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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.
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