Download ELEC 3105 Basic EM and Power Engineering

Compare to principle of
superposition applied to
determine the electric
field inside and outside
an infinite parallel plate
capacitor
ELEC 3105 BASIC EM AND POWER
ENGINEERING
Force and torque on Magnetic Dipole
Magnetic dipole =
product of current in loop
with surface area of loop
DEFINITION OF MAGNETIC DIPOLE
I = current in loop
A “S” = surface area of the loop
𝑛 = unit vector normal to loop surface - RHR
Magnetic moment
𝑚 vector in the direction of 𝑛
𝑚 = 𝐼𝐴 = 𝐼"𝑆"
Units of Am2
ẑ 
FORCE ON A MAGNETIC DIPOLE
Consider a circular ring of
current I placed at the end of a
solenoid as shown in the
figure. The current in the
solenoid produces a magnetic
field in which the current loop
is placed into. By postulates 1
and 2 of magnetic fields, the
current ring will be subjected
to a magnetic force.
out of page
into page
B
I
ẑ 
FORCE ON A MAGNETIC DIPOLE
F
Circular ring
F
out
out
F
down
F
out
F
B
F
down
I
Cancel in pairs around the ring
out
F
down
F
down
Will add in same direction on ring giving a net force.
ẑ 
FORCE ON A MAGNETIC DIPOLE
F
Circular ring
F
out
out
F
2r
down
F
down
F
B
F
down
I
Will add in same direction on ring giving a
net force.
down
Using postulate 1:
Gives:
We need for find Br
  
dF  I  Bd
Fdown  2rIBr

B
B
z
B
r
F
down
FORCE ON A MAGNETIC DIPOLE
z
Gaussian cylinder
B
z
We will relate Br to
z
z
Total magnetic flux through Gaussian
cylindrical surface must be zero. As
many magnetic field lines that enter the
surface, leave the surface. No magnetic
 
charges or monopoles.   B  0
Another important property of B
 
Recall
  B  0 everywhere
No net magnetic flux
through any closed
surface.
Closed surface S


 B  da  0
S
Using divergence theorem


   B dv
vol
0
vol
0
11
3-D view
FORCE ON A MAGNETIC DIPOLE
z
Flux through top:
r B  z  z   
2
z
top
z
Flux through side:
2rzB  
r
side
Flux through bottom:
 r B z   
2
z
bottom
  
side
top
bottom
0
3-D view
FORCE ON A MAGNETIC DIPOLE
  
side
top
z
0
bottom
z
2rzB  r B z  z   r B z   0
2
r
2
z
z
r B  z  z   B  z 
B 
2
z
z
z
r
r B
B 
2 z
z
r
We can now use this in our force on current ring expression
3-D view
ẑ 
FORCE ON A MAGNETIC DIPOLE
F
Circular ring
B
F
out
out
F
down
2r
I
F
down
We have found Br
r B
F  2rIB
B 
z
down
r
2 z
r

B
B
z
F
down
r B
 2rI
2 z
B
r
z
F
down
ẑ 
FORCE ON A MAGNETIC DIPOLE
Circular ring
F
F
out
out
2r
F
I
F
down
down
F
r B
 2rI
2 z
z
down
F
down
F
down
B
B
 A I
z
z

F
down
B
 r I
z
2
z
B
 m
z

B
B
z
z
B
F m
z Force pulls dipole into region
z
z
of stronger magnetic field
B
r
F
down
FORCE ON A MAGNETIC DIPOLE
z
In general
Fx  m
Fy  m
Fz  m
Bx
x
B y
y
Bz
z
Solenoid with axis along x
Solenoid with axis along y
Solenoid with axis along z
Form suggests some sort of
dot product with the del
operator
3-D view
FORCE ON A MAGNETIC DIPOLE
z
In general
 
F  m  B
 
F  m  B
 
F  m  B
x
x
y
y
z
z

  
F  m   B
3-D view
TORQUE ON A MAGNETIC DIPOLE

m

B

Side view
We will consider a dipole in a uniform magnetic field. We
can use any shape we want for the dipole. Here we will
select a square loop of wire.
I out of page
a
I into page
a
I
Wire loop
Top
view
TORQUE ON A MAGNETIC DIPOLE
a
a
2
a
a
I
Wire loop
Top
view
TORQUE ON A
MAGNETIC DIPOLE

m

F

Side view
Top
view
a

B

F
a
Pivot point
I
Pivot line
Torque attempts to align dipole
Wire loop
moment

m
with

B.
TORQUE ON A MAGNETIC DIPOLE

m

F

B

a
2
Side view

F
  
  r F
Total torque
  2F
Pivot point
a
sin  
2
F => Magnetic force on wire of length a
Torque attempts to align dipole
moment

m
with

B.
TORQUE ON A
MAGNETIC DIPOLE

m

F

a
2
Side view
  2F

F
a
sin  
2
Pivot point
F => Magnetic force on wire of length a
F  IBa
Then
Through postulate 1 for magnetic fields
  a 2 IB sin  

B
TORQUE ON A
MAGNETIC DIPOLE

m

F

B

a
2
Side view
  a 2 IB sin  
 
  m B sin  
  
  m B

F
Pivot point
a
a

m  a2 I
I
Wire loop
ELEC 3105 BASIC EM AND POWER ENGINEERING
Boundary conditions
Inductance
Magnetic energy
Principle of virtual work
21
NORMAL COMPONENT OF B FIELD AT BOUNDARY
 
 B  dA  0
S
Gaussian surface

B
2
Area A
B normal 
2

Interface

Very thin
T  0
2
1
B normal 
1

B
Area A
1
Net flux through a closed surface is zero.
B2 normal   B1 (normal )
The normal components of B are continuous across the interface
22
TANGENTIAL COMPONENT OF H FIELD AT BOUNDARY


H
 
H  d  I  0


B  H
2
H tangential
S
2
Square closed path
Length L

Interface

Very thin
2
1
Length L
T  0

H
H tangential
1
1
Integral of H around closed path is equal to the current enclosed (I = 0)
H 2 tangential  H1 ( tangential)
The tangential components of H are continuous across the interface
THIS BOUNDARY CONDITION ASSUMES NO SURFACE CURRENT AT THE INTERFACE.
23
TANGENTIAL COMPONENT OF H FIELD AT BOUNDARY

 
H  d  I

H
2
Square closed path
Very thin
2
H tangential
S
Interface


B  H
Length L

X
X
X
X
X
X

H
H tangential  H (tangential)
X
X
X
X
X
X
X
X
X

1
Length L
T  0
Surface
Current K2
1
X
2
H tangential
1
1
H 2 tangential  H1 ( tangential)  K
The tangential components of H are discontinuous across the interface
THIS BOUNDARY CONDITION ASSUMES A SURFACE CURRENT AT THE INTERFACE.
24
SUMMARY OF BOUNDARY CONDITIONS (GENERAL)
25
SUMMARY OF BOUNDARY CONDITIONS (CONDUCTORS)
26
SUMMARY OF BOUNDARY CONDITIONS (CONDUCTORS)
27
SELF INDUCTANCE
Introduction
A transformer is a device in which the current in one circuit induces an
EMF in a second circuit through the changing magnetic field.
B, H, and M relationship
17
To understand
how current in
one circuit
induced EMF in
another, we will
first examine how
a current in a
circuit can induce
an EMF in the
same circuit.
28
SELF INDUCTANCE
INDUCTANCE
SELF
Consider a single wire loop
Enclosed surface S
Current
 in loop produces a magnetic
field B , giving a flux through the
loop.

B

Bi
Biot-Savard Law
This expression is
The Biot-Savard Law
Consider a small segment of wire of overall length
d
Thus:
i
i
The Biot-Savard law applied to the small segment gives an
element of magnetic field dB at the point P.
 
 I d  rˆ21
dBr1   o
4 r21 2
i

dB
P
Magnetostatics
Same result as
postulate 2 for the
magnetic field
POSTULATE 2 FOR THE MAGNETIC FIELD

A current
element I d produces a magnetic

field B which at a distance R is given by:

  I  Rˆ
dB  o
d
4 R 2

dB

r
21
v
I
Units of {T,G,Wb/m2}
26
Lecture 15 slid 26
d
13
From Biot-Savard Law
WRITE:
   Li
29
SELF INDUCTANCE
Consider a single wire loop
Enclosed surface S
Current
 in loop produces a magnetic
field B , giving a flux through the
loop.

B
   Li
i
i
L is the self inductance of the loop
d
di
v
L
dt
dt
v
emf  

t
30
SELF INDUCTANCE
Consider a single wire loop
Enclosed surface S
Current
 in loop produces a magnetic
field B , giving a flux through the
loop.

B
   Li
i
It is difficult to compute L for a simple wire loop since the
magnetic field produced by the loop is not constant
across the surface of the loop.
A possible solution is to find B at center of loop and then
approximate:
B S
center
i
v
v
d
di
L
dt
dt
31
SELF INDUCTANCE
A simple example for the calculation of a self inductance is the long solenoid.
Magnetic field of a long solenoid
Current out of page
Axis of solenoid
Magnetic field of a long solenoid
Magnetic field of a long solenoid
In the vicinity of the point P

Bb  0
Current out of page
N : number of turns enclosed by length L
Current out of page
P
1
3
2
4
11
Current into page
3
5
Axis of solenoid
41
 
B  0
P
2
P
Infinite coil of wire carrying a current I
Evaluate B field here
5
4
resultant

B
Current into page
Expect B to lie along axis
of the solenoid
Implies that B field has no radial component. I.e. no component pointing towards42
or away from the solenoid axis.
Magnetic field of a FINITE solenoid
Magnetic field of a FINITE solenoid
Current out of page
NIr
dI 
d
L sin  
Axis of solenoid
a
sin   
r
B
dB 
P
Current into page
dB 
L
z
 a NIr
2
2r
finite solenoid start
2
3
La
Magnetic field of a FINITE solenoid
d

2 L

2
1
1 2 3 4 5
2
z
r 
 a NI
o
2r L
r
d
d
L
dB
We can now sum (integrate) the expression for
over the
angular extent of the coil. I.e. sum over all the rings of the finite
length solenoid.
 NI
cos    cos  
B 

 NI
2 L
B 
sin  d
 2
L
  NI
cos   cos zˆ
B
2 L
z
a
o
z
2
o
50
3
d 
 NI
•B is independent of distance from the axis of
the long solenoid as we are inside the
solenoid!
•B is uniform inside the long solenoid.
d

z
28
2
o
dB 
Evaluate B field here
Cross-section cut through solenoid axis
o
z
finite coil of wire carrying a current I
Radius of solenoid is a.
 dI a
L
infinite solenoid (36)
sub in
a
 o NI
1
2
o
sin  d
z
1
dB
o
1
z
34
2
35
32
SELF INDUCTANCE
Current out of page
Long solenoid of length


B
N turns of wire carrying current I
AREA
A
  NI
B

o

B is uniform over the cross-section of the solenoid
33
SELF INDUCTANCE
Long solenoid of length
  NI
B

o

B

AREA
A
Flux through one loop of area A
 
1
 NIA
o

34
SELF INDUCTANCE

Long solenoid of length

B
  NI
B

AREA
A
Flux through all N loops of solenoid
o
From
  LI
  N 
N
Then
L
1
 N IA
2
o

N A
2
o

35
SELF INDUCTANCE

Long solenoid of length
  NI
B

o
  LI
L
N A
2
o

AREA
A
Self inductance of a long
solenoid of N turns with a
current I in the windings. The
solenoid has cross-sectional
area A.
36
EXAMPLE: SELF INDUCTANCE
𝐼
Calculate the “self inductance”
per unit length for a segment of
a coax cable. Inner radius (a),
outer radius (b).

Example completed in class
37
Energy in Magnetic Field
Consider a long solenoid in order to develop a general expression for the energy stored in a magnetic field.
Magnetic field of a long solenoid
Current out of page
Axis of solenoid
Magnetic field of a long solenoid
Magnetic field of a long solenoid
In the vicinity of the point P

Bb  0
Current out of page
N : number of turns enclosed by length L
Current out of page
P
1
3
2
4
11
Current into page
3
5
Axis of solenoid
41
 
B  0
P
2
P
Infinite coil of wire carrying a current I
Evaluate B field here
5
4
resultant

B
Current into page
Expect B to lie along axis
of the solenoid
Implies that B field has no radial component. I.e. no component pointing towards42
or away from the solenoid axis.
Magnetic field of a FINITE solenoid
Magnetic field of a FINITE solenoid
Current out of page
NIr
dI 
d
L sin  
Axis of solenoid
a
sin   
r
B
dB 
P
Current into page
dB 
L
z
 a NIr
2
2r
finite solenoid start
2
3
La
Magnetic field of a FINITE solenoid
d

2 L

2
1
1 2 3 4 5
2
z
r 
 a NI
o
2r L
r
d
d
L
dB
We can now sum (integrate) the expression for
over the
angular extent of the coil. I.e. sum over all the rings of the finite
length solenoid.
 NI
cos    cos  
B 

 NI
2 L
B 
sin  d
 2
L
  NI
cos   cos zˆ
B
2 L
z
a
o
z
2
o
50
3
d 
 NI
•B is independent of distance from the axis of
the long solenoid as we are inside the
solenoid!
•B is uniform inside the long solenoid.
d

z
28
2
o
dB 
Evaluate B field here
Cross-section cut through solenoid axis
o
z
finite coil of wire carrying a current I
Radius of solenoid is a.
 dI a
L
infinite solenoid (36)
sub in
a
 o NI
1
2
o
sin  d
z
1
dB
o
1
z
34
2
35
38
Energy in Magnetic Field
Current out of page
May have core with
constant permeability
Long solenoid of length
 NI
B



AREA
A
N turns of wire carrying current I
Find work done by current source in building up magnetic field:
Power  V  I
V
d
dI
L
dt
dt
39
ENERGY
IN MAGNETIC
FIELD
Energy
in Magnetic
d
dI
V
L
dt
dt
dW
 Power  V  I
dt
THEN
I
W   LI dI 
Field
d
dW 
 I  dt
dt
THEN
0
THEN
LI
W
2
2
d
dW 
 I  dt
dt
Energy stored
40
ENERGY
IN MAGNETIC
FIELD
Energy
in Magnetic
LI
W
2
W
2
Field
N A
2
L

Energy stored
N AI
2
B
2
2
NI

For  core solenoid
1  N I 
W

 A 
2  

2
2
2
2
enclosed volume
1
W
B  A 
2
2
41
ENERGY
IN MAGNETIC
FIELD
Energy
in Magnetic
Field
Total magnetic energy stored in solenoid W  1 B 2  A 
2
W
Energy density
1
2
 B dv
2  vol
W
VOLUME
2
W
B

VOLUME
2
EXPRESSION
VALID
FOR
ALL
42
ENERGY
IN MAGNETIC
FIELD
Energy
in Magnetic
Energy in Magnetic Field
Energy in Magnetic Field
Current out of page
May have core with
constant permeability
Long solenoid of length

 NI
B


dW
 Power  V  I
dt
AREA
A
THEN
dW  L
I
W   LI dI 
N turns of wire carrying current I
d
dI
L
dt
dt
d
 I  dt
dt
THEN
dW  L
THEN
Find work done by current source in building up magnetic field:
d
dI
V
L
Power  V  I
dt
dt
Energy in Magnetic Field
V
0
W
LI
2
2
Field
Total magnetic energy stored in solenoid
Energy density
1
B  A 
2
2
W
VOLUME
2
d
 I  dt
dt
W
W
B

VOLUME
2
Expression
valid
for
all
Energy stored
Energy in Electric Field
43
Energy in Electric Field
For electric fields, we argued that the energy was really
stored in the potential energy of the charged particle’s
positions, since it would require that much energy to take
separate charges and form that distribution from a universe
with equally distributed charges.
Energy in Magnetic Field
This is harder to do for magnetic fields since there are no
magnetic charges. But one possible approach is to take
current loops enclosing zero area, and consider the forces
on the wires as we expand the loops so as to form the
current distributions which generate the magnetic field.
44
PRINCIPLE OF VIRTUAL WORK (MAGNETIC)
We can use the principle of virtual work to determine forces as we did for electric forces.
Gives correct magnitude
F
mag
U

s
mag
Energy stored in magnetic field
Position variable
Forces in Electrostatics
Conductor caries a surface charge of density  Find force
on plates of a parallel plate capacitor. Plate area A
x
U
d
L
2
xLS  yd 
y

F
F

E
d
s
oE 2
U  o E 2 xD

y
2
L
Force pulling metal insert into capacitor
Be very careful using the principle of virtual work
F
 o E 2 xD
2
45
PRINCIPLE OF VIRTUAL WORK (MAGNETIC)
Magnetic Relays
46
PRINCIPLE OF VIRTUAL WORK (MAGNETIC)
Use principle of virtual work to obtain
expression for the magnetic force on
the movable contact.
I
Magnetic Relays
GAP
Movable contact
V
Metal spring provides
restoring force when
current is zero
Example completed in class
47
MUTUAL INDUCTANCE
Enclosed surface S1

B
Enclosed surface S2


2
1
i
i
2
1
i
v 
v 
Loop 1
Loop 2
1
2
We shall consider two current loops close together.
48
MUTUAL INDUCTANCE

B
Suppose current i1 flows in loop 1,
creating a flux  in the loop and a
flux  in loop 2. We will set the
source current i2 zero for now.
1
12


1
S
S
1
2
2


   B  da
i
1
12
1
2
S2
v 
1
Loop 1
Loop 2
Integral over loop 2 surface
Magnetic field of loop 1 in the region of loop 2
Flux of loop 2 produced by current in loop 1
49
Now some math!!!!
MUTUAL INDUCTANCE


   B  da
Using magnetic vector potential
 

     A  da
Using Stoke’s theorem


   A  d
Using definition of magnetic vector potential

 i  d  
 
  d 
 
4  r 



d  d
Rearrange terms   i

4
r
12
1
2
S2
1
12
2
S2
12
2
1
2
o 1
12
2
1
1
2
21
o
12
1
1
2 1
21
50
2
MUTUAL INDUCTANCE



d  d
 i

4
r
o
12
1
1
2 1
  i M
12
1
2
21
12
Constant that depends on loop geometry
FLUX IN LOOP 2 DUE TO CURRENT IN LOOP 1
51
MUTUAL INDUCTANCE

B
Suppose current i2 flows in loop 2,
creating a flux  in the loop and a
flux  in loop 1. We will set the
source current i1 zero for now.
2
21


1
S
S
1
2
2
i
2


   B  da
21
2
1
S1
v 
1
Loop 1
Loop 2
Integral over loop 1 surface
Magnetic field of loop 2 in the region of loop 1
Flux of loop 1 produced by current in loop 2
Now some math!!!!
MUTUAL INDUCTANCE


   B  da
Using magnetic vector potential
 

     A  da
Using Stoke’s theorem


   A  d
Using definition of magnetic vector potential

 i  d  
 
  d 
 
4  r 



d  d
 i

Rearrange terms
4
r
21
2
1
S1
2
21
1
S1
21
1
2
1
o 2
21
1
2
2
1
12
o
21
2
2
1 2
1
12
53
MUTUAL INDUCTANCE



d  d
 i

4
r
o
21
2
2
1 2
 i M
21
2
1
12
21
Constant that depends on loop geometry
FLUX IN LOOP 1 DUE TO CURRENT IN LOOP 2
54
MUTUAL INDUCTANCE



d  d
 i

4
r
o
12
1
1
2 1
  i M
12
1



d  d
 i

4
r
o
2
21
21
12
2
Conclusion
M’s are geometrical factors
2
1
1 2
12
 i M
21
2
21
M  M M
12
21
MUTUAL INDUCTANCE BETWEEN LOOPS
55
MUTUAL INDUCTANCE
General result
Mutual Inductance
Enclosed surface S1

B
Enclosed surface S2
v 
1


d d
di
di

L
M
dt
dt
dt
dt
1
21
1
2
1
2
1
i
i
v 
2
1
2
i
v 
 v2 
Loop 1
Loop 2
d d
di
di

M
L
dt
dt
dt
dt
2
12
1
2
2
Sign convention
1
We shall consider two current loops close together.
Indicates v2 positive when v1 is positive

v
1

i
i
1
2
primary

v
2

56
ELEC 3105 BASIC EM AND POWER
ENGINEERING
START Extra
57
EXTRA ON INDUCTANCE
58
EXTRA ON INDUCTANCE
59
EXTRA ON INDUCTANCE
60
EXTRA ON INDUCTANCE
61
EXTRA ON INDUCTANCE
62
EXTRA ON INDUCTANCE
63
EXTRA ON INDUCTANCE
64