Chapter 34 Ampere’s Law is Wrong! Maxwell realized Ampere’s Law is not self-consistent This isn’t an experimental argument, but a theoretical one Consider.

Chapter 34

Ampere’s Law is Wrong!• Maxwell realized Ampere’s Law is not self-consistent

• This isn’t an experimental argument, but a theoretical one• Consider a parallel plate capacitor getting charged by a wire• Consider an Ampere surface between the plates• Consider an Ampere surface in front of plates• But they must give the same answer!

0 1 0d I B s

I I

0 2 0d I I B s• There must be something else that creates

B-fields

• Note that for the first surface, there is also an electric field accumulating in capacitor• Maybe electric fields?

• Take the time derivative of this formula• Speculate : This replaces I for first

surface

0 ?d I B s

0

QE

A

0 E Q 0

Ed dQ

dt dt

I

Ampere’s Law (New Recipe)• Is this self-consistent?• Consider two surfaces

with the same boundary

0 0Ed

d Idt

B s

10 1 0

Edd I

dt

B s 20 2 0

EdI

dt

1 2 0 2 0 1E E

dI I

dt

net 0 E

dI

dt

innet

dqI

dt

B

• Gauss’s Law for electric fields

0 inE q • This makes sense!

0 0 0Ed

d Idt

B s

I1

I2

E2

E1

CT – 1 For a charging capacitor, the total displacement current between the plates is equal to the total conduction current I in the wires. The capacitors in the diagram have circular plates of radius R. In (a), points A and B are each a distance d > R away from the line through the centers of the plates; in this case the magnetic field at A due to the conduction current is the same as that at B due to the displacement current. In (b), points P and Q are each a distance r < R away from the center line. Compared with the magnetic field at P, that at Q is

A. bigger. B. smaller. C. the same. D. need more information.

Maxwell’s EquationsWe now have four formulas that describe how to get electric and magnetic fields from charges and currents• Gauss’s Law• Gauss’s Law for Magnetism• Ampere’s Law (final version)• Faraday’s Law• Collectively, these are called

Maxwell’s Equations

in 0

0 in 0 0

ˆ

ˆ 0

S

S

E

B

dA q

dA

dd I

dtd

ddt

E n

B n

B s

E s

There is also a formula for forces on charges• Called Lorentz Force

All of electricity and magnetism is somewhere on this page

• Inside materials, these formulas are sometimes modified to include the effects of the reaction of the materials (like dielectric constants)

• These formulas commonly rewritten in derivative form

q F E v B

Wave solutions• We can solve Maxwell’s Equations(take my word for it) and come up with two “simple” differential equations.

0

0

, , , sin

, , , sin

x y z t kx t

x y z t kx t

E E

B B

• I could have used cosine instead, it makes no difference• I chose arbitrarily to make it move in the x-direction• These are waves where we have .

2 2

0 02 2

2 2

0 02 2

20 01

c

E E

x t

B B

x t

with

2

2

ff c

k

Show that

0

0

, , , sin

, , , sin

x y z t kx t

x y z t kx t

E E

B B

satisfies 2 2

0 02 2

2 2

0 02 2

20 01

c

E E

x t

B B

x t

with

On board

Wave Equations summarized:• Waves look like:• Related by:• Two independent solutions to these equations:

0

0

, , , sin

, , , sin

x y z t kx t

x y z t kx t

E E

B Bck

0 0

0 0

or

y z

z y

E cB

E cE

E0

B0

E0

B0

0 0E cB

• Note that E, B, and direction of travel are all mutually perpendicular

• The two solutions are called polarizations

• We describe polariza-tion by telling which way E-field points

• Note E B is in direction of motion

Understanding Directions for Waves• The wave can go in any direction you want• The electric field must be perpendicular to the wave direction• The magnetic field is perpendicular to both of them

• Recall: E B is in direction of motion

A wave has an electric field given by E = j E0 sin(kz – t). What does the magnetic field look like?A) B = i (E0/c) sin(kz - t) B) B = k (E0/c) sin(kz - t)C) B = - i (E0/c) sin(kz - t) D) B = - k (E0/c) sin(kz - t)

0 0E cB

• The magnitude of the wave is B0 = E0 / c• The wave is traveling in the z-direction, because of sin(kz - t).• The wave must be perpendicular to the E-field, so perpendicular to

j• The wave must be perpendicular to direction of motion, to k• It must be in either +i direction or –i direction• If in +i direction, then E B would be in direction j i = - k, wrong• So it had better be in the –i direction

CT – 2 A planar electromagnetic wave is propagating through space. Its electric field vector is given by E = Eo cos(kz – wt) i . Its magnetic field vector is A. B = Bo cos(kz – wt) j

B. B = Bo cos(ky – wt) k C. B = Bo cos(ky – wt) i D. B = Bo cos(kz – wt) k E. None of the above

The meaning of c:

• Waves traveling at constant speed• Keep track of where they vanish

0

0

, , , sin

, , , sin

x y z t kx t

x y z t kx t

E E

B Bck

• c is the velocity of these waves

0 0

1c

0kx t x tk

ct

83.00 10 m/sc

82.99792458 10 m/s

• This is the speed of light• Light is electromagnetic waves!• But there are also many other types of EM waves• The constant c is one of the most important fundamental constants of

the universe

Wavelength and wave number• The quantity k is called

the wave number• The wave repeats in time• It also repeats in space

0

0

sin

sin

kx t

kx t

E E

B B

ck 1

2

f T

f 2k

• EM waves most commonly described in terms of frequency or wavelengthc

k

2

2f

c f

• Some of these equations must be modified when inside a material

Ex- (Serway 34-19) In SI units, the electric field in an electromagnetic wave is described by Ey = 100 sin(1.00 x 107x -t). (a) Calculate the amplitude of the corresponding magnetic field. (b) Find the wavelength , (c) Find the frequency f. Also find an expression for the magnetic field.

Solve on Board

The Electromagnetic Spectrum• Different types of waves are classified

by their frequency (or wavelength)c f

Radio WavesMicrowavesInfraredVisibleUltravioletX-raysGamma Rays I

ncre

asin

gf I

ncre

asin

g

RedOrangeYellowGreenBlue

Violet

VermillionSaffron

ChartreuseTurquoise

Indigo

• Boundaries are arbitrary and overlap

• Visible is 380-740 nm

Which of the following waves has the highest speed in vacuum?A) InfraredB) OrangeC) GreenD) BlueE) It’s a tieF) Not enough info

Energy and the Poynting Vector • Let’s find the energy density in the wave

2102Eu E 2 21

0 02 sin kx t E 2 2 210 02 sinc B kx t

2 200

0 0

1sin

2B kx t

2

02Bu

B

220

0

sin2

Bkx t

2

20

0

sinB

u kx t

• Now let’s define the Poynting vector: 0

1

S E B

20 0

0

1sinS E B kx t

cu

0 0E cB

• It is energy density times the speed at which the wave is moving• It points in the direction energy is moving• It represents the flow of energy in a particular direction,

energy/area/time • Units: ucS

3

J m

m s

2

W

m

Intensity and the Poynting vector• The time-averaged Poynting vector is called the Intensity

• Power per unit area

S c u 2

20

0

sincB

kx t

20 02S cB

In Richard Williams’ lab, a laser can (briefly) produce 50 GW of power and be focused onto a region 1 m2

in area. How big are the electric and magnetic fields?

SA

P

10

26

5.0 10 W

10 m

22 25.0 10 W/m

020

2 SB

c

7 22 2

8

2 4 10 T m/A 5.0 10 W/m

3 10 m/s

8 24.2 10 T

0 20,000 TB 0 0E cB 120 6.1 10 V/mE

CT -3 At a fixed point, P, the electric and magnetic field vectors in an electromagnetic wave oscillate at angular frequency w. At what angular frequency does the Poynting vector oscillate at that point?

A. 2 w B. w C. w/2 D. 4 w

Momentum and Pressure• Light carries energy – can it carry momentum?

• Yes – but it’s hard to prove• p is the total momentum of a wave and U the total energy

• Suppose we have a wave, moving into a perfect absorber (black body)• As they are absorbed, they transfer momentum

• Intensity:

p U c

US A

t

• As waves hit the wall they transfer their momentum

cF

A

c pS

A t

P S ccP

• Pressure on a perfect absorber:• When a wave bounces off a mirror, the momentum is reversed

• The change in momentum is doubled• The pressure is doubled

2P S c

Cross-Section:• To calculate the power falling on an object, all that matters is

the light that hits it• Example, a rectangle parallel to the light feels no pressure

• Ask yourself: what area does the light see?• This is called the cross section

S P = F P=

If light of intensity S hits an absorbing sphere of radius a, what is the force on that sphere?A) a2S/c B) 2a2S/c C) 4a2S/c

P S c

• As viewed from any side, a sphere looks like a circle of radius a

• The cross section for a sphere, then, is a2

Sample Problem:A 150 W bulb is burning at 6% efficiency. What is the

force on a mirror square mirror 10 cm on a side 1 m away from the bulb perpendicular to the light hitting it?

• Light is distributed in all directions equally over the sphere of radius 1 m

2

0.06 150 W

4 mS

20.72 W/m

F P2 S

c

2

2

8

2 0.72 W/m0.1 m

3 10 m/s

1 m

114.8 10 NF

Ex- (Serway 34-41) A plane electromagnetic wave of intensity 6.00 W/m2 strikes a small pocket mirror, of area 40.0 cm2, held perpendicular to the approaching wave. (a) What momentum does the wave transfer to the mirror each second? (b) Find the force that the wave exerts on the mirror.

Solve on Board

Sources of EM waves• A charge at rest produces no EM waves

• There’s no magnetic field• A charge moving at uniform velocity produces no EM waves

• Obvious if you were moving with the charge• An accelerating charge produces electromagnetic waves

• Consider a current that changes suddenly• Current stops – magnetic field diminishes• Changing B-field produces E-field• Changing E-field produces B-field• You have a wave

+–

Simple Antennas• To produce long wavelength waves, easiest to use an antenna

• AC source plus two metal rods• Some charge accumulates on each rod• This creates an electric field• The charging involves a current

• This creates a magnetic field• It constantly reverses, creating a wave• Works best if each rod is ¼ of a

wavelength long• The power in any direction is

– – – – – –

++++++

2

2

sinS

r

distan

ce r

Common Sources of EM Radiation

Radio Waves AntennasMicrowaves Klystron, MagnetronInfrared Hot objectsVisible Outer electrons in atomsUltraviolet Inner electrons in atomsX-rays Accelerated electronsGamma Rays Nuclear reactions