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7/30/2019 TERM PAPER the Maxwell Equations
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The Maxwell equations
Introduction:-
One of Newton's great achievements was to show
that all of the phenomena of classical mechanics
can be deduced as consequences of three basic,
fundamental laws, namely Newton's laws of
motion. It was likewise one of Maxwell's great
achievements to show that all of the phenomena of
classical electricity and magnetism – all of the phenomena
discovered by Oersted, Ampère, Henry,
Faraday and others whose names are commemorated in
several electrical units – can be deduced as
consequences of four basic, fundamental equations. We
describe these four equations in this
chapter, and, in passing, we also mention Poisson's andLaplace's equations. We also show how
Maxwell's equations predict the existence of
electromagnetic waves that travel at a speed of 3 % 108
m s−1. This is the speed at which light is measured to
move, and one of the most important bases of
our belief that light is an electromagnetic wave.
Before embarking upon this, we may need a reminder of two mathematical theorems, as well as a
reminder of the differential equation that describes wave
motion.
The two mathematical theorems that we need to remind
ourselves of are:
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The surface integral of a vector field over a closed surface
is equal to the volume integral of its
divergence.
The line integral of a vector field around a closed planecurve is equal to the surface integral of its curl.
There are four basic equations, called Maxwell equations,
which form the axioms of electrodynamics. The so called
local forms of these equations are the following:
rot H = j + ∂D/∂t (1)
rot E = - ∂B/∂t (2)
div B = 0 (3)
div D = ρ (4)
Here rot (or curl in English literature) is the so called
vortex density, H is vector of the magnetic field strength, j
is the current density vector, ∂D/∂t is the time derivative of
the electric displacement vector D, E is the electric field
strength, ∂B/∂t is the time derivative of the magnetic
induction vector B, div is the so called source density and ρ
is the charge density.While the above local or differential forms are easy to
remember and useful in applications, they are not so easy to
understand as they use vector calculus to give spatial
derivatives of vector fields like rot H or div D. The global
or integral forms of the Maxwell equations are somewhat
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more complicated but at the same time they can be
understood without knowing vector calculus. They are
using path, surface, and volume integrals, however:
∫ G H•dr = I + IDISP (1)
∫ G E•dr = - ∂ΦB/∂t (2)
∫ A B•dA = 0 (3)
∫ A D•dA = ∫ V ρdV (4)
where
I is the electric current I = ∫ A
j•dA,
I
DISP
is the so called displacement I
DISP
= ∫ A (∂D/∂t)
•
dA,
and
ΦB is the flux of the magnetic induction B ΦB = ∫ A
B•dA.
It is important to realize that there are two variables to
describe the electric properties of the electromagnetic field
namely E and D, and also two variables for the magnetic
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properties of the field H and B. This is necessary when
some materials are present with oriented electric and
magnetic dipoles. If the electric dipole density is denoted
by P, and the magnetic dipole density by M, then we can
use the following definitions for D and B:
D = ε0E + P and,
B = µ0H + M .
Here ε0 and µ0 are the permittivity and the permeability of
the vacuum, respectively. If we are in vacuum (P = 0, M =0
) then the Maxwell equations can be written in the
following form:
rot H = j + ε0∂E /∂t (1)
rot E = - µ0∂H /∂t (2)
div H = 0 (3)
div E = ρ/ε0 (4)
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Thus we can see that in this case there are only one variable
for the electric field E, and another variable H for the
magnetic field. In other words the introduction of two more
variables D and B (or P and M ) is necessary only if wehave not only vacuum, but some material is also present.
To determine j, P, and M for a certain material we use the
so called material equations
j= j(E, Ei ), P = P (E), and M = M ( H ).
Here Ei includes all non electromagnetic forces. The
various functions in the material equations can be different
for each material, but they are often linear. In that case the
material equations are written in the following form: j= σ⋅(E+ Ei ), P = χe⋅ε0⋅(E), and M = χm⋅µ0⋅( H ),
where σ is the electric conductivity, χe is the electric and χm
is the magnetic susceptibility.
Thus the governing equations of electromagnetism
include the 4 Maxwell equations and the 3 material
equations. Finally one more equation is needed to establish
a connection with mechanics e.g.f = ρ⋅E + jxB
where f is the mechanical force density (force acting on the
unit volume). Another possibility to establish the
connection to mechanic is
ρEE = ½( E⋅D + H⋅B ),
where ρEE is the electromagnetic energy density, that is the
energy stored by the electric and magnetic fields in the unitvolume. (The concept of force and energy were developed
already in mechanics.)
Classification OF MAXWELL’S EQUATION:-
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i) Differential Form ii) Integral Form
Fundamental Postulate
A time-varying magnetic field B gives rise to an electric field E.
The fundamental postulate for the electrostatic case for the electric
field E i.e ∇ x E = 0 must therefore be replaced by
∇x E = - ∂B/∂t
We now have two curl equations and two divergence equation for
the time varying case i.e
∇ x E = -∂B/∂t Faraday's law
∇ x H = J Ampere's Law
∇.D = ρv Gauss' s Law
∇ .B = 0
Conservation of Charge
The principle of conservation of charge must be satisfied at all
times.
The mathematical expression of charge conservation is the
equation of continuity i.e
∇.J = -∂ρv/∂t
Is the set of four equations is consistent with the equation of
continuity in the time varying situation ?
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The answer is negative and this can be easily seen by taking the
divergence for curl of H above we obtain
∇.(∇ x H) = 0 = ∇.J
The divergence of a curl of H is always zero.(LHS)
∇.J = -∂ρv/∂t . (RHS)
LHS is not equal to the RHS. Something is wrong !!
Therefore this equation ∇.(∇ x H) = ∇.J is in general not true for
the time varying case.
How does Maxwell fix this ?
Add term ∂ρv/∂t to the divergence of J
( Recall that ∇.J = -∂ρv/∂t )
(∇.J +∂ρv/∂t ) = 0,
the divergence of the curl of H is now obeyed .i.e
∇.(∇ x H) = (∇.J+∂ρv/∂t ) = 0
L.H.S = R.H.S
Since ∇.D = ρv,
∇.(∇ x H) = (∇.J+∇.∂D/∂t ) = ∇.(J+∂D/∂t ) = 0
This implies that ∇ x H = J+∂D/∂t
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The equation for ∇ x H indicates that a time-varying electric field
will give rise to a magnetic field even in the absence of a free
current i.e J = 0.
The additional term ∂D/∂t is necessary to make the equation ∇ x
H = J+∂D/∂t consistent with the principle of the conservation of
charge.
The term ∂D/∂t is called the displacement current density.
The set of consistent equations with the modified
∇ x H is known as Maxwell's Equations.
Maxwell's Equations :
∇ x E = -∂B/∂t
∇ x H = J+∂D/∂t
∇.D = ρv
∇ .B = 0
Note that ρv is the volume density of free charges and J is the
density of free currents which may comprise convection current
ρvu and conduction current σE.
These equations can be used to explain and predict all
macroscopic electromagnetic phenomena.
These 4 equations are not entirely independent.
The two divergence equations ∇.D = ρv and ∇ .B can be derived
from the two curl equations ∇ x E = -∂B/∂t and ∇ x H = J+∂D/∂t
by making use of the equation of continuity.
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Curl of E can be used to derive the divergence of B
∇ x E = -∂B/∂t
Taking the divergence on both sides,
∇. (∇ x E) = ∇.( -∂B/∂t )
In the LHS, the divergence of the curl of any vector is always zero
0 = ∇.( -∂B/∂t )
Therefore ∇.B = 0
Curl of H can be used to derive the divergence of D
∇ x H = J+∂D/∂t
Taking the divergence on both sides,
∇. (∇ x H) = ∇.( J+∂D/∂t )
0 = ∇.J+∂∇.D/∂t )
Since ∇.J = -∂ρv/∂t
∂ρv/∂t = ∂∇.D/∂t )
Therefore, ∇.D = ρv
Displacement Current
The modified Ampere's Law in differential form is given by
∇ x H = J+∂D/∂t
If we take the surface integral of both sides over an arbitrary open
surface with contour C, we have
( ) sD
JsxH d t
d d s s s
... ∫ ∫ ∫ ∂
∂+=∇ s
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The surface integral of J is equal to the conduction current Ic
flowing through S, and the surface integral of ∇ x H can be
converted into a line integral of H over the contour C by invoking
the Stoke's Theorem.
Hence
∫ ∫ ∂
∂+=
sc
cd
t I d s
DH .. l
The second term on the right side has to have the same unit as the
current Ic i.e in amperes. This second term is also proportional to
the time derivative of the electric flux density D is called the
displacement current Id.
∫ ∫ ∂∂==
s sd d
t ds I s
DJd .. where Jd = ∂D/∂t
which represents the displacement current density.
Integral Form of Maxwell's Equations
The four Maxwell's equation are differential equations that are
valid at every point in space.
In explaining electromagnetic phenomena in a physical
environment, we must deal with finite objects of specified shapes
and boundaries. It is convenient to convert differential forms into
their integral-form equivalents.
We take the surface integral of both sides of the curl equations
over an open surface S with contour C and apply Stoke's Theorem
to obtain
∇ x E = -∂B/∂t
becomes
∫ −=∫ S C d
dt
d S
BE.dl .
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and
∇ x H = J+∂D/∂t
becomes
∫
+∫ =
S C d
dt
d S
DJH.dl .
Taking the volume integral of both sides of the divergence
equations over a volume with a closed surface S and using the
Divergence Theorem, we have
∇.D = ρv
becomes
∫ =∫ V V. dV d S ρ SD
and
∇.B = 0
becomes0. =∫ S
d SB
These set of four equations is the integral form of Maxwell'sEquation.
∫ −=∫ S C d
dt
d S
BE.dl .
∫
+∫ =
S C d
dt
d S
DJH.dl .
∫ =
∫ V V. dV d S ρ SD
0. =∫ S d SB
Derivation of MAXWELL’S Equation from
Farady Law:-
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Farady’s law and the Electromagnetic “force”
(EMF).
Faraday's law relates the electromotive force, , with the
change of the magnetic flux with time,
dt
d
c
Φ−=1
The electromagnetic force around the circuit through whichthe flux is passed is
∫ ∫ ⋅×∇=⋅=S C
dan E d E ˆ
We note that
( ) ( ) ∫ ∫ ∫ ⋅∂
∂
−×⋅=×⋅−=
Φ
− S S S dant
B
cvd Bcd xd Bdt
d
cdt
d
c ˆ
1111
Hence when the loop is not moving, we have
01
=∂∂
+×∇t
B
c E
The Maxwell equations in vacuum
J ct
E
c B
t
B
c E
B E
π
πρ
410
1
04
=∂
∂−×∇=
∂
∂+×∇
=⋅∇=⋅∇
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Implicit in those are the continuity equation,
0=∂
∂
+⋅∇ t J
ρ
and the Lorentz force
×+=
c
Bv E q F
The Maxwell equations in macroscopic media
f
f
J ct
D
c H
t
B
c E
B D
π
πρ
410
1
04
=∂
∂−×∇=
∂
∂+×∇
=⋅∇=⋅∇
with
M B H P E D π π 44 −=+=
Boundary conditions
Let the fields in two neighboring media be denoted by 1
and 2. Let the surface charge density on the surface
between the two media be σ and let the surface current
density there be K . Then the boundary conditions are
K c
H H nn E E
n B Bn D D π
πσ
4)(ˆ0ˆ)(
0ˆ)(4ˆ)(
1212
1212
=−×=×−
=⋅−=⋅−
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Scalar and vector potentials
Let us introduce the vector potential A such that A B ×∇= .
This ensures that 0=⋅∇ B . Then, from the Maxwell’s
equations, we have
01
=
∂∂
+×∇t
A
c E
and therefore we choose
Φ∇−=
∂∂
+
t
A
c E
1
These choices satisfy the two homogeneous Maxwell’s
equations. The other two equations determine the scalar
and the vector potentials,
J
ct c A
t
A
c A
At c
π
πρ
4)
1(
1
41
2
2
2
2
2
−=∂
Φ∂+⋅∇∇−
∂
∂−∇
−=⋅∇∂∂+Φ∇
We note that both the electric and the magnetic field are
unchanged by the transformation
t c
A A A
∂
Λ∂−Φ=Φ→Φ
Λ∇+=→
1'
'
and therefore we can choose the gauge
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01 =
∂Φ∂+⋅∇t c
A
to obtain the Maxwell’s equations in the form
J
ct
A
c A
t c
π
πρ
41
41
2
2
2
2
2
2
2
2
−=∂
∂−∇
−=∂
Φ∂−Φ∇
(Lorentz gauge)
But of course, we could have stayed with 0=⋅∇ A , and then
to obtain
J
ct ct
A
c A
π
πρ
4)
1(
1
4
2
2
2
2
2
−=∂
Φ∂∇−
∂
∂−∇
−=Φ∇
(Coulomb gauge)
Poynting theorem
The rate of work done by the electromagnetic fields on a
moving charge q is E vq ⋅ . For a system of current density,that rate will be
∫ ∫
∂
∂⋅−×∇⋅=⋅
t
D E H E cr d E J r d
)(4
1
π
Using
)()()( H E E H H E ×∇⋅−×∇⋅=×⋅∇
we find
∫ ∫
∂∂
⋅+∂
∂⋅+×⋅∇−=⋅
t
B H
t
D E H E cr d E J r d
)(
4
1
π
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Denoting the total energy density of the electromagnetic
field by U,
( ) H B D E U ⋅+⋅=π 8
1
we obtain the conservation law
E J S t
U ⋅−=⋅∇+
∂∂
total energy rate mechanical work
with the Poynting vector
H E
c
S ×= π 4
Exercise: A segment of a wire of length L and radius a,
carrying the current I. A voltage drop V is applied on the
wire. Find the Poynting vector.
Solution: We us cylindrical coordinates, placing the wire
along the z-direction. The magnetic field produced by the
current in the wire (on the surface of the wire) is
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ϕ ˆ2
ca
I B =
and the electric field is
z L
V E ˆ= ,
The Poynting vector at the surface of the wire is
ρ π π
ˆ24 L
V
a
I H E
cS =×= ,
which is the power supplied by the battery, IV P = , divided
by the surface area of the wire, aLπ 2 .
Derivation of Maxwell’s Equation from
Electromagnetic waves
In the absence of sources (i.e., no free charges or free
currents) the Maxwell’s
equations read
01
01
00
=∂
∂−×∇=
∂
∂+×∇
=⋅∇=⋅∇
t
D
c H
t
B
c E
B D
Let us describe the medium by its dielectric constant, E D ε =
, and magnetic permeability, H B µ = . We then have
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001
00
=∂
∂−×∇=
∂
∂+×∇
=⋅∇=⋅∇
t
E
c B
t
B
c E
B E
µε
We see that each of the filed components satisfy the wave
equation
01
2
2
2
2 =∂∂−∇t
u
vu
with the velocity
εµ
cv =
The solution is (in general)
t ir k it ir k i Be Aet r u
ω ω −⋅−⋅ +=),(
with the dispersion relation relating the wave-vector with
the frequency
cv
k ω
εµ ω
==
Let us now solve for the electromagnetic fields. We write
t ir nik
t ir nik
e Bt r B
e E t r E
ω
ω
−⋅
−⋅
=
=
ˆ
0
ˆ
0
),(
),(
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with the convention that the physical fields are the real
parts of the above expressions. From the homogeneous
Maxwell’s equations we have that
0ˆ
0ˆ
0
0
=⋅
=⋅
n B
n E
and therefore both the electric and the magnetic fields are
perpendicular to the direction of the wave propagation.
Such a wave is called "transverse wave". From the other
two Maxwell’s equations we have
00ˆ E n B ×= εµ
The Poynting vector, averaged over a period, is given by
the real part of
*
8 H E
cS
×=⟩⟨
π
Explanation:
⟩×+×+×+×⟨=⟩+×+⟨=⟩⟨ H E H E H E H E c
H H E E c
S
******
16)(
2
1)(
2
1
4 π π
⟩×+×⟨= H E H E c
**
16π
In our case
n E c
S ˆ||8
2
0
µ
ε
π =⟩⟨
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The energy density, averaged over a period is
⋅+⋅=⟩⟨ ** 1
16
1 B B E E u
µ ε
π
and in our case
2
0 ||8
E uπ
ε =⟩⟨
Electromagnetic waves in a conducting medium
( We consider the charge-free case.)
The Maxwell equations are
f
J ct
D
c H
t
B
c E
B E D
π
ε
410
100
=∂
∂−×∇=
∂
∂+×∇
=⋅∇=⋅∇=⋅∇
Using Ohm’s law, relating the current density to the
electric field via the conductivity
E J σ =
the fourth equation becomes
E ct
E
c B σ
π ε
µ
41 =∂∂−×∇
Therefore we now find
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04
2
2
22
2 =
∂∂
−
∂∂
−
∇
B
E
t c B
E
t c B
E
εµ πµσ
Using again a plane-wave solution, the dispersion law(relating the wave
vector to the frequency) is now
+=
ωε
πσ ω µε
41
2
22 i
ck
and the wave is damped.
Approximately, we find
1
42
)1(
142
>>+=
<<+=
ωε
πσ πωµσ
ωε
πσ σ
ε
µ π ω εµ
cik
ci
ck
In a good conductor (high conductivity, law
frequency), the electromagnetic
wave penetrates only over a length given by
πµωσ δ
2
c=
which is called "skin-depth".
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Exercise: A wave of frequency ω , polarized in the x-
direction, propagates
along the z-direction in a good conductor which
occupies the z>0 space. Find the current in theconductor.
Solution: Let us that 1, = µ ε . The wave is then given by
δ ω )1(
0ˆ i z t i
ee x E E −−−= ,
and the current density established in the conductor is
δ ω σ /)1(
0ˆ i z t i ee x E J −−−=
Let us consider a slab of dimensions dz h×× . The
current flowing in the slab (along the x-
direction) is
hdz ee E dI
i z t i δ ω σ
/)1(
0
−−−=
and the total current becomes
t ii z t i
ei
h E hdz ee E I
ω δ ω δ σ σ −−−−
∞
−==∫ 1
0
/)1(
0
0
Taking the real part to obtain the `true’ current, we find
2
)4
cos(
0
π ω
δ σ +
=t
h E I
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We see that the current is not in-phase with the field.
Waveguides
Consider a rectangular waveguide of cross section of
dimensions a and b and axis along the z-direction. In theabsence of charge and current distributions, the electric and
magnetic fields satisfy
01
2
2
2
2 =
∂∂
−
∇
H
E
t c H
E
When the wave propagates along the z-direction, and theelectric field is only in the x-y plane, then
z y x
y x
H z H y H x H
E y E x E
ˆˆˆ
ˆˆ
++=
+=
and all the z-dependence of the fields is in the form z ik g e .
Denoting
ck =ω
we have
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0)(22
2
2
2
2
=−+∂
∂+
∂
∂ z g
z z H k k y
H
x
H
From the Maxwell’s equation 01
=
∂
∂+×∇
t
B
c
E
we then find
x
g
y y
g
x H k
k E H
k
k E −== ,
and from his other equation 01 =∂∂−×∇t
D
c H
we have
0,0 =+∂∂
−=−∂∂
+ x g z
y y g z
x H ik x
H ikE H ik
y
H ikE
Therefore,
y
H H
k
k ik
x
H H
k
k ik
z y
g
g
z x
g
g
∂∂
=
−
∂∂
=
−
2
2
1
1
and it is sufficient to find z H to find all fields. Now we
solve for that component by variables separation. Writing
)()(),( 21 yh xh y x H z
=
we have
0,0 2
2
2
2
2
1
2
2
1
2
=+=+ hk dy
hd hk
dx
hd y x
with
7/30/2019 TERM PAPER the Maxwell Equations
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02222 =+−+ g y x k k k k
We now have the following solution
)coscossincoscossinsinsin( 4321 yk xk A yk xk A yk xk A yk xk Ae H y x y x y x y x
z ik
z g +++=
sincoscoscossinsincossin()][1( 4321
12
2k xk A yk xk A yk xk A yk xk Ae
k
k
k
kk i E x y x y x y x
z ik
g g
y
x g −+−−−= −
cossinsinsincoscossincos()][1(4321
12
2
yk xk A yk xk A yk xk A yk xk Aek
k
k
kk i E
y x y x y x y x
z ik
g g
x
y
g −−+−= −
Now we need to apply boundary conditions. Taking the
surface of the waveguide as
a conductor then 0= y E at x=0 and x=a gives 021 == A A
and
amk x /π =
where m is an integer. Taking 0= x E at y=0 and y=b
gives 031 == A A
7/30/2019 TERM PAPER the Maxwell Equations
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and
bnk y /π =
where n is an integer. Therefore the final solution is
)cos()cos(b
yn
a
xm Ae H
nm
mn
z ik
z
g π π ∑=
with
22
22
−
−=
b
n
a
mk k g
π π
limiting the allowed values of m and n!