Upload
oscar-brooks
View
231
Download
3
Tags:
Embed Size (px)
Citation preview
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.
2
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
5
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
Lecture 7
Ampere’s Law of Force (Cont’d)
• Experimental facts:– A short current-
carrying wire oriented perpendicular to a long current-carrying wire experiences no force.
8
I1
F12 = 0
I2
Lecture 7
Ampere’s Law of Force (Cont’d)
• 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
Ampere’s Law of Force (Cont’d)
• 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
Ampere’s Law of Force (Cont’d)
• 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
Ampere’s Law of Force (Cont’d)
• 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.
14
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
Magnetic Flux Density (Cont’d)
• 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
Magnetic Flux Density (Cont’d)
• 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
Magnetic Flux Density (Cont’d)
• 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.
21
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
The Biot-Savart Law (Cont’d)
23
lId
PR
r r
Lecture 7
The Biot-Savart Law (Cont’d)
• 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.
25
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.
29
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.
31
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.
32
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
Magnetic Flux Density of an Infinite Line Current Using Ampere’s Law (Cont’d)
(3) Evaluate the total current passing through the surface bounded by the Amperian path
35
S
encl sdJI
Lecture 7
Magnetic Flux Density of an Infinite Line Current Using Ampere’s Law (Cont’d)
36
Amperian path
IIencl
I
rx
y
Lecture 7
Magnetic Flux Density of an Infinite Line Current Using Ampere’s Law (Cont’d)
(4) For each Amperian path, evaluate the integral
37
BlldBC
2BldBC
magnitude of Bon Amperian
path.
lengthof Amperian
path.
Lecture 7
Magnetic Flux Density of an Infinite Line Current Using Ampere’s Law (Cont’d)
(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
41
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
Vector Magnetic Potential (Cont’d)
• The general form of the B-S law is
• Note that
44
V
vdR
RrJrB
30
4)(
3
1
R
R
R
Lecture 7
Vector Magnetic Potential (Cont’d)
• 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
Vector Magnetic Potential (Cont’d)
• Hence, we have
46
vd
R
rJrB
V
4
0
rA
Lecture 7
Vector Magnetic Potential (Cont’d)
• 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
Vector Magnetic Potential (Cont’d)
• 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.
48
Lecture 7
Vector Magnetic Potential (Cont’d)
• 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.
49
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).
50
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
Magnetic Dipole (Cont’d)
• 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
Magnetic Dipole (Cont’d)
• 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
Magnetostatic Boundary Conditions
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
Magnetostatic Boundary Conditions
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
84
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
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
86
– 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
87
Lecture 7
Diamagnetic Materials
• Note that mr = 1 + susceptibility
88
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
89
Lecture 7
Paramagnetic Materials
• Note that mr = 1 + susceptibility
90
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
91
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)
92
Lecture 7
Ferromagnetic Materials
• Curie Temperatures:– Iron: 770oC– Nickel: 354o C – Cobalt: 1115o C
93
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
94
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
95
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)
96
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
97
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)
98