physics 1304: lecture 17, pg 1 f()x x f(x x z y. physics 1304: lecture 17, pg 2 lecture outline l...

Physics 1304: Lecture 17, Pg 1

f( )x

x

f( x

x

z

y


Lecture Lecture OutlineOutline

Electromagnetic Waves: Experimental Ampere’s Law Is Incomplete: Displacement Current Review of Wave Properties Electromagnetic Waves: Theory

Maxwell’s Equations contain the wave equation! The velocity of electromagnetic waves = c The relationship between E and B in an e-m wave Energy in e-m waves: the Poynting vector

Text Reference: Chapter 34


Fields from Fields from Circuits?Circuits?

We have been focusing on what happens within the circuits we have been studying (eg currents, voltages, etc.)

What’s happening outside the circuits??

We know that:

» charges create electric fields and

» moving charges (currents) create magnetic fields. Can we detect these fields?Demo: Tesla coil & fluorescence bulbs… what’s this all about!


How Do We Understand? How Do We Understand? How can we understand the results?

DC current … no detected fieldsAC current … fields detected

Answer?Actually we do not yet have all that we need to know to understand this

phenomenon.One of our basic laws (Ampere’s Law) needs to be modified!!

J. C. Maxwell (~1860) realizes that the equations of electricity & magnetism are inconsistent with the conservation of charge!

It is probably not be obvious to you that these equations are inconsistent with conservation of charge, but there is a lack of symmetry here!


Ampere’s Law is the Ampere’s Law is the Culprit!Culprit!

Gauss’ Law: B dA 0

E dA

q 0

E d

d

dtB

B d I 0

both equations have the same symmetry (the difference is that magnetic charge does not exist!)

Ampere’s Law and Faraday’s Law:

If Ampere’s Law were correct, the right hand side of Faraday’s Law should be equal to zero -- since no magnetic current.Therefore(?), maybe there is a problem with Ampere’s Law.In fact, Maxwell proposes a modification of Ampere’s Law by adding another term (the “displacement” current) to the right hand side of the equation! ie


Maxwell’s Displacement Maxwell’s Displacement CurrentCurrent

Can we understand why this “displacement current” has the form it does?

• Consider applying Ampere’s Law to the current shown in the diagram.

• If the surface is chosen as 1, 2 or 4, the enclosed current = I

• If the surface is chosen as 3, the enclosed current = 0! (ie there is no current between the plates of the capacitor)

Big Idea: The added term is non-zero in this case, since the current I causes the charge Q on the capacitor to change in time which causes the Electric field in the region between the plates to change in time. The “displacement current” ID = 0 (dE/dt) in the region between the plates = the real current I in the wire.

circuit

Maxwell’s Displacement Maxwell’s Displacement CurrentCurrent

The Electric Field E between the plates of the capacitor is determined by the chrg Q on the plate of area A: E = Q/(A0)

What we want is a term that relates E to I without involving A. The answer: the time derivative of the electric flux!!

B d I ID 0( )

ddt

ddt

EAdQdt

IE

1 1

0 0

Therefore, if we want ID = I, we need to identify:

IddtD

E 0

In order to have for surface 2 to be equal to for surface 3, we want the displacement current in the region between the plates to be equal to the current in the wire.

B d

QdSEE0

1

Recall flux:


Lecture 22, ACT 1Lecture 22, ACT 1

At t=0, the switches in circuits I and II are closed and the capacitors start to charge.

What must be the relation between I and II such that the current in the circuits are the same at all times?

(a) I = II (b) I = 2II(c) impossible

• Suppose at time t, the currents flowing into C and 2C are identical and that 2C has twice the area of C as shown. Compare the magnetic fields at the points shown BI(ro/2) and BII(ro/2):

(a) BI < BII (b) BI = BII (c) BI > BII

1A

R

C

I

R/2

2C

I

I

II

I

II

1B ro/2ro

ro 2

I Iro/2

I II



At t=0, the switches in circuits I and II are closed and the capacitors start to charge.

What must be the relation between I and II such that the current in the circuits are the same at all times?

(a) I = II (b) I = 2II(c) impossible

1A

R

C

I

R/2

2C

I

I

II

I

II

• Both circuits have the same time constant = RC.• Therefore, if the currents are equal at any time, they are equal at all times.• The easiest time to evaluate the currents is at t=0.

IRI( )0

For circuit I: IR

II( )02

For circuit II:

• These currents are equal if: I II2


• Suppose at time t, the currents flowing into C and 2C are identical and that 2C has twice the area of C as shown. Compare the magnetic fields at the points shown BI(ro/2) and BII(ro/2):

(a) BI < BII (b) BI = BII (c) BI > BII

1B

• If the currents flowing to the capacitors are equal, the total displacement currents are also equal.• The displacement currents are distributed uniformly throughout the volume between the two plates.• Therefore, when we apply our new form of Ampere’s Law to find the magnetic field, the fraction of the displacement current with r<ro/2 is smaller in case II than it is in case I. Therefore, BII < BI.

• Here are the equations behind these words:

ro/2ro

ro 2

I Iro/2

I II


On to On to Waves!!Waves!!

Note the symmetry of Maxwell’s Equations in free space, i.e. when no charges are present

h is the variable that is changing in space (x) and time (t). v is the velocity of the wave

We can now predict the existence of electromagnetic waves. Why? The wave equation is contained in these equations. Do you remember the wave equation????

Review of Waves from Phys 1303Review of Waves from Phys 1303

k 2

v fk

22

fT

h x t A kx t, cos h

x

A

A = amplitude = wavelengthf = frequencyv = speedk = wave number

2

2 2

2

2

1h

x v

h

tThe one-dimensional wave equation:

h x t h x vt h x vt( , ) ( ) ( ) 1 2

has a general solution of the form:

where h1 represents a wave traveling in the +x direction and h2 represents a wave traveling in the -x direction.

A specific solution for harmonic waves traveling in the +x direction is:



Snapshots of a wave with angular frequency are shown at 3 times:

Which of the following expressions describes this wave?

(a) y = sin(kx-t) (b) y = sin(kx+t) (c) y = cos(kx+t)

(a) +x direction (b) -x direction • In what direction is this wave traveling?2B

2A

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

x

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

x

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

xx

y

0

0

0

0

t 2

t 0

t






2A

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

x

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

x

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

xx

y

0

0

0

0

t 2

t 0

t

• The t=0 snapshot at t=0, y = sinkx• At t=/2 and x=0, (a) y = sin(-/2) = -1• At t=/2 and x=0, (b) y = sin(+/2) = +1






• In what direction is this wave traveling?

(a) +x direction (b) -x direction

2A

2B

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

x

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

x

x ..,0.0r1

nr1

0 2 4 61

0

1

h( )x

xx

y

0

0

0

0

t 2

t 0

t

• We claim this wave moves in the -x direction.• The orange dot marks a point of constant phase.

• It clearly moves in the -x direction as time increases!!


Step 1 Assume we have a plane wave propagating in z (ie E, B not functions of x or y)

E dddt

B

E z E z x x zB

tx xy

2 1

Ez

z x x zB

tx y

Ez

B

tx y

4-step Plane Wave Derivation

E E kz tx 0 sin( )Example: does this

x

z

y

z1 z2

Ex Ex

Z

x

By

Step 2 Apply Faraday’s Law to infinitesimal loop in x-z plane


B dddt

E 0 0

t

EzyyzBzB x

0o2y1y

t

Ezyyz

z

Bx

00y

B

zEt

y x 0 0

2x

2

00y

2

t

E

zt

B

2x

2y

2

z

E

tz

B

2

2 0 0

2

2

E

z

E

tx x

4-step Plane Wave Derivation

x

z

y

z1 z2

By

Z

yBy

Ex

Step 3 Apply Ampere’s Law to an infinitesimal loop in the y-z plane:

Step 4 Combine results from steps 2 and 3 to eliminate By

Velocity of Electromagnetic Velocity of Electromagnetic WavesWaves

We derived the wave equation for Ex:

2

2 0 0

2

2

E

z

E

tx x

cs/m1000.3v 8

Therefore, we now know the velocity of electromagnetic waves in free space:

Putting in the measured values for 0 & 0, we get:

This value is identical to the measured speed of light! We identify light as an electromagnetic wave.

How is B related to How is B related to E?E?

We derived the wave eqn for Ex:

)tkzcos(kEz

E

t

B0

xy

)tkzsin(Ek

dt)tkzcos(kEB 00y

• By is in phase with Ex

• B0 = E0 / c

2x

2

22x

2

t

E

c

1

z

E

2y

2

22y

2

t

B

c

1

z

B

We could have also derived for By:

How are Ex and By related in phase and magnitude?

E E kz tx 0 sin( ) kc

Consider the harmonic solution: where

(Result from step 2)

E & B in Electromagnetic E & B in Electromagnetic WaveWave

Plane Harmonic Wave:

E E kz tx 0 sin( )B B kz ty 0 sin( )

kcE cB0 0where:

x

z

y

where are the unit vectors in the (E,B) directions. , e bNothing special about (Ex,By); eg could have (Ey,-Bx)

sNote: the direction of propagation is given by the cross product

s e b

s e b e b s b s e Note cyclical relation:



Suppose the electric field in an e-m wave is given by:

(a) + z direction (b) -z direction

• Which of the following expressions describes the magnetic field associated with this wave?

3B

E jE kz t cos( )0

(a) Bx = -(Eo/c)cos(kz + t) (b) Bx = +(Eo/c)cos(kz - t) (c) Bx = +(Eo/c)sin(kz - t)

3A In what direction is this wave traveling ?




In what direction is this wave traveling ?

(a) + z direction (b) -z direction3A

E jE kz t cos( )0

• To determine the direction, set phase = 0: kz t 0 zk

t

• Therefore wave moves in + z direction!• Another way: cos( ) cos( ) E E kz to cos( )



In what direction is this wave traveling ?

(a) + z direction (b) -z direction

• Which of the following expressions describes the magnetic field associated with this wave?

(a) Bx = -(Eo/c)cos(kz + t) (b) Bx = +(Eo/c)cos(kz - t) (c) Bx = +(Eo/c)sin(kz - t)

3A

3B

E jE kz t cos( )0

• B is in phase with E and has direction determined from: b s e • At t=0, z=0, Ey = -Eo

• Therefore at t=0, z=0, ( ) b s e k j i B i

E

ckz t cos( )0

Energy in Electromagnetic Energy in Electromagnetic WavesWaves

Electromagnetic waves contain energy. We know already expressions for the energy density stored in E and B fields:

u EE 12 0

2 uB

B 12

2

0u

E

cE uB E 1

212

2

20

02

u u u EE B 02

I c EE

c

02

2

0

• The Intensity of a wave is defined as the average power transmitted per unit area = average energy density times wave velocity:

• For ease in calculation, define Z0 as: Z c0 0 377 I

E 2

377

In an e-m wave, B = E/c

Therefore, the total energy density in an e-m wave = u, where

The Poynting The Poynting VectorVector

The direction of the propagation of the electromagnetic wave is given by:

SEB E

cEZ

E 0

2

0

2

0

2

377

S

E B 0

SEZ

EZ

kz tEZ

2

0

02

0

2 02

0

12

sin ( )

s e b

SE

I2

377

This wave carries energy. This energy transport is defined by the Poynting vector S as:

The direction of S is the direction of propagation of the waveThe magnitude of S is directly related to the energy being transported by the wave:

The intensity for harmonic waves is then given by:

physics 1304: lecture 17, pg 1 f()x x f(x x z y. physics 1304: lecture 17, pg 2 lecture outline l...

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