Chapter 5 Magnetostatics

5.1 The Lorentz Force Law

5.2 The Biot-Savart Law

5.3 The Divergence and Curl of

5.4 Magnetic Vector Potential

B

5.1.1 Magnetic Fields

Charges induce electric field

FF E

Q

3q

1q2q

Source charges Test charge

Q

)( Bv

5.1.1

Ex.1 Cyclotron motion

relativistic cyclotron frequency

EM wave

relativistic electron cyclotron maser

microwave

light laser

5.1.2 Magnetic Force

Lorentz Force Law

moment

cyclotron frequency

( )magF Q v B

[ ( )]F Q E v B

magF QvB

2vcentripetal c RF mvw m

p mv QBR QB

c mw

~ eBw wce me

1.1

cQB

wm

~1960 EM : maser [ 1959 J.Schneider ; A.V. Gaponov]

ES : space [1958 R.Q. Twiss ]

(1976) K.R. Chu & J.L. Hirshfield : physics in gyrotron/plasma1978 C.S. Wu & L.C. Lee : EM in space ( )

1986 K.R. Chen & K.R. Chu : ES in gyrotron

relativistic ion cyclotron instability

1993 K.R. Chen ES in Lab. plasma [fusion ( EM ? Lab. & space plasmas ?

)]

5.1.2 (2)

~ 1.01

~ ici

i

N eBw w

m

~ 1.00093

Ex.2 Cycloid Motion

assume

5.1.2 (3)

E

zvˆB zF qv By

( 0) 0 ( ) zv t v t t v qE t

( )dv

m q E v Bdt

y zdy dzv vdt dt

ˆˆE Ez B Bx

2 2

2 2( )

d y dydz d zm q B m q E Bdt dtdt dt

( )c cEy w z z w yB

y yd ydy V V V const

( ) ( )z y zz V V t V t

1 2

2 1

sin cos

sin cos

y c z y c c

z c y z c c

V w V V C w t C w t

V w V V C w t C w t

5.1.2 (4)

)(t )(t )(t

( )Ey c z z c y ydBV w V V w V V

Eyd BV

2

1

( 0) 0

( 0) 0 0

Ey y y yd B

z z z

v t v V V C

v t v V C

(1 cos )

sin

Ey cB

Ez cB

v w t

v w t

3

4

( sin )

cos

EcwB

EcwB

y wt w t C

z w t C

5.1.2 (5)

3

4

( 0) 0 0

( 0) 0 EwB

y t C

z t C

( sin ) ( sin )

(1 cos ) (1 cos )

c

c

Ec c c cw B

Ec cw B

y w t w t R w t w t

z w t R w t

2 2 2cy Rw t z R R

2 2 2y z R

sin

cosc c

c

y y Rw t y R w t

z z R z R w t

.E B drift

the cause of Hall effect

Magnetic forces do not work

5.1.2 (6)

for

( )

0

mag mag

Q dl vdt

dW F dl

Q v B vdt

moving

5.1.3 Currents

The current in a wire is the charge per unit time passinga given point.

Amperes 1A = 1 C/S

The magnetic force on a segment of current-carrying wire

( )

v tI v A

t

I v

magF dl v B I B dl

I dl B

magF I dl B

I Idl

surface current density

the current per unit length-perpendicular-to-flow

The magnetic force on a surface current is(mobile)

5.1.3 (2)

dIKdl

:K v surface charge density

magF da v B K B da

da Q

volume current density

The current per unit area-perpendicular-to-flow

The magnetic force on a volume current is

5.1.3 (3)

dIJda

:J v volume charge density

magF d v B

( )d Q

magF J B d

Ex. 3

Sol.

Ex. 4(a) what is J ? (uniform I)

Sol.

5.1.3 (4)

If ( ) ( ) 0

what is I ?

mag gF F

magF IBa mg

mgI

Ba

2IJR

I B

I

I B

I B

B

(b) For J = kr, find the total current in the wire.

Sol.

5.1.3 (5)

dr rd

32

0

22

3

R

I Jda kr rdrd

kRk r dr

~ ~ ~q dl da d

1

~ ~ ~n

i ii line surface volume

q v Idl Kda Jd

J relation?

Continuity equation

(charge conservation)

5.1.3 (6)

s sI Jda J da

s V

ddt tV V

J da J d

d d

Jt

5.2.1 Steady Currents

Stationary charges constant electric field: electrostaticsSteady currents constant magnetic field: magnetostatics

I No time dependence

0 0Jt

024

v RB q

R

5.2.2 The Magnetic Field of a Steady Current

Biot-Savart Law: for a steady line current

Permeability of free space

Biot-Savart Law for surface currents

Biot-Savart Law for volume currents

for a moving point charge not steady current

02

02

( )4

4

I RB p dl

R

dl RI

R

( )

1 1

tesla T

Ntesla

A m

70 2

4 10N

A

4

:

1 10

cgs gauss

T gauss

02

02

4

4

K RB da

RJ R

B dR

5.2.2 (2)

Solution:

In the case of an infinite wire,

( ) ?B P

dl

z

R

ˆ sin cosdl R dl dl

2

2 2 21 cos

tan tan , cos ,cos

z zl z dl z d d

R R z

2

1

2

1

20 0

2 2 2

0 02 1

cos( )cos

4 4 cos

cos (sin sin )4 4

dl R zB I I d

R zI I

dz z

1 2, ;2 2

0

2

IB

z

5.2.2 (3)

Force?

The field at (2) due to is 1I

The force at (2) due to is 1I

The force per unit length is

0 1 ( )2

IB

d

0 12( )

2

IF I dl

d

0 1 2

2

I If

d

02

02

ˆcos

ˆcos4

cosˆ( )2

4

B dB z

dlI z

IRz

12 2 2cos ( )

RR z

5.2.2 (4)

2

?B

32

20

2 2ˆ

2 ( )

I RB z

R z

0 0

0

2 2

I IB dl dl dl

R RI

5.3.1 an example: Straight-Line Currents

I

R

zr

BB

0

2

IB

R

0 ˆ ˆˆ ˆ2

IB dl drr rd dzz

R

20 000

1

2 2

I IB dl r d d I

r

2 1

1 20d d

0 ;enc encB dl I I J da

( )B da

0 J da

0B J

5.3.2 The Divergence and Curl ofBiot-savart law

00

B

( , , )

( , , )

B x y z

J x y z

ˆ ˆ( ) ( )

ˆ ( )

ˆ ˆ ˆ

ˆ ˆ ˆ

R x x i y y j

z z k

d dx dy dz

i x j y z z

i x j y z z

( ) ( ) ( ) ( ) ( )A C C A A C A C C A

x x

2

ˆ( )

RJ

R

02

ˆ

4

J RB d

R

��������������

02

2 2 2

ˆ( ) ,

4ˆ ˆ ˆ

( ) ( ) ( )

RB J d d dx dy dz

R

R R RJ J J

R R R

0 00B

( , , )J x y z

02

ˆ( )

4

RB J d

R

3

2 2 2 2 2

ˆ4 ( )

ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( ) ( )

R

R R R R RJ J J J J

R R R R R

0 for steady current

To where

Ampere’s law in differential form

5.3.2

2 3 3 3

ˆˆˆ ˆR x x y y z z

i j kR R R R

3 3 3

3 3

( ) ( ) ( )

( )( ) [ ] ( )( )

[ ] 0volume surface

fA f A A f

x x x x x xJ J J

R R Rx x x x

J d J daR R

0J

300( )4 ( ) ( )

4B J r r r d J r

0 ( )B J r

5.3.3 Applications of Ampere’s LawAmpere’s Law in differential form

Ampere’s Law in integral form

Electrostatics: Coulomb GaussMagnetostatics: Bio-Savart AmpereThe standard current configurations which can be handled by Ampere's law:

1. Infinite straight lines2. Infinite planes3. Infinite solenoids4. Toroid

0B J

0 0( ) encB da B dl J da I

0 encB dl I

5.3.3 (2)

B ?

Ex.7

symmetry

Ex.8

Bl2

r

0 0

0

2

2

encB dl I I

B dl B r

IB

r

0 0

ˆ( )[ ]

enc

B z j

B dl I Kl

0

0

ˆ 02

ˆ 02

Kj for zB

Kj for z

Ex.9

loop 1.

loop 2.

5.3.3 (3)

r

0(2 ) 0

0

encB dl B r I

B

Bz

01[ ( ) ( )] 0

( ) ( )

encB dl B a B b L I

B a B b

0 02

0 ˆ

0

encB dl BL I NIL

NIz insideB

outside

Ex.10

Solution:

5.3.3 (4)

0

0ˆ

( ) 20

enc

nIinside

B dr I B r routside

p

5.3.3 (5)

( ) ?B p 0

34

I RdB dl

R

0 0( ,0, )

' ( cos , sin , )

p x z

r r r z

0 0' ( cos , sin , )

ˆ ˆ ( cos , sin , )

( 0)r z r r z

R p r x r r z z

I I r I z I I I

I

ˆ ˆˆ( ) ( )I R y B in

0 0

0 0

0 0 0

ˆ

ˆ ˆ ˆ[ ] [ ] [ ]

ˆ ˆ[ sin ( ) ( sin )] [ ( cos )

ˆcos ( )] [ cos ( sin ) sin ( cos )]

ˆ ˆ{sin [ ( ) ]} { cos [ (

ijk i j

y z z y z x x z x y y x

r z z

r r r

r z z z r

I R I R k

x I R I R y I R I R z I R I R

x I z z I r y I x r

I z z z I r I x r

x I z z I r y I x I r I z z

0

)]}

ˆ[ sin ]rz I x ˆ ˆand components cancel out sin from ' and "x z r r

5.4.1 The Vector Potential

E.S. : constVV ]0[ const

M.S. : ? AA

a constant-like vector

]0[

function

A

is a vector potential in magnetostatics

If there is that , can we find a function to obtain with

A

0 A

0 A

?AA

Gauge transformation

0E

0B

B A

E V

20( ) ( )B A A A J

5.4.1 (2)

Ampere’s Law

if

20 A A

2 ( )A

1

4

Ad

R

20A J

0

4

JA d

R

( ) 0J

0 0 0ˆ

;4 4

IJ KA d d A da

R R R R

2

0V

1

4V d

R

5.4.1 (3)

Example 11

Solution:

For surface integration over easier 1

A spiring spheres

( ) ?A P

0( )4

KA P da

12 2 22, ( 2 cos ) , sin

ˆˆ ˆ

sin 0 cos

sin cos sin sin cos

K v R S RS da R d d

v r

i j k

R R R

ˆ( , )v v r

5.4.1 (4)

ˆ ˆ[ (cos sin sin ) (cos sin cos sin cos )

ˆ(sin sin sin ) ]

R i j

k

2 2

0 0sin cos 0d d

30

2 2 1/ 20

sin cos sin ˆ( )2 ( 2 cos )

RA P d j

R S RS

( cos )u 2 2

2 21 2 2 2 2

2 2 1/ 2 2 21 2

1[ ( )]

( 2 ) 2

R S RS

R S RS

udu y R S dy

R S RSu R S

2 2 2 2 21( 2 ; ( ); )

2

ydyy R S RSu u R S y du

RS RS

5.4.1 (5)

2 2

2 22 2 2 2

2 2 2

1[ ( )]

2

R S RS

R S RSy R S dy

R S

22 2

2 2[ ( )]

32

R S

R S

y yR S

R S

2 2 2 22 2

2 2 2 22 2

[ 2 3( )]6

[ 2 3( )]6

R SR S RS R S

R SR S

R S RS R SR S

2 2 2 22 21

[ ( ) ( )( )]3

R S R S RS R S R S RSR S

5.4.1 (6)

2 2 2 22 21

[( )( ) ( )( )]3

R S R S RS R S R S RSR S

2 22 21

[2 2 ( )]3

RRS S R SR S

22

3

S

R

2 22 21

[2 2 ( )]3

SRS R R SR S

22

3

R

S

if R > S

if R < S

5.4.1 (7)

for

inside,

outside,

for

inside,

outside,

in

30 sinˆ( ) ( )

2

RA P j

22

3

S

R

22

3

R

S

R S

R S

S

ˆsinS j 0 ( )3

RS

40

3( )

3

RS

S

R S

R Sr

z ˆS rr

0

40

2

ˆsin ( )3

( , , )sin

ˆ ( )3

Rr r R

A rR

r Rr

( )A P

1 1 ˆˆ(sin ) ( )sin

B A A r rAr r r

5.4.1 (8)

B

Note:

ˆ( )A A

00 0

2 2 2ˆˆ ˆ(cos sin )3 3 3

Rr R z R

is uniform inside the spherical shell

r R

5.4.1 (9)

adBadAdA

=

0

02

ˆ2

ˆ2

NIr

ANI R

r

?A

12Example N turns per unit length

:sol

00 0

inenc

out

B NIB d I

B

20

20

( )

( )

NI r r RA d B da

NI R r R

2A rˆ( )

r R

r R

5.4.2 Summary and Magnetostaic Boundary Conditions

V E

dτΡ

ΡΒ

πΑ

24

1

0A

A B

00 2

ˆ0

4

J Rrecall B B J B d

2

ˆ1

4

B RA d

0

ˆ( )a bE E n

( )a bE E

0 0

( )above below

enc

B d B B

I K

5.4.2 (2)

0above belowB B K

B.C. for B

0B da

above belowB B

0 ˆ( )above belowB B K n

5.4.2 (3)

B.C. for A

0 above belowA A A

( )a bV V

A B

0 ˆ( ) ( ) ( )above belowA A K n

0ˆ ˆ ˆ( ) ( ) ( )above belown A n A K nn n

0

a bV V

n n

0above belowA A

Kn n

5.4.3 Multipole Expansion of the Vector Potential

linecurrent

=0

0

4

JA d

R

0 1

4

IdR

2 2 1/2 1/21 1 1 1

(1 cos )( 2 cos ) [1 2( )cos ]

r

R r rr r rr r r r

r r

0 02

1 1 1(1 cos ) [ cos ]

4 4

I r IA dl dl r dl

r r r r

monopole dipole

5.4.3(2)

momentdipolemagnetic

0 02 2

ˆ( ) cos ( )4 4

dipI I

A r r dl r r dlr r

ˆ ˆ ˆ( ) ( ) ( )r r dr r r dr dr r r

( )dl dr

ˆ ˆ ˆ( ( ) ( ) ( )d r r r r r dr dr r r ˆ2 ( )r r dr

02

1ˆ( ) ( )

24dip

IA r r r dr

r

02

ˆ( )

4dipm r

A rr

( )

2

Im r dl

m Ia1

( )2

a r dl

5.4.3(3)

?m

2 2 ˆˆm IS j IS k solution:

Ex. 13

5.4.3(4)

Field of a “pure” magnetic dipole Field of a “physical” magnetic dipole

0

0

2

2

ˆ( )

4

sin ˆ =4

dipm r

A rrm

r

1 1 ˆˆ(sin ) ( )sin

B A A r rAr r r

03

ˆˆ(2cos sin )4

mr

r