Chapter VI Magnetics Fields in Matter

Magnetization

As mentioned in the previous chapter, all magnetic phenomena are due to

electric charge in motion.

In atomic scale these charges are electrons orbiting around nuclei and

electrons spinning about their axes.

For macroscopic scale these current loops are so small that we treated as

magnetic dipoles.

In the normal case, these dipoles cancel each other because of their random

orientations.

When an external magnetic field is applied, these dipoles aligned in some

direction and so the material becomes magnetized.

Paramagnetic materials: Are materials that acquire a magnetization parallel to

the external magnetic field B.

Diamagnetic materials: Are materials that acquire a magnetization Opposite

to the external magnetic field B.

Ferromagnetic materials: Are materials that retain their magnetization even

after the external magnetic field B is removed.

Torques and Forces on Magnetic Dipoles

It is known that the electric dipole experiences a torque in an electric field. In

analogues, the magnetic dipole is expected to experience a torque under the

action of external magnetic field.

Consider a rectangular loop of sides a and b that is tilt at an angle from the z-

axis as shown. The magnetic field is applied along the z-axis.

For the sides a we have for the back side

m

x

y

z

B

I

a b

Fa

Fa

Fb

Fb

iIaBkkjIaBBlIFmˆˆ)ˆˆ(

And for the front side we have

iIaBkkjIaBBlIFmˆˆ)ˆˆ(

So the two forces cancel out and

since they are act on the same axis

they don’t make any torque.

The sides b are both along the x-axis, so the forces are along the y-axis but

opposite to each other.

jIaBkiIbBBlIFmˆˆˆ

Although these two forces cancel each other, they have a net torque given as

xabIBxFa

xa

Fa

FN

ˆsinˆsin

ˆsin2

sin2

)1(BmN

Is the magnetic dipole moment. It direction is determined by the Hand-Right

rule.

)2(AImwith

m

y

z

B

a

Fb

Fb

The torque is tend to align these dipoles along the direction of the magnetic

field. It is this torque that accounts for Paramagetism.

It is known that the magnetic force acting on a loop is given by

BldIF

If the magnetic field is uniform we have

Due to the spinning of electrons around its own axis, each electron can be

considered as a magnetic dipole, so Paramagnetism is a universal phenomena.

But it is not the case if the magnetic field is nonuniform.

Considering a circular loop near the end of a short

solenoid, as shown in the figure.

The magnetic field here has two components: The z-

components (Bsin) and the radial components

(Bcos), is the angle between the field the radial axis

.

The z-components constitute uniform field and so have

no force. The radial components give a force by

zIRBsBRIF ˆcos2ˆcosˆ2

For infinitesimal loops the force can be written as

)3(BmF

0 BldIF

Effect of a Magnetic Field on Atomic Orbits.

Let us assume that the electron moves around the

nucleus in a circular orbit. This motion results a

current given as

R

ev

T

eI

2

The orbital dipole moment is given by

If the attractive force between the electron and the

nucleus then we have

)4(ˆ21 zevRAIm

In the presence of a magnetic field there is

an additional force so we have

)5(4

2

2

2

R

vm

R

ee

o

)6(4

2

2

2

R

vm

R

eBve e

o

Substituting Eq.(5) into Eq.(6) we get

R

vm

R

vmBve ee

22

vvvvR

m

R

vvmBve e

e

22

Assuming the change is very small, i.e.,

vvvvvv 2&0

)7(2 em

eRBv

That is, when the magnetic field turned on the electron speed up.

From Eq.(4) it is clear that the change in the orbital speed means a chanhe in

the dipole moment as

)8(4

ˆ22

21 B

m

RezvRem

e

Note that the change of magnetic dipole moment is opposite to the direction of

the magnetic field. This is the mechanism responsible for Diamagnetism.

The Field of a Magnetized Object:

Let us calculate the magnetic field produced by a magnetized material.

Let us calculate the magnetic field produced

by a magnetized material with magnetization

M (magnetic dipole moment per unit volume).

From the last section of the previous chapter,

we have for the vector potential of a single

dipole moment

)9(ˆ

4 2

mA o

dipole

2

ˆ

4

dMAor o

dipole

The total vector potential of the object is, then

)10(ˆ)(

4 2

d

rMA o

Using the identity 2

ˆ1

)11(1

4

drMA o

Now using the identity

AAA

)12(

1

4

d

rMdrMA o

Sv

SdAdA

Recalling the divergence theorem

Now replacing vectorconstantais, BBAbyA

)13( Sv

SdBAdBA

ABBAABBABut

ABCBACCBAand

)14( Sv

SdABdAB

Eq.(13) becomes

vectorconstantanyisBut B

)15( Sv

SdAdA

Using Eq.(15) Eq.(12) becomes

)16(11

4

adrMdrMA

S

o

Defining the volume current density (current per unit area) Jb and the surface

current density (current per unit length) Kb as

)17(MJb

)18(n̂MKb

Eq.(16) becomes

)19(

4

S

bbo rKd

rJA

This means that, Instead of using Eq.(10), the vector potential and hence the

magnetic field is the same as would be produced by a volume current Jb

through the material plus a surface current Kb on the boundary.

dvAneI

It should be noted that using the relation between current and velocity, Eq.(25)

of the previous chapter, we can show that

)20(dd vveV

NJ

)21(dd vveA

NK

Example 6.1 Find the magnetic field of a

uniformly magnetized sphere.

Solution: Since the magnetization M is uniform

0 MJb

ˆsinˆ MnMKb

SdrKd

rJA oo

44

One can find the vector potential using

But we can realise that the magnetism of the sphere is causing current to flow

along the surface of the sphere. This is the same as of a spinning spherical

shell, the moving charges would form a surface current. Such a shell would

have ˆsinRK

Therefore, the field of a magnetized sphere is identical to that of a spinning

spherical shell with

MR

Rrr

R

RrrR

Ao

o

ˆsin

3

ˆsin3

2

4

Referring back to Example 5.11 we have

Rrr

MR

RrrM

Ao

o

ˆsin

3

ˆsin3

2

3

22

22

sin3

00

ˆsinˆˆ

sin

1

sin

ˆsinˆˆ

sin

1

rM

r

rrr

rArrAA

r

rrr

rAB

or

inin

To find the m.field B we have

MzMrM

B ooo

in

32

32 ˆˆsinˆcos

3

2 Uniform magnetic field

ˆsinˆcos2

4

sin

400

ˆsinˆˆ

sin

13

2

2

r

r

m

r

m

r

rrr

rAB o

o

outout

22

3ˆ

4ˆsin

3 r

rm

r

MRA oo

For outside region the vector potential looks like

Which exactly like a pure dipole. The magnetic field is now

MRm 3

34with Is the magnetic dipole moment

Physical Interpretation of Bound Currents

Consider a thin slab of uniform magnetized

material where the magnetic dipoles are

represented by tiny current loops.

It is clear that all the internal currents

cancel out, while at the edge there is no

adjacent loop to do the canceling.

It is clear that all the internal currents cancel

out, while at the edge there is no adjacent

loop to do the canceling. His means that the

slab is equivalent to a single ribbon of

current I flowing around the boundary.

Now, suppose that each current loop has

area a and thickness t. Its dipole moment is

atMaIm

bKt

IM

Or in vector notation we have

nMKb ˆ

When the magnetization is not uniform the internal currents no longer cancel.

Consider two adjacent blocks of

magnetized material. The current in the

right block is greater than that of the left

block. On the surface where they join there

is a net current in the x-direction given by.

dydzy

MdzyMdyyMI z

zx

The corresponding volume current density

is then

)22(y

MJ z

xb

Similarly, a magnetization along the y-axis

would give a current density by

)23(z

MJ

yxb

From Eqs.(22 & 23) we get

)24(z

M

y

MJ

yzxb

Or in general we can write

)25(MJb

The Auxiliary Field H

For Magnetized material the net current is the sum of the free current and the

bound current, i.e.,

)26(bf JJJ

)27(rJB o

Recalling Ampere’s law from Chapter 5 we have

From the last equation it is clear that 0 bJ

Which agree with conservation principle, as stated by the continuity equation.

rJMB

fo

Defining the auxiliary field (the magnetic field strength) H as

)28(MB

Ho

Ampere’s law now becomes

)29(rJH f

)30(fencIldH

Or in integral form we have

From Eqs.(26 & 25) Ampere’s law reads

MrJB

fo

Example 6.2 Find the m.f a distance s from a

long, straight, magnetized wire carrying a steady

current I.

Solution. For s>R, let us choose a circular loop

of radius s around the wire. Ampere’;s law is

R I

s

fencIldH

By symmetry we see that, like B, H must be constant in magnitude and tangent

to the loop at every point on the loop. Then

IsHdlH )2(

s

IH

2

Using Eq.(28) and noting that the magnetization M=0 outside the wire we

conclude that

s

IHMHB ooo

2

For a point inside the wire the circular loop in

this case has a radius sR.

R I s

To find the current enclosed by the loop, we note that

22 s

I

R

IJ

fencf

2

2

R

sIIenc

fencIldH

22 R

IsH

The magnetic field inside can’t be found because we have no information about

the magnetization inside the wire.

The Boundary Value Problems and the Boundary Conditions.

Recalling the fundamental two equations concerning the magnetic fields,

)31(0 B

)32(rJH f

Substituting for B from Eq.(28), Eq.(31) becomes

0MHo

)33(MH

In regions of space where there is no free currents, Eqs.(32 & 33) becomes

)34(0 MHH

Because the curl of H is zero

)35(mH

Where m is defined as a scalar magnetic potential. Taking the divergence of

Eq.(25) we get

)36(2mH

From Eq.(33) we get

)37(2 Mm

Poisson Eq. of magnetosatics

If the divergence of magnetization is zero (either M=0 or M is uniform) we get

)38(02 m Laplace’s Eq. of magnetosatics

To study the B.C. at the surface of two

regions in space we first apply Gauss’s

law of magnetism for the pillbox shown.

Region I

Region II

n1

n2 0. SdB

0ˆˆ 2211 nBanBa

nnnBut ˆˆˆ 21 )39(21 nn BB

That is, the component of the m.field normal to the interface is continuous

across the interface

Now applying Ampere’s law to the closed

loop shown, we get

Region I

Region II

lKlHlH f 2211

lllBut

21 lKlHH f

21

Where Kf is the surface current density normal to the plane of the rectangle.

The ;last equation can be written as

fencIldH

)40(ˆ21 nKHH ft

Equations (39) and (40) can be combined in a single equation as

)41(ˆ 21 fKHHn

The tangential components of H is discontinuous at the interface when a

surface current density exists on the interface.

Example Let us resolve example 6.1 as a

boundary value problem.

Solution: Since the magnetization M is uniform

inside and zero outside so Laplace’s Eq. is

satisfied in both regions. The general solution of

Laplace’s equation is

01

cos),(l

llll

lm Pr

BrAr

The B.C. are: inr 0at (i) 0at (ii) outr

From the second condition we have

0

1cos),(

lll

lout P

r

Br

And from the first condition we have

0

cos),(l

ll

lin PrAr

Now since there is no free surface current on the surface of the sphere we have

021 HH

From Eq.(35) we conclude that, as a third B.C

),(),( RR outin

01

0

coscosl

lll

ll

ll P

R

BPRA

)1(12

1

l

lllll

l RABR

BRA

From the continuity of the normal components of B we have

),(),( RBRB outrinr

But from Eq.(28) and Eq.(35)we have

cosM

rM

rBMHB ororo

coscoscos1 00

1

02

MPRAlPR

Bl

ll

ll

lll

l

Substituting fro Bl from Eq.(1), the last equation gives

coscoscos10

1

0

1 MPRAlPRAll

ll

ll

ll

l

coscos120

1 MPRAll

ll

l

10&3 1 lforAMA l

331

131

1 & MRBMA

cos),(31 Mrrin

2

331 cos

),(r

MRrout

mH

Now from Eq.(35) we have

ˆsinˆcos2),(3

3

31 r

r

RMrHout

kMrHinˆ),(

31

But from Eq.(28) and knowing that M=o outside, w get

kMkMHB oinoinˆˆ

32

ˆsinˆcos2),(3

3

31 r

r

RMrHB ooutoout

Linear and Nonlinear Media For most substances the magnetization is proportional to the magnetic field, i.e.,

)42(HM m

The dimensionless constant m is called the magnetic susceptibility. Materials that obey Eq.(43) are called linear.

For paramagnetic materials m is +ve, while for diamagnetic materials m is –ve.

From Eq.(28) we have )43(1 HMHB moo

)44(HMHB o

)45(1 mowith

Is called the magnetic permeability of the material. It is clear that in vacuum = o and so m =0

Example 6.3 An infinite solenoid with n

turns per unit length and current I is

filled with linear material of

susceptibility . Find the magnetic field

inside the solenoid.

Solution. The m.field for a long

solenoid is zero outside and uniform

inside. To calculate the internal

magnetic field, we can apply Ampere’s

law to the rectangular loop abcd of

length l and width w.

encfIldH

encf

a

d

d

c

c

b

b

a

IldHldHldHldH

For the portions bc and da H is to dl and for the portion cd H=0, then only

the 1st integral survives

encf

b

a

IldH

For the portion ab H is to dl and its magnitude is constant along it

encfIHl

But the current crossing the surface enclosed by the loop is I multiplied by the

No. of turns bounded by the loop

knIIl

NHNIHl ˆ

With n is the No. of turns per unit length. Now using Eq.(43) we get

knIHB momoˆ11

Note that if the solenoid is filled with vacuum m =0 and we recover the well-known value for a long solenoid.

If the material is paramagnetic the field is enhanced, while if the material is

diamagnetic the field is reduced.

Let us calculate the surface bound current. We have from eq.(18) and eq.(42)

ˆˆˆˆˆ mmmb nIsknInHnMK

This means that Kb is in the same direction as I for paramagnetic materials and

in opposite direction to I for diamagnetic materials.

This explain why the m.field enhanced for paramagnetic materials and reduced

for diamagnetic materials.

Ferromagnetism

Unlike paramagnets and diamagnets, ferromagnets, which are not linear,

require no external fields to sustain the magnetization. Iron, nickel, cobalt are

the most common examples of ferromagnets.

In a ferromagnet, each dipole “like” to point in the same direction as its

neighbors. All the spins point the same way.

If one magnify a piece of iron and see the

individual dipoles as tiny arrows, it would look

something like shown in the figure with all the

spins pointing the same way.

This alignment occurs in small regions called domains, each

domain contains many dipoles. The domains themselves

are randomly oriented

Within the domain, the magnetization is intense, but in a bulk sample the

material will usually be demagnetized because the many domains will be

randomly oriented with respect to one another.

A small externally imposed magnetic field can cause the magnetic domains to

line up with each other.

(a) Starting at zero the material follows at first a non-linear magnetization curve

and reaches the saturation level, when all the magnetic domains are aligned

with the direction of a field;

(b) Reducing the applied field there is a partial return to randomly oriented

domains (demagnetization). Even with zero field there is some residual

(remanent) magnetization. The material is now a permanent magnet.

c) In order to demagnetize a ferromagnetic material the strong magnetic field of

the opposite direction (called coercive field, Hc ) has to be applied. A saturation

level will be reached in the other direction.

d) To complete the story, turn the field on again in the positive sense: M returns

to zero and eventually to the forward saturation point.

Hysteresis Loop

When a ferromagnetic material is

magnetized in one direction, it will not

relax back to zero magnetization when

the imposed magnetizing field is removed.

If a m.field is applied to the material,

its magnetization will trace out a loop

called a hysteresis loop.

As a final remark concerning ferromagnetism, it should be noted that all

ferromagnets have a maximum temperature where the ferromagnetic property

disappears as a result of thermal agitation. This temperature is called the Curie

temperature.

Above this temperature the alignment is destroyed and the material becomes

paramagnetic. Fortunately the Curie temperature for iron is 770o, which is high

enough for practical use.