Download Lecture12_Static Magnetic Fields 2

Electromagnetic Fields Lecture-12:
Static Magnetic Fields 2
Chadi Abou-Rjeily
Department of Electrical and Computer Engineering
Lebanese American University
[email protected]
March 23, 2017
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetization and Equivalent Current Densities (1)
A magnetic dipole corresponds to a small circular loop of
radius b ≪ 1 that carries a current I.
The magnetic dipole moment is defined as:
~ = I(πb 2 )~az
m
This corresponds to a vector:
whose magnitude is the product of the current and the area
of the loop.
whose direction is the direction of the thumb as the fingers
of the right hand follow the direction of the current.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetization and Equivalent Current Densities (2)
All materials are composed of atoms with a positively
charged nucleus and a number of orbiting negatively
charged electrons.
The orbiting electrons cause circulating currents and form
microscopic magnetic dipoles.
In the absence of an external magnetic field, the magnetic
dipoles of the atoms of most materials (except permanent
magnets) have random orientations resulting in no net
magnetic moment.
The application of an external magnetic field causes an
alignment of the magnetic moments of the spinning
electrons.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetization and Equivalent Current Densities (3)
To study the change in the magnetic flux density caused by
the presence of a magnetic material, we define the
~ by:
magnetization vector M
~ = lim
M
∆v →0
Pn∆v
k =1
∆v
~k
m
(A/m)
~k is the magnetic dipole moment of the k -th atom.
m
n is the number of atoms per unit volume.
~
From the above definition, we can observe that M
corresponds to the volume density of magnetic dipole
moment.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetization and Equivalent Current Densities (4)
~ is equivalent to both a
It can be proven that the effect of M
~
volume current density Jm and a surface current density
~Jms such that:
~
~Jm = ∇ × M
(A/m2 )
~Jms = M
~ × ~an
(A/m)
~an is the unit outward normal from the surface.
Consider the case of an external
magnetic field that caused the atomic
circulating currents to align with it as
shown in the adjacent figure.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetization and Equivalent Current Densities (5)
~ × ~an correctly predicts a surface
The relation ~Jms = M
current flow in the upward direction on the right surface
and in the downward direction on the left surface.
~ is uniform inside the material, the currents of the
If M
neighboring atomic dipoles that flow in opposite directions
will cancel everywhere leaving no net currents in the
~ =0
interior. This is predicted by the relation ~Jm = ∇ × M
~
when M does not depend on the spatial coordinates.
~ has space variations, the internal
On the other hand, if M
atomic currents do not completely cancel resulting in a net
volume density ~Jm .
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Field Intensity and Relative Permeability (1)
~ satisfies the relation:
In free space, B
~ = µ0~J
∇×B
⇒
1
~ = ~J
∇×B
µ0
In magnetic materials, the equivalent magnetization
volume current density ~Jm must be taken into consideration
~
when evaluating the curl of B:
⇒
1
~ = ~J + ~Jm = ~J + ∇ × M
~
∇×B
µ0
!
~
B
~ = ~J
∇×
−M
µ0
~ by:
We define the magnetic field intensity H
~
~ = B −M
~
H
µ0
Chadi Abou-Rjeily
(A/m)
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Field Intensity and Relative Permeability (2)
Consequently, the curl postulate can be written as:
~ = ~J
∇×H
(A/m2 )
~J is the volume density of free currents (in A/m2 ).
Taking the surface integral of both sides results in:
Z
Z
~ ~s =
~J.d~s
(∇ × H).d
S
S
Applying Stokes’s theorem:
I
~ ~l = I
H.d
(A)
C
C is the closed contour bounding the surface S and I is the
total free current passing through S.
The last relation corresponds to another form of Ampere’s
circuital law.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Field Intensity and Relative Permeability (3)
When the magnetic properties of the medium are linear
and isotropic, the magnetization is directly proportional to
the magnetic field intensity:
~
~ = χm H
M
χm is a dimensionless quantity called magnetic
susceptibility.
Consequently:
~ = µ0 H
~ +M
~
B
~ = µ0 µr H
~ = µH
~
= µ0 (1 + χm )H
where:
µ r = 1 + χm =
µ
µ0
µr is a dimensionless quantity known as the relative
permeability.
µ = µ0 µr is the absolute permeability (or permeability) of
the medium in H/m.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Boundary Conditions (1)
~ and H
~
In this part, we determine the relations verified by B
at the interface of two media having different magnetic
properties.
We have seen in lecture-10 that the normal component of
a divergence-less field is continuous across an interface.
~ = 0:
Consequently, since ∇.B
B1n = B2n
(T)
Note that a similar relation was obtained for the current
density ~J since ∇.~J = 0.
A similar result was also obtained for the electric flux
density in the absence of free charges since in this case
~ = ρ = 0.
∇.D
~ 1 = µ1 H
~ 2 = µ2 H
~ 1 and B
~ 2 , the last equation can be
Since B
written as:
µ1 H1n = µ2 H2n
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Boundary Conditions (2)
To determine the boundary conditions on the tangential
component of the field, construct a closed contour abcda
where the sides ab and cd are parallel to the interface and
bc = ad = ∆h → 0.
Applying Ampere’s circuital law on the contour:
I
~ ~l = I
H.d
abcda
−→ ~
−−→
~ 1 .−
⇒ H
∆w + H
2 .(−∆w) = Jsn ∆w
⇒ H1t − H2t = Jsn
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Boundary Conditions (3)
Note that in the previous equation Jsn is the surface current
density on the interface normal to the contour C:
Its direction is determined by the right-hand rule.
In the previous example, the positive direction of Jsn is out
of the paper.
~
The boundary condition on the normal component of H
can be written in a more concise way as:
~1 − H
~ 2 ) = ~Js
~an2 × (H
(A/m)
~an2 is the outward unit normal from medium 2.
When the conductivities of both media are finite, currents
are defined by volume current densities and free surface
currents do not exist on the interface. In this case, ~Js = 0
~ is continuous across
and the tangential component of H
the interface.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Boundary Conditions (4)
Example: Consider two magnetic media
with permeabilities µ1 and µ2 . The
magnetic field intensity in medium 1 at
point P1 has a magnitude H1 and makes
an angle α1 with the normal. Determine
~ at
the magnitude and the direction of H
point P2 in medium 2.
~ implies that:
The continuity of the normal component of B
µ2 H2 cos α2 = µ1 H1 cos α1
Since neither of the media is a conductor, the tangential
~ is continuous:
component of H
H2 sin α2 = H1 sin α1
Dividing the above equations results in:
tan α2
µ2
=
tan α1
µ1
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Boundary Conditions (5)
~ 2 can be determined from:
The magnitude of H
q
q
2 + H2 =
H2 = H2t
(H2 sin α2 )2 + (H2 cos α2 )2
2n
"
2 #1/2
µ1
2
H2 = H1 sin α1 +
cos α1
µ2
If medium 1 is nonmagnetic (like air) and medium 2 is
ferromagnetic (like iron):
µ2 ≫ µ1 and α2 ≈ π/2.
Consequently, the magnetic field in a ferromagnetic
medium runs almost parallel to the interface.
If medium 1 is ferromagnetic and medium 2 is
nonmagnetic:
µ1 ≫ µ2 and α2 ≈ 0.
Consequently, if a magnetic field originates in a
ferromagnetic medium, the flux lines will emerge into air in
a direction almost normal to the interface.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (1)
In electric circuits, we are required to find the voltages
across and the currents in various branches of an electric
network.
There is an analogous class of problems dealing with
magnetic circuits. In this type of problems, we are
interested in the magnetic fluxes and magnetic field
intensities in various parts of a circuit caused by windings
carrying currents around ferromagnetic cores.
The analysis of these circuits is based on Ampere’s law:
I
~ ~l = NI = ϑm
H.d
C
N is the number of turns of a winding carrying a current I.
ϑm is called the magnetomotive force (mmf). It is
analogous to electromotive force (emf) in electrical circuits.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (2)
Example: Assume that N turns are
wound around a toroidal core of a
ferromagnetic material with permeability
µ. The core has a mean radius r0 , a
circular cross section of radius a ≪ r0
and a narrow air gap of length lg . A
steady current I0 flows in the wire.
~ f and H
~ f in the ferromagnetic
Determine B
~
~
core and Bg and Hg in the air gap.
Applying Ampere’s circuital law around a circular path C of
radius r0 :
I
~ ~l = NI0
H.d
C
If flux leakage is neglected, the same total flux will flow in
both the ferromagnetic core and the air gap.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (3)
Given that the magnetic field flows in the ~aφ direction and
~ along an interface:
following from the continuity of B
~f = B
~ g = B~aφ
B
In the ferromagnetic core:
~ f = B ~aφ
H
µ
In the air gap:
~ g = B ~aφ
H
µ0
Ampere’s circuital law implies that:
B
B
(2πr0 − lg ) + lg = NI0
µ
µ0
µ0 µNI0
~f = B
~g =
~aφ
⇒ B
µ0 (2πr0 − lg ) + µlg
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (4)
~ f and H
~ g are given by:
Consequently, H
µ0 NI0
~aφ
µ0 (2πr0 − lg ) + µlg
µNI0
~g =
~aφ
H
µ0 (2πr0 − lg ) + µlg
~f =
H
Note that:
Hg
µ
=
>1
Hf
µ0
and the magnetic field intensity in the air gap is much
stronger than that in the ferromagnetic core.
Note that the magnitude of the magnetic flux density can
be written as:
B=
µ0 µNI0
NI0
=
µ0 (2πr0 − lg ) + µlg
(2πr0 − lg )/µ + lg /µ0
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (5)
Since the radius of the cross section is much smaller than
~ is approximately constant in the core. Consequently,
r0 , B
the magnetic flux in the circuit can be approximated by:
NI0
Φ ≃ BS =
(2πr0 − lg )/(µS) + lg /(µ0 S)
S is the cross section of the core.
The above equation can be written as:
ϑm
Φ=
Rf + Rg
Rf is the reluctance of the ferromagnetic core and Rg is
the reluctance of the air gap (in H−1 ):
Rf =
2πr0 − lg
l
= f
µS
µS
;
Rg =
lg
µ0 S
lf = 2πr0 − lg is the length of the ferromagnetic core.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (6)
Note the analogy of the reluctance with the resistance of a
straight piece of homogeneous material of length l, uniform
cross section S and conductivity σ:
R=
l
σS
Consequently, magnetic circuits can be analyzed by the
same techniques used in electric circuits:
The mmf ϑm is analogous to the emf ϑ.
The magnetic flux φ is analogous to the electric current I.
The reluctance R is analogous to the resistance R.
The permeability µ is analogous to the conductivity σ.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (7)
Similar to Kirchhoff’s voltage law, we can write for any
closed path in a magnetic circuit:
X
X
Nj Ij =
Rk Φk
j
k
Similar to Kirchhoff’s current law, we can write for any
junction in a magnetic circuit:
X
Φj = 0
j
The equivalent magnetic circuit and the analogous electric
circuit for a toroid with an air gap is shown below:
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (8)
Example: Consider the magnetic circuit shown in the figure
below. Steady currents I1 and I2 flow in windings of N1 and N2
turns, respectively, on the outside legs of the ferromagnetic
core. The core has a cross section S and permeability µ.
Determine the magnetic flux in the center leg.
Figure (b) shows the equivalent magnetic circuit. Two
sources of mmf’s N1 I1 and N2 I2 are shown with proper
polarities.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2
Magnetic Circuits (9)
The reluctances are computed on the basis of average
path lengths:
l1
R1 =
µS
l2
R2 =
µS
l3
R3 =
µS
The two loop equations are:
Loop 1:
N1 I1 = (R1 + R3 )Φ1 + R1 Φ2
Loop 2 :
N1 I1 − N2 I2 = R1 Φ1 + (R1 + R2 )Φ2
Solving the above equations results in:
Φ1 =
R2 N1 I1 + R1 N2 I2
R1 R2 + R1 R3 + R2 R3
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-12: Static Magnetic Fields 2