5. MAGNETOSTATICS Applied EM by Ulaby, Michielssen and Ravaioli.
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Transcript of 5. MAGNETOSTATICS Applied EM by Ulaby, Michielssen and Ravaioli.
5. MAGNETOSTATICS Applied EM by Ulaby, Michielssen and Ravaioli
Chapter Outline
Maxwell’s Equations Magnetic Forces and Torques The total electromagnetic force, known as Lorentz
force Biot- Savart’s law Gauss’s law for magnetism Ampere’s law for magnetism Magnetic Field and Flux Vector magnetic potential Properties of 3 different types of material Boundary conditions between two different media Self inductance and mutual inductance Magnetic energy
Chapter 5 Overview
Course Outcome 3 (CO3)
Ability to analyze the concept of electric current density and boundary conditions, magnetic flux and magnetic flux density in a steady magnetic field and the basic laws of magnetic fields.
Maxwell’s equations
Where;
E = electric field intensityD = electric flux densityρv = electric charge density
per unit volumeH = magnetic field intensityB = magnetic flux densityt
t
DHH
0B
BE
D v
Maxwell’s equations:
Maxwell’s equations
For static case, ∂/∂t = 0. Maxwell’s equations is reduced to:
Electrostatics Magnetostatics
0
E
D vJH
B
0
Electric vs Magnetic Comparison
Electric & Magnetic Forces
Electromagnetic (Lorentz) force
Magnetic force
Magnetic Force on a Current Element
Differential force dFm on a differential current I dl:
I
Magnetic Force
q
N m BuF q
Magnetic Forces and Torques The electric force Fe per unit charge acting
on a test charge placed at a point in space with electric field E.
When a charged particle moving with a velocity u passing through that point in space, the magnetic force Fm is exerted on that charged particle.
where B = magnetic flux density (Cm/s or Tesla T)
N m BuF q
Magnetic Forces and Torques If a charged particle is in the presence of
both an electric field E and magnetic field B, the total electromagnetic force acting on it is:
q BuEF
me FFF
;EF qe BuF qm
BuEFFF qqme
force)(Lorentz
Magnetic Force on a Current- Carrying Conductor
For closed circuit of contour C carrying I , total magnetic force Fm is:
In a uniform magnetic field, Fm is zero for a closed circuit.
C
m dI N BlF N m BuF q
Magnetic Force on a Current- Carrying Conductor
On a line segment, Fm is proportional to the vector between the end points.
BF Im
Example 1
The semicircular conductor shown carries a current I. The closed circuit is exposed to a uniform magnetic field .
Determine (a) the magnetic force F1 on the straight section of the wire and
(b) the force F2 on the curved section.
0yBB
C
m dI N BlF
BF Im
Solution to Example 1
a) the magnetic force F1 on the straight section of the wire
N 2ˆˆ2ˆ 001 IrBBIr zyxF
0ˆ
2ˆ
, Using
B
r
Im
yB
x
BF
N 2ˆ
)11(ˆ
cosˆsin
:direction ˆ- in the is d ofproduct the
section. curved on the F2 force the F2 b)
02
0
00
0
0
0
2
IrB
IrB
IrBBdrI
dI
zF
z
z
BlF
zBl
))(sin(
ˆ
0
rdl
ByB o
Torque
d = moment armF = forceT = torque
Magnetic Torque on Current Loop
No forces on arms 2 and 4 ( because I and B are parallel, or anti-parallel)
Magnetic torque:
Area of Loop
Inclined Loop
For a loop with N turns and whose surface normal is at angle theta relative to B direction:
The Biot–Savart’s Law
The Biot–Savart law is used to compute the magnetic field generated by a steady current, i.e. a continual flow of charges, for example through a wire
Biot–Savart’s law states that:
where:
dH = differential magnetic field dl = differential length
A/m R
dd
2Rl
4
1H
Biot-Savart Law
Magnetic field induced by a differential current:
For the entire length:
Magnetic Field due to Current Densities
Example 2
Determine the magnetic field at the apex O of the pie-shaped loop as shown. Ignore the contributions to the field due to the current in the small arcs near O.
O A
C O
A C
= dl
= -dl
0
?
• For segment AC, dl is in φ direction,
• Using Biot- Savart’s law:
add zRl aR
radians in is wherea
H
a
ad
a
ad
4
1z
z
4
1z
4
1H
22
A/m ˆ
4
12
l R
d RlH
Example 5-2: Magnetic Field of Linear Conductor
Cont.
Example 5-2: Magnetic Field of Linear Conductor
Magnetic Field of Long Conductor
Example 5-3: Magnetic Field of a Loop
Cont.
dH is in the r–z plane , and therefore it hascomponents dHr and dHz
z-components of the magnetic fields due to dl and dl’ add because they are in the same direction, but their r-components cancel
Hence for element dl:
Magnitude of field due to dl is
Example 5-3:Magnetic Field of a Loop (cont.)
For the entire loop:
Magnetic Dipole
Because a circular loop exhibits a magnetic field pattern similar to the electric field of an electric dipole, it is called a magnetic dipole
Forces on Parallel Conductors
Parallel wires attract if their currents are in the same direction, and repel if currents are in opposite directions
Gauss’s Law for Magnetism Gauss’s law for magnetism states
that:
Magnetic field lines always form continuous closed loops.
form)(integral0form)ial(different0 S
dsBB
0 BsD
Ampère’s Law
Ampere’s law states that: true for an infinite length of conductor
C
law sAmpere' IdlH
Ampere’s law for magnetism
true for an infinite length of conductor
I, +az
HdlC, +aø
r
Internal Magnetic Field of Long ConductorFor r < a
Cont.
External Magnetic Field of Long Conductor
For r > a
Magnetic Field of Toroid
Applying Ampere’s law over contour C:
The magnetic field outside the toroid is zero. Why?
Ampere’s law states that the line integral of H around a closed contour C is equal to the current traversing the surface bounded by the contour.
Magnetic Flux
The amount of magnetic flux, φ in Webers from magnetic field passing through a surface is found in a manner analogous to finding electric flux:
SB d
a
2H
r
I
a
2B 0 r
I
HB 0
Example 4
An infinite length coaxial cable with inner
conductor radius of 0.01m and outer conductor
radius of 0.05m carrying a current of 2.5A
exists along the z axis in the + az direction.
Find the flux passing through the region between
two conductors with height of 2 m in free space.
Solution to Example 4
1) inner conductor radius = r1 0.01m
2) outer conductor radius = r2 0.05m
3) current of 2.5A (in the +az direction)
4) Flux radius = 2m
z
xy
a
2HB 00 r
I
Sd
r
I
a20
Iaz=2.5A
r1 r2
Flux,z
SB d
aø
Solution to Example 4
where dS is in the aø direction.
So,
Therefore,
aS drdzd
Wb I
adrdzar
I
SdB
z r
60
2
0
05.0
01.0
0
1061.101.0
05.0ln
2
2
2
Magnetic Vector Potential A
Electrostatics
Magnetostatics
Vector Magnetic Potential
For any vector of vector magnetic potential A:
We are able to derive: .
Vector Poisson’s equation is given as:
where
0A
2Wb/mA B
J2 A
Wb/m ''4 '
dvR
JA
v
Magnetic Properties of Materials Magnetization in a material is associated
with atomic current loops generated by two principal mechanisms: Orbital motions of the electrons around the
nucleus, i.e orbital magnetic moment, mo
Electron spin about its own axis, i.e spin magnetic moment, ms
Magnetic Properties of Materials
Magnetization vector M is defined as
where = magnetic susceptibility (dimensionless)
Magnetic permeability is defined as:
and to define the magnetic properties in term of relative permeability is defined as:
Magnetic Permeability
m
HM m
H/m 10 m mH 104 where 70
mr 10
HB
metals have a very weak and negative susceptibility ( ) to magnetic field
slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed
Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.
Magnetic Materials - Diamagnetic
m
Magnetic Materials - Paramagnetic
Paramagnetic materials have a small and positive susceptibilities to magnetic fields.
slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed.
Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.
)( m
However, the absolute susceptibilities value of both materials is in the order 10-5. Thus, can be ignored. Hence, we have Magnetic permeability:
Diamagnetic and paramagnetic materials include dielectric materials and most metals.
Magnetic Materials – Diamagnetic, Paramagnetic
m
0or 1 r
Magnetic Materials – Ferromagnetic Materials Ferromagnetic materials is characterized
by magnetized domain - a microscopic region within which the magnetic moments of all its atoms are aligned parallel to each other.
Hysteresis – “to lag behind”. It determines how easy/hard for a magnetic material to be magnetized and demagnetized.
Process of Magnetic Hysteresis
material is magnetizedand can serve as permanent magnet!
material is demagnetize
B
Magnetic Hysteresis of Ferromagnetic Materials Comparison of hysteresis curves for (a) a hard
and (b) a soft ferromagnetic material is shown.
Hard magnetic material- cannot be easily magnetized & demagnetized by an
external magnetic field.
Soft magnetic material – easily magnetized & demagnetized.
Magnetic Hysteresis
Boundary Conditions
Magnetic boundary conditions Boundary condition related to normal
components of the electric field;
By analogy, application of Gauss’s law for magnetism, we get first boundary condition:
Since , For linear, isotropic media, the first
boundary condition which is related to H;
SnnSDDQd 21 sD
nnSBBd 21 0 sB
nn HH 2211
HB
z
xy
stt JHH 12
By applying Ampere’s law
C
IdlH
Magnetic boundary conditions
The result is generalized to a vector form:
Where However, surface currents can exist only
on the surfaces of perfect conductors and perfect superconductors (infinite conductivities).
Hence, at the interface between media with finite conductivities, Js=0. Thus:
sJHHn 212ˆ
2 medium fromaway pointing vector normal theis ˆ 2n
tt HH 21
Example
tt HH 21
))(( 0 r
xy (plane)
• Solution:1)H1t = H2t thus, H2t = 6ax + 2ay
2)Hn1 = 3az,
but, Hn2 = ??
6000μ0(3az) = 3000 μ0(Hn2)
Hn2 = 6az
thus, H2 =6ax + 2ay + 6az
))(( 0 r
μr1 = 6000 ; μr2 = 3000 ;
Inductance
An inductor is the magnetic analogue of an electrical capacitor.
Capacitor can store electric energy in the electric field present in the medium between its conducting surfaces.
Inductor can store magnetic energy in the volume comprising the inductors.
)(1
1212 H
IL
INDUCTANCEstore magnetic
energy
H I
L
32 J/m 2
1H
v
Ww m
m
Solenoid
Inside the solenoid:
Inductance
Example of an inductor is a solenoid - a coil consisting of multiple turns of wire wound in a helical geometry around a cylindrical core.
Magnetic Field in a Solenoid For one cross section of
solenoid,
When l >a, θ1≈−90° and θ2≈90°,
Where, N=nl =total number of turns
over the length l
12 sinsin2
ˆ
nIzB
l
NInI
al with solenoidlong For
zzB
1/
Self Inductance
The self-inductance of a circuit is used to
describe the reaction of the circuit to a
changing current in the circuit,
(The ratio of the magnetic flux to the
current)
The self-inductance of a circuit is used to
describe the reaction of the circuit to a
changing current in the circuit,
(The ratio of the magnetic flux to the
current)
Self Inductance
Self-inductance of any conducting structure is the ratio of the magnetic flux linkage, Λ to the current I flowing through the structure.
Magnetic flux linkage, Λ is the total magnetic flux linking a given conducting structure.
H I
L
(Wb) 2
IS
l
NN
Self Inductance
Magnetic flux, linking a surface S is given by:
In a solenoid with uniform magnetic field, the flux linking a single loop is:
Wb
S
dsB
loop the of area sectional-cross Swhere
ISl
N
dsIl
NS
zz
Wb d flux, Magnetic•
S sB
ISl
N
dsIl
NS
zz
l
NIzB
Sd
S s
Self Inductance – magnetic flux in solenoid
Self Inductance
Magnetic flux, linking a surface S is given by:
In a solenoid with uniform magnetic field, the flux linking a single loop is:
Wb S
dsB
loop the of area sectional-cross Swhere
ISl
N
dsIl
NS
zz
Self Inductance
For a solenoid:
For two conductor configuration:
solenoid S2
l
NL
S
dIII
L sB 1
Self Inductance for a solenoid
H I
L
ISl
NNN
(Wb) IS
l
N
2
Thus,
HI
ISl
N
I
L
2
Sl
NL
2
Mutual Inductance
Mutual inductance – produced by magnetic coupling between two different conducting structures.
Mutual Inductance Magnetic field B1 generated by current I1 results
in a flux Φ12 through loop 2:
If loop 2 consists of N2 turns all coupled by B1 in exactly the same way, the total magnetic flux linkage through loop 2 due to B1 is:
2 112
SdSB
2 1212212
SdNN SB
Mutual Inductance
Hence, the mutual inductance:
H dI
N
IL
s
2
sB11
2
1
1212
2 112
SdSB
2 1212212
SdNN SB
Inductance
Magnetic Flux
Flux Linkage
Inductance
Solenoid
Magnetic Energy
Consider an inductor with an inductance L connected to a current source.
The current I flowing through the inductor is increased from zero to a final value I.
The energy expended in building up the current in the inductor:
i.e the magnetic energy stored in the inductor
2
02
1LIidiLivdtpdtW
l
m
Magnetic Energy
Magnetic energy density (for solenoid):
i.e magnetic energy per unit volume
Magnetic energy in magnetic field:
322
1J/m H
v
Ww m
m
32
1J/m BdvHW
vm
The magnetic field in the region S between the two conductors is approximately
Example 5-7: Inductance of Coaxial Cable
Total magnetic flux through S:
Inductance per unit length:
Magnetic Energy Density
Magnetic field in the insulating material is
The magnetic energy stored in thecoaxial cable is
Summary