MAGNETIC FIELDS IN MATTER
Magnetization
When considering the electric ¤elds in matter we needed to think about
the atomic structure of the material. Applied electric ¤elds tend to distort
the atoms and/or molecules by stretching and rotating them. The electron
orbits in atoms constitute little current loops, and so they form little magnetic
~ ; so
dipoles. As electric ¤elds align electric dipoles with a torque ~¿ = ~p £ E
~ (This is
also magnetic ¤elds align magnetic dipoles with a torque ~¿ = m
~ £ B:
how compass needles work.) In most materials, thermal agitation ensures that
the atomic dipoles are randomly oriented, but applied ¤elds tend to align the
~ The material is then magnetized. However, the result of
dipoles parallel to B:
the alognment is quite di¤erent from what we found with electric ¤elds. The
electric dipoles create a charge layer on the surface of a dielectric, and that layer
produces a ¤eld within the material that is opposite the applied ¤eld. Thus the
result is a net ¤eld inside the material that is smaller than the applied ¤eld.
Aligned magnetic dipoles produce a surface current layer (LB Figure 29.25)
and the magnetic ¤eld inside this current loop is parallel to the applied ¤eld.
Thus the net result is a ¤eld inside the material that is greater than the applied
¤eld. The e¤ect is called paramagnetism.
But this is not yet the whole story. Changing magnetic ¤elds produce electric
¤elds through Faradayzs law, and those electric ¤elds e¤ect the atomic currents.
Letzs see how it goes.
We model a simple atomic magnetic moment as an electron orbiting with
radius r at speed v: The magnetic moment is
e
m = IA = ¼r2
T
where T is the period 2¼r=v: Thus
e
m = rv
2
Now letzs worry about directions. The current is opposite the velocity, because
z
the charge is negative, and so if ~! = ! z^ = v^
; then
r
e
m
~ = ¡ rv^z
2
The ¢ux of external magnetic ¤eld through this loop is
~ e xt ¢ z^
© e xt = ¼r2 B
and the changing ¢ux produces an emf
I
~ ¢ d~l = ¡ d©e xt = ¡¼r2 d B
~ e x t ¢ z^
"=
E
dt
dt
~ e xt = 0, the centripetal force on the electron orbit is purely
If we start with B
electric:
mv 2
Ze
F~ c = ¡
^r = ¡e
^r
(1)
r
4¼"0 r2
1
The induced electric ¤eld is given by
2¼rE Á = ¡¼r2
d ~
Be x t ¢ z^
dt
~ ¢ z^ is increasing, E µ will be negative, F Á will be positive and the particlezs
So if B
~ ¢ ^z is decreasing, Eµ will be positive, F Á
speed will increase. However, if B
will be negative and the particlezs speed will decrease. So after time ¢t; the
speed will have changed by ¢v; and we also have an additional force component
~
F~ m ag = ¡e~v £ B
2
Fr = ¡
Ze 2
m (v + ¢v)
mv 2
2mv¢v
¡ evBz = ¡
=¡
¡
4¼"0 r
r
r
r
where we ignore the square of the small change ¢v, as usual. The ¤rst terms
on each side are equal, by (1). Thus
evBz
=
¢v
=
2mv¢v
r
eBz r
2m
As we concluded above, ¢v is positive if Bz is positive. This change in speed
also changes the magnetic moment:
e
m
~ + ¢~
m = ¡ r (v + ¢v) z^
2
So
¢m
~ =¡
er
2
µ
eBz r
2m
¶
z^ = ¡
e2 r 2 ~
Be x t
4m
~ That means that increasing the
The change in m
~ is opposite the change in B:
~
external B tends to decrease the internal ¤eld due to the dipoles. This e¤ect is
called diamagnetism.
Both these e¤ects happen at the same time. Diamagnetic e¤ects are usually
weak, and show up in materials in which the intrinsic atomic magnetic moments
are small (these are usually atoms with an even number of electrons, like Helium)
or materials that are hot, because the thermal agitation dirupts the alignment
of the dipoles. What this means is that the response of any given material to an
applied magnetic ¤eld will depend on that materialzs properties: in some cases
the internal ¤eld is increased (these are paramagnetic materials) and in others,
decreased (these are diamagnetic materials). In both cases the e¤ect is small.
A very interesting class of materials, the ferromagnetic materials, are able
to maintain the alignment of the dipoles even when the applied ¤eld is removed,
thus forming a permanent magnet. The details are due to quantum mechanical
interactions, and we wonzt discuss them here. Each ferrormagnetic material has
a characteristic temperature, called the Curie temperature, Above that temperature, thermal e¤ects disrupt the alignment and a permanent magnet cannot be
2
formed. There are additional very interesting features of these materials, such
as hysteresis, which make them more di¢cult to analyze. More later.
As with dielectric polarization, we de¤ne the magnetization of a magnetic
~ as the dipole moment per unit volume,
material M
~ = n~
M
m
Dielectrics are drawn into stronger electric ¤eld regions. Similarly, paramagnetic
materials are drawn into stronger magnetic ¤eld regions, while diamagnetic
materials are repelled from such regions.
Fields in magnetic materials.
~ to aid in our
With dielctrics we found it useful to introduce a new ¤eld D
~ The
analysis. Similarly here we shall ¤nd it useful to introduce a new ¤eld H:
~ is magnetic induction, while H
~ is called the magnetic ¤eld.
proper name for B
(Gri¢ths seems to hate this language, and I do too, so for the most part wezll
~ the magnetic ¤eld. Please use context to be
ignore it, and continue to call B
sure you know what is meant.) So letzs see how this goes.
We start with the magnetic vector potential due to one dipole.
~ £ r^
~ = ¹0 m
A
4¼ r2
~ due to a magnetized object:
and from this we can compute A
Z ~
^
M £R
~ (~r) = ¹0
A
d¿ 0
2
4¼
R
~ = ~r ¡ ~r0 : We may rewrite the integrand as follows:
where, as usual, R
µ
¶
Z
~ (~r) = ¹0
~ £ r
~ 0 1 d¿ 0
A
M
4¼
R
à !
#
Z "
~
~0£M
~
0
¹0
M
r
~
=
¡r £
+
d¿ 0
4¼
R
R
"Z
#
Z ~0
~
~
¹0
M
r
£
M
~0 +
=
£ dA
d¿ 0
4¼
R
s R
(2)
We may identify the bound current density
~j b ou n d = r
~ £M
~
(3)
and bound surface current density
~ b o un d = M
~ £n
K
^
~ becomes
so that A
~ (~r) = ¹0
A
4¼
"Z
s
Z ~
~ b ou n d
K
j b ou n d 0
0
dA +
d¿
R
R
3
(4)
#
Equation (2) may be used for any magnetized object, including permanent magnets.
~ = M z^ inside, and
Suppose we have a uniformly magnetized sphere, with M
~ is constant except at the
zero outside, of course. Then ~jb ou n d = 0; because M
~
surface. But we take care of the surface with K:
^
~ b o un d = M
~ £ r^ = M sin µ Á
K
Now this looks just like our rotating, uniformly charged spherical shell, and the
¤eld is identical if we replace ¾!a with M: The internal ¤eld is uniform and the
external ¤eld is a pure dipole.
~ in t = 2 ¹ M
~
B
3 0
and the dipole moment is
m
~ =
4¼ 3 ~
~ £ volume of sphere
a M =M
3
This exactly what we might have expected.
~ b o u nd : For any volume
We can use this result to motivate our expression for K
V;
~V
m
~ =M
Let us look at a slab of area A and thickness t; so that
m = M At = IA
Thus the e¤ective "loop"current is I = M t and thus the surface current is
K = I=t = M : Putting back the vectors to get the directions gives equation (4).
Now we take the total current and divide it into two parts: the bound current
due to magnetization and the free current due to moving charges,
~j = ~j b ou n d + ~jfr ee
Amperezs law in di¤erential form is
³
´
h³
´
i
~ £B
~ = ¹0 ~jb o un d + ~j fre e = ¹0 r
~ £M
~ + ~jf re e
r
Put the curl terms together to get
³
´
~ £ B
~ ¡¹ M
~ = ¹ ~j fre e
r
0
0
~ by
Now de¤ne the ¤eld H
and we get
~ ´ 1 B
~ ¡M
~
H
¹0
(5)
~ £H
~ = ~jfre e
r
(6)
4
~ which is
We can then deduce the integral form of Amperezs law for H;
I
Z
~ ¢ d~l = Ifr ee = ~jf re e ¢ n
H
^ dA
This is a very useful relation in situations with su¢cient symmetry.
~ satis¤es a second equation that we can get from its de¤nition (5). Taking
H
the divergence, we get
~ ¢H
~ ´ 1 r
~ ¢B
~¡r
~ ¢M
~
r
¹0
~ ¢B
~ = 0, so
But r
~ ¢H
~ ´ ¡r
~ ¢M
~
r
(7)
This can be very useful in situations where ~jf re e = 0: For then equation (6)
~ £H
~ = 0; and so we may express H
~ as the gradient of a scalar
becomes r
function Izll call Â: If we write
~ = ¡rÂ
~
H
equation (7) becomes Poissonzs equation for Â:
~ ¢M
~
r 2Â = r
This problem is exactly analgous to an electrostatic potential problem, where
~ ¢M
~ acts like a magnetic charge density.
¡r
~ and H:
~ The ¤eld
Now we can see some important di¤erences between B
~ form closed loops, because r
~ ¢B
~ = 0: But the ¤eld lines of H
~ begin
lines of B
~ ¢M
~ 6= 0:
and end on "magnetic charge" where r
Letzs look again at the magnetized sphere. Inside the sphere
~
~ in = 1 B
~ ¡M
~ = 1 2¹ M
~ ¡M
~ = ¡M
H
0
¹0
¹0 3
3
~ diverge from the surface of the sphere where r
~ ¢M
~ 6= 0:
Thus the lines of H
~
Boundary conditions for H
Starting with the curl equation, and integrating around a rectangle across
the boundary, we have
5
Z ³
´
~ £H
~ ¢ dA
~ =
r
I
~ ¢ d~l =
H
³
´ ¡ ¢
~1 ¡H
~ 2 ¢ ¡^t =
H
Since ^t = ^n £ N^ ; we get
Z
~j fre e ¢ dA
Z
^
N ¢ ~j fre e dh w
~ fre e
N^ ¢ K
³
´
~1 ¡ H
~2 = K
~ f re e
n
^£ H
(8)
Then integrating (7) over a pillbox, we get
Z
Z
~
~
~ ¢M
~ dV
r ¢ HdV ´ ¡ r
Z
Z
~ ¢ dA
~ = ¡ M
~ ¢ dA
~
H
~ is zero on one side of the surface (a frequent occurrence) we get
Now if M
³
´
~ ou t ¡ H
~ in ¢ n
~ ¢n
H
^=M
^
(9)
~ ¢n
~
M
¹ acts like a surface "magnetic charge density" in producing H:
~
Suppose a bar magnet has a uniform magnetization M running along its
~ ¢M
~ is zero everywhere except at the surface of the magnet
length. Then r
~ ¢n
(where it is in¤nite). M
^ is also zero except at the ends of the magnet. Thus
~ ¢ ^n is
we have no "magnetic charge density" except at the two ends, where M
positive on one end and negative on the other. Thus we have a physical dipole.
~ runs from the positive "magnetic charge" at one end to
Inside the magnet H
the negative "magnetic charge" on the other. Outside we have a dipole-type
¤eld.
~ =¹ H
~ +M
~ forms closed loops. Outside there is no M
~ and the B
~
Now B
0
~
~
¤eld lines follow the H lines exactly. But the B lines form closed loops that
~ and H
~ are opposite each other.
close inside the magnet, where B
~ and H
~ :
Letzs summarize the boundary conditions for B
³
~1 ¡ H
~2
n
^£ H
´
Bn or ma l is continuous always
~ f re e ;
= K
~ t an ge ntia l is continuous if K
~ fr ee = 0
H
Linear Media
In LIH materials, the magnetization is proportional to the net ¤eld:
~ = ÂmH
~
M
6
Here Âm is the magnetic susceptibility. Unlike the electric susceptibility, Â m
can take either sign: it is positive in paramagnetic materials and negative in
diamagnetic materials. But in both cases it is almost always very small, Â m ¿
1: For LIH materials,
³
´
~ =¹ H
~ +M
~ = ¹ (1 + Â ) H
~ = ¹H
~
B
0
0
m
¹ is the permeability of the material. For most LIH materials where this formulation makes sense, ¹ ' ¹0 :
For these materials the bound current is
³
´
~ £M
~ =r
~ £ Â H
~
~jb ou n d = r
m
and because of the IH part of LIH, we can move the susceptibilty through the
derivative to get
~jb ou n d = Â m r
~ £H
~ = Âm~j fre e
If there is no free current density inside the material, all the bound current will
be at the surface.
Feromagnetism
In a ferromagnetic material, the atomic dipoles tend to align with each other,
even in the absence of applied ¤elds. the dipoles align in regions called domains.
Normally the domains are oriented randomly. When an external ¤eld is applied,
the dipoles at the edge of a domain feel an additonal torque, and the net e¤ect is
to cause the domains with magnetization parallel to the applied ¤eld to grow. If
the applied ¤eld is strong enough, the other domains actually rotate to align with
the applied ¤eld. When all the domains are aligned, the material is saturated
~ has its maximum value.
and B
To magnetize the material we set up a coil so that we get an (almost) uniform
~ (Remember it is H
~ that is directly related to the free currents.) As the
H.
current is increased, H increases and so does the magnetization. But once the
material is saturated, M does not increase further. Then B can increase only
slowly, since
³
´
~ =¹ H
~ +M
~
B
0
with M À H: If we now begin to turn the current (and thus H) down, the
~ reverses do we get a substantial decrease
domains remain aligned Only when H
~ : Thus the relation between B
~ and H
~ is not at all linear. Worse, the value
in M
~ does not depend only on H;
~ but also on the historyw what has happened
of B
previously. This phenomenon is called hysteresis. The graph of B versus H
looks like this:
7
This is a hysteresis loop (Diagram from Reitz,Milford and Christy). . (See
LB page 942 for more examples.)
There are two kinds of ferromagnetic materials. Materials with a fat hysteresis loop are hard magnetic materials that are used to make permanent magnets.
Such materials include cobalt steel and the alloy called alnico. Materials that
have a thin hysteresis loop (such as ferrites) are soft magnetic materials used
in transformer cores. A H value of about 80 A/m is su¢cient to saturate iron,
which is a soft magnetic material.
Magnetic shielding
One of the important uses of magnetic materials is for magnetic shieldingw
creating a volume of space where magnetic ¤elds are reduced to almost zero.
As an example, consider an in¤nitely long cylindrical shell with internal and
external radii a and b; and large magnetic permeability ¹: We place the z¡axis
along the axis of the cylinder. The shell is placed in a uniform external ¤eld
~ 0 = B0 x^ : Letzs ¤nd the magnetic ¤eld everywhere.
B
~ as the gradient of a scalar potential  that satis¤es
We may express H
Laplacezs equation everywhere outside the shell, and in the hole. In the shell
itself
~ ¢H
~ ´ ¡r
~ ¢M
~
r
But
~ = ¹H
~
B
8
and
³
´
~ ¢B
~ =0=r
~ ¢ ¹H
~ = ¹r
~ ¢H
~
r
So
r2 Â = 0
everywhere except at the edges of the shell.
We may express the solution in cylindrical coordinates with no z¡dependence.
Now we write three solutions, one valid in each of the three regions. In the hole,
the solution must be ¤nite as r ! 0; so
Âin =
1
X
rn (An sin nÁ + Bn cos nÁ)
n=1
In the exterior, the solution must give us the applied ¤eld as r ! 1; so
Âo u t = ¡
1
X
B0
r cos Á +
r¡n (Cn sin nÁ + Dn cos nÁ)
¹0
n=1
In the shell itself, we need all the terms:
Âsh e ll =
1
X
¡
n=1
¢
r ¡n + ®n r n (E n sin nÁ + F n cos nÁ)
Now we apply the boundary conditions.
Normal B is continuous at each boundary.
¡B0 cos Á¡¹0
1
X
nb¡ n¡1 (Cn sin nÁ + Dn cos nÁ) = ¹
n=1
n= 1
and
¹0
1
X
1
X
¡
nan¡ 1 (An sin nÁ + Bn cos nÁ) = ¹
n=1
1
X
¡
n=1
¢
¡nb ¡n¡1 + n® n b n¡ 1 (E n sin nÁ + F n cos nÁ)
¢
¡na¡n¡1 + n®n an¡1 (En sin nÁ + F n cos nÁ)
~ is continuous at the boundaries:
Tangential H
1
1
X
X
¡
¢
B0
b sin Á+
nb ¡n (C n cos nÁ ¡ Dn sin nÁ) =
n b ¡n + ®n bn (E n cos nÁ ¡ F n sin nÁ)
¹0
n=1
n= 1
and
1
X
n=1
1
X
¡
¢
n a¡n + ®n an (E n cos nÁ ¡ Fn sin nÁ) =
nan (An cos nÁ ¡ Bn sin nÁ)
n=1
We have seen this kind of thing before. The external ¤eld imposes the n = 1;
cosine modew all the others are zero.Verify this for yourself. So looking at n = 1;
we have the 4 equations:
¡
¢
¡B0 ¡ ¹0 b ¡2 D1 = ¹ ¡b¡ 2 + ®1 F1
(10)
9
¡
¢
¹0 B1 = ¹ ¡a¡ 2 + ®1 F1
¡
¢
B0
b + b ¡1 (¡D1 ) = b ¡1 + ® 1 b (¡F 1 )
¹0
and
¡
¢
a¡1 + ® 1 a (¡F1 ) = ¡aB1
(11)
(12)
(13)
for the four unknowns B1 ; D1 ; F1 and ®1 : We can already see from equation
(10) that the coe¢cient F 1 has to be of order ¹0 =¹: and from (13) B1 will be
of roder F1 : This is the source of the shielding.
Going at it systematically, we get from (13)
¡
¢
B1 = a¡2 + ®1 F 1
Then putting this result into (11) we
¡
¢
¹ ¡a¡ 2 + ®1 F1
¢
¹ ¡ ¡2
¡a + ®1
¹0
µ
¶
¹
®1
¡1
¹0
have
®1
Then putting (14) into (10) we have
¡
¢
= ¹0 a¡ 2 + ®1 F1
¡
¢
= a¡ 2 + ®1
µ
¶
¹
= a¡2 1 +
¹0
¹
+
¹
0
= a¡2
¹ ¡ ¹0
(14)
µ
¶
b2 ¹ + ¹0
¡B0 b2 ¡ ¹0 D1 = ¹ ¡1 + 2
F1
a ¹ ¡ ¹0
and puting (14) into (12) we get
µ
¶
b2 ¹ + ¹0
B0 b ¡ ¹0 D1 = ¡¹0 1 + 2
F1
a ¹ ¡ ¹0
2
Subtracting these two, we get
Ã
2
2B0 b =
F1 =
Adding gives
!
F1
2B0 a2 b 2 (¹ ¡ ¹0)
a2 (¹ ¡ ¹0 )2 ¡ b 2 (¹ + ¹0 )2
(15)
µ
¶
b2
2¹0 D1 = F 1 (¹ + ¹0 ) 1 ¡ 2
a
so
D1
2
b2 (¹ + ¹0 )
¹ ¡ ¹0 ¡ 2
a ¹ ¡ ¹0
µ
b2
1¡ 2
a
¶
2B0 a2 b 2 (¹ ¡ ¹0 )
=
(¹ + ¹0 )
2¹0
=
¢
B0 ¡ 2
a ¡ b2
¹0
a2 (¹ ¡ ¹0 )2 ¡ b2 (¹ + ¹0 )2
10
a2 (¹ ¡ ¹0 )2 ¡ b2 (¹ + ¹0 )2
¡
¢
b2 ¹2 ¡ ¹20
(16)
and ¤nally
B1
¡
¢
a¡2 + ® 1 F1
µ
¶
1
2¹
2B0 a2 b 2 (¹ ¡ ¹0 )
a2 ¹ ¡ ¹0 a2 (¹ ¡ ¹0 )2 ¡ b2 (¹ + ¹0 )2
=
=
4¹B0 b2
=
(17)
a2 (¹ ¡ ¹0 )2 ¡ b 2 (¹ + ¹0 )2
Thus
4¹B0 b2
Âin =
r cos Á
a2 (¹ ¡ ¹0 ) 2 ¡ b 2 (¹ + ¹0 )2
"¡
#
¢
¡
¢
B0
a2 ¡ b 2
b 2 ¹2 ¡ ¹20
Âou t =
cos Á
¡r
2
2
¹0
r
a2 (¹ ¡ ¹0 ) ¡ b 2 (¹ + ¹0 )
µ
¶
(¹ ¡ ¹0 )
r
2a2 b2
Âsh e ll = B0
+ (¹ + ¹0 ) 2
cos Á
r
a
a 2 (¹ ¡ ¹0 )2 ¡ b 2 (¹ + ¹0 )2
~ =B
~ 0 everywhere if ¹ ! ¹ :
Verify that we get back B
0
Letzs see what happens as ¹=¹0 becomes very large:
B 0 4b2
r cos Á
¹ a2 ¡ b 2
·
¸
B0 b 2
Âou t !
¡ r cos Á
¹0 r
µ
¶
B0 1
r
2a2 b 2
 sh e ll !
+ 2
cos Á
¹ r
a
a2 ¡ b 2
 in !
The ¤elds in the cavity become
B0 4b 2
~ in = ¡ rÂ
~
H
x^
in !
¹ b 2 ¡ a2
and
2
~ in = ¹0 H
~ in ! ¹0 B 0 4b ^x
B
2
¹
b ¡ a2
Both are very small, so the shell has e¤ectively shielded the cavity from the
external ¤elds. Within the shell
Ã
!
2
^Á
B
2b
cos
Á
r
^
+
sin
Á
0
2
~ s he ll =
H
^x ¡ a
¹ b 2 ¡ a2
r2
which is small, but
~ s he ll
B
2b 2
= B0 2
b ¡ a2
Ã
cos Á r^ + sin Á ^Á
x^ ¡ a
r2
2
which is comparable to B0 :
11
!