BmF

Force on an ideal magnetic dipole with dipole moment m in

a B-field B is given by

A simple proof of this is to consider a square current loop

with side-length , as shown below

BmF

0

By considering the forces on the

four sides of the square, we can

prove that, when

(Proof in assignment)

N B

BmN

The torque on in a uniform magnetic field is given by

Consider a rectangular loop with sides

a, b and making an angle with B.

Without loss of generality, let B point

along the z-direction and the loop is

tilted from the z-axis towards the y-axis

by an angle .

Proof:

x

z

y

a b

I

m

x

z

y

m

a b

I

x

x

xN

ˆ sin

ˆ sin

ˆ sin

mB

IabB

aF

IbBF

In vector form: BmN

Magnetic Field in Matters

Paramagnetism

? ?

? ?

? ?

? ?

?

?

Inside matters, there are a lot of tiny currents due to the

electrons orbiting around the nuclei and intrinsic spins.

BmN The torque, , however, is non-zero & tends to align

m in the same direction of B

This effect is called paramagnetism.

What is Paramagnetism?

The scale of these small “current loops” are so small

that the applied B-field can be considered uniform.

Bm

0 BmF

is constant and so In a uniform B-field,

Diamagnetism

Magnetic Field in Matters

What is Diamagnetism?

This effect is called diamagnetism.

The induced net magnetic dipole moment is opposite to the

field.

m B

v

RT

2

where R is the radius of the orbit.

For electrons orbiting around nuclei, in addition to the

paramagnetic effect, there is another less significant effect

due to an external B-field.

A full, rigorous treatment of this effect requires

quantum mechanics. Here is a qualitative analysis:

Let the speed of an orbiting electron be v, then the period

Effect of Magnetic field on Atomic Orbits

R

ev

T

eI

2

2 ˆ

1ˆ

2

I

I R

evR

m a

z

z

This yields a steady current

The orbital dipole moment is therefore

R

vm

R

ee

2

2

2

04

1

Then, in addition to the electric force

between the electron and the nucleus,

there is a magnetic force.

Suppose now a B-field zB ˆ B is applied.

In the absence of the B-field, the centripetal

force is contributed by the electric force

alone, so

B

v

R

vmBve

R

ee

2

2

2

04

1

vvv

0)(2

2

22

22

vR

mv

R

vmveB

R

vmveBvv

R

m

R

vmveB

R

vm

ee

e

e

ee

With the magnetic field, the velocity of the

electron is changed from v to , so that

Let

B

zm

zm

em

Re

Rve

evR

4

ˆ 2

1

ˆ 2

1

22

Therefore, the electron speeds up or slows down,

depending on the direction of the applied field.

v

v

em

eRBv

2

Assume that is so small that the second-order term

can be ignored, then

0B B, i.e. is along the When

z-direction, it speeds up.

0B B, i.e. is along the When

-ve z-direction, it slows down.

22 ( ) 0e em v meBv v v

R R

Ferromagnetism

Magnetic Field in Matters

What is Ferromagnetism?

• “Forzen in” magnetic dipoles

• Not due to external fields, but sustained by

interaction between nearby dipoles

• Quantum mechanics Dipoles “like” to

point in the same direction as their

neighbors.

• Emphatically non-linear

Magnetization

Magnetic Field in Matters

MThe magnetization of an object, , is defined as the

amount of magnetic dipole moment per unit volume

M Magnetic dipole moment per unit volume.

What is Magnetization?

The Field of a

Magnetized Object

Magnetic Field in Matters

0

2( )

4

ˆ

mA r

rr

0

2

( )( )

4

ˆd

M rA r

rr

The vector potential A of a single ideal

magnetic dipole moment m is

So, for a magnetized object with

magnetization M

Similar to the case in electric polarization, the potential can

be re-written in another form which gives better physical

interpretation.

0

0

1 ( ) ( ) ( )

4

1 1 1 ( ) ( ) ( ) ( )

1 1 ( ) ( ) ( )

4

d

d

A r M r

M r M r M r

A r M r M r

r

r r r

r r

0

2

( )( )

4

ˆd

M rA r

rr

Recall:

1 1( ) ( )

s

d d

M r M r ar r

0 0ˆ( ) ( )

( ) 4 4

d da

M r M r nA r

r r

Consider the second term in the integral. We shall proof

later that

where S is the surface of the object.

n̂ adwhere is the normal unit vector of the area element .

0 0

s

( ) ( )( ) d

4 4v

da

b bJ r K rA r

r r

MJb

nMKbˆ

bJ bK

The potential is equivalent to that due to a volume current

density and surface current density , i.e.,

where

is the volume bound current density

and

is the surface bound current density.

0 0ˆ( ) ( )

( ) 4 4

d da

M r M r nA r

r r

v

( ) 1 ( )

s

d d

M r

M r ar r

sv

d ''' d acvcv

sv

dd vaccvvc'''' ][

BAABBA

BACCBA

To prove:

Consider a vector field v and a constant vector c . By divergence theorem,

where we have used the vector identities:

0' c

sv

dd ''' avcvc

v s

dd '''avv

Since c is a constant vector,

Because c is an arbitrary constant vector, therefore,

'1( )v M r

rThe proof is completed by setting

sv

dd vaccvvc'''' ][

Example:

Find the magnetic field of a uniformly magnetized sphere.

'

ˆ ˆ

b

b

J M 0

K M n M r

Solution:

Hence, '0

'0

ˆ( )

4

ˆ

4

da

da

M rA r

rM

r

r

3

0

3

30

3

0

2

4( )

4 3

4

4 3

ˆ

4

R

r

Rr

r

A r M r

M r

m r

3

3

4R

Mm

For r >R ,

where

is the total dipole moment of the sphere.

It can be shown that

(supplementary notes)

3

3

4if

ˆ 3

4if

3

S

r R

daR

r Rr

rr

r rr

3

3

4R

Mm

mrrmB ˆˆ31

4 3

0

r

So, outside the sphere, the vector potential, and hence the B-

field, is exactly the same as if all the dipole moments were at

the center, giving rise to an ideal dipole

Accordingly, the B-field outside the sphere is

at the origin.

It can be shown that

(supplementary notes)

3

3

4if

ˆ 3

4if

3

S

r R

daR

r Rr

rr

r rr

0 04( )

4 3 3

A r M r M r

0 3

B A

M r

MrrMrMMrrM

3

M r M r M r

M M r

ˆ ˆ ˆ

2

x y zM M M x y zx y z

M r x y z M

M r M

For r <R,

Since M is a constant vector,

But

And

0

2

3 B M

Physical Interpretation

of Bound Currents

Magnetic Field in Matters

Consider a small area element on the surface of a magnetized

object with length b and width t.

sint

Consider a very small volume element

as shown above, with length a,

width b and thickness .

The region is so small that M can

be taken as constant.

There are tiny current loops inside the volume due to

magnetization.

Iabm

sinMabtIab

sinMtI

Inside the volume, the current at a point due to adjacent loops

cancel each other.

where I is the surface current.

The net result is a current flowing on the surface.

The magnetic dipole moment is therefore

But the net dipole moment must be the same as that of the

total contribution of all the dipoles inside the volume.

sinMt

IK

nMK ˆ

By the definition of surface current density,

)sin( MtI

Physical Interpretation of volume bound current:

When M is not uniform, the current carried by adjacent

loops cannot cancel each other.

For example, consider the x-component of the currents.

Since Mx is responsible for currents flowing on the plane

parallel to the y-z plane, we need not consider Mx

Since Mx is responsible for currents flowing on the plane parallel

to the y-z plane, we need not consider Mx

zM

zM

0

y

M z

dxdydzyMdxdyyI z )()(

dzyMyI z )()(

If

If Mz is non-uniform in the y-direction, i.e.

the current on the surface of the two adjacent volume elements

are of different magnitudes and cannot cancel each other

Similarly, dzdyyMdyyI z )()(

.

dzyMdyyMyIdyyII zz )]()([)()(

( ) ( )( ) z z z

x

M y dy M y MI

dydz dy y

bJ

By definition of volume current density,

The net current flowing in the x-direction is

The variation of in the z-direction can also give rise to a

current in the x-direction. yM

dxdydzzMdxdzzI y )()(

dyzMzI y )()(

dydzzMdzzI y )()( Similarly,

( ) ( ) [ ( ) ( )]y yI I z dz I z M z dz M z dy

( ) ( )( )

y y y

x

M z dz M z MI

dydz dz z

bJ

By definition of volume current density,

.

The net current flowing in the x-direction is

( )yz

x

MM

y z

bJ

( ) x zy

M M

z x

bJ

( )yx

z

MM

y z

bJ

Similarly, one can prove that

Taking into account the two contributions,

MJb

The Auxiliary Field H

Magnetic Field in Matters

bf JJJ

JB 0

bf JJ 00

MJ f 00

fJMB

0

1

Inside matters,

By Ampere’s law,

J

fJ

bJ

: Total current density

: Free current density

: Bound current density (due to magnetization)

MBH 0

1

fJH

encfd H l I

Define the H-field,

(Ampere’s law of H)

In integral form,

encfIwhere is the free current enclosed.

A long copper rod of radius

R carries a uniformly

distributed (free) current I .

Find H inside and outside

the rod.

Example:

R

I

The argument is similar to that of a solenoid.

Answer:

First, the H field has no radial component.

And the field depends only on s, but not on z and .

)()( bHaH zz 0, ba

0 zH

In addition, if one considers the

amperian loop 1, it can be shown

that Hz is constant.

However, we know that H = 0 at infinity,

φ̂Therefore, the H field only has the -component.

This is true no matter where the

loop is located, because unlike

the case of a solenoid, the

current is now flowing in the

z-direction and loop 1 always encloses no current.

( ) 2H s s I

ˆ2

I

s H φ

2 2

2 2( ) 2

s sH s s I I

R R

2ˆ

2

Is

R H φ

Now consider the circular amperian

loop 2 coaxial with the axis of the rod.

By Ampere’s law,

For s > R,

For s < R,

0M

HB 0

0 ˆ2

I

s

φ

Since M is unknown, the B-field inside the rod cannot be

determined. However, outside the rod,

Rs ,for

MBH

0

1

M

JB 0 0 B

0

1

H B

fJ J

fJH JB 0Note, that although looks similarly to

0 BHowever, while,

Both and

have been employed to determine the B-field.

It is not valid to perform the simple substitution,

,which may not vanish.

Linear Media

Magnetic Field in Matters

BM

MBH 0

1

M

1

1

0k

HMk

k

0

0

1

For some materials, the magnetization is directly proportional

to the applied field.

BM kLet ,

i.e., M is also proportional to H.

m

HM m

HMHB )1()( 00 m

H

By convention, the magnetic susceptibility is defined by,

)1(0 m where is called the permeability of the material

Recall that 0 is called the permeability of free space

m

1

0

is called the relative permeability Besides,

0

1

H B

fJ J

BH

1

01

BH

Note: Recall that we cannot simply substitute

MH Since may be non-zero,

In linear media,

BBH

11

It seems that the substitution becomes possible.

In fact, this is true if the entire space is filled by a single medium.

In other words,

may be non-zero.

is position dependent, then However, if

is constant in the entire space.

For example:

In this case, we have to match the boundary conditions.

0 H on the boundary!!!

Boundary Conditions

Magnetic Field in Matters

MH

SS

dd aMaH

)( belowabovebelowabove

MMHH

fKConsider a surface with free current density

Consider a pillbox Gaussian surface as shown.

or in integral form,

Since

we have,

nKBB ˆ0

//

below

//

above

nKHH ˆ//

below

//

above f

For the parallel components, since

fJH JB 0is similar to

Recall that we have derived from the Ampere’s law that,

Therefore, similarly

0fK

//

below

//

above HH

BH

1

//

below

below

//

above

above

belowabove

11BB

BB

In particular, if

then

In linear media,

At the interface of two linear media,