Download Module I: Electromagnetic waves - Lecture 4: Energy in electric and

Module I: Electromagnetic waves
Lecture 4: Energy in electric and magnetic fields
Amol Dighe
TIFR, Mumbai
Outline
Energy in static electric field
Energy in static magnetic field
Energy of EM waves
Coming up...
Energy in static electric field
Energy in static magnetic field
Energy of EM waves
Work done in vacuum
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Work done in increasing charge density by δρ(~x):
Z
Z
~
δW = φ(~x) δρ(~x)dV = 0 φ(~x)δ(∇ · E)dV
(1)
Integrate by parts:
δW
δW
Z
~
∇ · (φδ E)dV
− 0
Z
~ · δ EdV
~
E
= 0
= 0
Z
~
(∇φ) · (δ E)dV
(2)
Total work done
W =
0
2
Z
~ 2 dV
E
(3)
Work done in a dielectric
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Work done in increasing charge density by δρ(~x):
Z
Z
~
δW = φ(~x) δρ(~x)dV = φ(~x)δ(∇ · D)dV
(4)
Integrate by parts:
δW
=
δW
=
Z
~
∇ · (φδ D)dV
−
Z
~ · δ DdV
~
E
Z
~ = E)
~
Only for a linear dielectric (D
Z
1 ~ ~
W =
E · DdV
2
~
(∇φ) · (δ D)dV
(5)
(6)
Matching with energy of collection of discrete charges
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For N charges qi , we expect that the energy is
0 X
qi φi
U=
2 n
(7)
where φi is the potential at the i th charge qi due to the others.
~ i : electric field due to charge i
Let E
R 2
~ dV , we get
I Using U = (0 /2) E
Z
X
0
~2 +
~i · E
~ j )dV
U =
(Σi E
E
(8)
i
2
i6=j
Z
0 X ~
= U0 +
Ei · (−∇φi )dV
(9)
2
i
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The first term, U0 , does not depend on the distribution of
charges, it is the “self-energy”.
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The second term matches with what we expect.
Coming up...
Energy in static electric field
Energy in static magnetic field
Energy of EM waves
Differences compared to the electric field
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A steady current has to be maintained, for which energy should
continue to be supplied by an external electric field, Eext .
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When this current is increased by a small amount, a back-EMF
Eind is induced, against which work needs to be done. (This is
the analog of work done against a repulsive force when bringing
a charge from infinity.)
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This work done will be stored in the magnetic field produced.
The rest of the energy supplied by the external EMF will be
dissipated as heat in the conductor.
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~ ext + E
~ ind ) ⇒ ~J2 = σ(E
~ ext · ~J + E
~ ind · ~J)
Total current ~J = σ(E
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Rate of energy input:
dUin
~ ext · ~JdV =
=E
dt
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Z ~2
Z
J
~ ind · ~JdV
dV − E
σ
(10)
The first term is rate of energy dissipation as heat, the second
term is rate of storage of energy in the magnetic field
Energy stored in static magnetic field
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~ =E
~ ind )
Rate of energy storage in magnetic field: (E
Z
Z
dUm
~ · ~JdV = − E
~ · (∇ × H)dV
~
= − E
dt
Z
Z
~ · (∇ × E)dV
~
~ × H)dV
~
= − H
+ ∇ · (E
Z
=
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~ · (∂ B/∂t)dV
~
H
Incremental energy stored: δUm =
(11)
(12)
R
~ · δ BdV
~
H
~ is linear in B
~
Only if H
Z
Z
1 ~
1 ~ ~
~
Um =
H · BdV =
H · (∇ × A)dV
2
2
Z
Z
1
~ · (∇ × H)dV
~
~ × A)dV
~
=
A
− ∇ · (H
2
Z
1 ~ ~
A · JdV
=
2
(13)
(14)
~ ×H
~
A note about E
R
H
~
~ × H)dV
~
~ × H)
~ · dS
∇ · (E
= (E
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We neglected the term
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This is indeed valid for static electromagnetic fields
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But for time-dependent fields, we shall see later that the leading
~ and H
~ go as 1/r , so over the surface of a sphere with
terms in E
large radius r , the integral actually will have a constant nonzero
vale.
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This will be the energy radiated away at infinity due to the
~ ≡E
~ ×H
~ will be defined as the Poynting
changing currents. N
vector, which gives the rate of loss of energy through radiation.
Recap of topics covered in the lecture
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Energy stored in static electric fields: in vacuum and in a
dielectric
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Energy stored in a static magnetic field
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Note: some of the results are only valid when the material media
have linear properties.
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Energy stored and transported by an EM wave
Coming up...
Energy in static electric field
Energy in static magnetic field
Energy of EM waves
Quadratic quantities and factors of 2
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In Electrodynamics, for convenience, we often use notation
involving complex numbers (mainly exponentials), e.g.
~ =E
~ 0 ei(kx−ωt) ,
E
~ = −i B
~ 0 ei(kx−ωt)
B
(15)
when we actually want to represent
~
E
~
B
~ 0 cos(kx − ωt) = Re(E
~ 0 ei(kx−ωt) )
= E
~ 0 sin(kx − ωt) = Re(i B
~ 0 ei(kx−ωt) )
= B
(16)
(17)
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While performing calculations in complex notation and taking the
real part of the final answer works as long as we are dealing with
~ or B,
~ one has to be careful while dealing
quantities linear in E
with quadratic (or higher order) quantities.
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For example, in the complex notation above,
~ 2 i = h|E
~ ∗ · E|i
~ = |E
~ 0 |2
h|E|
(18)
while the actual answer should be (using real notation)
~ 0 |2
~ 2 i = |E
~ 0 |2 hcos2 (kx − ωt)i = 1 |E
h|E|
2
(19)
Energy density stored in EM fields
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We have already seen that the energy stored in electric field is
~ 2 (we showed this result for a static field). When
Ue = (1/2)0 |E|
the electric field represents a propagating wave, then taking into
account the “factor of 2” for averaged quadratic quantities, we get
hUe i =
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(20)
~ 2 /µ0 (we
The energy stored in magnetic field is Um = (1/2)|B|
showed it for a static magnetic field). For a propagating wave,
~ = |/˛ω||E|.
~ Including the “factor of 2”, we get
|B|
hUm i =
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1 ~ 2
0 |E0 |
4
~ 0 |2
1 |B
1 ||˛2 ~ 2
1 ~ 2
=
|E0 | = 0 |E
0|
4 µ0
4 ω 2 µ0
4
(21)
For a plane EM wave, energy stored in electric and magnetic
field is equal. The total energy of an EM wave is
hUi = hUe i + hUm i =
1 ~ 2
0 |E0 |
2
(22)
Energy transported by the EM fields
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The rate of energy transport is given by the Poynting vector,
~ =E
~ ×H
~
N
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(23)
The time-averaged value of this quantity is
~ =
h|N|i
1 ~ ||˛ ~
1
1 ~ ~
|E0 |.|H0 | = |E
|E0 | =
0|
2
2
ωµ
2
r
0 ~ 2
|E0 |
µ0
(24)
Compared with the rate ofpenergy consumption in a conductor,
~ 0 |2 , the quantity 0 /µ0 is termed the conductance of
(1/2)σ|E
vacuum
p
I Similarly,
/µ is the conductance of a medium through which
an EM wave propagates
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Recap of topics covered in this lecture
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Energy stored in static electric and magnetic fields
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Energy transported by the EM field