Lecture 7 EEE 403 Lecture 7: Magnetostatics: Ampere’s Law Of Force; Magnetic Flux Density; Lorentz Force; Biot-savart Law; Applications Of Ampere’s Law

Lecture 7

EEE 403

Lecture 7: Magnetostatics: Ampere’s Law Of Force; Magnetic Flux Density; Lorentz Force; Biot-savart Law; Applications Of Ampere’s Law In Integral Form; Vector Magnetic Potential; Magnetic Dipole;

Magnetic Flux

1

Lecture 7

Lecture 7 Objectives

• To begin our study of magnetostatics with Ampere’s law of force; magnetic flux density; Lorentz force; Biot-Savart law; applications of Ampere’s law in integral form; vector magnetic potential; magnetic dipole; and magnetic flux.

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Lecture 7

Overview of Electromagnetics

3

Maxwell’sequations

Fundamental laws of classical electromagnetics

Special cases

Electro-statics

Magneto-statics

Electro-magnetic

waves

Kirchoff’s Laws

Statics: 0t

d

Geometric Optics

TransmissionLine

Theory

CircuitTheory

Input from other

disciplines

Lecture 7

Magnetostatics

• Magnetostatics is the branch of electromagnetics dealing with the effects of electric charges in steady motion (i.e, steady current or DC).

• The fundamental law of magnetostatics is Ampere’s law of force.

• Ampere’s law of force is analogous to Coulomb’s law in electrostatics.

4

Lecture 7

Magnetostatics (Cont’d)

• In magnetostatics, the magnetic field is produced by steady currents. The magnetostatic field does not allow for– inductive coupling between circuits– coupling between electric and magnetic fields

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Lecture 7

Ampere’s Law of Force

• Ampere’s law of force is the “law of action” between current carrying circuits.

• Ampere’s law of force gives the magnetic force between two current carrying circuits in an otherwise empty universe.

• Ampere’s law of force involves complete circuits since current must flow in closed loops.

6

Lecture 7

Ampere’s Law of Force (Cont’d)

• Experimental facts:– Two parallel wires

carrying current in the same direction attract.

– Two parallel wires carrying current in the opposite directions repel.

7

I1 I2

F12F21

I1 I2

F12F21

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• Experimental facts:– A short current-

carrying wire oriented perpendicular to a long current-carrying wire experiences no force.

8

I1

F12 = 0

I2

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• Experimental facts:– The magnitude of the force is inversely

proportional to the distance squared.– The magnitude of the force is proportional to the

product of the currents carried by the two wires.

9

Lecture 7

Ampere’s Law of Force (Cont’d)• The direction of the force established by

the experimental facts can be mathematically represented by

10

1212

ˆˆˆˆ 12 RF aaaa

unit vector in direction of force on

I2 due to I1

unit vector in direction of I2 from I1

unit vector in direction of current I1

unit vector in direction of current I2

Lecture 7


• The force acting on a current element I2 dl2 by a current element I1 dl1 is given by

11

2

12

1122012

12ˆ

4 R

aldIldIF R

Permeability of free spacem0 = 4p 10-7 F/m

Lecture 7


• The total force acting on a circuit C2 having a current I2 by a circuit C1 having current I1 is given by

12

2 1

12

212

1221012

ˆ

4 C C

R

R

aldldIIF

Lecture 7


• The force on C1 due to C2 is equal in magnitude but opposite in direction to the force on C2 due to C1.

13

1221 FF

Lecture 7

Magnetic Flux Density

• Ampere’s force law describes an “action at a distance” analogous to Coulomb’s law.

• In Coulomb’s law, it was useful to introduce the concept of an electric field to describe the interaction between the charges.

• In Ampere’s law, we can define an appropriate field that may be regarded as the means by which currents exert force on each other.

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Lecture 7

Magnetic Flux Density (Cont’d)

• The magnetic flux density can be introduced by writing

15

2

2 1

12

1222

212

1102212

ˆ

4

C

C C

R

BldI

R

aldIldIF

Lecture 7


• where

16

1

12

212

11012

ˆ

4 C

R

R

aldIB

the magnetic flux density at the location of dl2 due to the current I1 in C1

Lecture 7


• Suppose that an infinitesimal current element Idl is immersed in a region of magnetic flux density B. The current element experiences a force dF given by

17

BlIdFd

Lecture 7


• The total force exerted on a circuit C carrying current I that is immersed in a magnetic flux density B is given by

18

C

BldIF

Lecture 7

Force on a Moving Charge

• A moving point charge placed in a magnetic field experiences a force given by

19

BvQ

The force experienced by the point charge is in the direction into the paper.

BvQF m vQlId

Lecture 7

Lorentz Force• If a point charge is moving in a region where both

electric and magnetic fields exist, then it experiences a total force given by

• The Lorentz force equation is useful for determining the equation of motion for electrons in electromagnetic deflection systems such as CRTs.

20

BvEqFFF me

Lecture 7

The Biot-Savart Law

• The Biot-Savart law gives us the B-field arising at a specified point P from a given current distribution.

• It is a fundamental law of magnetostatics.

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Lecture 7

The Biot-Savart Law (Cont’d)

• The contribution to the B-field at a point P from a differential current element Idl’ is given by

22

30

4)(

R

RldIrBd

Lecture 7


23

lId

PR

r r

Lecture 7


• The total magnetic flux at the point P due to the entire circuit C is given by

24

C R

RldIrB

30

4)(

Lecture 7

Types of Current Distributions

• Line current density (current) - occurs for infinitesimally thin filamentary bodies (i.e., wires of negligible diameter).

• Surface current density (current per unit width) - occurs when body is perfectly conducting.

• Volume current density (current per unit cross sectional area) - most general.

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Lecture 7

The Biot-Savart Law (Cont’d)• For a surface distribution of current, the B-S law

becomes

• For a volume distribution of current, the B-S law becomes

26

S

s sdR

RrJrB

30

4)(

V

vdR

RrJrB

30

4)(

Lecture 7

Ampere’s Circuital Law in Integral Form

• Ampere’s Circuital Law in integral form states that “the circulation of the magnetic flux density in free space is proportional to the total current through the surface bounding the path over which the circulation is computed.”

27

encl

C

IldB 0

Lecture 7

Ampere’s Circuital Law in Integral Form (Cont’d)

28

By convention, dS is taken to be in the

direction defined by the right-hand rule applied

to dl.

S

encl sdJI

Since volume currentdensity is the most

general, we can write Iencl in this way.

S

dl

dS

Lecture 7

Ampere’s Law and Gauss’s Law

• Just as Gauss’s law follows from Coulomb’s law, so Ampere’s circuital law follows from Ampere’s force law.

• Just as Gauss’s law can be used to derive the electrostatic field from symmetric charge distributions, so Ampere’s law can be used to derive the magnetostatic field from symmetric current distributions.

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Lecture 7

Applications of Ampere’s Law

• Ampere’s law in integral form is an integral equation for the unknown magnetic flux density resulting from a given current distribution.

30

encl

C

IldB 0known

unknown

Lecture 7

Applications of Ampere’s Law (Cont’d)

• In general, solutions to integral equations must be obtained using numerical techniques.

• However, for certain symmetric current distributions closed form solutions to Ampere’s law can be obtained.

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Lecture 7

Applications of Ampere’s Law (Cont’d)

• Closed form solution to Ampere’s law relies on our ability to construct a suitable family of Amperian paths.

• An Amperian path is a closed contour to which the magnetic flux density is tangential and over which equal to a constant value.

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Lecture 7

Magnetic Flux Density of an Infinite Line Current Using Ampere’s Law

Consider an infinite line current along the z-axis carrying current in the +z-direction:

33

I

Lecture 7

Magnetic Flux Density of an Infinite Line Current Using Ampere’s Law (Cont’d)

(1) Assume from symmetry and the right-hand rule the form of the field

(2) Construct a family of Amperian paths

34

BaB ˆ

circles of radius r where

0

Lecture 7


(3) Evaluate the total current passing through the surface bounded by the Amperian path

35

S

encl sdJI

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36

Amperian path

IIencl

I

rx

y

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(4) For each Amperian path, evaluate the integral

37

BlldBC

2BldBC

magnitude of Bon Amperian

path.

lengthof Amperian

path.

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(5) Solve for B on each Amperian path

38

l

IB encl0

2ˆ 0IaB

Lecture 7

Applying Stokes’s Theorem to Ampere’s Law

39

S

encl

SC

sdJI

sdBldB

00

Because the above must hold for any surface S, we must have

JB 0 Differential formof Ampere’s Law

Lecture 7

Ampere’s Law in Differential Form

• Ampere’s law in differential form implies that the B-field is conservative outside of regions where current is flowing.

40

Lecture 7

Fundamental Postulates of Magnetostatics

• Ampere’s law in differential form

• No isolated magnetic charges

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JB 0

0 B B is solenoidal

Lecture 7

Vector Magnetic Potential

• Vector identity: “the divergence of the curl of any vector field is identically zero.”

• Corollary: “If the divergence of a vector field is identically zero, then that vector field can be written as the curl of some vector potential field.”

42

0 A

Lecture 7

Vector Magnetic Potential (Cont’d)

• Since the magnetic flux density is solenoidal, it can be written as the curl of a vector field called the vector magnetic potential.

43

ABB 0

Lecture 7


• The general form of the B-S law is

• Note that

44

V

vdR

RrJrB

30

4)(

3

1

R

R

R

Lecture 7


• Furthermore, note that the del operator operates only on the unprimed coordinates so that

45

R

rJ

rJR

RrJ

R

RrJ

1

13

Lecture 7


• Hence, we have

46

vd

R

rJrB

V

4

0

rA

Lecture 7


• For a surface distribution of current, the vector magnetic potential is given by

• For a line current, the vector magnetic potential is given by

47

sd

R

rJrA

S

s

4

)( 0

L R

ldIrA

4

)( 0

Lecture 7


• In some cases, it is easier to evaluate the vector magnetic potential and then use B = A, rather than to use the B-S law to directly find B.

• In some ways, the vector magnetic potential A is analogous to the scalar electric potential V.

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Lecture 7


• In classical physics, the vector magnetic potential is viewed as an auxiliary function with no physical meaning.

• However, there are phenomena in quantum mechanics that suggest that the vector magnetic potential is a real (i.e., measurable) field.

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Lecture 7

Magnetic Dipole

• A magnetic dipole comprises a small current carrying loop.

• The point charge (charge monopole) is the simplest source of electrostatic field. The magnetic dipole is the simplest source of magnetostatic field. There is no such thing as a magnetic monopole (at least as far as classical physics is concerned).

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Lecture 7

Magnetic Dipole (Cont’d)

• The magnetic dipole is analogous to the electric dipole.

• Just as the electric dipole is useful in helping us to understand the behavior of dielectric materials, so the magnetic dipole is useful in helping us to understand the behavior of magnetic materials.

51

Lecture 7


• Consider a small circular loop of radius b carrying a steady current I. Assume that the wire radius has a negligible cross-section.

52

x

y

b

Lecture 7

Magnetic Dipole (Cont’d)• The vector magnetic potential is

evaluated for R >> b as

53

sin4

ˆ

sincosˆsinˆ

4

cossin1cosˆsinˆ

4

ˆ

4)(

2

20

20

2

0 20

2

0

0

r

bIa

r

baa

Ib

dr

b

raa

Ib

R

bdaIrA

yx

yx

Lecture 7


• The magnetic flux density is evaluated for R >> b as

54

sinˆcos2ˆ4

23

0 aabIr

AB r

Lecture 7

Magnetic Dipole (Cont’d)• Recall electric dipole

• The electric field due to the electric charge dipole and the magnetic field due to the magnetic dipole are dual quantities.

55

sinˆcos2ˆ

4 30

aar

pE r

Qdp moment dipole electric

Lecture 7

Magnetic Dipole Moment

• The magnetic dipole moment can be defined as

56

2ˆ bIam z

Direction of the dipole moment is determined by the direction of current using the right-hand rule.

Magnitude of the dipole moment is the product of the current and the area of the loop.

Lecture 7

Magnetic Dipole Moment (Cont’d)

• We can write the vector magnetic potential in terms of the magnetic dipole moment as

• We can write the B field in terms of the magnetic dipole moment as

57

20

20

4

ˆ

4

sinˆ

r

am

r

maA r

rmaam

rB r

1

4sinˆcos2ˆ

40

30

Lecture 7

Divergence of B-Field

• The B-field is solenoidal, i.e. the divergence of the B-field is identically equal to zero:

• Physically, this means that magnetic charges (monopoles) do not exist.

• A magnetic charge can be viewed as an isolated magnetic pole.

58

0 B

Lecture 7

Divergence of B-Field (Cont’d)

• No matter how small the magnetic is divided, it always has a north pole and a south pole.

• The elementary source of magnetic field is a magnetic dipole.

59

N

S

NS

NS

I

N

S

Lecture 7

Magnetic Flux

• The magnetic flux crossing an open surface S is given by

S

sdB

60

S

B

C

Wb/m2Wb

Lecture 7

Magnetic Flux (Cont’d)• From the divergence theorem, we have

• Hence, the net magnetic flux leaving any closed surface is zero. This is another manifestation of the fact that there are no magnetic charges.

61

000 SV

sdBdvBB

Lecture 7

Magnetic Flux and Vector Magnetic Potential

• The magnetic flux across an open surface may be evaluated in terms of the vector magnetic potential using Stokes’s theorem:

62

C

SS

ldA

sdAsdB

Lecture 7

Electromagnetic Force

The first term in the Lorentz Force Equation represents the electric force Fe acting on a charge q within an electric field is given by.

e qF E

The electromagnetic force is given by Lorentz Force Equation (After Dutch physicist Hendrik Antoon Lorentz (1853 – 1928))

q F E u B

The electric force is in the direction of the electric field.

The Lorentz force equation is quite useful in determining the paths charged particles will take as they move through electric and magnetic fields. If we also know the particle mass, m, the force is related to acceleration by the equation.mF a

Lecture 7

Since the magnetic force is at right angles to the magnetic field, the work done by the magnetic field is given by

cos 90 0W d FdL F L

Magnetic Force

The magnetic force is at right angles to the magnetic field. The magnetic force requires that the charged particle be in motion.

It should be noted that since the magnetic force acts in a direction normal to the particle velocity, the acceleration is normal to the velocity and the magnitude of the velocity vector is unaffected.

The second term in the Lorentz Force Equation represents magnetic force Fm(N) on a moving charge q(C) is given by

m q F u B

where the velocity of the charge is u (m/sec) within a field of magnetic flux density B (Wb/m2). The units are confirmed by using the equivalences Wb=(V)(sec) and J=(N)(m)=(C)(V).

Lecture 7

Magnetic ForceD3.10: At a particular instant in time, in a region of space where E = 0 and B = 3ay Wb/m2, a 2 kg particle of charge 1 C moves with velocity 2ax m/sec. What is the particle’s acceleration due to the magnetic field?

2sec

13 3

22

mq

m

x y za u B a a a

To calculate the units: 2 2 2

sec

sec sec sec

C m Wb kg m N m J V m

kg m N J C V Wb

P3.33: A 10. nC charge with velocity 100. m/sec in the z direction enters a region where the electric field intensity is 800. V/m ax and the magnetic flux density 12.0 Wb/m2 ay. Determine the force vector acting on the charge.

92

10 10 800 100 12 4x z xy

V m Wbq x C N

m s m

F E u B a a a

a

Given: q= 10 nC, u = 100 az (m/sec), E = 800 ax V/m, B = 12.0 ay Wb/m2.

Given: q= 1 nC, m = 2 kg, u = 2 ax (m/sec), E = 0, B = 3 ay Wb/m2.

mF aNewtons’ Second Law Lorentz Force Equation

q q F E u B u BEquating

Lecture 7

Magnetic Force on a current ElementConsider a line conducting current in the presence of a magnetic field. We wish to find the resulting force on the line. We can look at a small, differential segment dQ of charge moving with velocity u, and can calculate the differential force on this charge from

d dQ F u B

The velocity can also be written

d

dt

Lu

dQd d

dt F L B

Therefore

Now, since dQ/dt (in C/sec) corresponds to the current I in the line, we have

d Id F L B(often referred to as the motor equation)

We can use to find the force from a collection of current elements, using the integral

12 2 2 1 .I d F L B

dQ

u

segment

velocity

Lecture 7

Magnetic Force – An infinite current Element

Let’s consider a line of current I in the +az direction on the z-axis. For current element IdLa, we have

a a z .Id IdL z a

12 2 2 1d .I d F L B

The magnetic flux density B1 for an infinite length line of current is

11

2o I

B a11

2

I

H a1 1oB H

We know this element produces magnetic field, but the field cannot exert magnetic force on the element producing it. As an analogy, consider that the electric field of a point charge can exert no electric force on itself.What about the field from a second current element IdLb on this line? From Biot-Savart’s Law, we see that the cross product in this particular case will be zero, since IdL and aR will be in the same direction. So, we can say that a straight line of current exerts no magnetic force on itself.

Lecture 7

Magnetic Force – Two current Elements

By inspection of the figure we see that ρ = y and a = -ax. Inserting this in the above equation and considering that dL2 = dzaz, we have

Now let us consider a second line of current parallel to the first. The force dF12 from the magnetic field of line 1 acting on a differential section of line 2 is

12 2 2 1d I d F L B

The magnetic flux density B1 for an infinite length line of current is recalled from equation to be

11

2o I

B a

a = -ax

ρ = y

1 112 2 2 1 2 2

2 2o o

z z x

I II d I dz I dz

F a a a aL B -

1 212

2o

y

I I

ydz

F a

Lecture 7

Magnetic Force on a current ElementTo find the total force on a length L of line 2 from the field of line 1, we must integrate dF12 from +L to 0. We are integrating in this direction to account for the direction of the current.

0

1 212

1 2

2

2

o

L

o

I Idz

y

I I L

y

y

y

F a

a

This gives us a repulsive force. Had we instead been seeking F21, the magnetic force acting on line 1 from the field of line 2, we would have found F21 = -F12.

Conclusion:1) Two parallel lines with current in opposite directions experience a force of repulsion. 2) For a pair of parallel lines with current in the same direction, a force of attraction would result.

a = -ax

ρ = y

Lecture 7

Magnetic Force on a current ElementIn the more general case where the two lines are not parallel, or not straight, we could use the Law of Biot-Savart to find B1 and arrive at

2 1 1212 2 1 2

124o

d dI I

R

L L aF

This equation is known as Ampere’s Law of Force between a pair of current carrying circuits and is analogous to Coulomb’s law of force between a pair of charges.

Lecture 7

Magnetic ForceD3.11: A pair of parallel infinite length lines each carry current I = 2A in the same direction. Determine the magnitude of the force per unit length between the two lines if their separation distance is (a) 10 cm, (b)100 cm. Is the force repulsive or attractive? (Ans: (a) 8 mN/m, (b) 0.8 mN/m, attractive)

1 212

2o I I L

y

yF a

1 212

2o I I

yL

y

Fa

Case (a) y = 10 cm

Magnetic force between two current elements when current flow is in the same direction

Magnetic force per unit length

12

7

2

(4 10 )(2)(2)

2 (10 10 )8 N/m

L

y

Fa

Case (a) y = 10 cm

12

7

2

(4 10 )(2)(2)

2 (100 10 )0.8 N/m

L

y

Fa

Lecture 7

Magnetic Materials

and Boundary Conditions

Lecture 7

Magnetic Materials

Material mr

Diamagnetic bismuthgoldsilver

copperwater

0.999830.999860.99998

0.9999910.999991

Paramagnetic airaluminumplatinum

1.00000041.000021.0003

Ferromagnetic(nonlinear)

cobaltnickel

iron (99.8% pure)iron (99.96% pure)Mo/Ni superalloy

250600

5000280,000

1,000,000

The degree to which a material can influence the magnetic field is given by its relative permeability,r, analogous to relative permittivity r for dielectrics.

In free space (a vacuum), r = 1 and there is no effect on the field.

We know that current through a coil of wire will produce a magnetic field akin to that of a bar magnet.

We also know that we can greatly enhance the field by wrapping the wire around an iron core. The iron is considered a magnetic material since it can influence, in this case amplify, the magnetic field.

Relative permeabilities for a variety of materials.

Lecture 7

In the presence of an external magnetic field, a magnetic material gets magnetized (similar to an iron core). This property is referred to as magnetization M defined as

(1 ) =o m o r B H H H

where is the material’s permeability, related to free space permittivity by the factor r, called the relative permeability.

1r m

Magnetic Flux Density

Where

mM H

where m (“chi”) is the material’s magnetic susceptibility.

The total magnetic flux density inside the magnetic material including the effect of magnetization M in the presence of an external magnetic field H can be written as

+o o B H M

mM HSubstituting

Lecture 7

Magnetostatic Boundary ConditionsWill use Ampere’s circuital law and Gauss’s law to derive normal and tangential boundary conditions for magnetostatics.

Ampere’s circuit law:

encd IH L

.encI KdW K w The current enclosed by the path is

We can break up the circulation of H into four integrals:

.b c d a

a b c d

d d K w H L H L

T1 T T T1

0

.b w

a

d H dL H w

H L a a

0 / 2

N1 N N N2 N N N1 N2

/ 2 0 2

c h

b h

hd H dL H dL H H

H L a a a a

Path 1

Path 3

Path

2Path

4

Path 1:

Path 2:

Lecture 7

0 / 2

N2 N N N1 N N N1 N2

/ 2 0 2

a h

d h

hd H dL H dL H H

H L a a a a

T1 T2H H K

Now combining our results (i.e., Path 1 + Path 2 + Path 3 + Path 4), we obtain

A more general expression for the first magnetostatic boundary condition can be written as

21 1 2 a H H K

where a21 is a unit vector normal going from media 2 to media 1.

Magnetostatic Boundary Conditions0

T2 T T T2 .d

c w

d H dL H w

H L a a Path 3:

Path 4:

T1 T2d wH H H L encI KdW K w Equating

Tangential BC:

encd IH LACL:

Lecture 7

The tangential magnetic field intensity is continuous across the boundary when the surface current density is zero.

o r B HWe know that

Important Note:

Special Case: If the surface current density K = 0, we get

Magnetostatic Boundary Conditions

T1 T2H HIf K = 0

T1 T2H H K

1 2

T1 T2

o o

B B

Using the above relation, we obtain

T1 T2H H

The tangential component of the magnetic flux density B is not continuous across the boundary.

Therefore, we can say that T1 T2B B

o r

BH(or)

Lecture 7


Gauss’s Law for Magnetostatic fields:

= 0dB S

To find the second boundary condition, we center a Gaussian pillbox across the interface as shown in Figure.

We can shrink h such that the flux out of the side of the pillbox is negligible. Then we have

N1 N N N 2 N N

N1 N 2

( )

0.

d B dS B dS

B B S

B S a a a a

N1 N 2 .B BNormal BC:

Lecture 7


Thus, we see that the normal component of the magnetic flux density must be continuous across the boundary.

We know that

Important Note:

Using the above relation, we obtain

The normal component of the magnetic field intensity is not continuous across the boundary (but the magnetic flux density is continuous).

Therefore, we can say that

N1 N2B BNormal BC:

o r B H

1 2N1 N2o oH H N1 N2B B

N1 N2H H

Lecture 7

Magnetostatic Boundary ConditionsExample 3.11: The magnetic field intensity is given as H1 = 6ax + 2ay + 3az (A/m) in a medium with r1 = 6000 that exists for z < 0. We want to find H2 in a medium with r2 = 3000 for z >0.

Step (a) and (b): The first step is to break H1 into its normal component (a) and its tangential component (b). Step (c): With no current at the interface, the tangential component is the same on both sides of the boundary. Step (d): Next, we find BN1 by multiplying HN1 by the permeability in medium 1. Step (e): This normal component B is the same on both sides of the boundary. Step (f): Then we can find HN2 by dividing BN2 by the permeability of medium 2. Step (g): The last step is to sum the fields .

Lecture 7

Magnetization and Permeability

• M can be considered the magnetic field intensity due to the dipole moments when an external field H is applied

• Hence, the total magnetic field intensity inside the material is M+H

• The magnetic field density inside the material is B=mo(M+H)

• But M depends on H, Define the magnetic susceptibility m . Hence M= m H

• Hence, we get B=mo(m H +H) = B=moH (m +1)

• Finally define the relative permeability mr = (m +1)

• B=mo mr H= m H

81

Lecture 7

The Nature of Magnetic Materials

• Materials have a different behavior in magnetic fields

• Accurate description requires quantum theory• However, simple atomic model (central

nucleus surrounded by electrons) is enough for us– We can also say that B tries to make the Magnetic

Dipole Moment m m in the same direction of B

82

Lecture 7

Magnetic Dipole Moments in Atom

• There are 3 magnetic dipole moments:1. Moment due to rotation of the electrons2. Moment due to the spin of the electrons3. Moment due to the spin of the nucleus

• The 3 rotations are 3 loop currents• The first two are much more effective• Electron spin is in pairs, in two opposite direction– Hence, a net moment due to electron spin occurs only

when there is an un-filled shell (or orbit) • The combination of moment decide the magnetic

characteristics of the material

83

Lecture 7

Type of Magnetic Materials

• We will study 6 types:– Diamagnetic– Paramagnetic– Ferromagnetic– Anti-ferromagnetic– Ferrimagnetic– Super paramagnetic

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Lecture 7

Diamagnetic Materials• Without an external magnetic field,

diamagnetic materials have no net magnetic field

• With an external magnetic field, they generate a small magnetic field in the opposite direction

• The value of this opposite field depends on the external field and the diamagnetic material

• Most materials are diamagnetic (with different parameters)

• We will see that the relative permeability

mr but 1 85

http://upload.wikimedia.org/wikipedia/commons/c/c9/Diamagnetic_graphite_levitation.jpg

Lecture 7

Why Diamagnetic Materials, 1• Each atom has zero total Magnetic Dipole Moment – No torque due to external field and do not add any field

• But if some electrons have their magnetic dipole moment with the external field

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– External field will cause a small outward force on electrons, which adds to their centrifugal force

– Electrons cannot leave shells to next shell (not enough energy)

– Coulombs attraction force with nucleus is the same– To stay in same orbit, centrifugal force must go down.

Hence, velocity reduces– The magnetic dipole moment of the atom decreases– Net magnetic dipole moment of the atom is created,

opposite to B

Lecture 7

Why Diamagnetic Materials, 2• Also if some electrons have their magnetic dipole moment opposite to the

external field– External field will cause a small inward force on electrons, which reduces

the centrifugal force– Electrons cannot leave shells to next shell (not enough energy)– Coulombs attraction force with nucleus is the same– To stay in same orbit, centrifugal force must increase– The magnetic dipole moment of the atom increases– Net magnetic dipole moment of the atom is created, opposite to B

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Lecture 7

Diamagnetic Materials

• Note that mr = 1 + susceptibility

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Material Susceptibility x 10-5

Bismuth -16.6

Carbon (diamond) -2.1

Carbon (graphite) -1.6

Copper -1.0

Superconductor -105

Lecture 7

Paramagnetic Materials• Atoms have a small net magnetic dipole moment• The random orientation of atoms make the average dipole

moment in the material zero• Without an external field, there is no magnetic property• When an external magnetic field is applied there is a small

torque on atoms and they become aligned with the field • Hence, inside the material, atoms add their own field to the

external field• The diamagnetism due to orbiting electrons is also acting• If the net effect is an increase in the field B, the material is

called paramagnetic

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Lecture 7

Paramagnetic Materials

• Note that mr = 1 + susceptibility

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Material Susceptibility x 10-5

Iron oxide (FeO) 720

Iron amonium alum 66

Uranium 40

Platinum 26

Tungsten 6.8

Lecture 7

Ferromagnetic Materials• Atoms have large dipole moment, they affect

each other• Interaction among the atoms causes their

magnetic dipole moments to align within regions, called domains

• Each domain have a strong magnetic dipole moment

• A ferromagnetic material that was never magnetized before will have magnetic dipole moments in many directions

• The average effect is cancellation. The net effect is zero

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Lecture 7

Ferromagnetic Materials• When an external magnetic field B is applied

the domains with magnetic dipole moment in the same direction of B increase their size at the expense of other domains

• The internal magnetic field increases significantly

• When the external magnetic field is removed a residual magnetic dipole moment stay, causing the permanent magnet

• The only ferromagnetic material at room temperature are Iron, Nickel and Cobalt

• They loose ferromagnetism at temperature > Curie temperature (which is 770o C for iron)

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Lecture 7

Ferromagnetic Materials

• Curie Temperatures:– Iron: 770oC– Nickel: 354o C – Cobalt: 1115o C

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Medium Relative Permeability μr

Mu Metal 20,000

Permalloy 8000

Electrical Steel 4000

Ferrite (nickel zinc) 16-640

Ferrite (manganese zinc) >640

Steel 100

Nickel 100-600

Lecture 7

Antiferromagnetic Materials

• Atoms have a net dipole moment• However, the material is such that atoms dipole moments

line-up in opposite direction• Net dipole moment is zero• No much difference when an external magnetic field is

present• Phenomena occurs at temperature well below room

temperature• No engineering importance at present time

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http://upload.wikimedia.org/wikipedia/commons/5/5d/Antiferromagnetic_ordering.svg

Lecture 7

Ferrimagnetic Materials

• Similar to antiferromagnetic materials, atoms dipole moments line-up in opposite direction

• However, the dipole moments are not equal. Hence, there is a net dipole moment

• Ferrimagnetic materials behave like ferrormagnetic materials, but the magnetic field increase is not as large

• Effect disappear above Curie temperature

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http://upload.wikimedia.org/wikipedia/commons/8/81/Ferrimagnetic_ordering.svg

Lecture 7

Ferrimagnetic Materials

• The main advantage is that they have high resistance. Hence can be used as the core of transformers, specially at high frequency

• Also used in loop antennas in AM radios• In this case the losses Eddy current are much

smaller than iron core• Example material: Iron Oxide (Fe3O4) and

Nickel Ferrite (NiFe2O4)

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Lecture 7

Super Paramagnetic Materials

• Composed of Ferromagnetic particles inside non-ferromagnetic materials

• Domains occur but can not expand when exposed to external field

• Used in magnetic tapes for audio and video tape recording

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Lecture 7

Magnetization and Permeability• Now let us discuss the magnetic effect of

magnetic material in a quantitative manner• Let us call the current inside the material due

to electron orbit, electron spin and atom spin by the bound current Ib

• The material includes many dipole moments m (units A m2) that add-up

• Define the Magnetization M as the magnetic dipole moment per unit volume

• M has a unit A/m (which is similar to the units of H)

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Documents

Lecture 7 EEE 403 Lecture 7: Magnetostatics: Ampere’s Law Of Force; Magnetic Flux Density; Lorentz Force; Biot-savart Law; Applications Of Ampere’s Law