Chapter 32 Maxwell’s Equations and Electromagnetic Wavess Eqn.pdf · MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 4 ∫ ∫µ µ i i o o C C S B dl = J dA= I Something

Chapter 32

Maxwell’s Equations and

Electromagnetic Waves

MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 2

Maxwell’s Equations and EM Waves

• Maxwell’s Displacement Current

• Maxwell’s Equations

• The EM Wave Equation

• Electromagnetic Radiation


µ µ∫ ∫� �� i i� o o C

C S

B dl = J dA = I

Something is Missing From Ampere’s Law

The surface S in the integral above can be

any surface whose boundary is C.

If the surface S2 is chosen for

use in the above integral the

result will be that the magnetic

field around C is zero. But there

is current flowing through the

wire so we know there is a

magnetic field present.


µ µ∫ ∫� �� i i� o o C

C S

B dl = J dA = I

Something is Missing From Ampere’s Law

The surface S2 has the same boundary as S1 but there is no current

passing through S2. The charge is accumulating on the capacitor .

Maxwell noticed this deficiency in

Ampere’s law and fixed it by

defining the Displacement Current Id.

He began by taking surfaces S1 and

S2, putting them together and treating

them like one closed surface S


µ µ∫ ∫� �� i i� o o C

C S

B dl = J dA = I

The Displacement Current

Charge is building up on the disk within the closed surface S.

Therefore there are electric field lines, E, that are crossing the

surface S. We can use Gauss’s Law here.

∂ ∂

∂ ∂

∂

∂

∫��i�

enclosed

e

S o

e

d

o o

e

d o

Qφ = E dA =

ε

φ 1 Q 1 = = I

t ε t ε

φI =ε

t


∫ ∫� �� i i� o o C

C S

B dl = µ J dA = µ I

The Displacement Current

Maxwell fixed the problem with Ampere’s law by adding

another current to the right hand side of the equation below

∂

∂e

d o

φI =ε

t

∂

∂∫ ∫� �� i i�

e

o o C o d o C o 0

C S

φB dl = µ J dA = µ I + µ I = µ I + µ ε

t


Displacement Current Example

In calculating the displacement

current we will be making the

approximation that the electric

field is everywhere uniform.

This requires that the plate

separation be much smaller than

R, the radius of the plate.

The surface S must not extend

past the edge of the capacitor

plates. So r must be less than R.


Displacement Current Example

In calculating the displacement

current we will need to compute

the electric flux across the

surface S.e

d o

dφI =ε

dt

ˆ∫�ie

S

φ = E ndA = EA

o o o

Qσ QAE = = =ε ε ε A

( )

d o o o

o

d EA dE d Q dQI = ε = ε A = ε A =

dt dt dt ε A dt


B-Field from the Displacement Current

In calculating the B-Field from the

displacement current we will be

making the same approximations

that were made in the last example:

the electric field is everywhere

uniform.

∂

∂∫��i�

e

o C o 0

C

φB dl = µ I + µ ε

t

( )∫��i�

C

B dl = B 2πr

There is no current through S so IC is zero

( )∫��i�

eo o

C

dφB dl = B 2πr = 0 + µ ε

dt


B-Field from the Displacement Current

( )∫��i�

eo o

C

dφB dl = B 2πr = 0 + µ ε

dt

2 2

e

o

σφ = AE = πr E = πr

ε2

2 2

e 2 2

o o o

σ Q Qrφ = πr = πr =

ε ε πR ε R

The size of S will vary so φe will depend on r

( )

=

2 2

o o o2 2

o

o o

2 2

d Qr r dQB 2πr = µ ε = µ

dt dt ε R R

µ µr dQ rB = I

2π dt 2πR R


Maxwell’s Equations



( )∂

∂∫��i�

e

o C d o C o 0

C

φB dl = µ I + I = µ I + µ ε

tAmpere’s Law

Faraday’s Law

∫� insiden

0S

QE dA =

εGauss’s Law

∫� n

S

B dA = 0 No name - there are no

magnetic monopoles

[ ] [ ]∂ ∂

∂ ∂∫ ∫��i�

m n

C

φ B= E dl = 0 - = 0 - dA

t tε



( )∂

∂∫ ∫��i�

n

o C d o C o 0

C S

EB dl = µ I + I = µ I + µ ε dA

tAmpere’s Law

Faraday’s Law

∫ ∫� �inside

n

S V0 0

Q1E dA = ρdV =

ε εGauss’s Law

∫� n

S


magnetic monopoles

∂ ∂

∂ ∂∫ ∫��i�

m n

C

φ B= E dl = - = - dA

t tε


EM Wave Equation


Conservative Forces and Potentials

from Vector Analysis

Work around a closed loop = 0

Stokes Theorem

Therefore a potential function V exists for a conservative force.

( )

( )

( )

⋅

⋅ = ∇ ⋅

∇ ⋅ ⇒ ∇

∇ ∇ ∇

∫

∫ ∫

∫

��

��

� � � ��

� � � �

�

�

C

C S

S

W = F dl = 0

F dl × F da

× F da = 0 × F = 0

F = - V since × V = 0


Vector Analysis

ψ��

φ and are scalar functions

F and G are vector functions

∇

∇ ⋅

∇

�

� � � �

� � � �

φ= grad φ= gradient of φ

F = div F = divergence of F

× F = curl F = curl of F


Vector Analysis

Gradient

Divergence

Curl

ˆˆ ˆφ φ φ∂ ∂ ∂

∇∂ ∂ ∂

�φ= i + j + k

x y z

∂∂ ∂∇ ⋅

∂ ∂ ∂

� �yx z

FF FF = + +

x y z

ˆˆ ˆ

∂ ∂ ∂∇ ×

∂ ∂ ∂

� �

x y z

i j k

F =x y z

F F F


Vector Identities

( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

∇ ∇ ∇

∇ ∇ ∇ ∇ ∇

∇ ∇ ∇

� � �

� � � � �� i i i

� � �� i i i

fg = f g + g f

A B = B A+ A B + B× × A + A× × B

fA = f A+ f A

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( )

∇ ∇ ∇

∇ ∇ ∇

∇ ∇ ∇ ∇ ∇

∇ ∇ ∇ ∇ ∇

� � �� i i i

� � ��

� � � � �� i i i i

� � �� i

2

A× B = B × A - A × B

× fA = f × A+ f × A

× A× B = B A - A B + B A - A B

× × A = A - A


Vector Identities

( ) ˆ

ˆ

ˆ

∂ ∂ ∂∇ ∂ ∂ ∂

∂ ∂ ∂ + ∂ ∂ ∂

∂ ∂ ∂+ ∂ ∂ ∂

� � �i

x x xx y z

y y y

x y z

z z zx y z

B B BA B = A + A + A i

x y z

B B BA + A + A j

x y z

B B BA + A + A k

x y z


Maxwell’s Equations:

Integral Form to Differential Form

Stokes Theorem

Divergence Theorem

ˆ∇∫ ∫�� i i�

C S

E dl = × E ndA

ˆ ∇∫ ∫� � �i i�

S V

F ndA = FdV



∂

∂∫ ∫��i�

n

o C o 0

C S

EB dl = µ I + µ ε dA

tAmpere’s Law

Faraday’s Law

∫ ∫� �n

S V0

1E dA = ρdV

εGauss’s Law

∫� n

S


magnetic monopoles

∂

∂∫ ∫��i�

n

C

BE dl = - dA

t



ˆ =∫ ∫ ∫�i� � �n

S S V0

1E dA = E ndA ρdV

εGauss’s Law

Use the Divergence Theorem to recast the surface

integral into a volume integral

ˆ ρ∇ =∫ ∫ ∫� � �i i�

oS V V

1E ndA = EdV dV

ε

ρ

ρ

∇ =

∇ =

∫� �i

� �i

oV

o

E - dV 0ε

E - 0ε



∂

∂∫ ∫��i�

n

o C o 0

C S

EB dl = µ I + µ ε dA

tAmpere’s Law

Faraday’s Law

∫ ∫� �n

S V0

1E dA = ρdV

εGauss’s Law

∫� n

S

B dA = 0

∂

∂∫ ∫��i�

n

C

BE dl = - dA

t

ρ∇� �i

o

E =ε

∇ =� �iB 0

∂∇ ×

∂

�� B

E + = 0t

0 0µ ε

∂∇ ×

∂

��

o m

1 EB - = µ J

t

Integral form Differential form


Wave Eqn from Maxwell’s Eqn

The differential form of Maxwell’s equations brings out

the symmetry and non-symmetry of the E and B fields

We will use the following vector identity with the E-field

( )∇ ∇ ∇ ∇ ∇� � ��

i2

× × A = A - A

( )∇ ∇ ∇ ∇ ∇� � � � � � �

i2

× × E = E - E ∂∇ ×

∂

�� B

E + = 0t

( ) ∂∇ ∇ ∇ ∇

∂

� � � � � �i

2E - E = × B

t

( ) ∂ ∂

∇ ∇ ∇ + ∂ ∂

��

i2

f

EE - E = µJ εµ

t t


Wave Eqn from Maxwell’s Eqn

( ) ∂ ∂

∇ ∇ ∇ + ∂ ∂

��

i2

f

EE - E = µJ εµ

t t

These are the source

terms

ρ

ε

∂∂ ∇ + ∇

∂ ∂

�� 2

f2

2

JEE -εµ = µ

t t

∂∇

∂

∂∇

∂

��

��

22

o o 2

22

2 2

o o

EE - ε µ = 0

t

1 E 1E - = 0; where c =

c t ε µ

In free space there are no sources

This is the form of a wave equation

is the speed of light


Solutions of the Wave Equation

In free space the solutions of the wave equations show

that E and B are in phase.

These equations describe plane waves that are uniform

through out any plane perpendicular to the x-axis.

( )

x xo

y yo

E E= sin kx -ωt

B B

2π 2πk = ; ω= = 2πf

λ T


Plane Polarized Waves

( )

x xo

y yo

E E= sin kx -ωt

B B

2π 2πk = ; ω= = 2πf

λ T

Examining the E and B components

show that this represents a plane

polarized wave.

The E vector is oriented in the x

direction and the B vector is

oriented in the y direction.


Relationships Between E and B Vectors

oo o o

Ek 1B = E = E =

ω c c

E = cB

The Poynting Vector describes the propagation of the

electromagnetic energy � ��

o

E× BS =

µ

With E in the x-direction and B in the y-direction the

energy flows in the z-direction.


Relationships Between E and B Vectors



o

E× BS =

µ

( ) ( )

( )

ˆ ˆ

ˆ

� �

� �

o o

2

o o

E× B = E sin kx -ωt i× B sin kx -ωt j

E × B = E B sin kx -ωt k

The energy is proportional to E and B and is flowing

in the z-direction, perpendicular to E and B.


The Principle of Invariance

The laws of Physics should be the same for all

non-accelerated observers.

Einstein’s fundamental postulate of relativity can be stated:

“It is physically impossible to detect the uniform motion of a

frame of reference from observations made entirely within

that frame.”


The Principle of Invariance

The laws of Physics should be the same for all

non-accelerated observers.

If two observers watch the motion of an object from

two different inertial reference systems (no

acceleration), moving at a relative velocity v, they

should find the same laws of Physics

F1 = m1a1 and F2=m2a2


Galilean and Lorentz Transformations



The inertial reference frames are related by a Galilean transformation.

Newton’s laws are invariant under these transformations but not Maxwell’s

Equations

Prior to Einstein’s Theory of Special Relativity it was determined that a

Lorentz transformation kept Maxwell’s equation invariant. However, no one

knew exactly what they meant.




Electromagnetic Radiation



These are all different forms of

electromagnetic radiation.

Anytime you accelerate or

decelerate a charged particle it

gives off electromagnetic radiation.

Electrons circulating about their

nuclei don't give off radiation

unless they change energy levels.

Thermal motion gives off continuous EM radiation. Example –

Infrared radiation which peaks below the visible spectrum.

Electromagnetic Radiation


Electric Dipole Radiation


Electric Dipole Radiation


Dipole Antenna - Radiation Distribution

Note the different orientation

of the angle measurement∝2

2

sinθI(θ)

r



2

o 2

sin θI(θ) = I

r

In these problems you will need to determine the value of

Io or else take a ratio so that the Io factor will cancel out.



2

o 2

sin θI(r,θ) = I

r

(a.) Find I1 at r1 = 10m and θ = 90o

(b.) Find I2 at r2 = 30m and θ = 90o

(c.) Ratio of I2 / I1

=2

o o1 o 2

I1I = I(r = 10,θ = 90 ) = I

10 100

=2

o o2 o 2

I1I = I(r = 30,θ = 90 ) = I

30 900

(a.)

(b.)

(c.)o

2

o1

II 1900= =

II 9100



Electric - Dipole Antenna

http://www.austincc.edu/mmcgraw/physics_simulations.htm


http://www.falstad.com/mathphysics.html

Oscillating Ring Antenna


http://www.falstad.com/mathphysics.html

Oscillating Ring Pair Antenna


Electric - Dipole Antenna

Plane wave – Far from source antenna - “Far Field”


Magnetic - Loop Antenna

Plane wave – Far from source antenna - “Far Field”



( )

( )

( )

( )

∂

∂

∂

∂

∂

∂

∂ ∂

2m

2

rms

o

o

o

o rmsrms

rms

rms

d BAdφ B= - = - = - πr

dt dt t

B= πr

t

B = B sin kx -ωt

B= -ωB cos kx -ωt

t

ωBB= ωB -cos kx -ωt = = ωB

t 2

ε

ε

c

∂=

∂

= =

2 2

rms

rms

22 2 2rms

rms rms

rms

rms

B= πr πr ωB

t

E 2π= πr ωB πr ω r fE

c

ε

ε

Find εrms ?Find εrms ?



22

rmsrms

2π= r f E

cε

rms

r = 10.0cm

N = 1

E = 0.150 V/m

f : (a.) 600 kHz; (b.) 60.0 MHz

( ) ( ) =2

2 3

rms

2π= 0.10 600x10 0.150 59.2µV

cε(a.)

(b.) ( ) ( ) =2

2 6

rms

2π= 0.10 60x10 0.150 5.92mV

cε


Energy and Momentum in an

Electromagnetic Wave


Energy and Momentum in an

Electromagnetic Wave




o

E× BS =

µ

avg avg

avg avg

avg

avg

U u LAP = = = u Ac

∆t L c

P I = = u c

A

Uavg is the total energy and uavg is the energy density.

I is the intensity, the average power per unit area.

E-M Energy and Momentum



22

e o m

o

1 Bu = ε E and u =

2 2µ

These are the electric and magnetic energy densities

Since E = cB

( )2

2 22

m o e2

o o o

E cB E 1u = = = = ε E = u

2µ 2µ 22µ c

Therefore the energy density can be expressed in different ways.

=2

2

e m o

o o

B EBu = u + u = ε E =

µ µ c



The energy density

=2

2

e m o

o o

B EBu = u + u = ε E =

µ µ c

�rms rms o o

avg avg

o o

E B E B1I = u c = = = S

µ 2 µ

The intensity I is the energy/(m2 sec) = power/m2;

� ��

o

E× BS =

µ

This is the Poynting vector, its magnitude is the intensity.


Radiation Pressure pr

=

ir avg

2 2

o o rms rms o or 2

o o o o

Momentum Ip = = u

Unit Area Unit Time c

E B E B E BIp = = = = =

c 2µ c µ c 2µ c 2µ


Radiation Pressure pr - Example

A lightbulb emits spherically symmetric electromagnetic waves in al

directions. Assume 50W of electromagnetic radiation is emitted. Find (a)

the intensity, (b) the radiation pressure, (c) the electric and magnetic field

magnitudes at 3.0m from the bulb.

The energy spreads out uniformly over a sphere of radius r.

The surface area of the sphere is 4πr2.

2 2

Power 50 WI = Intensity = = = 0.442

Area 4πr m(a.)

(b.)-9

r 8

I 0.442p = = = 1.47x10 Pa

c 3.0x10


Radiation Pressure pr - Example

A lightbulb emits spherically symmetric electromagnetic waves in al

directions. Assume 50W of electromagnetic radiation is emitted. Find (a)

the intensity, (b) the radiation pressure, (c) the electric and magnetic field

magnitudes at 3.0m from the bulb.

(c.) Remember

2 2

o or o o2

o o

E BIp = = = and E = cB

c 2µ c 2µ

( )( )( )

( )

-9 -7

o o r

-8

o

8 -8

0 o

0

B = 2µ p = 1.47x10 2 4πx10

B = 6.08x10 T

E = cB = 3.0x10 6.08x10

VE = 18.2

m


Extra Slides


Documents

Chapter 32 Maxwell’s Equations and Electromagnetic Wavess Eqn.pdf · MFMcGraw-PHY 2426 Chap32-Maxwell's Eqn-Revised: 6/24/2012 4 ∫ ∫µ µ i i o o C C S B dl = J dA= I Something