ × π e ).
Taking relation ∇2 π e = −k02 π e −
P
ϵ0
from Eq. (9), we obtain
E = ∇ × ∇ × πe −
P
ϵ0 .
(12)
Problem 3.3:
For the time-harmonic magnetic field in a region of free space:
H(t) = ax H0 sin (ky y) cos (ωt − βz)
(13)
(a) Find the corresponding vector phasor H:
H(t) = ax H0 sin (ky y)e−jβz .
2
(14)
(b) Determine the electric field E in phasor form:
ax
0
Hx
∇×H=
ay
az
∂
∂y
∂
∂z
0
0
∂Hx
x
= ay ∂H
∂z − az ∂y
= −ay jβH0 sin (ky y)e−jβz − az ky H0 cos (ky y)e−jβz
(15)
= jωϵ0 E ⇒
−jβz
jky H0
0
E = − ay βH
.
ωϵ0 sin (ky y) + az ωϵ0 cos (ky y) e
(c) Determine the constant β.
ax
∇×E= 0
0
ay
az
∂
∂y
∂
∂z
Ey
Ez
z
= ax ( ∂E
∂y −
= ax −
jky2 H0
ωϵ0
= ax −
j(ky2 +β 2 )H0 sin (ky y)e−jβz
ωϵ0
sin (ky y) −
jβ 2 H0
ωϵ0
sin (ky y) e−jβz
= −jωµ0 H ⇒
k2 +β 2
H = ax ωy2 µ0 ϵ0 H0 sin (ky y)e−jβz ⇒
β=
q
ω 2 µ0 ϵ0 − ky2 .
3
∂Ey
∂z )
(16)