1

PHY481 - Lecture 29: Magnetic materials

Griﬃths: Chapter 6

Magnetic materials

Extension of Maxwell’s equations to treat magnetic ﬁelds inside and outside magnetic materials is achieved in

manner that in some ways is like the treatment of the dielectric response of materials - through analysis of dipoles, in

the magnetic case of course we consider magnetic dipoles. To treat a large number of aligned electric dipoles inside

a material using Maxwell’s equations, we introduced the polarization ~

P, the electric dipole density. To treat a large

number of aligned magnetic dipoles using Maxwell’s equations we introduce an analogous quantity, the magnetization

~

Mwhich is the magnetic moment density. In the dielectric case, atoms and molecules have either an induced or

intrinsic electric dipole moment. In a similar way, atoms and molecules may have an intrinsic magnetic moment

and/or a magnetic moment induced by an applied magnetic ﬁeld.

Atomic origins of magnetic properties

The magnetic properties of atoms comes from a combination of the orbital motion of the electrons about the

nucleus and the intrinsic magnetic moment of electrons, protons and neutrons. The orbital response produces a

magnetic moment that opposes an applied magnetic ﬁeld, while the spin part typically produces a magnetic moment

that enhances an applied magnetic ﬁeld.

The fundamental quantum of electron magnetic moment - the Bohr magneton

If a single electron moves in a circular orbit at speed vand with radius r, then the current is

I=e

2πr v(1)

The magnetic moment, morbital is then,

morbital =Iπr2=−1

2evr =−e

2me

L(2)

where L=mevr is the angular momentum of the electron. This relation is usually written as,

~morbital =gL~

L(3)

where gL=−e

2meis called the gyromagnetic ratio.

According to Bohr theory, the angular momentum is quantized, so that,

L=l¯h(4)

where l= 0,1,2... is an integer. We then have,

morbital =−e

2me

¯hl =−mBl(5)

where we have deﬁned the Bohr magneton,

mB=e¯h

2me

= 9.27 ×10−24Am2(6)

In addition, the electron carries an intrinsic magnetic moment due to its spin. The intrinsic magnetic moment of the

electron has a value very close to mB.

The total magnetic moment is found by a vector addition of the orbital and spin contributions. This sum is called

~m0. The magnetization used in Maxwell’s equations are the density of aligned moments ~m0.

Response of electron spins and electron orbits to an applied magnetic ﬁeld

Above we showed that an electron in circular orbit may have orbital or spin contributions, however in many

situations neither of these contributions is large. The orbital contributions may be small in s-states or if there are

electrons in each of the possible high angular momentum states, such as p-states. Moreover, in most situations, the

response of the electron spin to an applied ﬁeld depends on whether the atom or molecule has paired or unpaired

electrons in each orbital. If the electrons are paired, the spin response is weak as the paired electrons must be broken

up in order for the spins to align with an applied ﬁeld. In contrast if there are unpaired electrons in an atom or

2

molecule, the unpaired electrons can easily line up with an applied ﬁeld. Paramagnetism and ferromagnetism occur

when electron spins line up in a common direction. In the case of paramagnetism, an applied ﬁeld is used to line up

the spins. In ferromagnetic materials the spins can remain lined up even when an applied ﬁeld is switched oﬀ.

Orbital contributions to the magnetic moment are almost always zero until the atom or molecule is placed in an

applied ﬁeld. The qualitative eﬀect is that an applied magnetic ﬁeld produces a diagmagnetic eﬀect where the induced

dipole moment is opposite to the original direction of the dipole in a circular orbit. This can be seen by considering

the relationship between radius and velocity for the Bohr atom, ie

mv2

r=ke2

r2; or vr =ke2

mv (7)

Now consider applying a small magnetic ﬁeld that does not signiﬁcantly alter this relationship, though of course a

magnetic force q~v ∧~

Bdoes occur. The orbital moment is given by −evr/2 so increasing the product vr leads to an

increase in the orbital moment. As seen from the equation above, smaller velocities (larger orbits) leads to larger

orbital magnetic moment of course this is consistent with Bohr theory as vr is proportional to the magnitude of the

angular momentum.

However the most strongest diagmagnetism occurs in superconductors where the eﬀect can be perfect so as to screen

out an applied ﬁeld completely. In these materials superconducting screening currents are set up to completely cancel

an applied magnetic ﬁeld. In that sense they are qualitatively similar to the perfect dielectric behavior of metals

where static charges perfectly screen out an applied electric ﬁeld. As we shall see below there are some important

diﬀerences that are not evident in these qualitative analogies.

The bottom line is that most atoms and molecules that have unpaired electrons are likely to be dominated by

paramagnetic response, while those that have an even number of electrons, assuming that the electrons are paired,

are typically diamagnetic. Superconductors are a special case where diamagnetism can be very strong.

Maxwell’s equations for magnetic materials

We would like to extend Maxwell’s equations to treat the magnetic ﬁelds inside and outside magnetic materials

so we would like to replace the magnetization with current sources. It turns out that the relations between bound

currents and the magnetization are,

~

Kb=~

M∧~n;~

jb=~

∇ ∧ ~

M(8)

If the magnetization inside a material is a constant, only surface currents are required to produce the magnetization,

while if the magnetization varies in space, a bulk bound current density is also needed. The physical reasoning leading

to these observations is constructed by taking local current loops and showing that if the local current loop density

is a constant only a surface current remains. The formal calculation of these results is as follows.

The vector potential due to a magnetized material is found by adding up the vector potential contribution from

each part of the material, so that,

~

A(~r) = µ0

4πZ~

M(~r0)∧(~r −~r0)

|~r −~r0|3dτ 0=µ0

4πZ~

M(~r0)∧~

∇0(1

|~r −~r0|)dτ 0(9)

Product rule 7 of Griﬃths,

~

∇(f~

A) = f(~

∇ ∧ ~

A)−~

A∧~

∇f(10)

leads to,

~

A(~r) = µ0

4πZ(1

|~r −~r0|~

∇0∧~

M(~r0))dτ 0−µ0

4πZ~

∇0∧(~

M(~r0)

|~r −~r0|)dτ 0(11)

Using the vector identity (see Problem 1.60b of Griﬃths)

Z(~

∇ ∧ ~

A)dτ0=−I~

A∧d~a (12)

we ﬁnally have,

~

A(~r) = µ0

4πZ(1

|~r −~r0|~

∇0∧~

M(~r0))dτ 0+µ0

4πI1

|~r −~r0|(~

M(~r0)∧ˆn0)da0(13)

3

The ﬁrst term is a superposition formula for bulk currents, provided ~

jb=~

∇∧ ~

M, while the second integral corresponds

to a superposition of surface currents provided we deﬁne ~

Kb=~

M∧ˆn.

Incorporating bound currents into Ampere’s law

If we have a problem where we know the magnetization, we can solve for the magnetic ﬁelds using Ampere’s law

where,

~

∇ ∧ ~

B=µ0(~

jf+~

jb) = µ0(~

jf+~

∇ ∧ ~

M) = µ0(~

∇ ∧ ~

H+~

∇ ∧ ~

M) (14)

where we deﬁned ~

Hto be the magnetic intensity and it obeys,

~

∇ ∧ ~

H=~

jfso that I~

H·d~

l=if.(15)

From this equation it is seen that Amperian contours can be used to relate ifto the magnetic intensity (or Auxilliary

ﬁeld) ~

H. The magnetic intensity and magnetization have units Amp per meter (A/m) as can be seen from the

deﬁnition of ~

H. Once the magnetic intensity has been found, the magnetic ﬁeld is simply,

~

B=µ0(~

H+~

M) (16)

A uniformly magnetized cylinder with length >> radius

We take the magnetization to be uniform, ~

M=M0ˆzand to be oriented along the axis of the cylinder. There are no

free currents, so there is no gain in using the magnetic intensity (H= 0 as if= 0) and ~

B=µ0~

M. The magnetization

is uniform, so the curl of the magnetization is zero, therefore the only bound currents are the surface currents ~

Kb.

Since the magnetic ﬁeld and magnetization are uniform, we can think of the problem as being similar to the the

uniform ﬁeld inside a solenoid, so we expect to ﬁnd bound surface currents ﬂowing around the cylinder. We ﬁnd the

suface currents through,

~

Kb=M0ˆz∧ˆs=M0ˆ

φ(17)

This result applies in central regions of the magnetized cylinder. If we now consider a point ~r which is well removed

from the cylinder, i.e. r >> l, then we can treat the cylinder as a dipole with dipole moment ~mcyl =V olume ×~

M,

and then ﬁnd the ﬁeld using the dipole formula,

~

B(~r) = µ0

4π

3(~mcyl ·ˆr)ˆr−~m

r3(18)

As a check, lets use Ampere’s law to ﬁnd the magnetic ﬁeld inside the cylinder. An amperian loop leads to

Bl =µ0Kbl, so that ~

B=µ0Kb.

A Uniformly magnetized sphere of radius R

Take the magnetization to be, ~

M=M0ˆz, so that the bound currents are now,

~

Kb=~

M∧ˆr=M0sinθ ˆ

φ(19)

Assume that the magnetic ﬁeld inside is uniform and along the ˆzdirection, and that outside it is like a dipole, so

that,

~

Bint =B0ˆz; and ~

Bext =µ0

4π

3(~msphere ·ˆr)ˆr−~m

r3(20)

The boundary conditions on the magnetic ﬁeld at the sphere surface are, Bext

t−Bint

t=µ0Kb,Bext

n=Bint

n. The

latter equation gives,

B0ˆz·ˆr=B0cosθ =µ0

4π

2~msphere ·ˆr)

R3=µ0

4π

2mspherecosθ

R3(21)

From this expression, we ﬁnd that

B0=µ0

4π

2M04πR3

3R3=2µ0M0

3(22)

4

and hence,

~

Binside =2µ0M0

3ˆz(23)

It is easy to show that the boundary condition on the tangential component is also satisﬁed, so we have the correct

solution. Notice that this solution diﬀers by a factor of two from the analagous solution for a uniformly polarization

sphere.

##### Document information

Uploaded by:
kapor

Views: 1312

Downloads :
0

University:
Dhirubhai Ambani Institute of Information and Communication Technology

Subject:
Electricity and Magnetism

Upload date:
21/02/2013