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8 Power Considerations and the
Poynting Vector
Electromagnetic waves carry energy (the capacity to do work) through
space. At any point in space, the flow of energy can be described by a
power density vector P, which specifies both the power density in Watts
per square metre, and the direction of flow. The vector P is called the
Poynting vector, and is a simple cross product of electric and magnetic field
vectors,
P=E×H
The units of E are Vm−1 and the units of H are Am−1 and thus the units
of P are VAm−2 or Wm−2. The total power passing through a surface S1 is
obtained by integration over S1 , i.e.
Z
P · dS Watts
W =
S1
The power dP passing through an elemental surface dS with normal n̂ is
dP = |P| cosθ dS = P · n̂ dS = P · dS
To understand why P = E × H represents power flow, we consider the
rate of energy loss though from a volume V enclosed by closed surface S,
and relate this to the electric and magnetic fields. If U is the total energy
in V , the rate of energy loss is
I
Z
dU
−
= P · dS =
∇ · PdV (by Divergence theorem)
dt
V
We re-express the term ∇ · P by substituting P = E × H, and making use
8-1
of a vector identities1 and Maxwell’s curl equations:
∇ · P = ∇ · (E × H)
= (∇ × E) · H − (∇ × H) · E
∂H
∂E
= −µ
· H − (J + ε ) · E
∂t
∂t
1 ∂E · E
1 ∂(H · H)
= −µ
−ε
−J·E
2
∂t
2 ∂t
∂ 1
∂ 1
= − ( µH 2 ) − ( εE 2) − J · E
∂t 2
∂t 2
Thus
I
P · dS =
Z
∇ · PdV
VZ
Z
Z
∂ 1 2
∂ 1
2
( µH )dV −
( εE )dV −
J · E dV
= −
∂t
2
∂t
2
VZ
VZ
ZV
1
1
∂
∂
( µH 2 )dV −
( εE 2)dV −
= −
J · E dV
∂t V 2
∂t
2
V
V
Z
∂
∂
J · E dV
= − UM − UE −
∂t
∂t
V
where
R
• UM = V ( 21 µH 2 )dV is the energy in the magnetic field. The quantity
1
2
−3
2 µH is the magnetic energy density in Jm .
R
• UE = V ( 12 εE 2)dV is the energy stored in the electric field. The quantity 21 εE 2 is the electric energy density in Jm−3.
R
• V J · E dV is a term which represents either power dissipated through
ohmic losses or power generated by a source inside V.
If representing power dissipated, the term J · E can be re-expressed as
J · E = σE · E = σE 2 in Wm−3.
H
Thus we conclude that E × H · dS represents the outflow of power from a
volume, and P = E × H is the power density vector.
1
Identity 1: ∇ · (E × H) = (∇ × E) · H − (∇ × H) · E
1 ∂(H·H)
Identity 2: ∂H
∂t · H = 2
∂t
∂H
∂t
x
· H = Hx ∂H
∂t + Hy
∂Hy
∂t
z
+ Hz ∂H
∂t =
2
1 ∂Hx
2 ∂t
+
2
1 ∂Hy
2 ∂t
8-2
+
2
1 ∂Hz
2 ∂t
Example of Electric Energy stored in a Capacitor
As an example of energy stored in a field, let us consider the energy stored
in the electric field of a parallel plate capacitor, of surface area A, gap d,
and charged to a voltage V. The capacitance is approximately C ≈ εA
d if
the gap is small compared to the plate dimensions. The electric field is
concentrated in the gap between the plates, and has strength of E = V /d.
The energy stored in the electric field is
Z
1
UE =
( εE 2)dV
V 2
1 V2
≈ ε 2 Ad
2 d
1 εA 2
=
V
2 d
1
≈ CV2
2
a familiar result from circuit theory.
8.1 Power in a Sinusoidal Plane Wave
A sinusoidal electromagnetic plane wave propagating in the ẑ direction is
described by the EM pair
E(z, t) = Ex (z, t)x̂
H(z, t) = Hy (z, t)ŷ
where
Ex (z, t) = E1 cos(ωt − kz + ψ1 )
Hy (z, t) =
Ex (z, t) E1
=
cos(ωt − kz + ψ1)
η
η
The power density is
π
E2
P = E × H = Ex Hy sin ẑ = 1 cos2(ωt − kz + ψ1 )ẑ
2
η
8-3
The time averaged power density is
P = |E × H| =
E12 2
E2 1 1
1 E12
cos ( ) = 1 ( + cos2 (2ωt − 2kz + 2ψ1) =
Wm−2
η
η 2 2
2 η
8.1.1 Poynting Vector for Complex Notation
Phasor notation is commonly used for treatment of steady state sinusoidal
signals. The phasor representation of the sinusoidal wave EM plane wave is
Ẽ = Ẽx x̂ = E1ejψ1 e−jkz x̂
E1 jψ1 −jkz
ŷ
e e
H̃ = H̃y ŷ =
η
The Poynting vector for complex phasor representation is defined as
1
P̃ = Ẽ × H̃∗
2
such that the real part of P̃ is the time averaged power density, i.e. Re{P̃} =
P = E × H. To see why the factor of 1/2 is required, we proceeding as
before, but substituting the phasor forms of Maxwell’s curl equations:
I
∗
(Ẽ × H̃ ) · dS =
=
=
=
Z
ZV
∇ · (Ẽ × H̃∗ ) dV
(∇ × Ẽ) · H̃∗ − (∇ × H̃∗ ) · Ẽ dV
ZV
(−jωµH̃) · H̃∗ − (J̃∗ − jωεẼ∗ ) · Ẽ dV
V
−jωµH̃ · H̃∗ + jωεẼ · Ẽ∗ − Ẽ · J̃∗ dV
ZV
In the non-phasor case, J · E represents the instantaneous dissipated (or
generated) power density in Wm−3 - in other words J · E is a function of
time. Here Ẽ · J̃∗ = σ Ẽ · Ẽ∗ equals twice the time averaged power density
(= 12 σE 2 ). Hence P̃ is defined as P̃ = 12 Ẽ × H̃∗ in order that the real part
equals time averaged power flow in Wm−2.
8-4
If we substitute the phasor representations for a plane wave,
1
Ẽ × H̃∗
2
1
1
= E1 ejψ1 e−jkz E1e−jψ1 ejkz
2
η
2
1 E1
=
Wm−2
2 η
P̃ =
which agrees with the result previously obtained with the real signal representation.
8.2 Power density from Radiating Antennas
An isotropic antenna is an idealised radiating source which radiates power
uniformly in all directions. The power density at a distance r from the
source is
Pt
P =
Wm−2
2
4πr
where Pt is the power in Watts flowing from the source and 4πr 2 is the
surface area of a spherical shell.
For any physical antenna, the energy is focused in a particular direction,
and hence the power density increases compared to an isotropic radiator.
The increase in power density is specified by a parameter called the gain of
the antenna, which is defined as the factor by which the power density is
increased above that of an isotropic radiator. The power density in position
(r, θ, φ) is given by
Pt
Wm−2
P = G(θ, φ)
2
4πr
where G(θ, φ) is the gain pattern, which is a function of angle.
Example
An antenna radiates 100 Watts. Calculate the electric field intensity and
the magnetic field intensity at a distance r = 1000 m from the antenna
assuming (i) an isotropic radiator and (ii) a horn antenna with a peak gain
8-5
of 10. Assume that the polarization is linear and the field is locally approximated as a plane wave.
Solution
(i) For an isotropic radiator, the power density
100
−6
Wm2.
P = 4π1000
2 = 8.0 × 10
2
The power density for a plane wave is P = 12 Eη which implies
√
√
E = 2ηP = 2 377 8.0 × 10−6 = 0.08 Vm−1.
−4
Am−1.
H = Eη = 0.08
377 = 2.1 × 10
100
−5
(ii) For the horn antenna with a gain of 10, P = 10 4π1000
2 = 8.0 × 10
Wm2.
√
√
E = 2ηP = 2 377 8.0 × 10−5 = 0.25 Vm−1.
−4
Am−1.
H = Eη = 0.25
377 = 6.5 × 10
8.3 Power Flow in a Simple Circuit
Consider the circuit shown below consisting of a resistor connected to a
battery via a pair of wires. The resistor is made of a cylinder of resistive
material, placed between two metal plates as illustrated.
H
Metal plate
I
H into page
Battery
J
P
E
V
Resistor
Length d
E
P
P
Ez
Hφ
I
H
The magnetic field lines circulate around the wires and resistor as illustrated (apply right hand rule for direction). The electric field lines extend
from the wire at the positive voltage down to negative voltage wire. The
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power flow at any point in space is described by P = E×H. Below is shown
a top view of the circuit. The Poynting vector points from left to right.
P
TOP VIEW
(plane of battery
and resistor)
P
P
H
P=E×H
r
E into page
P
Integration surface
for calculating
total power entering load.
P
P
Consider the region local to the resistor. The electric field between the
plates is vertically directed (z-direction), and is given by
V
IR
=
d
d
and the magnetic field surrounding the resistor is
Ez =
I
2πr
The power flow at any point in space is described by the cross product
Hφ =
P = E×H
= −Ez Hφ sin 90 r̂
IR I
= −
r̂
d 2πr
I 2R
r̂
= −
2πrd
which is a vector which points radially inwards as illustrated 2. From this
we observe that
• the power flows from the outside inwards towards the centre axis of the
column.
2
You should verify the direction E × H by considering the cross product at several points in space.
Note that r̂ points radially outwards from the centre axis.
8-7
• the conductor will get warm; i.e. molecules vibrate more vigorously as
electromagnetic energy is converted into kinetic energy.
• energy is re-radiated by the mechanisms of
– thermal radiation, also known as black body radiation (so called
because objects which are dark in colour radiate energy effectively
in the optical frequency range; dark objects are also good absorbers
in the optical band).
– convection through the surrounding air.
– conduction - i.e. through objects which are physically in contact
with the resistor, e.g. metal plates and wires.
The total power entering the resistor can be calculated by integrating the
Poynting vector over a closed surface enclosing the resistor. Since E ≈ 0
in the top metal plates, we need only consider the contribution through the
side walls of a cylinder of radius r (see Top View figure), which has surface
area = circumf erence × length = 2πr × d:
I
I 2R
P = P · dS ≈ −
2πrd = I 2 R Watts
2πrd
which is in agreement with circuit theory.
Question: How much power flows through the wires to the load?
Answer: For perfect wires, σ → ∞, and E → 0 within the wires. Thus
P = E × H → 0, within the wires, and we conclude that NO power flows
through the wires. From a field theory perspective, the power flows through
the air. The power (i.e. the capacity to do work) is carried in the EM field,
which travels from source to load. Why then do we need wires? The pair of
wires serve to guide the wave from source to load. Transmission line theory
is used to predict the behaviour of such guided waves when the length of
the transmission line is comparable to the wavelength.
8-8
8.3.1 Transmission Lines for 50 Hz AC Mains Electricity Supply
Wires pairs are used for getting EM power from a power station to a home,
which may be many kilometres away. At 50 Hz, however, λ = c/f =
3 × 108/50 = 6 × 106 m = 6000km, and so circuit theory with appropriate
lumped element models is more applicable for describing power transfer even over considerable distances like across a country.
Below is depicted a cross section through a pair of wires: the conductor on
the left is carrying current I into the page (in the direction of the load) and
the conductor on the right is carrying the return current out of the page.
Note the directions of the electric and magnetic field lines at this instant in
time. The Poynting vector P = E × H points into the page, in the direction
of the load. Since the current is alternating “AC”, what happens to the
Poynting vector when the current reverses direction? (Answer: E × H still
points into the page - you should check this yourself).
H
H
E
I
into page
P =E×H
points into the page
E
I
out of page
H
E
8.3.2 Transmission lines for Microwave Circuits
In microwave circuits, the wavelength (e.g. at 10 GHz, λ = 3 cm) is often
comparable to circuit board dimensions, and so wires carrying signals must
be treated as transmission lines. A common method of circuit construction is to use a ground plane on the underside of the printed circuit board
8-9
(PCB), and the signals are routed via tracks on the top side (see figure).
The wavelength of the guided wave and the characteristic impedance of the
guiding structure are functions of the dielectric constant of the material, the
width of the track, and the thickness of the PCB. For standard fibreglass
PCB, a transmission line with a 50 Ohm characteristic impedance can be
made by etching a track of about 2mm wide on the top side; the underside
is a ground plane. Increasing the width of the line reduces the characteristic
impedance; reducing the track width increases the characteristic impedance
of the line.
Substrate material
fibreglass
or special
low loss
microwave board
e.g. RT Duroid.
Copper track
H
E
Ground plane
(copper)
8-10