dielectric

44
3 The effect of dielectric and magnetic media on electric and magnetic fields 3.1 Polarization The terrestrial environment is characterized by dielectric media (e.g., air, water) which are, for the most part, electrically neutral, since they are made up of neutral atoms and molecules. However, if these atoms and molecules are placed in an electric field they tend to polarize. Suppose that when a given neutral molecule is placed in an electric field E the centre of charge of its constituent electrons (whose total charge is -q ) is displaced by a distance -r with respect to the centre of charge of its constituent atomic nuclei. The dipole moment of the molecule is defined p = q r. If there are N such molecules per unit volume then the electric polarization P (i.e., the dipole moment per unit volume) is given by P = N p. More generally, P (r)= X i N i hp i i, (3.1) where hp i i is the average dipole moment of the ith type of molecule in the vicinity of point r, and N i is the average number of such molecules per unit volume at r. It is easily demonstrated that any divergence of the polarization field P (r) gives rise to an effective charge density ρ b in the medium. In fact, ρ b = -∇· P . (3.2) This charge density is attributable to bound charges (i.e., charges which arise from the polarization of neutral atoms), and is usually distinguished from the charge density ρ f due to free charges, which represents a net surplus or deficit of electrons in the medium. Thus, the total charge density ρ in the medium is ρ = ρ f + ρ b . (3.3) It must be emphasized that both terms in this equation represent real physical charge. Nevertheless, it is useful to make the distinction between bound and free charges, especially when it comes to working out the energy associated with electric fields in dielectric media. 84

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Page 1: Dielectric

3 The effect of dielectric and magnetic media on

electric and magnetic fields

3.1 Polarization

The terrestrial environment is characterized by dielectric media (e.g., air, water)which are, for the most part, electrically neutral, since they are made up of neutralatoms and molecules. However, if these atoms and molecules are placed in anelectric field they tend to polarize. Suppose that when a given neutral moleculeis placed in an electric field E the centre of charge of its constituent electrons(whose total charge is −q) is displaced by a distance −r with respect to the centreof charge of its constituent atomic nuclei. The dipole moment of the molecule isdefined p = q r. If there are N such molecules per unit volume then the electricpolarization P (i.e., the dipole moment per unit volume) is given by P = Np.More generally,

P (r) =∑

i

Ni〈pi〉, (3.1)

where 〈pi〉 is the average dipole moment of the ith type of molecule in the vicinityof point r, and Ni is the average number of such molecules per unit volume at r.

It is easily demonstrated that any divergence of the polarization field P (r)gives rise to an effective charge density ρb in the medium. In fact,

ρb = −∇·P . (3.2)

This charge density is attributable to bound charges (i.e., charges which arisefrom the polarization of neutral atoms), and is usually distinguished from thecharge density ρf due to free charges, which represents a net surplus or deficit ofelectrons in the medium. Thus, the total charge density ρ in the medium is

ρ = ρf + ρb. (3.3)

It must be emphasized that both terms in this equation represent real physicalcharge. Nevertheless, it is useful to make the distinction between bound andfree charges, especially when it comes to working out the energy associated withelectric fields in dielectric media.

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Page 2: Dielectric

Gauss’ law takes the differential form

∇·E =ρ

ε0=

ρf + ρb

ε0. (3.4)

This expression can be rearranged to give

∇·D = ρf , (3.5)

whereD = ε0E + P (3.6)

is termed the electric displacement, and has the same dimensions as P (dipolemoment per unit volume). The divergence theorem tells us that

S

D ·dS =

V

ρf dV. (3.7)

In other words, the flux of D out of some closed surface S is equal to the totalfree charge enclosed within that surface. Unlike the electric field E (which is theforce acting on unit charge) or the polarization P (the dipole moment per unitvolume), the electric displacement D has no clear physical meaning. The onlyreason for introducing it is that it enables us to calculate fields in the presenceof dielectric materials without first having to know the distribution of polarizedcharges. However, this is only possible if we have a constitutive relation connectingE and D. It is conventional to assume that the induced polarization P is directlyproportional to the electric field E, so that

P = ε0χeE, (3.8)

where χe is termed the electric susceptibility of the medium. It follows that

D = ε0εE, (3.9)

whereε = 1 + χe (3.10)

is termed the dielectric constant or relative permittivity of the medium. (Likewise,ε0 is termed the permittivity of free space.) It follows from Eqs. (3.5) and (3.9)that

∇·E =ρf

ε0ε. (3.11)

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Page 3: Dielectric

Thus, the electric fields produced by free charges in a dielectric medium are anal-ogous to those produced by the same charges in a vacuum, except that they arereduced by a factor ε. This reduction can be understood in terms of a polarizationof the atoms or molecules of the dielectric medium that produces electric fieldsin opposition to that of given charge. One immediate consequence is that thecapacitance of a capacitor is increased by a factor ε if the empty space betweenthe electrodes is filled with a dielectric medium of dielectric constant ε (assumingthat fringing fields can be neglected).

It must be understood that Eqs. (3.8)–(3.11) are just an approximation whichis generally found to hold under terrestrial conditions (provided that the fields arenot too large) for isotropic media. For anisotropic media (e.g., crystals) Eq. (3.9)generalizes to

D = ε0 ε·E, (3.12)

where ε is a second rank tensor known as the dielectric tensor. For strong elec-tric fields D ceases to vary linearly with E. Indeed, for sufficiently strong elec-tric fields neutral molecules are disrupted and the whole concept of a dielectricmedium becomes meaningless.

3.2 Boundary conditions for E and D

When the space near a set of charges contains dielectric material of non-uniformdielectric constant then the electric field no longer has the same form as in vac-uum. Suppose, for example, that the space is occupied by two dielectric mediawhose uniform dielectric constants are ε1 and ε2. What are the matching condi-tions on E and D at the boundary between the two media?

Imagine a Gaussian pill-box enclosing part of the boundary surface betweenthe two media. The thickness of the pill-box is allowed to tend towards zero, sothat the only contribution to the outward flux of D comes from its flat faces.These faces are parallel to the bounding surface and lie in each of the two media.Their outward normals are dS1 (in medium 1) and dS2, where dS1 = −dS2.Assuming that there is no free charge inside the disk (which is reasonable in the

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Page 4: Dielectric

limit where the volume of the disk tends towards zero), then Eq. (3.7) yields

D1 ·dS1 + D2 ·dS2 = 0, (3.13)

where D1 is the electric displacement in medium 1 at the boundary with medium2, etc. The above equation can be rewritten

(D2 −D1)·n21 = 0, (3.14)

where n21 is the normal to the boundary surface, directed from medium 1 tomedium 2. If the fields and charges are non time varying then Maxwell’s equa-tions yield ∇∧E = 0, which give the familiar boundary condition (obtained byintegrating around a small loop which straddles the boundary surface)

(E2 −E1) ∧ n21 = 0. (3.15)

In other word, the normal component of the electric displacement and the tan-gential component of the electric field are both continuous across any boundarybetween two dielectric materials.

3.3 Boundary value problems with dielectrics - I

Consider a point charge q embedded in a semi-infinite dielectric ε1 a distance daway from a plane interface which separates the first medium from another semi-infinite dielectric ε2. The interface is assumed to coincide with the plane z = 0.We need to find solutions to the equations

ε1∇·E =ρ

ε0(3.16)

for z > 0,ε2∇·E = 0 (3.17)

for z < 0, and∇∧E = 0 (3.18)

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Page 5: Dielectric

AA

q q

P

/

/ d d

R12 R

ε ε12

z = 0

z ->

everywhere, subject to the boundary conditions at z = 0 that

ε1Ez(z = 0+) = ε2Ez(z = 0−), (3.19a)

Ex(z = 0+) = Ex(z = 0−), (3.19b)

Ey(z = 0+) = Ey(z = 0−). (3.19c)

In order to solve this problem we will employ a slightly modified form of thewell known method of images. Since ∇∧E = 0 everywhere, the electric field canbe written in terms of a scalar potential. So, E = −∇φ. Consider the regionz > 0. Let us assume that the scalar potential in this region is the same as thatobtained when the whole of space is filled with the dielectric ε1 and, in addition tothe real charge q at position A, there is a second charge q′ at the image positionA′ (see diagram). If this is the case then the potential at some point P in theregion z > 0 is given by

φ(z > 0) =1

4πε0ε1

(

q

R1+

q′

R2

)

, (3.20)

where R1 =√

ρ2 + (d− z)2 and R2 =√

ρ2 + (d + z)2, when written in terms ofcylindrical polar coordinates (ρ, ϕ, z). Note that the potential (3.20) clearly is a

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Page 6: Dielectric

solution of Eq. (3.16) in the region z > 0. It gives ∇·E = 0, with the appropriatesingularity at the position of the point charge q.

Consider the region z < 0. Let us assume that the scalar potential in thisregion is the same as that obtained when the whole of space is filled with thedielectric ε2 and a charge q′′ is located at the point A. If this is the case then thepotential in this region is given by

φ(z < 0) =1

4πε0ε2

q′′

R1. (3.21)

The above potential is clearly a solution of Eq. (3.17) in the region z < 0. It gives∇·E = 0, with no singularities.

It now remains to choose q′ and q′′ in such a manner that the boundaryconditions (3.19) are satisfied. The boundary conditions (3.19b) and (3.19c) areobviously satisfied if the scalar potential is continuous at the interface betweenthe two dielectric media:

φ(z = 0+) = φ(z = 0−). (3.22)

The boundary condition (3.19a) implies a jump in the normal derivative of thescalar potential across the interface:

ε1∂φ(z = 0+)

∂z= ε2

∂φ(z = 0−)

∂z. (3.23)

The first matching condition yields

q + q′

ε1=

q′′

ε2, (3.24)

whereas the second yieldsq − q′ = q′′. (3.25)

Here, use has been made of

∂z

(

1

R1

)

z=0

= −∂

∂z

(

1

R2

)

z=0

=d

(ρ2 + d2)3/2. (3.26)

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Page 7: Dielectric

Equations (3.24) and (3.25) imply that

q′ = −

(

ε2 − ε1ε2 + ε1

)

q, (3.27a)

q′′ =

(

2ε2ε2 + ε1

)

q. (3.27b)

The polarization charge density is given by ρb = −∇·P , However, insideeither dielectric P = ε0χeE, so ∇·P = ε0χe∇·E = 0, except at the point chargeq. Thus, there is zero polarization charge density in either dielectric medium. Atthe interface χe takes a discontinuous jump,

∆χe = ε1 − ε2. (3.28)

This implies that there is a polarization charge sheet on the interface betweenthe two dielectric media. In fact,

σpol = −(P2 − P1)·n21, (3.29)

where n21 is a unit normal to the interface pointing from medium 1 to medium2 (i.e., along the positive z-axis). Since

Pi = ε0(εi − 1)E = −ε0(εi − 1)∇φ (3.30)

in either medium, it is easy to demonstrate that

σpol = −q

ε2 − ε1ε1(ε2 + ε1)

d

(ρ2 + d2)3/2. (3.31)

In the limit ε2 � ε1, the dielectric ε2 behaves like a conducting medium (i.e.,E → 0 in the region z < 0), and the polarization surface charge density on theinterface approaches that obtained in the case when the plane z = 0 coincideswith a conducting surface.

The above method can clearly be generalized to deal with problems involvingmany point charges in the presence of many different dielectric media whoseinterfaces form parallel planes.

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Page 8: Dielectric

z = 0 z = b

E EE0 01

ε

3.4 Boundary value problems with dielectrics - II

Consider a plane slab of dielectric ε lying between z = 0 and z = b. Suppose thatthis slab is placed in a uniform z-directed electric field of strength E0. What isthe field strength E1 inside the slab?

Since there are no free charges and this is a one-dimensional problem, it isclear from Eq. (3.5) that the electric displacement D is the same in both thedielectric slab and the vacuum region which surrounds it. In the vacuum regionD = ε0E0, whereas D = ε0ε E1 in the dielectric. It follows that

E1 =E0

ε. (3.32)

In other words, the electric field inside the slab is reduced by polarization charges.As before, there is zero polarization charge density inside the dielectric. However,there is a uniform polarization charge sheet on both surfaces of the slab. It iseasily demonstrated that

σpol(z = b) = −σpol(z = 0) = ε0 E0ε− 1

ε. (3.33)

In the limit ε � 1, the slab acts like a conductor and E1 → 0.

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Let us now generalize this result. Consider a dielectric medium whose dielec-tric constant ε varies with z. The medium is assumed to be of finite extent andis surrounded by a vacuum, so that ε(z) → 1 as |z| → ∞. Suppose that thisdielectric is placed in a uniform z-directed electric field E0. What is the fieldE(z) inside the dielectric?

We know that the electric displacement inside the dielectric is given by D(z) =ε0 ε(z) E(z). We also know from Eq. (3.5) that, since there are no free chargesand this is a one-dimensional problem,

dD(z)

dz= ε0

d[ε(z)E(z)]

dz= 0. (3.34)

Furthermore, E(z) → E0 as |z| → ∞. It follows that

E(z) =E0

ε(z). (3.35)

Thus, the electric field is inversely proportional to the dielectric constant insidethe dielectric medium. The polarization charge density inside the dielectric isgiven by

ρb = ε0dE(z)

dz= ε0 E0

d

dz

[

1

ε(z)

]

. (3.36)

3.5 Boundary value problems with dielectrics - III

Suppose that a dielectric sphere of radius a and dielectric constant ε is placed ina z-directed electric field of strength E0 (in the absence of the sphere). What isthe electric field inside and around the sphere?

Since this is a static problem we can write E = −∇φ. There are no freecharges, so Eqs. (3.5) and (3.9) imply that

∇2φ = 0 (3.37)

everywhere. The boundary conditions (3.14) and (3.15) reduce to

ε∂φ

∂r

r=a−=

∂φ

∂r

r=a+

, (3.38a)

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Page 10: Dielectric

∂φ

∂θ

r=a−=

∂φ

∂θ

r=a+

. (3.38b)

Furthermore,φ(r, θ, ϕ) → −E0 r cos θ (3.39)

as r → 0. Here, (r, θ, ϕ) are spherical polar coordinates centred on the sphere.

Let us search for an axisymmetric solution, φ = φ(r, θ). Since the solutionsto Poisson’s equation are unique, we know that if we can find such a solutionwhich satisfies all of the boundary conditions then we can be sure that this is thecorrect solution. Equation (3.37) reduces to

1

r

∂2(rφ)

∂r2+

1

r2 sin θ

∂θ

(

sin θ∂φ

∂θ

)

= 0. (3.40)

Straightforward separation of the variables yields

φ(r, θ) =∞∑

l=0

(Al rl + Bl r

−(l+1))Pl(cos θ), (3.41)

where l is a non-negative integer, the Al and Bl are arbitrary constants, and Pl(x)is a solution to Legendre’s equation,

d

dx

[

(1− x2)dPl

dx

]

+ l(l + 1) Pl = 0, (3.42)

which is single-valued, finite, and continuous in the interval −1 ≤ x ≤ +1. Itcan be demonstrated that Eq. (3.42) only possesses such solutions when l takesan integer value. The Pl(x) are known as Legendre polynomials (since they arepolynomials of order l in x), and are specified by Rodrigues’ formula

Pl(x) =1

2ll!

dl

dxl(x2 − 1)l. (3.43)

Since Eq. (3.42) is a Sturm-Liouville type equation, and the Legendre polynomi-als satisfy Sturm-Liouville type boundary conditions at x = ±1, it immediately

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follows that the Pl(cos θ) are orthogonal functions which form a complete set inθ-space. The orthogonality relation can be written

∫ 1

−1

Pl′(x)Pl(x) dx =2

2l + 1δll′ . (3.44)

The Legendre polynomials form a complete set of angular functions, and itis easily demonstrated that the rl and the r−(l+1) form a complete set of radialfunctions. It follows that Eq. (3.41), with the Al and Bl unspecified, represents acompletely general axisymmetric solution of Eq. (3.37) which is well behaved inθ-space. We now need to find values of the Al and Bl which are consistent withthe boundary conditions.

Let us divide space into the regions r ≤ a and r > a. In the former region

φ(r, θ) =

∞∑

l=0

Al rl Pl(cos θ), (3.45)

where we have rejected the r−(l+1) radial solutions because they diverge unphys-ically as r → 0. In the latter region

φ(r, θ) =∞∑

l=0

(Bl rl + Cl r

−(l+1) )Pl(cos θ). (3.46)

However, it is clear from the boundary condition (3.39), and Eq. (3.43), that theonly non-vanishing Bl is B1 = −E0. This follows since P1(cos θ) = cos θ. Theboundary condition (3.38b) (which integrates to give φ(r = a−) = φ(r = a+) fora potential which is well behaved in θ-space) gives

A1 = −E0 +C1

a3, (3.47)

and

Al =Cl

a2l+1(3.48)

for l 6= 1. Note that it is appropriate to match the coefficients of the Pl(cos θ)since these functions are orthogonal. The boundary condition (3.38a) yields

εA1 = −E0 − 2C1

a3, (3.49)

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Page 12: Dielectric

and

ε l Al = −(l + 1)Cl

a2l+1(3.50)

for l 6= 1. Equations (3.48) and (3.50) give Al = Cl = 0 for l 6= 1. Equations (3.47)and (3.49) reduce to

A1 = −

(

3

2 + ε

)

E0, (3.51a)

C1 =

(

ε− 1

ε + 2

)

a3E0. (3.51b)

The solution is therefore

φ = −

(

3

2 + ε

)

E0 r cos θ (3.52)

for r ≤ a, and

φ = −E0 r cos θ +

(

ε− 1

ε + 2

)

E0a3

r2cos θ (3.53)

for r > a.

Equation (3.52) is the potential of a uniform z-directed electric field of strength

E1 =3

2 + εE0. (3.54)

Note that E1 < E0 provided that ε > 1. Thus, the electric field strength is re-duced inside the dielectric sphere due to partial shielding by polarization charges.Outside the sphere the potential is equivalent to that of the applied field E0 plusthe field of a point electric dipole, located at the origin and pointing in the z-direction, whose dipole moment is

p = 4πε0

(

ε− 1

ε + 2

)

a3E0. (3.55)

This dipole moment can be interpreted as the volume integral of the polarizationP over the sphere. The polarization is

P = ε0(ε− 1) E1 z = 3ε0

(

ε− 1

ε + 2

)

E0 z. (3.56)

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Page 13: Dielectric

Since the polarization is uniform there is zero polarization charge density insidethe sphere. However, there is a polarization charge sheet on the surface of thesphere whose density is given by σpol = P ·r (see Eq. (3.29) ). It follows that

σpol = 3ε0

(

ε− 1

ε + 2

)

E0 cos θ. (3.57)

The problem of a dielectric cavity of radius a in a dielectric medium withdielectric constant ε and with an applied electric field E0 parallel to the z-axiscan be treated in much the same manner as that of a dielectric sphere. In fact, itis easily demonstrated that the results for the cavity can be obtained from thosefor the sphere by making the transformation ε → 1/ε. Thus, the field inside thecavity is uniform, parallel to the z-axis, and of magnitude

E1 =3ε

2ε + 1E0. (3.58)

Note that E1 > E0 provided that ε > 1. The field outside the cavity is theoriginal field plus that of a z-directed dipole, located at the origin, whose dipolemoment is

p = −4πε0

(

ε− 1

2ε + 1

)

a3E0. (3.59)

Here, the negative sign implies that the dipole points in the opposite direction tothe external field.

3.6 The energy density within a dielectric medium

Consider a system of free charges embedded in a dielectric medium. The increasein the total energy when a small amount of free charge δρf is added to the systemis given by

δU =

φ δρf d3r, (3.60)

where the integral is taken over all space and φ(r) is the electrostatic potential.Here, it is assumed that the original charges and the dielectric are held fixed, so

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that no mechanical work is performed. It follows from Eq. (3.5) that

δU =

φ∇·δD d3r, (3.61)

where δD is the change in the electric displacement associated with the chargeincrement. Now the above equation can also be written

δU =

∇·(φ δD) d3r −

∇φ·δD d3r, (3.62)

giving

δU =

φ δD ·dS −

∇φ·δD d3r, (3.63)

where use has been made of Gauss’s theorem. If the dielectric medium is of finitespatial extent then we can neglect the surface term to give

δU = −

∇φ·δD d3r =

E ·δD d3r. (3.64)

This energy increment cannot be integrated unless E is a known function of D.Let us adopt the conventional approach and assume that D = ε0εE, where thedielectric constant ε is independent of the electric field. The change in energyassociated with taking the displacement field from zero to D(r) at all points inspace is given by

U =

D

0

δU =

D

0

E ·δD d3r, (3.65)

or

U =

∫ ∫ E

0

ε0ε δ(E2)

2d3r =

1

2

ε0εE2 d3r, (3.66)

which reduces to

U =1

2

E ·D d3r. (3.67)

Thus, the electrostatic energy density inside a dielectric is given by

W =E ·D

2. (3.68)

This is a standard result which is often quoted in textbooks. Nevertheless, it isimportant to realize that the above formula is only valid in dielectric media inwhich the electric displacement D varies linearly with the electric field E.

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3.7 The force density within a dielectric medium

Equation (3.67) was derived by considering a virtual process in which true chargesare added to a system of charges and dielectrics which are held fixed, so that nomechanical work is done against physical displacements. Let us now considera different virtual process in which the physical coordinates of the charges anddielectric are given a virtual displacement δr at each point in space, but no freecharges are added to the system. Since we are dealing with a conservative system,the energy expression (3.67) can still be employed, despite the fact that it wasderived in terms of another virtual process. The variation in the total electrostaticenergy δU when the system undergoes a virtual displacement δr is related to theelectrostatic force density f acting within the dielectric medium via

δU = −

f ·δr d3r. (3.69)

If the medium is moving with a velocity field u then the rate at which electrostaticenergy is drained from the E and D fields is given by

dU

dt= −

f ·u d3r. (3.70)

Let us now consider the energy increment due to both a change δρf in thefree charge distribution and a change δε in the dielectric constant, caused by thedisplacements. From Eq. (3.67)

δU =1

2ε0

[

D2δ(1/ε) + 2D ·δD/ε]

d3r, (3.71)

or

δU = −ε02

E2 δε d3r +

E ·δD d3r. (3.72)

Here, the first term represents the energy increment due to the change in dielec-tric constant associated with the virtual displacements, whereas the second termcorresponds to the energy increment caused by displacements of the free charges.The second term can be written

E ·δD d3r = −

∇φ·δD d3r =

φ∇·δD d3r =

φ δρf d3r, (3.73)

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where surface terms have been neglected. Thus, Eq. (3.72) implies that

dU

dt=

∫(

φ∂ρf

∂t−

ε02

E2 ∂ε

∂t

)

d3r. (3.74)

In order to arrive at an expression for the force density f we need to expressthe time derivatives ∂ρ/∂t and ∂ε/∂t in terms of the velocity field u. This can beachieved by adopting a dielectric equation of state; i.e., a relation which gives thedependence of the dielectric constant ε on the mass density ρm. Let us assumethat ε(ρm) is a known function. It follows that

Dt=

dρm

Dρm

Dt, (3.75)

whereD

Dt≡

∂t+ u·∇ (3.76)

is the total time derivative (i.e., the time derivative in a frame of reference which islocally co-moving with the dielectric.) The hydrodynamic equation of continuityof the dielectric is

∂ρm

∂t+∇·(ρmu) = 0, (3.77)

which implies thatDρm

Dt= −ρm∇·u. (3.78)

It follows that∂ε

∂t= −

dρmρm∇·u− u·∇ε. (3.79)

The conservation equation for the free charges is written

∂ρf

∂t+∇·(ρfu) = 0. (3.80)

Thus, we can express Eq. (3.74) in the form

dU

dt=

∫[

−φ∇·(ρfu) +ε02

E2 dε

dρmρm∇·u +

(ε02

E2∇ε)

·u

]

d3r. (3.81)

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Integrating the first term by parts and neglecting any surface contributions, weobtain

φ∇·(ρfu) d3r =

ρf ∇φ·u d3r. (3.82)

Likewise,∫

ε02

E2 dε

dρmρm∇·u d3r = −

ε02∇

(

E2 dε

dρmρm

)

·u d3r. (3.83)

Thus, Eq. (3.81) becomes

dU

dt=

∫[

−ρfE +ε02

E2∇ε−ε02∇

(

E2 dε

dρmρm

)]

· u d3r. (3.84)

Comparing with Eq. (3.70), we see that the force density inside the dielectric isgiven by

f = ρfE −ε02

E2∇ε +ε02∇

(

E2 dε

dρmρm

)

. (3.85)

The first term in the above equation is the standard electrostatic force density.The second term represents a force which appears whenever an inhomogeneousdielectric is placed in an electric field. The last term, known as the electrostrictionterm, gives a force acting on a dielectric in an inhomogeneous electric field. Notethat the magnitude of the electrostriction force depends explicitly on the dielectricequation of state of the material, through dε/dρm. The electrostriction term giveszero net force acting on any finite region of dielectric if we can integrate over alarge enough portion of the dielectric that its extremities lie in a field free region.For this reason the term is frequently omitted, since in the calculation of the totalforces acting on dielectric bodies it usually does not contribute. Note, however,that if the electrostriction term is omitted an incorrect pressure variation withinthe dielectric is obtained, even though the total force is given correctly.

3.8 The Clausius-Mossotti relation

Let us now investigate what a dielectric equation of state actually looks like.Suppose that a dielectric medium is made up of identical molecules which develop

100

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a dipole momentp = αε0E (3.86)

when placed in an electric field E. The constant α is called the molecular polar-izability. If N is the number density of such molecules then the polarization ofthe medium is

P = Np = Nαε0E, (3.87)

or

P =NA ρmα

Mε0E, (3.88)

where ρm is the mass density, NA is Avogadro’s number, and M is the molecularweight. But, how does the electric field experienced by an individual moleculerelate to the average electric field in the medium? This is not a trivial questionsince we expect the electric field to vary strongly (on atomic length-scales) insidethe dielectric.

Suppose that the dielectric is polarized with a mean electric field E0 which isuniform (on macroscopic length-scales) and directed along the z-axis. Considerone of the molecules which constitute the dielectric. Let us draw a sphere of ra-dius a about this particular molecule. This is intended to represent the boundarybetween the microscopic and the macroscopic range of phenomena affecting themolecule. We shall treat the dielectric outside the sphere as a continuous mediumand the dielectric inside the sphere as a collection of polarized molecules. Ac-cording to Eq. (3.29) there is a polarization surface charge of magnitude

σpol = −P cos θ (3.89)

on the inside of the sphere, where (r, θ, ϕ) are spherical polar coordinates, andP = P z = ε0(ε − 1)E0 z is the uniform polarization of the dielectric. Themagnitude of Ez at the molecule due to the surface charge is

Ez = −1

4πε0

σpol cos θ

a2dS, (3.90)

where dS = 2πa2 sin θ dθ is a surface element of the sphere. It follows that

Ez =P

2ε0

∫ π

0

cos2 θ sin θ dθ =P

3ε0. (3.91)

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It is easily demonstrated that Eθ = Eϕ = 0 at the molecule. Thus, the field atthe molecule due to the surface charges on the sphere is

E =P

3ε0. (3.92)

The field due to the individual molecules within the sphere is obtained bysumming over the dipole fields of these molecules. The electric field at a distancer from a dipole p is

E = −1

4πε0

[

p

r3−

3(p·r)r

r5

]

. (3.93)

It is assumed that the dipole moment of each molecule within the sphere is thesame, and also that the molecules are evenly distributed throughout the sphere.This being the case, the value of Ez at the molecule due to all of the othermolecules within in the sphere,

Ez = −1

4πε0

mols

[

pz

r3−

3(px xz + py yz + pz z2)

r5

]

, (3.94)

is zero, since∑

mols

x2 =∑

mols

y2 =∑

mols

z2 =1

3

mols

r2 (3.95)

and∑

mols

xy =∑

mols

yz =∑

mols

zx = 0. (3.96)

It is easily seen that Eθ = Eϕ = 0. Hence, the electric field at the molecule dueto the other molecules within the sphere vanishes.

It is clear that the net electric field seen by an individual molecule is

E = E0 +P

3ε0. (3.97)

This is larger than the average electric field E0 in the dielectric. The aboveanalysis indicates that this effect is ascribable to the long range (rather than the

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short range) interactions of the molecule with the other molecules in the medium.Making use of Eq. (3.88) and the definition P = ε0(ε− 1)E0, we obtain

ε− 1

ε + 2=

NA ρmα

3M. (3.98)

This is called the Clausius-Mossotti relation. This formula is found to work prettywell for a wide class of dielectric liquids and gases. The Clausius-Mossotti relationyields

dρm=

(ε− 1)(ε + 2)

3ρm. (3.99)

3.9 Dielectric liquids in electrostatic fields

Consider the behaviour of an uncharged dielectric liquid placed in an electrostaticfield. If p is the pressure in the liquid when in equilibrium with the electrostaticforce density f , then force balance requires that

∇p = f . (3.100)

It follows from Eq. (3.85) that

∇p = −ε02

E2∇ε +ε02∇

(

E2 dε

dρmρm

)

=ε0ρm

2∇

(

E2 dε

dρm

)

. (3.101)

We can integrate this equation to give

∫ p2

p1

dp

ρm=

ε02

([

E2 dε

dρm

]

2

[

E2 dε

dρm

]

1

)

, (3.102)

where 1 and 2 refer to two general points in the liquid. Here, it is assumed thatthe liquid possesses an equation of state, so that p = p(ρm). If the liquid isessentially incompressible (ρm ' constant) then

p2 − p1 =ε0ρm

2

[

E2 dε

dρm

]2

1

. (3.103)

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Finally, if the liquid obeys the Clausius-Mossotti relation then

p2 − p1 =

[

ε0E2

2

(ε− 1)(ε + 2)

3

]2

1

. (3.104)

According to Eqs. (3.54) and (3.104), if a sphere of dielectric liquid is placedin a uniform electric field E0 then the pressure inside the liquid takes the constantvalue

p =3

2ε0E

20

ε− 1

ε + 2. (3.105)

It is clear that the electrostatic forces acting on the dielectric are all concentratedat the edge of the sphere and are directed radially inwards; i.e., the dielectric iscompressed by the external electric field. This is a somewhat surprising resultsince the electrostatic forces acting on a rigid conducting sphere are concentratedat the edge of the sphere but are directed radially outwards. We might expectthese two cases to give the same result in the limit ε → ∞. The reason thatthis does not occur is because a dielectric liquid is slightly compressible and is,therefore, subject to an electrostriction force. There is no electrostriction forcefor the case of a completely rigid body. In fact, the force density inside a rigiddielectric (for which ∇·u = 0) is given by Eq. (3.85) with the third term (theelectrostriction term) missing. It is easily seen that the force exerted by an electricfield on a rigid dielectric is directed outwards and approaches that exerted on arigid conductor in the limit ε → 0.

As is well known, when a pair of charged (parallel plane) capacitor plates aredipped into a dielectric liquid the liquid is drawn up between the plates to someextent. Let us examine this effect. We can, without loss of generality, assumethat the transition from dielectric to vacuum takes place in a continuous manner.Consider the electrostatic pressure difference between a point A lying just abovethe surface of the liquid in between the plates and a point B lying just above thesurface of the liquid well away from the capacitor where E = 0. The pressuredifference is given by

pA − pB = −

∫ B

A

f ·dl. (3.106)

Note, however, that the Clausius-Mossotti relation yields dε/dρm = 0 at both Aand B, since ε = 1 in a vacuum (see Eq. (3.99) ). Thus, it is clear from Eq. (3.85)

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that the electrostriction term makes no contribution to the line integral (3.106).It follows that

pA − pB =ε02

∫ B

A

E2∇ε·dl. (3.107)

The only contribution to this integral comes from the vacuum/dielectric interfacein the vicinity of point A (since ε is constant inside the liquid, and E = 0 in thevicinity of point B). Suppose that the electric field at point A has normal andtangential (to the surface) components En and Et, respectively. Making use ofthe boundary conditions that Et and εEn are constant across a vacuum/dielectricinterface, we obtain

pA − pB =ε02

[

E 2t (ε− 1) + ε2E 2

n (ε)

∫ ε

1

ε2

]

, (3.108)

giving

pA − pB =ε0 (ε− 1)

2

[

E 2t +

E 2n

ε

]

. (3.109)

This electrostatic pressure difference can be equated to the hydrostatic pressuredifference ρm g h to determine the height h that the liquid rises between the plates.At first sight, the above analysis appears to suggest that the dielectric liquidis drawn upward by a surface force acting on the vacuum/dielectric interfacein the region between the plates. In fact, this is far from being the case. Abrief examination of Eq. (3.104) shows that this surface force is actually directeddownwards. According to Eq. (3.85), the force which causes the liquid to risebetween the plates is a volume force which develops in the region of non-uniformelectric field at the base of the capacitor, where the field splays out betweenthe plates. Thus, although we can determine the height to which the fluid risesbetween the plates without reference to the electrostriction force, it is, somewhatparadoxically, this force which is actually responsible for supporting the liquidagainst gravity.

Let us consider another paradox concerning the electrostatic forces exerted ina dielectric medium. Suppose that we have two charges embedded in a uniformdielectric ε. The electric field generated by each charge is the same as that invacuum, except that it is reduced by a factor ε. Therefore, we expect that theforce exerted by one charge on another is the same as that in vacuum, except

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that it is also reduced by a factor ε. Let us examine how this reduction inforce comes about. Consider a simple example. Suppose that we take a parallelplate capacitor and insert a block of solid dielectric between the plates. Suppose,further, that there is a small vacuum gap between the faces of the block and eachof the capacitor plates. Let ±σ be the surface charge densities on each of thecapacitor plates, and let±σp be the polarization charge densities which develop onthe outer faces of the intervening dielectric block. The two layers of polarizationcharge produce equal and opposite electric fields on each plate, and their effectstherefore cancel each other. Thus, from the point of view of electrical interactionalone there would appear to be no change in the force exerted by one capacitorplate on the other when a dielectric slab is placed between them (assuming thatσ remains constant during this process). That is, the force per unit area (whichis attractive) remains

fs =σ2

2ε0. (3.110)

However, in experiments in which a capacitor is submerged in a dielectric liquidthe force per unit area exerted by one plate on another is observed to decrease to

fs =σ2

2ε0ε. (3.111)

This apparent paradox can be explained by taking into account the differencein liquid pressure in the field filled space between the plates and the field freeregion outside the capacitor. This pressure difference is balanced by internalelastic forces in the case of the solid dielectric discussed earlier, but is transmittedto the plates in the case of the liquid. We can compute the pressure differencebetween a point A on the inside surface of one of the capacitor plates and a pointB on the outside surface of the same plate using Eq. (3.107). If we neglect endeffects then the electric field is normal to the plates in the region between theplates and is zero everywhere else. Thus, the only contribution to the line integral(3.107) comes from the plate/dielectric interface in the vicinity of point A. UsingEq. (3.109), we find that

pA − pB =ε02

(

1−1

ε

)

E2 =σ2

2ε0

(

1−1

ε

)

, (3.112)

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Page 24: Dielectric

where E is the normal field strength between the plates in the absence of dielectric.The sum of this pressure force and the purely electrical force (3.110) yields a netattractive force per unit area

fs =σ2

2ε0ε(3.113)

acting between the plates. Thus, any decrease in the forces exerted by chargeson one another when they are immersed or embedded in some dielectric mediumcan only be understood in terms of mechanical forces transmitted between thesecharges by the medium itself.

3.10 Magnetization

All matter is built up out of atoms, and each atom consists of electrons in mo-tion. The currents associated with this motion are termed atomic currents. Eachatomic current is a tiny closed circuit of atomic dimensions, and may thereforebe appropriately described as a magnetic dipole. If the atomic currents of a givenatom all flow in the same plane then the atomic dipole moment is directed normalto the plane (in the sense given by the right-hand rule) and its magnitude is theproduct of the total circulating current and the area of the current loop. Moregenerally, if j(r) is the atomic current density at the point r then the magneticmoment of the atom is

m =1

2

r ∧ j d3r, (3.114)

where the integral is over the volume of the atom. If there are N such atoms ormolecules per unit volume then the magnetization M (i.e., the magnetic dipolemoment per unit volume) is given by M = Nm. More generally,

M(r) =∑

i

Ni〈mi〉, (3.115)

where 〈mi〉 is the average magnetic dipole moment of the ith type of molecule inthe vicinity of point r, and Ni is the average number of such molecules per unitvolume at r.

Consider a general medium which is made up of molecules which are polar-izable and possess a net magnetic moment. It is easily demonstrated that any

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circulation in the magnetization field M(r) gives rise to an effective current den-sity jm in the medium. In fact,

jm = ∇∧M . (3.116)

This current density is called the magnetization current density, and is usuallydistinguished from the true current density jt, which represents the convectionof free charges in the medium. In fact, there is a third type of current called apolarization current, which is due to the apparent convection of bound charges.It is easily demonstrated that the polarization current density jp is given by

jp =∂P

∂t. (3.117)

Thus, the total current density j in the medium is given by

j = jt +∇∧M +∂P

∂t. (3.118)

It must be emphasized that all terms on the right-hand side of this equationrepresent real physical currents, although only the first term is due to the motionof real charges (over more than atomic dimensions).

The Ampere-Maxwell equation takes the form

∇∧B = µ0 j + µ0ε0∂E

∂t, (3.119)

which can also be written

∇∧B = µ0 jt + µ0∇∧M + µ0∂D

∂t, (3.120)

where use has been made of the definition D = ε0E + P . The above expressioncan be rearranged to give

∇∧H = jt +∂D

∂t, (3.121)

where

H =B

µ0−M (3.122)

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is termed the magnetic intensity, and has the same dimensions as M (i.e., mag-netic dipole moment per unit volume). In a steady-state situation, Stokes’s the-orem tell us that

C

H ·dl =

S

jt ·dS. (3.123)

In other words, the line integral of H around some closed curve is equal tothe flux of true current through any surface attached to that curve. Unlike themagnetic field B (which specifies the force e v ∧B acting on a charge e movingwith velocity v) or the magnetization M (the magnetic dipole moment per unitvolume), the magnetic intensity H has no clear physical meaning. The onlyreason for introducing it is that it enables us to calculate fields in the presence ofmagnetic materials without first having to know the distribution of magnetizationcurrents. However, this is only possible if we possess a constitutive relationconnecting B and H.

3.11 Magnetic susceptibility and permeability

In a large class of materials there exists an approximately linear relationshipbetween M and H. If the material is isotropic then

M = χmH, (3.124)

where χm is called the magnetic susceptibility. If χm is positive the material iscalled paramagnetic, and the magnetic field is strengthened by the presence ofthe material. If χm is negative then the material is diamagnetic and the magneticfield is weakened in the presence of the material. The magnetic susceptibilitiesof paramagnetic and diamagnetic materials are generally extremely small. A fewsample values are given in Table 1.10

A linear relationship between M and H also implies a linear relationshipbetween B and H. In fact, we can write

B = µH, (3.125)

10Data obtained from the Handbook of Chemistry and Physics, Chemical Rubber Company

Press, Baca Raton, FL

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Material χm

Aluminium 2.3× 10−5

Copper −0.98× 10−5

Diamond −2.2× 10−5

Tungsten 6.8× 10−5

Hydrogen (1 atm) −0.21× 10−8

Oxygen (1 atm) 209.0× 10−8

Nitrogen (1 atm) −0.50× 10−8

Table 1: Magnetic susceptibilities of some paramagnetic and diamagnetic mate-rials at room temperature

whereµ = µ0(1 + χm) (3.126)

is termed the magnetic permeability of the material in question. (Likewise, µ0

is termed the permeability of free space.) It is clear from Table 1 that the per-meabilities of common diamagnetic and paramagnetic materials do not differsubstantially from that of free space. In fact, to all intents and purposes themagnetic properties of such materials can be safely neglected (i.e., µ = µ0).

3.12 Ferromagnetism

There is, however, a third class of magnetic materials called ferromagnetic ma-terials. Such materials are characterized by a possible permanent magnetization,and generally have a profound effect on magnetic fields (i.e., µ/µ0 � 1). Unfor-tunately, ferromagnetic materials do not exhibit a linear dependence between M

and H or B and H, so that we cannot employ Eqs. (3.124) and (3.125) withconstant values of χm and µ. It is still expedient to use Eq. (3.125) as the defi-nition of µ, with µ = µ(H), however this practice can lead to difficulties undercertain circumstances. The permeability of a ferromagnetic material, as definedby Eq. (3.125), can vary through the entire range of possible values from zeroto infinity, and may be either positive or negative. The most sensible approachis to consider each problem involving ferromagnetic materials separately, try to

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Figure 2: Magnetization curve and relative permeability of commercial iron (an-nealed)

determine which region of the B-H diagram is important for the particular casein hand, and then make approximations appropriate to this region.

First, let us consider an unmagnetized sample of ferromagnetic material. Ifthe magnetic intensity, which is initially zero, is increased monotonically, thenthe B-H relationship traces out a curve such as that shown in Fig. 2. This iscalled a magnetization curve. It is evident that the permeabilities µ derived fromthe curve (according to the rule µ = B/H) are always positive, and show a widerange of values. The maximum permeability occurs at the “knee” of the curve.In some materials this maximum permeability is as large as 105 µ0. The reasonfor the knee in the curve is that the magnetization M reaches a maximum valuein the material, so that

B = µ0(H + M) (3.127)

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Page 29: Dielectric

-H H

B

R

C

Figure 3: Typical hysteresis loop of a ferromagnetic material

continues to increase at large H only because of the µ0H term. The maximumvalue of M is called the saturation magnetization of the material.

Next, consider a ferromagnetic sample magnetized by the above procedure.If the magnetic intensity H is decreased, the B-H relation does not follow backdown the curve of Fig. 2, but instead moves along a new curve, shown in Fig. 3,to the point R. The magnetization, once established, does not disappear withthe removal of H. In fact, it takes a reversed magnetic intensity to reduce themagnetization to zero. If H continues to build up in the reversed direction, thenM (and hence B) becomes increasingly negative. Finally, when H increasesagain the operating point follows the lower curve of Fig. 3. Thus, the B-H curvefor increasing H is quite different to that for decreasing H. This phenomenon isknown as hysteresis.

The curve of Fig. 3 is called the hysteresis loop of the material in question. Thevalue of B at the point R is called the retentivity or remanence. The magnitudeof H at the point C is called the coercivity. It is evident that µ is negative in the

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second and fourth quadrants of the diagram and positive in the first and thirdquadrants. The shape of the hysteresis loop depends not only on the nature of theferromagnetic material but also on the maximum value of H to which the materialis subjected. However, once this maximum value, Hmax, becomes sufficient toproduce saturation in the material the hysteresis loop does not change shape withany further increase in Hmax.

Ferromagnetic materials are used either to channel magnetic flux (e.g., aroundtransformer circuits) or as sources of magnetic field (permanent magnets). Foruse as a permanent magnet, the material is first magnetized by placing it in astrong magnetic field. However, once the magnet is removed from the externalfield it is subject to a demagnetizing H. Thus, it is vitally important that apermanent magnet should possess both a large remanence and a large coercivity.As will become clear later on, it is generally a good idea for the ferromagneticmaterials used to channel magnetic flux around transformer circuits to possesssmall remanences and small coercivities.

3.13 Boundary conditions for B and H

What are the matching conditions for B and H at the boundary between twomedia? The governing equations for a steady state situation are

∇·B = 0, (3.128)

and∇∧H = jt. (3.129)

Integrating Eq. (3.128) over a Gaussian pill-box enclosing part of the boundarysurface between the two media gives

(B2 −B1)·n21 = 0, (3.130)

where n21 is the unit normal to this surface directed from medium 1 to medium2. Integrating Eq. (3.129) around a small loop which straddles the boundarysurface yields

(H2 −H1) ∧ n21 = 0, (3.131)

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assuming that there is no true current sheet flowing in this surface. In general,there is a magnetization current sheet flowing in the boundary surface whosedensity is given by

Jm = n21 ∧ (M2 −M1), (3.132)

where M1 is the magnetization in medium 1 at the boundary, etc. It is clear thatthe normal component of the magnetic field and the tangential component of themagnetic intensity are both continuous across any boundary between magneticmaterials.

3.14 Permanent ferromagnets

Let us consider the magnetic field generated by a distribution of permanent ferro-magnets. Let us suppose that the magnets in question are sufficiently “hard” thattheir magnetization is essentially independent of the applied field for moderatefield strengths. Such magnets can be treated as if they contain a fixed, specifiedmagnetization M(r).

Let us assume that there are no true currents in the problem, so that jt = 0.Let us also assume that we are dealing with a steady state situation. Under thesecircumstances Eq. (3.121) reduces to

∇∧H = 0. (3.133)

It follows that we can writeH = −∇φm, (3.134)

where φm is called the magnetic scalar potential. Now, we know that

∇·B = µ0∇·(H + M) = 0. (3.135)

Equations (3.134) and (3.135) combine to give

∇2φm = −ρm, (3.136)

whereρm = −∇·M . (3.137)

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Thus, the magnetostatic field H is determined by Poisson’s equation. We canthink of ρm as an effective magnetic charge density. Of course, this magneticcharge has no physical reality. We have only introduced it in order to makethe problem of the steady magnetic field generated by a set of permanent mag-nets look formally the same as that of the steady electric field generated by adistribution of charges.

The unique solution of Poisson’s equation, subject to sensible boundary con-ditions at infinity, is well known:

φm(r) =1

ρm(r′)

|r − r′|d3r′. (3.138)

This yields

φm(r) = −1

∇′ ·M(r′)

|r − r′|d3r′. (3.139)

If the magnetization field M(r) is well behaved and localized we can integrateby parts to obtain

φm(r) =1

M(r′)·∇′(

1

|r − r′|

)

d3r′. (3.140)

Now

∇′(

1

|r − r′|

)

= −∇

(

1

|r − r′|

)

, (3.141)

so our expression for the magnetic potential can be written

φm(r) = −1

4π∇·

M(r′)

|r − r′|d3r′. (3.142)

Far from the region of non-vanishing magnetization the potential reduces to

φm(r) ' −∇

(

1

4πr

)

·

M(r′) d3r′ 'm· r

4πr3, (3.143)

where m =∫

M d3r is the total magnetic moment of the distribution. Thisis the scalar potential of a dipole. Thus, an arbitrary localized distribution of

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magnetization asymptotically produces a dipole magnetic field whose strength isdetermined by the net magnetic moment of the distribution.

It is often a good approximation to treat the magnetization field M(r) as adiscontinuous quantity. In other words, M(r) is specified throughout the “hard”ferromagnets in question, and suddenly falls to zero at the boundaries of thesemagnets. Integrating Eq. (3.137) over a Gaussian pill-box which straddles oneof these boundaries leads to the conclusion that there is an effective magneticsurface charge density,

σm = n·M , (3.144)

on the surface of the ferromagnets, where M is the surface magnetization, andn is a unit outward directed normal to the surface. Under these circumstancesEq. (3.139) yields

φm(r) = −1

V

∇′ ·M(r′)

|r − r′|d3r′ +

1

S

M(r′)·dS′

|r − r′|, (3.145)

where V represents the volume occupied by the magnets and S is the boundingsurface to V . Here, dS is an outward directed volume element to S. It is clearthat Eq. (3.145) consists of a volume integral involving the volume magneticcharges ρm = −∇·M and a surface integral involving the surface magnetic chargesσm = n·M . If the magnetization is uniform throughout the volume V then thefirst term in the above expression vanishes and only the surface integral makes acontribution.

We can also write B = ∇ ∧A in order to satisfy ∇·B = 0 automatically. Itfollows from Eqs. (3.121) and (3.122) that

∇∧H = ∇∧ (B/µ0 −M) = 0, (3.146)

which gives∇2A = −µ0 jm, (3.147)

since jm = ∇ ∧ M . The unique solution to Eq. (3.147), subject to sensibleboundary conditions at infinity, is very well known:

A(r) =µ0

jm(r′)

|r − r′|d3r′. (3.148)

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Thus,

A(r) =µ0

∇′ ∧M(r′)

|r − r′|d3r′. (3.149)

If the magnetization field is discontinuous it is necessary to add a surfaceintegral to the above expression. It is straightforward to show that

A(r) =µ0

V

∇′ ∧M(r′)

|r − r′|d3r′ +

µ0

S

M(r′) ∧ dS′

|r − r′|. (3.150)

It is clear that the above expression consists of a volume integral involving thevolume magnetization currents jm = ∇ ∧ M and a surface integral involvingthe surface magnetization currents Jm = M ∧ n (see Eq. (3.132) ). If the mag-netization field is uniform throughout V then only the surface integral makes acontribution.

3.15 A uniformly magnetized sphere

Consider a sphere of radius a, with a uniform permanent magnetization M =M0 z, surrounded by a vacuum region. The simplest way of solving this problemis in terms of the scalar magnetic potential introduced in Eq. (3.134). FromEqs. (3.136) and (3.137), it is clear that φm satisfies Laplace’s equation,

∇2φm = 0, (3.151)

since there is zero volume magnetic charge density in a vacuum or a uniformlymagnetized magnetic medium. However, according to Eq. (3.144), there is amagnetic surface charge density,

σm = r ·M = M0 cos θ, (3.152)

on the surface of the sphere. One of the matching conditions at the surface ofthe sphere is that the tangential component of H must be continuous. It followsfrom Eq. (3.134) that the scalar magnetic potential must be continuous at r = a,so that

φm(r = a+) = φm(r = a−). (3.153)

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Integrating Eq. (3.136) over a Gaussian pill-box straddling the surface of thesphere yields

[

∂φm

∂r

]r=a+

r=a−

= −σm = −M0 cos θ. (3.154)

In other words, the magnetic charge sheet on the surface of the sphere gives riseto a discontinuity in the radial gradient of the magnetic scalar potential at r = a.

The most general axisymmetric solution to Eq. (3.151) which satisfies physicalboundary conditions at r = a and r = ∞ is

φm(r, θ) =

∞∑

l=0

Al rlPl(cos θ) (3.155)

for r < a, and

φm(r, θ) =

∞∑

l=0

Bl r−(l+1)Pl(cos θ) (3.156)

for r ≥ a. The boundary condition (3.153) yields

Bl = Al a2l+1 (3.157)

for all l. The boundary condition (3.154) gives

−(l + 1) Bl

al+2− lAl a

l−1 = −M0 δl1 (3.158)

for all l, since Pl(cos θ) = cos θ. It follows that

Al = Bl = 0 (3.159)

for l 6= 1, and

A1 =M0

3, (3.160a)

B1 =M0a

3

3. (3.160b)

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Thus,

φm(r, θ) =M0a

2

3

r

a2cos θ (3.161)

for r < a, and

φm(r, θ) =M0a

2

3

a

r2cos θ (3.162)

for r ≥ a. Since there is a uniqueness theorem associated with Poisson’s equation,we can be sure that this axisymmetric potential is the only solution to the problemwhich satisfies physical boundary conditions at r = 0 and infinity.

In the vacuum region outside the sphere

B = µ0H = −µ0∇φm. (3.163)

It is easily demonstrated from Eq. (3.162) that

B(r > a) =µ0

[

−m

r3+

3(m·r) r

r5

]

, (3.164)

where

m =4

3πa3M . (3.165)

This, of course, is the magnetic field of a magnetic dipole m. Not surprisingly,the net dipole moment of the sphere is equal to the integral of the magnetizationM (which is the dipole moment per unit volume) over the volume of the sphere.

Inside the sphere we have H = −∇φm and B = µ0(H + M), giving

H = −M

3, (3.166)

and

B =2

3µ0M . (3.167)

Thus, both the H and B fields are uniform inside the sphere. Note that themagnetic intensity is oppositely directed to the magnetization. In other words,the H field acts to demagnetize the sphere. How successful it is at achieving

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operating pointBR

−µ0

µ= -2

HC

B-H curve

-H

B

00

µ0

Figure 4: Schematic demagnetization curve for a permanent magnet

this depends on the shape of the hysteresis curve in the negative H and posi-tive B quadrant. This curve is sometimes called the demagnetization curve ofthe magnetic material which makes up the sphere. Figure 4 shows a schematicdemagnetization curve. The curve is characterized by two quantities: the reten-tivity BR (i.e., the residual magnetic field strength at zero magnetic intensity)and the coercivity µ0Hc (i.e., the negative magnetic intensity required to demag-netize the material: this quantity is conventionally multiplied by µ0 to give it theunits of magnetic field strength). The operating point (i.e., the values of B andµ0H inside the sphere) is obtained from the intersection of the demagnetizationcurve and the curve B = µH. It is clear from Eqs. (3.166) and (3.167) that

µ = −2 µ0 (3.168)

for a uniformly magnetized sphere in the absence of external fields. The magne-tization inside the sphere is easily calculated once the operating point has beendetermined. In fact, M0 = B − µ0H. It is clear from Fig. 4 that for a magneticmaterial to be a good permanent magnet it must possess both a large retentivityand a large coercivity. A material with a large retentivity but a small coercivity

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is unable to retain a significant magnetization in the absence of a strong externalmagnetizing field.

3.16 A soft iron sphere in a uniform magnetic field

The opposite extreme to a “hard” ferromagnetic material, which can maintain alarge remnant magnetization in the absence of external fields, is a “soft” ferro-magnetic material, for which the remnant magnetization is relatively small. Letus consider a somewhat idealized situation in which the remnant magnetizationis negligible. In this situation there is no hysteresis, so the B-H relation for thematerial reduces to

B = µ(B) H, (3.169)

where µ(B) is a single valued function. The most commonly occurring “soft”ferromagnetic material is soft iron (i.e., annealed, low impurity iron).

Consider a sphere of soft iron placed in an initially uniform external fieldB0 = B0 z. The µ0H and B fields inside the sphere are most easily obtainedby taking the solutions (3.166) and (3.167) (which are still valid in this case),and superimposing on them the uniform field B0. We are justified in doing thisbecause the equations which govern magnetostatic problems are linear. Thus,inside the sphere we have

µ0H = B0 −1

3µ0M , (3.170a)

B = B0 +2

3µ0M . (3.170b)

Combining Eqs. (3.169) and (3.170) yields

µ0M = 3

(

µ− µ0

µ + 2µ0

)

B0, (3.171)

with

B =

(

µ + 2µ0

)

B0, (3.172)

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where, in general, µ = µ(B). Clearly, for a highly permeable material (i.e.,µ/µ0 � 1, which is certainly the case for soft iron) the magnetic field strengthinside the sphere is approximately three times that of the externally applied field.In other words, the magnetic field is amplified inside the sphere.

The amplification of the magnetic field by a factor three in the high perme-ability limit is specific to a sphere. It can be shown that for elongated objects(e.g., rods), aligned along the direction of the external field, the amplificationfactor can be considerably larger than this.

It is important to realize that the magnetization inside a ferromagnetic mate-rial cannot increase without limit. The maximum possible value of M is called thesaturation magnetization, and is usually denoted Ms. Most ferromagnetic mate-rials saturate when they are placed in external magnetic fields whose strengthsare greater than, or of order, one tesla. Suppose that our soft iron sphere firstattains the saturation magnetization when the unperturbed external magneticfield strength is Bs. It follows from (3.170b) and (3.171) (with µ � µ0) that

B = B0 + 2Bs (3.173)

inside the sphere, for B0 > Bs. In this case, the field amplification factor is

B

B0= 1 + 2

Bs

B0. (3.174)

Thus, for B0 � Bs the amplification factor approaches unity. We conclude thatif a ferromagnetic material is placed in an external field which greatly exceedsthat required to cause saturation then the material effectively loses its magneticproperties, so that µ ' µ0. Clearly, it is very important to avoid saturating thesoft magnets used to channel magnetic flux around transformer circuits. This setsan upper limit on the magnetic field strengths which can occur in such circuits.

3.17 Magnetic shielding

There are many situations, particularly in experimental physics, where it is de-sirable to shield a certain region from magnetic fields. This can be achieved by

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surrounding the region in question by a material of high permeability. It is vitallyimportant that a material used as a magnetic shield does not develop a permanentmagnetization in the presence of external fields, otherwise the material itself maybecome a source of magnetic fields. The most effective commercially availablemagnetic shielding material is called Mumetal, and is an alloy of 5% Copper, 2%Chromium, 77% Nickel, and 16% Iron. The maximum permeability of Mumetalis about 105 µ0. This material also possesses a particularly low retentivity andcoercivity. Unfortunately, Mumetal is extremely expensive. Let us investigatehow much of this material is actually required to shield a given region from anexternal magnetic field.

Consider a spherical shell of magnetic shielding, made up of material of per-meability µ, placed in a formerly uniform magnetic field B0 = B0 z. Supposethat the inner radius of the shell is a and the outer radius is b. Since there areno free currents in the problem, we can write H = −∇φm. Furthermore, sinceB = µH and ∇·B = 0, it is clear that the magnetic scalar potential satisfiesLaplace’s equation, ∇2φm = 0, throughout all space. The boundary conditionsare that the potential must be well behaved at r = 0 and r → ∞, and also thatthe tangential and the normal components of H and B, respectively, must becontinuous at r = a and r = b. The boundary conditions on H merely imply thatthe scalar potential φm must be continuous at r = a and r = b. The boundaryconditions on B yield

µ0∂φm(r = a−)

∂r= µ

∂φm(r = a+)

∂r, (3.175a)

µ0∂φm(r = b+)

∂r= µ

∂φm(r = b−)

∂r. (3.175b)

Let us try the following test solution for the magnetic potential:

φm = −B0

µ0r cos θ +

α

r2cos θ (3.176)

for r > b,

φm =(

β r +γ

r2

)

cos θ (3.177)

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for b ≥ r ≥ a, andφm = δ r cos θ (3.178)

for r < a. This potential is certainly a solution of Laplace’s equation throughoutspace. It yields the uniform magnetic field B0 as r → ∞, and satisfies physicalboundary conditions at r = 0 and infinity. Since there is a uniqueness theoremassociated with Poisson’s equation, we can be certain that this potential is thecorrect solution to the problem provided that the arbitrary constants α, β, etc.can be adjusted in such a manner that the boundary conditions at r = a andr = b are also satisfied.

The continuity of φm at r = a and r = b requires that

β a +γ

a2= δ a, (3.179)

and

β b +γ

b2= −

B0

µ0b +

α

b2. (3.180)

The boundary conditions (3.175) yield

µ0 δ = µ

(

β −2γ

a3

)

, (3.181)

and

µ0

(

−B0

µ0−

b3

)

= µ

(

β −2γ

b3

)

. (3.182)

It follows that

µ0 α =

[

(2µ + µ0)(µ− µ0)

(2µ + µ0)(µ + 2µ0)− 2(a3/b3)(µ− µ0)2

]

(b3 − a3)B0,

(3.183a)

µ0 β = −

[

3(2µ + µ0)µ0

(2µ + µ0)(µ + 2µ0)− 2(a3/b3)(µ− µ0)2

]

B0, (3.183b)

µ0 γ = −

[

3(µ− µ0)µ0

(2µ + µ0)(µ + 2µ0)− 2(a3/b3)(µ− µ0)2

]

a3B0, (3.183c)

µ0 δ = −

[

9µµ0

(2µ + µ0)(µ + 2µ0)− 2(a3/b3)(µ− µ0)2

]

B0. (3.183d)

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Consider the limit of a thin, high permeability shell for which b = a + d,d/a � 1, and µ/µ0 � 1. In this limit, the field inside the shell is given by

B '3

2

µ0

µ

a

dB0. (3.184)

Thus, if µ ' 105µ0 for Mumetal, then we can reduce the magnetic field strengthinside the shell by almost a factor of 1000 using a shell whose thickness is only1/100 th of its radius. Clearly, a little Mumetal goes a long way! Note, however,that as the external field strength, B0, is increased, the Mumetal shell eventuallysaturates, and µ/µ0 gradually falls to unity. Thus, extremely strong magneticfields (typically, B0

>∼ 1 tesla) are hardly shielded at all by Mumetal, or similar

magnetic materials.

3.18 Magnetic energy

Consider an electrical conductor. Suppose that a battery with an electromotivefield E′ is feeding energy into this conductor. The energy is either dissipated asheat or is used to generate a magnetic field. Ohm’s law inside the conductor gives

jt = σ(E + E′), (3.185)

where jt is the true current density, σ is the conductivity, and E is the inductiveelectric field. Taking the scalar product with jt, we obtain

E′ ·jt =j 2t

σ−E ·jt. (3.186)

The left-hand side of this equation represents the rate at which the battery doeswork on the conductor. The first term on the right-hand side is the rate of Jouleheating inside the conductor. We tentatively identify the remaining term withthe rate at which energy is fed into the magnetic field. If all fields are quasi-stationary (i.e., slowly varying) then the displacement current can be neglected,and the Ampere-Maxwell equation reduces to ∇ ∧ H = jt. Substituting this

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expression into Eq. (3.186) and integrating over all space, we get∫

E′ ·(∇∧H) d3r =

(∇∧H)2

σd3r −

E ·(∇∧H) d3r. (3.187)

The last term can be integrated by parts using the relation

∇·(E ∧H) = H ·(∇∧E)−E ·(∇∧H). (3.188)

The divergence theorem plus the Faraday-Maxwell equation yield∫

E ·(∇∧H) d3r = −

H ·∂B

∂td3r −

(E ∧H)·dS. (3.189)

Since E ∧H falls off at least as fast as 1/r5 in electrostatic and quasi-stationarymagnetic fields (1/r2 comes from electric monopole fields, and 1/r3 from magneticdipole fields), the surface integral in the above expression can be neglected. Ofcourse, this is not the case for radiation fields, for which E and H fall off like1/r. Thus, the constraint of “quasi-stationarity” effectively means that the fieldsvary sufficiently slowly that any radiation fields can be neglected.

The total power expended by the battery can now be written∫

E′ ·(∇∧H) d3r =

(∇∧H)2

σd3r +

H ·∂B

∂td3r. (3.190)

The first term on the right-hand side has already been identified as the energyloss rate due to Joule heating. The last term is obviously the rate at which energyis fed into the magnetic field. The variation δU in the magnetic field energy cantherefore be written

δU =

H ·δB d3r. (3.191)

This result is analogous to the result (3.64) for the variation in the energy of anelectrostatic field.

In order to make Eq. (3.191) integrable, we must assume a functional relation-ship between H and B. For a medium which magnetizes linearly the integrationcan be carried out in much the same manner as Eq. (3.67), to give

U =1

2

H ·B d3r. (3.192)

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Thus, the magnetostatic energy density inside a linear magnetic material is givenby

W =H ·B

2. (3.193)

Unfortunately, most interesting magnetic materials, such as ferromagnets, exhibita nonlinear relationship between H and B. For such materials, Eq. (3.191) canonly be integrated between definite states, and the result, in general, depends onthe past history of the sample. For ferromagnets, the integral of Eq. (3.191) hasa finite, non-zero value when B is integrated around a complete magnetizationcycle. This cyclic energy loss is given by

∆U =

∫ ∮

H · dB d3r. (3.194)

In other words, the energy expended per unit volume when a magnetic materialis carried through a magnetization cycle is equal to the area of its hysteresis loopas plotted in a graph of B against H. Thus, it is particularly important to ensurethat the magnetic materials used to form transformer cores possess hysteresisloops with comparatively small areas, otherwise the transformers are likely to beextremely inefficient.

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