(1) Chapter 5

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Chapter Five

Magnetostatics

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8/12/2019 (1) Chapter 5

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Biot and Savart law

dr

)(1

)(4

0

3

unitsGaussianc

korSIk

r

rdkIBd

==

=

r

lr

Orsted (1819) : Wires carry electric currents produced deflections of

permanent Magnetic dipoles placed in their neighborhood. Currents

were sources of magnetic flux density

Biot and savart law (1820): Laws relating the magnetic induction to

the currents (experimental laws)

I

d

p

outwards

(1)

ra

rdI

rrrdIBd 2030 4

),

sin(4

lllr

==

( )

a

I

a

IB

a

da

IB

24

2

4

00

2/322

0

==

+=

r

l

lr

Compare witha

E

21

0=r


8/12/2019 (1) Chapter 5

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B

I

a p

rld

( )[ ]

( ) 2/3220

2/322

0

4

)()(

4za

iadzI

za

jakzkdzIBd

+=

+

+=

r

( )

ia

IB

za

dzia

IB

2

4

0

2/322

0

=

+=

r

r

x

y

z


8/12/2019 (1) Chapter 5

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Amperes law (1820-1825)

( )BdIFd l = 11 11 ldI

Did not deal with the relation between currents and magnetic

induction, but with the force

Current elements In the presence of a magnetic

induction

12r#1

#2

Magnetic field at #1 is

=12

12220

4 r

rdIB r

lr

1 2

=

1 2 12

1221210

4 c c r

rddIIF r

llr

(2)

).().()().().()(sin

211212121221 lr

lrrr

lr

lrr

lr

lr

ddrrddrddBACCABCBAgU

==

Equation (2) becomes

= 1 2 1 2

312

21123

12

2121210 ).().(

4c c c c r

ddr

r

drdIIF

r

lr

lr

r

lrr

lr


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( ) danr

danAdr

rdr

rd

ssc

cc

.1

..1

1

.

).(

0

121

12

1213

12

121

1

11

=

==

=

=

434 21

r

rrrrl

321

r

r

r

r

lr

l

r

= 1 23

12

1221

21

0

12

).(

4c c r

rdd

IIF r

llr

(3)

Two long parallel straight wires

Force per unit length

21 II

z

dy

;11 BdIFd l = +== )()(

2 2220

IIi

aIB

r

( )( )

=

=

)(

)(

2

2

2

2210

1210

Ij

Ij

d

II

d

Fd

ikdd

IFd

l

r

lr

Attractive

repulsive


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Current density )(rJ

== rdrBrJFrdJId 33 )()(;l (4)

Total torque == rdBJrFrN 3)(rr

Differential equation of magnetostatics and Amperes law:

From equation (1)

rd

rr

rrrrErd

rr

rrrJrB

r

rdIrB

=

=

=

3

3

0

3

3

0

3

0

)()(

4

1)(

)()(

4)(

4)(

rr

rrrr

rr

rrrrrr

r

rlrr

BandB

---(5)

434 21

rrr

rrrr

rr

r

rr

rrr

rr

rrrrr

r=

+

=

=

0

30

)(1

)(1)(

1)(

4)(

rJrr

rJrrrr

rJ

rdrr

rJrB

acts on r

rrrJ rr

rr

r

= 1)(

0)(

.4

.

)(

4)(

30

30

=

=

=

rdrr

rJB

rdrr

rJrB

rr

rrrrrr

rr

rrr

(6)

00 == EB


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aaa rrr 2)( =

)(4 rr

)(

1

)(4 030

rJrdrrrJB

rr

rr

rrrrrr

+

=

rdrr

rJBrdrr

rJrB

=

=

30

30

)(

4

)(

4)( rr

rrrr

rr

rrr

rdrr

rJrd

rr

rJB

= 32030 )(

4

)(

4 rr

r

rr

rrrrr

rdrr

rJrdrr

rJB

= 32030

1)(

4

1)(

4 rr

rr

rr

rrrrrr

)()()(

4 0

330 rJrdrr

rJrd

rr

rJB

rr

rr

r

rr

rrrrr

+

=

!0 after integration

From the continuity equation 00 ==+ JJ

trrrr

(no charge)

)(0 rJB =0

= E

rr

Differential form of Ampere"s law

(7)


8/12/2019 (1) Chapter 5

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Integral form of Amperes law:

=ss

danJdanB 0

Using Stock"s theorem =sc

danJdB 0l

The total current I passing through the closedcontour c

IdB

c

0= l

n

s

da

ld

Amperes law

=

s

qdanE

0

r

(8)

Problem 5.6.A cylindrical conductor of radius a has a hole of

radius b bored parallel to, and centered a distance d, from the

cylinder axis (d+ b

8/12/2019 (1) Chapter 5

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Cylinder 1

Ampere"s law IdBc

0= l 10112 Ir =

1

101

2 r

IB

= 3

r

rdkIBd

r

lr =

The direction of B1 is in the direction of )( 11 rkrd l

or in the direction of

)(2

121

101 rk

rB rr =

At point p inside the hole

Cylinder 2

)(2

22

2

202 rk

r

IB

rr=

The current density is uniform throughout he remaining metal of the

cylinder

)( 222

2

22

1

1

ba

I

rr

I

==

2222

222

22

22221

122

21

1 ;ba

I

r

I

ba

rII

ba

I

r

I

ba

rII

=

=

=

=

principle of linear superposition 21 BBB +=

)(

)(

2 22210

ba

rrkIB

=r

jikdirr ;21 ==

jba

dB

)(2 22

0

=

r


8/12/2019 (1) Chapter 5

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u1 u2

u3

Unit vectors in curvilinear coordinates:

x

y

z

),,(),,(),,(

),( 3,21

zrrrzyxr

uuurr

rrr===

=

The tangent to u1 curve at point p isp

Lr ;

1

11

1 ur

uru

u

r

=

Spherical coordinates:

cossincossinsin

cossinsinsin

/

/

cossinsincossin

22

jir

jrri

r

r

rkrjrizkyjxir

+=+

+=

=

++=++=

r

r

( )

( )1121

101

11111

1

101

1

101

2

)sin(sin;cos

;cossin2

2

xjyirIB

rrryrx

jir

I

r

IB

+=

===

+==

r

r

21

202

2

rB =

r( )

222

2

20 2

xjyir

I+=


8/12/2019 (1) Chapter 5

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( ))()()(2

2121220 xxjyyi

abB +

=

r

dxxyy == 2121 ;0 d

jab

dB

)(2 220

=

rp

x2

x1

dxx = 21

+= 21 BBB

Scalar potential:

)(0 rJB = 0=

0=JIf In the region of interest 0= MB =

02 == MB 02 = Laplace equation for magnetostatics.

Same technique as in electrostatics.

Vector potential: General method

0= B Divergence of any curl is zero (also curl of any grade is zero)

ABA == 0)(

Vector potential


8/12/2019 (1) Chapter 5

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From equation (5)

+

=

=

r

rr

r

rr

rrrrd

rr

rJArd

rr

rJrB

3030 )(

4

)(

4)( (9)

For a given magnetic induction B , the vector potential can be freely

transformed according to

AA rrrr

+ (Gauge transformation)(Bunchanged)

JAA

JArJBrrrrr

02

00

)(

)(

=

==

Choose # to make 0= (Coulomb gauge)

JA 02 = )(

0

2

=

rdrr

rJA

= 30 )(

4 rr

r

))(

4

1)(( 3

0

rdrr

rr

=

rr

r

(10)With # constant


8/12/2019 (1) Chapter 5

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Example: Circular current loop

ld

. JcomponlyJ =

2

)()(cos)(

a

arNrJ

=

r

NIa

drdra

arNaI

rJdId

dxx

=

=

=

=

0

1)(

0

2

2

3

2sin)(cos)(

2

1

1

444 3444 21

l

2

)()(cos)(

a

arIrJ

=

r

JJ=

From the curvilinear coordinates:

cossincossin JjJiJji +=+=

From equation (9): rdrr

rJ

A

= 30 )(

4 rr

r

For purpose of calculation choose the observation point

on the x- z axes ($=0)

)cos(sinsincoscoscos

cos,sinsin,cossin

cos.

+=

===

++==

rzryrx

zzyyxxrrrr


8/12/2019 (1) Chapter 5

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++

+= ddrdr

rrrr

arji

a

IrA sin

cossinsincos(cos2

)()(coscossin

4)( 2

22

0rr

( )

+

+= d

raar

jia

a

IrA

2

0

22

20

cossin2

cossin

4)(

rr

The azimuthal integration is symmetric about "=0, the x-component of

J does not contribute. This leaves only the y-component, which is

+

= d

raara

IrA

2

022

0

cossin2

cos

4),(

The integral can be solved by elliptic integral

( ) +== cossinsincoscos0 rrrr

Expansion of rr 1

, point on the z-axis

=

=

=

=

0)(cos)(

11

l

ll

rr Pr

r

rrr

where r) is the smaller (larger) of rr & , and

is the angle between rr &

= >= 0

)(11l

lrr

rrr

),(),(12

14

1

0*1 m

mm

rYY

rrr l

l

l

lll

l

lrr

= = +

>+=

+

+

+

+

+

+

+ az

a

(b) r=a

( ) 2/3222

02/3

2

0

420

0

20

2)(1

1

2

2.4

1.5

2

31

2

)!(2

!)!12()1(

2

az

aIB

a

ra

I

a

r

a

r

a

I

a

r

n

n

a

IB

z

n

n

nn

r

+=

+

=

+

=

+=

=

L

(%=0)

)0( => az

)(1

rArr

B=

rrara == >< ;)(

=

+

+

++

+

=

0

22

12

112

0)12(

)(cos)!1(2

!)!12()1(

4

1

n

n

n

nn

n

r

anP

n

nIa

rB

= +

+

++

=0

12

12

112

0 )(cos)!1(2

!)!12()1(

4),(

nn

n

nnn

r

aP

n

nIarrA


8/12/2019 (1) Chapter 5

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n

nnn

n

r

a

rP

n

nIaB

2

03

112

20 1)(cos

)!1(2

!)!12()1(

4

+

+=

=+

rrar == >< ,

n

nnn

n

nn

n

nnn

nn

n

nnn

ra

rnP

nnIaB

r

anP

n

nIaB

r

anPn

nIar

B

2

03

112

20

032

12

112

0

022

12

112

0

)12()(cos)!1(2!)!12()1(

4

)12()(cos

)!1(2

!)!12()1(

4

)12()(cos)!1(2

!)!12()1(4

1

+

+=

+

+

=

++

=

=

+

= +

+

+

= +

+

+

= +

+

++

=

022

121

120 )(cos

)!1(2

!)!12()1(

4),(

nn

n

nn

n

a

rP

n

nIarA

= +

+

++

=0

22

221

120 )(cos

)!1(2

!)!12()1(

4),(

nn

n

nn

n

a

rP

n

nIarrA

(b) r=a

( ) =

++

++

+=

022

12112

0 )22()(cos)!1(2

!)!12()1(

4 nn

nnn

n

a

rnP

n

nIarA

r

)(1

rArr

B

=

= ++

+

+

=0

22

21

120 )22()(cos

)!1(2

!)!12()1(

4 nn

n

nnn

a

rnP

n

nIaB


8/12/2019 (1) Chapter 5

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( ) ( )

= ++

+

+

++= 0 22

21

120 )22(

)(cos)!1(2

12/!!12

)1(4 nn

n

nnn

a

rn

Pn

nnIa

B

( )

=+

++

++=

0

112

2

3

20 )(cos

1

)12(

)22(

)!1(2

!!12)1(

4 nn

n

nn P

a

r

an

n

n

nIaB

arrr == >< ,

8/12/2019 (1) Chapter 5

23/28

See Haliday and Resnick and Walker pp.942

ld

r

a

d

=

=

30

30

44 r

rdIB

r

rdIB

llr

2

0

4 r

dIdB

l

=

cossin dBkdBjBd +=

0 after integration (from symmetry)

r

a

r

dIk

r

dIkB

==

2

0

2

0

4

cos

4

llr

{

2/322

20

2

2/3220

)(2

)(4

az

aIkB

daz

aIkB

a

+=

+=

r

lr


8/12/2019 (1) Chapter 5

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For %=0 (point is on the z-axis)

z

r

=3

0

4 r

rdIB

lr

==+ zkrzkr y

x

x

y

sincos ajaiyjxi

+=+=

sincos ajaizkr =

)cossin( jiaddad +==l

)coscossincos(

)sincos)(cossin(

22

kziakzjad

ajaizkjiadrd

++=

+=l

{

( ) 2/3222

03

20

2

2

0

0

2

0

0

2

03

0

2

2

sincos4

az

aIkk

r

IaB

dkdzidzjarIaB

+==

+=

r

434 21434 21

r


8/12/2019 (1) Chapter 5

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Faraday"s law of induction (1831)

Time varying field: Time &varying magnetic fields give rise to electric fields

and vice versa- electromagnetic fields rather than electric or magnetic fields.

Faradays law of induction:

Experimental observations

A transient current is induced in a circuit if

(a) The study current flowing in an adjacent circuit is tuned on or off

(b) The adjacent circuit with a steady current flowing is moved relative

to the first circuit

(c) A permanent magnet is thrust into or out of the circuit.

Transient current flow as being due to a changing magnetic flux

linked by the circuit


8/12/2019 (1) Chapter 5

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S

n

da

The magnetic flux linking the circuit is

=s

danBF

The changing flux induced an electric field around

the circuit, the line integral of which is called

electromotive force. So, if is the electric field at

the element of the circuit C, then the electromotive

force around the circuit is

ld

C

=C

dE l

Faraday"s observation

dt

dFk= Faraday"s law

-ve sign (Lenz law) The induced current ( and magnetic flux)

is such a direction as to appose the change of flux through the circuit

=sc

danBdt

dkldE .

rrr(1)

=c

k 11 SI units

Gaussian units

v

v

C

),,,( tzyxB

danBvdant

B

danBzt

z

yt

y

xt

xdan

t

BdanB

dt

d

s s

s ss

)(

)(.

+

=

+

+

+

=

rrrr

rr

vBBvvBFrom321

0

=


8/12/2019 (1) Chapter 5

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( )[ ]

( )

+

=

s

dvB

ss

c

danvBdant

BdanBdt

d

44 344 21

rrrr

lrrr

.

(Stockes theorem)

From equation (1)

( )[ ]

= dan

t

BldBvE

c

.rrrr

(2)

This is an equivalent statement of Faraday"s law applied to the

moving circuit C

=sc

danBdt

dkldE .

rrr

If is the electric field in the laboratory (the circuit C and surface

S as instantaneously at a certain position in space in the

laboratory)

Applying Faraday"s law (equation 1) to that fixed circuit, we find

=

c

dant

BldE .rr

The electric field in the moving coordinate system is

BvEE +=

= c sdan

tBdE l

rr(3)

Different interpretation:


8/12/2019 (1) Chapter 5

28/28

Since the circuit C and the surface S are arbitrary, then the

integrand must vanish at all points

0=

+ tB

E

rr(4) For electrostatics 0=

The differential form of Faraday"s law

Differential form:

0)(0 =

==

+ dantB

danEdant

BdE

c

rrlrr

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