TERM PAPER the Maxwell Equations

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:



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



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



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)



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:-



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 ?



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



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.



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



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 .



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:-



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

=∂

∂−×∇=

∂

∂+×∇

=⋅∇=⋅∇



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

=−×=×−

=⋅−=⋅−



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



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



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



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



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

),(

),(



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 =⟩⟨



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



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".



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



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



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



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



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!

TERM PAPER the Maxwell Equations

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