Electromagnetism, Prior

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Electromagnetism

Christopher R Prior

ASTeC Intense Beams Group

Rutherford Appleton Laboratory

Fellow and Tutor in Mathematics

Trinity College, Oxford

Thursday, 25 October 2012



2

Contents

• Review of Maxwell’s equations and Lorentz Force Law• Motion of a charged particle under constant

Electromagnetic fields• Relativistic transformations of fields

• Electromagnetic waves – Waves in vacuo – Waves in conducting medium

• Waves in a uniform conducting guide – Simple example TE 01 mode

– Propagation constant, cut-off frequency – Group velocity, phase velocity – Illustrations




3

Reading

• J.D. Jackson: Classical Electrodynamics (Wiley, 1998)

• H.D. Young, R.A. Freedman & L. Ford: University Physics(with Modern Physics) (Addison-Wesley,2007)

• P.C. Clemmow: Electromagnetic Theory (CUP, 1973)

• Feynmann Lectures on Physics (Basic Books, 2011)

• W.K.H. Panofsky & M.N. Phillips: Classical Electricity and

Magnetism (Addison-Wesley, 2005)• G.L. Pollack & D.R. Stump: Electromagnetism (Addison-

Wesley, 2001)




Gradient is normal to

surface = constant.

For a scalar function (x,y,z , t ),

gradient: =∂∂ x

, ∂∂ y

, ∂∂ z

For a vector F = F 1 , F 2 , F 3 :

divergence: · F = ∂ F 1∂ x

+ ∂ F 2∂ y

+ ∂ F 3∂ z

curl: F =∂ F 3∂ y

− ∂ F 2

∂ z, ∂ F 1

∂ z−

∂ F 3∂ x

, ∂ F 2

∂ x−

∂ F 1∂ y

4

Vector Calculus




· F G = G · F − F · G

φ = 0 , · F = 0

( F ) = ( · F ) − 2 F

Stokes’ Theorem

Basic Vector Calculus

5

Divergence or Gauss’Theorem

Closed surface S, volume V,outward pointing normal

S

F · d

S = C

F · d r

V

·

F d V = S

F · d

S

nd

S = n d S

Oriented

boundary C




What is Electromagnetism?

• The study of Maxwell’s equations , devised in 1863 torepresent the relationships between electric and magneticfields in the presence of electric charges and currents,whether steady or rapidly fluctuating, in a vacuum or inmatter.

• The equations represent one of the most elegant andconcise way to describe the fundamentals of electricity andmagnetism. They pull together in a consistent way earlierresults known from the work of Gauss, Faraday, Ampère,Biot, Savart and others.

• Remarkably, Maxwell’s equations are perfectly consistentwith the transformations of special relativity.

6




Maxwell’s EquationsRelate Electric and Magnetic fields generated bycharge and current distributions.

7

E = electric eld D = electric displacement

H = magnetic eld B = magnetic ux density

ρ = electric charge density j = current density

µ 0 = permeability of free space, 4 π 10− 7

0 = permittivity of free space, 8 .854 10 − 12

c = speed of light, 2 .99792458 10 8

In vacuum:

D = 0 E,

B = µ 0

H, 0 µ 0 c

2

= 1

· D = ρ

· B = 0

E = −∂ B∂ t

H = j + ∂ D∂ t




E = q 4 π 0

rr 3

= sphere

E · d S = q 4 π 0

sphere

d S r 2

= q

0

Equivalent to Gauss’ Flux Theorem:

The flux of electric field out of a closed region is proportional to the total

electric charge Q enclosed within the surface. A point charge q generates an electric field:

Maxwell’s 1 st Equation· E =

ρ0

· E = ρ

0 V

· E d V = S

E · d S = 1

0 V

ρ d V = Q

0

Area integral gives a measure of the net charge enclosed; divergence ofthe electric field gives the density of the sources. 8




Gauss’ law for magnetism:

The net magnetic ux out of any

closed surface is zero. Surround amagnetic dipole with a closed surface.The magnetic ux directed inwardtowards the south pole will equal theux outward from the north pole.

If there were a magnetic monopolesource, this would give a non-zerointegral.

Maxwell’s 2 nd Equation

Gauss’ law for magnetism is then astatement that

There are no magnetic monopoles 9

·

B = 0

B · d

S = 0

·

B = 0




Equivalent to Faraday’s Law of Induction:

(for a xed circuit C)

The electromotive force round a circuit

is proportional to the rate of change of ux of

magnetic eld through the circuit.

Maxwell’s 3 rd Equation

! "= l d E !!

#

!! "=# S d B!!

10

S

E · d S = − S

∂ B

∂ t · d S

C

E · d l = −d

d t S

B · d S = −d Φ

d t

E = −∂ B

∂ t

Faraday’s Law is the basis for electric generators. It also

forms the basis for inductors and transformers.

Michael Faraday




B = µ0 I

4 π d l rr 3

Maxwell’s 4 th Equation

Ampère

11

Biot

B = µ 0 j +

1

c2

∂ E ∂ t

Originates from Ampère’s (Circuital) Law :

Satised by the eld for a steady line current(Biot-Savart Law, 1820):

B = µ 0

j

C

B · d

l = S

B · d

S = µ0 S

j · d

S = µ0 I

B =µ 0 I 2 π r

For a straight line current




• pply mpere to sur ace 1 a atdisk): the line integral of B =µ0 I .

• Applied to surface 2, line integralis zero since no current penetratesthe deformed surface.

• In a capacitor,

E = Q

0 A and I =

dQdt

= 0 AdE dt

,

so there is a current density j d = 0

∂ E ∂ t

.

12

Displacement Current

Surface 1 Surface 2

Closed loop

Current I

• Faraday : vary B-field, generate E-field

• Maxwell : varying E-field should then produce a B-field, but not covered by Ampère’s Law.

B = µ0 ( j + j d ) = µ0 j + 0 µ 0

∂ E ∂ t




13

Consistency with Charge Conservation

Charge conservation is implicit in Maxwell’s Equations

j · d S = −

d

d t ρ d V

· j d V = − ∂ρ∂ t

d V

· j + ∂ρ∂ t = 0

B = µ0 j +1

c2

∂ E ∂ t

= · B = µ0 · j + 1c2 ∂

∂ t· E

= 0 = · j + 0 µ 0∂ ∂ t

ρ0

= 0 = · j +∂ρ

∂ t




In vacuum:

D = 0 E, B = µ 0 H, 0 µ 0 c2 = 1

Source-free equations:

· B = 0

E + ∂ B

∂ t = 0

Source equations:

· E = ρ0

B − 1c2

∂ B∂ t

= µ0 j

Equivalent integral form (useful forsimple geometries):

E · d S = 1

0 ρ dV

B · d S = 0

E · d l = − ddt B · d S = − dΦ

dt

B

· d

l = µ0

j d

S +

1c2

ddt

E

· d

S

Maxwell’s Equations in Vacuum

14




Also from E = − ∂ B∂ t

B = µ0 j +

1c2

∂ E ∂ t

then gives current density necessary tosustain the elds

Example: Calculate E from B

E · d

l = −d

d t

B · d S

!"#

>

<=

0

00

0

sin

r r

r r t B B

z

$

r

z

15

2 π rE θ = −d

d tπ r 2 B 0 sin ω t = − π r 2 B 0 ω cos ω t

= E θ = −1

2B 0 ω r cos ω t

r < r 0

2 π rE θ = −d

d tπ r 2

0 B 0 sin ω t = − π r 20 B 0 ω cos ω t

= E θ = −ω r 2

0 B 0

2 r cos ω t

r > r 0




E · d l = −d t B · d S

= 2 π rE θ = −d Φ

d t

The BetatronMagnetic

flux, !

Generated E -field

r

Particles accelerated by the rotational electriceld generated by a time-varying magnetic eld

16

−mv 2

r = evB = B = −

per

=∂

∂ t B (r, t ) = −

1er

d pdt = −

E r =

12π r 2

dΦ

dt

= B (r, t ) = 1

21

π r 2 B dS

B-field on orbit needs to be one half the average B over the circle. This imposes a limit on the energythat can be achieved. Nevertheless the constant radius principle is attractive for high energy circularaccelerators.

For circular motion at a constant radius:




Lorentz Force Law

• Thought of as a supplement to Maxwell’s equations but actuallyimplicit in relativistic formulation, gives force on a charged particlemoving in an electromagnetic field:

• For continuous distributions, use force density:

• Relativistic equation of motion

– 4-vector form:

– 3-vector component: Energy component:

F =dP dτ

= γ v · f

c , f = γ

1

cdE dt

, d pdt

d

d tm 0 γ v = f = q E + v B v · f =

d E d t

= m 0 c2 d γ d t

17

f = q

E +

v

B

f d = ρ E + j B




d t m 0 γ v = f = q E + v B = q v Bd

d tm 0 γ c2 = v · f = q v · v B = 0

Motion in Constant Magnetic Fields

• From energy equation, ! is constant

• From momentum equation,

18

No acceleration with a magnetic eld

|

v | constant and |

v | constant

= |

v | also constant

= |

v | is constant




Motion in Constant Magnetic Field

Constant magnetic eld givesuniform spiral about B with

constant energy.

19

dt (m 0 γ v) = q v B

=d vdt

= q m 0 γ

v B

=v 2

ρ = q

m 0 γ v B

= circular motion with radius ρ = m 0 γ v

qB

at an angular frequency ω = v

ρ =

qBm 0 γ

= qB

m

Magnetic Rigidity

B ρ =

m 0 γ vq

=

pq




d

d tm 0 γ v = f = q E + v B =

d

d tm 0 γ v = q E

20

Motion in Constant Electric Field

Constant E-eld gives uniform acceleration instraight line

Solution is γ v = q E m 0

t

Then γ 2 = 1 +γ vc

2

= γ = 1 +q Etm 0 c

2

If E = ( E, 0, 0), dx

dt =

(γ v)γ

= x = x 0 + m 0 c2

qE 1 +qEtm 0 c

2

− 1

≈ x 0 + 1

2

qE

m 0t 2 for qE m 0 c

Energy gain is m 0 c2 (γ − 1) = qE (x − x 0 )




• According to observer O in frame F, particle has velocity , fields are andand Lorentz force is

• In Frame F ’ , particle is at rest and force is

• Assume measurements give same charge and force, so

• Point charge q at rest in F:

• See a current in F’

, giving a field

• Suggests

E = q 4 π 0

rr 3 , B = 0

v

E

B

Relativistic Transformations of E and B

f = q

E

q = q and

E =

E +

v ×

B

B = −

µ 0 q 4 π

v × rr 3 = −

1

c2 v × E

B =

B −1

c 2

v ×

E

21

f = q

E +

v

B

R o u g h i d

e a




h d l



ω ∆ t − k ∆ x = 0

v p

=∆ x

∆ t

=ω

k

Superposition of plane waves. Whileshape is relatively undistorted, pulsetravels with the Group Velocity

Phase and Group Velocities

[ ]! "

"#

# dk ek A kxt k i )()( $

Plane wave has constantphase at peaks

sin(ω t − kx )ω t − kx = 1

2π

vg

=

d ω

dk

23




24

Wave Packet Structure

• Phase velocities of individual plane waves makingup the wave packet are different,

• The wave packet will then disperse with time




N f El i W



Nature of Electromagnetic Waves• A general plane wave with angular frequency ! travelling in the

direction of the wave vector has the form

• Phase number of waves and so is a Lorentzinvariant.

• Apply Maxwell’s equations:

• Waves are transverse to the direction of propagation; and aremutually perpendicular

k

E =

E 0e i ( ω t − k · x ) ,

B =

B 0 e i ( ω t − k · x )

↔ − i k∂

∂ t ↔ i ω

26

ω t − k · x = 2 π ×

E,

B

k

· E = 0 = · B ←→ k · E = 0 = k · B

E = −

∂ B

∂ t ←→

k E = ω

B


Plane Electromagnetic Wa e



Plane Electromagnetic Wave

27Thursday, 25 October 2012

Plane Electromagnetic Waves



= speed of electromagnetic waves in vacuum is ω

k = c

Plane Electromagnetic Waves

!

" #

! $

2Frequency

k

2Wavelength

=

= !

Reminder: The fact that is aninvariant tells us that

is a Lorentz 4-vector, the 4-Frequency vector.Deduce frequency transforms as

Λ =

ω

c ,

k

ω t − k · x

ω = γ (ω − v · k ) = ω c − v

c + v

B =

1

c 2

∂ E

∂ t

k B = −

ω

c 2

E

Combined with k E = ω B =| E || B |

= ω

k =

kc 2

ω

28


Waves in a Conducting Medium



Waves in a Conducting Medium

• (Ohm’s Law) For a medium of conductivity ! ,

• Modified Maxwell:

• Putconduction

currentdisplacement

current

Dissipationfactor

j = σ E

H = j +∂ E ∂ t

= σ E +∂ E ∂ t

− i k H = σ E + i ω E

D =

σ

ω

E =

E 0e i (ω t − k · x ) ,

B =

B 0e i ( ω t − k · x )

29

40

8-

120

7

1057.21.2,103:Teflon

10,108.5:Copper !

"=#="=

=#="=

D

D

$ $ %

$ $ %


Attenuation in a Good Conductor



− i k H = σ E + i ω E k H = i σ E − ω E = ( i σ − ω ) E

ombine with E = −∂ B∂ t

= k E = ω µ H

= k ( k E ) = ω µ k H = ω µ(iσ − ω ) E

= ( k · E ) k − k2

E = ω µ(iσ − ω ) E

= k2 = ω µ(− iσ + ω ) since k · E = 0

Attenuation in a Good Conductor

For a good conductor, D 1, σ

ω

, k2

≈ − iω

µσ

= k ≈ ω µσ

2 (1 − i) =

1δ

(1 − i) where δ = 2ω µσ

is the skin-depth

Wave-form is: e i (ω t − kx ) = e i ( ω t − (1 − i )x ) = e− x e i ( ω t − x )

30


Charge Density in a Conducting Material



• Inside a conductor (Ohm’s law)

• Continuity equation is

• Solution is

• Charge density decays exponentially with time. For a verygood conductor, charge flows instantly to the surface toform a surface current density and (for time varying fields)a surface current. Inside a perfect conductor:

Charge Density in a Conducting Material j = σ E

ρ = ρ 0 e− σ t/

32

∂ρ∂ t

+ · j = 0 ∂ρ

∂ t + σ · E = 0

∂ρ∂ t

+ σ ρ = 0 .

(σ → ∞ )

E = H = 0


A Uniform Perfectly Conducting Guide



E = −∂ B∂ t

= − iω µ H

H = ∂ D

∂ t = iω E

=

2 E = ( · E ) − E

= iω µ H

= − ω 2 µ E

2 + ω 2 µ E

H = 0

A Uniform Perfectly Conducting Guide

Hollow metallic cylinder with perfectly conducting boundarysurfaces

Maxwell’s equations with time dependence e i! t are:

Assume )(

)(

),(),,,(

),(),,,( z t i

z t i

e y x H t z y x H

e y x E t z y x E ! "

! "

#

#

=

=

!!

!!

! is the propagation constant

Can solve for the fields completelyin terms of E

z and H

z

zx

y

33

Then 2t + ω2 µ + γ 2

E H

= 0

Helmholtz Equation


A simple model: “Parallel Plate Waveguide”



To satisfy oun ary con itions: E = 0 on x = 0 an x = a.

= E = A sin Kx, with K = K n ≡n π

a , n integer

E = (0 , 1, 0)E (x)e i ω t − γ z where E satises

2t E =

d2 E dx 2 = − K 2 E, K 2 = ω2 µ + γ 2 .

with solution E = A cos Kx or A sin Kx

Propagation constant is

γ = K 2n − ω

2 µ = n π

a 1 − ω

ω c

2

, ω c = K n√ µ

34

A simple model: Parallel Plate Waveguide

Transport between two infinite conducting plates (TE 01 mode):

z

x

y

x = 0 x = a


Cut-off Frequency "



γ =

n π

a 1

− ω

ω c

2

, E = sin

n π x

a ei ω t − γ z

, ω c =

n π

a√ µ

35

Cut off Frequency, c

! " <" c gives real solution for # , so attenuationonly. No wave propagates: cut-off modes.

! " >" c gives purely imaginary solution for # ,and a wave propagates without attenuation.

! For a given frequency " only a finite number of

modes can propagate .

For given frequency, convenient tochoose a s.t. only n=1 modeoccurs.

γ = ik, k = √ µ ω2

− ω2c

12 = ω √ µ 1

−ω

2c

ω2

12

ω > ω c =

π

a √ µ= n <

ω

π√ µ





k = √ µ ω 2

−ω 2

c

1

2 < ω √ µ

λ =2 π

k >

2 π

ω √ µ,

k 2 = µ ω2

−ω

2

c

= vg =

ω

d k =

kω µ

< 1

√ µ

37


v p =

ω

k >

1

√ µ

# free-space wavelength

# larger than free-spacevelocity

# smaller than free-space velocity

• Wave number

• Wavelength

• Phase velocity

• Group velocity


Calculation of Wave Properties



Calculation of Wave Properties• If cm, cut-off frequency of lowest order mode is

• At 7 GHz, only the n=1 mode propagates and

a = 3

f c = ω c

2 π=

1

2 a√ µ ≈3 × 10 8

2 × 0 .03 ≈ 5 GHz ω c = n π

a√ µ

c38

k = √ µ ω 2

− ω 2c

12

≈ 2π (72

− 52 ) 12 ×109 / 3 × 108 = 103 m − 1

λ = 2π

k ≈ 6 cm

v p = ω

k = 4.3 × 108 ms− 1 > c

vg = kω µ

= 2.1 × 108 ms− 1 < c


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