Download Exercise 5 14 October 2016

ELEC–E4130
Electromagnetic Fields
Exercise 5
Fall 2016
14 October 2016
Solution
1. Define:
(a) Poynting theorem
(b) Group velocity
(c) Energy velocity
Solution
(a) Poynting’s theorem is a statement of conservation of energy for the electromagnetic
field:
H
R
R 1
R 1
d
d
− area (E × H) · ds = vol J · E dv + dt
D
·
E
dv
+
dt vol 2 B · H dv
vol 2
On the right-hand side, the first integral is the total (but instantaneous) ohmic power
dissipated within the volume. The second integral is the total energy stored in the
electric field, and the third integral is the stored energy in the magnetic field. Since
time derivatives are taken of the second and third integrals, those results give the time
rates of increase of energy stored within the volume, or the instantaneous power going
to increase the stored energy. The sum of the expressions on the right must therefore
be the total power flowing into this volume, and so the total power flowing out of the
volume is:
H
W
area (E × H) · dS
where the integral is over the closed surface surrounding the volume. The cross product E × H is known as the Poynting vector, S,
S=E×H
W/m2
(b) Group velocity is the velocity of propagation of the wave-packet envelope (of a group
of frequencies). A general relation between the group and phase velocities is:
up
ug =
ω dup
1− u
p dω
(c) Energy velocity is the ration of the time-average propagated power to the time-average
stored energy per unit guide length:
uen = (PWz )0 av ,
av
where the time-average power (Pz )av is equal to the time-average Poynting vector
Pav integrated over the guide cross section:
R
(Pz )av = S Pav · ds,
0 is the sum of the time-average
and the time-average stored energy per unit length Wav
stored energy density (we )av and the time-average stored magnetic energy density
(wm )av integrated over the guide cross section:
R
0 =
Wav
S [(we )av + (wm )av ]ds.
2. What do we mean when we say that incident wave is (a) TM wave, and (b) TE wave.
Solution
In the figure below, two cases are shown that differ by the choice of electric field orientation.
In (a), the E field is polarized in the plane of the page, with H therefore perpendicular
to the page and pointing outward. In this illustration, the plane of the page is also the
plane of incidence, which is more precisely defined as the plane spanned by the incident
k vector and the normal to the surface. With E lying in the plane of incidence, the wave
is said to have parallel polarization or to be p-polarized (E is parallel to the incidence
plane). Note that although H is perpendicular to the incidence plane, it lies parallel (or
transverse) to the interface. Consequently, another name for this type of polarization is
transverse magnetic, or TM polarization.
Figure (b) shows the situation in which the field directions have been rotated by 90 deg.
Now H lies in the plane of incidence, whereas E is perpendicular to the plane. Because
E is used to define polarization, the configuration is called perpendicular polarization, or
is said to be s-polarized. E is also parallel to the interface, and so the case is also called
transverse electric, or TE polarization.
3. The figure below shows a side and a top view of a solenoid carrying current I with electric
and magnetic fields E and B at time t. In the solenoid, the current I is:
(a) Increasing in time.
(b) Constant in time.
(c) Decreasing in time.
Explain.
Solution The correct answer is (c): the current I is decreasing in time.
Assume that a unit vector is pointing out of page. The line integral of the CCW oriented
electric field is positive. In accordance with Faraday’s law:
H
R
d
E · dL = − dt
S B · dS,
where the fingers of our right hand indicate the direction of the closed path, and our
thumb indicates the direction of dS. Since left-hand side of this equality is positive then
the positive magnetic flux must be decreasing. Since B is proportional to I, the current
must be also decreasing.
4. In the lecture you studied the Rayleigh formula in terms of frequency. It can be also
dup
represented in view ug = up + β
. Starting from the given expression, please prove its
dβ
dup
form in terms of lambda ug = up - λ
.
dλ
Solution
ug = up + β
up −
dup
dβ
= up + β
dup dλ
dλ dβ
= up + β
2π
dup d( β )
dλ dβ
= up + β
dup −2π
dλ β 2
= up + β(− βλ )
dup
dλ
=
du
λ dλp
5. A standing electromagnetic wave is established in a vacuum in the x-direction, the electric
component is: E = E0 cos (ωt) cos (kx). Find the magnetic component B(x,t). Show an
approximate picture of the distribution of the electric and magnetic wave components (E
and B) at t = 0 and t = T/4, where T = 2π/ω is the period of oscillations. Find the
Poynting vector and its average value over one period.
Solution
Let’s choose coordinate axes in such way that k is along x and E is along y. Then we can
find B(x, t) through known electric field using Maxwell’s equation:
∇ × E = − ∂B
∂t
ax ay az ∂
∂Ey
∂Ey
∂
∂ ∂
∇×E = ∂x ∂y
∂z = −ax ∂z +az ∂x = az ∂x (E0 cos(ωt)cos(kx)) = −az E0 kcos(ωt)sin(kx)
0 E
0
y
R
B = az E0 kcos(ωt)sin(kx)dt = az Ec0 sin(kx)sin(ωt)
E2
S = E × H = ax 4µ00 c sin(2ωt)sin(2kx)
RT
hSi = T1 0 Sdt = 0