elementos de electromagnetismo sadiku 2 edicion

Lecture 26
97.315 Basic E&M and Power Engineering Topic: Inductance
TITLE
SELF INDUCTANCE
ENERGY STORED IN A
MAGNETIC FIELD
MUTUAL INDUCTANCE

Lecture 26
OUTLINE
• Theory
• Self inductance
• Examples of calculation
• Self inductance of a long solenoid
• Theory
• Energy stored in a magnetic field
• Mutual Inductance
• Examples of calculation
• Inductance calculations
• Assignment
• References
• Summary
Walk in the park

Lecture 26
THEORY
SELF INDUCTANCE
A transformer is a device in which the current in
one circuit induces an EMF in a second circuit
through the changing magnetic field.
Introduction
Lecture 23
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.
Lecture 23
97.315 Basic E&M and Power Engineering Topic: Magnetization
THEORY
THEORY
B, H, AND M RELATIONSHIP
R
NI
B o
o


2

I
V
voltmeter
An arrangement to measure the magnetic field
inside a toroid. The subscript Bo denotes that the
interior of the toroid is void of magnetic material.

Lecture 26
THEORY
SELF INDUCTANCE
Consider a single wire loop

 v
i
i
B

Enclosed surface S
Current in loop produces a
magnetic field , giving a
flux through the loop.
B

From Biot-Savard Law i
B 

Thus: i


WRITE: Li



Lecture 21
97.315 Basic E&M and Power Engineering Topic: Biot-Savard
REVIEW
REVIEW
BIOT-SAVARD LAW
  2
21
21
1
ˆ
4 r
r
d
I
r
B
d o



 



Consider a small segment of wire of overall length
I

d
P
21
r


d
B
d

Same result as
postulate 2 for the
magnetic field
Lecture 16
The Biot-Savard law applied to the small segment gives an
element of magnetic field at the point P.
B
d

21
r̂
Lecture 16
97.315 Basic E&M and Power Engineering Topic: H,B BASICS
THEORY
THEORY
Magnetostatics
Postulate 2 for the magnetic field
A current element produces a magnetic
field which at a distance R is given by:



d
R
R
I
B
d o
2
ˆ
4






d
I
B

Units of {T, G, Wb/m2}

d
I I
B
d
 R
R̂

Lecture 26
THEORY
SELF INDUCTANCE

 v
i
i
B

Enclosed surface S
B

Li



L is the self inductance of the loop
dt
di
L
dt
d
v 



t
emf






Lecture 26
THEORY
SELF INDUCTANCE

 v
i
i
B

Enclosed surface S
B

Li



dt
di
L
dt
d
v 



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:
S
Bcenter



Lecture 26
THEORY
SELF INDUCTANCE
A simple example for the calculation of a self inductance is the the long solenoid.
Lecture 21
Lecture 17
Lecture 17
97.315 Basic E&M and Power Engineering Topic: Ampere's Law
EXAMPLE
EXAMPLE
Example (Question)
Obtain an expression for the electric field at a point
inside a long solenoid.
Current out of page I
Current into page I
Infinite coil of wire carrying a current I
Axis of solenoid
P
Evaluate B field here


spring
Lecture 17
EXAMPLE
EXAMPLE
Example (Solution)
Obtain an expression for the electric field at a point inside a long
solenoid.
1 2 3 4 5
P
P
3
1
2 4 5
1
resultant
Expect B to lie along
axis of the solenoid
B

1
B
d

2
B
d

3
B
d

4
B
d

5
B
d

Blow up
of region
about
point P
Fields produced at P
Lecture 17
EXAMPLE
EXAMPLE
Example (Solution)
Obtain an expression for the electric field at a point inside a long solenoid.
0

b
B

P
L
NI
B o


N : number of turns enclosed by length L
•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.
• Direction of B from right hand rule
x
L
NI
B o
ˆ


 x̂
Current out of page
Current into page
END
Lecture 21
EXAMPLE
EXAMPLE
Example (Question)
Obtain an expression for the magnetic along the axis of a long but finite
length solenoid. See figure for dimensions.
1 2 3 4 5
Axis of solenoid
Current out of page
z
dB

d
Segment of the solenoid coil
r


d
 


sin
rd
d 

arc length 
rd

a
 
r
a


sin

Develop a few
relations
Lecture 21
EXAMPLE
EXAMPLE
In Lecture 17 we examined the magnetic field
inside an infinitely long solenoid. We found that
no magnetic field existed on the outside of the
solenoid and that inside the magnetic field was
uniform and directed along the axis.
Example (Question)
Obtain an expression for the magnetic along the axis of
a long but finite length solenoid. See figure for dimensions.
Current out of page
Current into page
finite
finite coil of wire carrying a current I
Axis of solenoid
P
a
Radius of solenoid is a.
Cross-section cut through solenoid axis
L
Lecture 21
EXAMPLE
EXAMPLE
Example (Question)
Obtain an expression for the magnetic along the axis of a long but finite length
solenoid. See figure for dimensions.
L

d
90
1


180
2


z
   
 
180
cos
90
cos
2


L
NI
B o
z

z
L
NI
B o
ˆ
2



L
NI
B o
z
2


Magnetic field is ½ that of center
END

Lecture 26
THEORY
SELF INDUCTANCE
Current out of page
N turns of wire carrying current I
is constant over the cross-section of the solenoid
B

B

AREA
A
Long solenoid of length

 NI
B o




Lecture 26
THEORY
SELF INDUCTANCE
B
 AREA
A

 NI
B o



Flux through one loop of area A

NIA
o


1
is constant over the cross-section of the solenoid
B

BA

1

Lecture 26
THEORY
SELF INDUCTANCE
AREA
A

 NI
B o



Flux through all N loops of solenoid

IA
N
N o
N
2
1





B

From LI


Then

A
N
L o
2



Lecture 26
THEORY
SELF INDUCTANCE
AREA
A

 NI
B o



LI



A
N
L o
2


Self inductance of a long
solenoid of N turns with a current
I in the windings. The solenoid
has cross-sectional area A.

Lecture 26
THEORY
Lecture 21
Lecture 17
Lecture 17
EXAMPLE
EXAMPLE
Example (Question)
Obtain an expression for the electric field at a point
inside a long solenoid.
Current out of page I
Current into page I
Infinite coil of wire carrying a current I
Axis of solenoid
P


spring
Lecture 17
EXAMPLE
EXAMPLE
Example (Solution)
Obtain an expression for the electric field at a point inside a long
solenoid.
1 2 3 4 5
P
P
3
1
2 4 5
1
resultant
Expect B to lie along
axis of the solenoid
B

1
B
d

2
B
d

3
B
d

4
B
d

5
B
d

Blow up
of region
about
point P
Fields produced at P
Lecture 17
EXAMPLE
EXAMPLE
Example (Solution)
Obtain an expression for the electric field at a point inside a long solenoid.
0

b
B

P
L
NI
B o


N : number of turns enclosed by length L
•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.
• Direction of B from right hand rule
x
L
NI
B o
ˆ


 x̂
Current out of page
Current into page
END
Lecture 21
EXAMPLE
EXAMPLE
Example (Question)
Obtain an expression for the magnetic along the axis of a long but finite
length solenoid. See figure for dimensions.
1 2 3 4 5
Axis of solenoid
Current out of page
z
dB

d
Segment of the solenoid coil
r


d
 


sin
rd
d 

arc length 
rd

a
 
r
a


sin

Develop a few
relations
Lecture 21
EXAMPLE
EXAMPLE
In Lecture 17 we examined the magnetic field
inside an infinitely long solenoid. We found that
no magnetic field existed on the outside of the
solenoid and that inside the magnetic field was
uniform and directed along the axis.
Example (Question)
Obtain an expression for the magnetic along the axis of
a long but finite length solenoid. See figure for dimensions.
Current out of page
Current into page
finite
finite coil of wire carrying a current I
Axis of solenoid
P
a
Radius of solenoid is a.
Cross-section cut through solenoid axis
L
Lecture 21
EXAMPLE
EXAMPLE
Example (Question)
Obtain an expression for the magnetic along the axis of a long but finite length
solenoid. See figure for dimensions.
L

d
90
1


180
2


z
   
 
180
cos
90
cos
2


L
NI
B o
z

z
L
NI
B o
ˆ
2



L
NI
B o
z
2


Magnetic field is ½ that of center
END
Consider a long solenoid in order to develop a general expression for the energy stored in a
magnetic field.
ENERGY IN MAGNETIC FIELD

Lecture 26
THEORY
Current out of page
AREA
A

 NI
B



May have core with
constant permeability 
Find work done by current source in building up magnetic field:
I
V
Power 

dt
dI
L
dt
d
V 



Lecture 26
THEORY
dt
dI
L
dt
d
V 


I
V
Power
dt
dW



THEN
dt
I
dt
d
dW 



THEN
t
d
I
t
d
d
dW 



 


I
I
d
I
L
W
0
THEN
2
2
LI
W 
Energy stored

Lecture 26
THEORY
2
2
LI
W 
Energy stored 
A
N
L
2



 NI
B


For  core solenoid

2
2
2
AI
N
W


 


A
I
N
W 





 2
2
2
2
2
1 

 

A
B
W 2
2
1


enclosed volume
of solenoid
For long solenoid

Lecture 26
THEORY

2
2
B
VOLUME
W

Energy density
VOLUME
W
 

A
B
W 2
2
1


Total magnetic energy stored in solenoid


vol
dv
B
W 2
2
1
Energy density

2
2
B

EXPRESSION
VALID
FOR
ALL

Lecture 26
THEORY
Energy in Magnetic Field
Lecture 26
Lecture 26
THEORY
THEORY

2
2
B
VOLUME
W

Energy density
VOLUME
W
 

A
B
W 2
2
1


Total magnetic energy stored in solenoid


vol
dv
B
W 2
2
1
Energy density

2
2
B

EXPRESSION
VALID
FOR
ALL
Lecture 26
THEORY
THEORY
dt
dI
L
dt
d
V 


I
V
Power
dt
dW



THEN
dt
I
dt
d
dW 



THEN
t
d
I
t
d
d
dW 



 


I
I
d
I
L
W
0
THEN
2
2
LI
W 
Energy stored
Lecture 26
THEORY
THEORY
Current out of page
AREA
A

 NI
B



May have core with
constant permeability

I
V
Power 
 dt
dI
L
dt
d
V 



Lecture
97.315 Basic E&M and Power Engineering Topic: Poisson’s equ.
TEXT
TEXT
Reference (8) page 172
Lecture 26
THEORY
Lecture 6
Energy in Electric Field
Lecture
THEORY
THEORY
+ -
+Q -Q
A
Q
E
o
o
s






Consider a capacitor at potential difference
V and of charge +Q , -Q on the plates.
Area of plates (A) and spacing (D)
Energy stored in the capacitor: 2
2
2
CV
QV
U 

But:
 
AD
E
D
V
AD
D
AV
CV
U o
o
o
2
2
2
2
2
2
2
2













D
A
 
plates
between
volume
2
2
E
U o


Energy stored in electric field
V
D
A
C o

 and D
V
E 
Lecture
THEORY
THEORY
In general for any volume where electric field exists:
Energy stored is:


Volume
o
dv
E
U 2
2

Potential energy stored in electrostatic field
Energy stored in electric field

Lecture 26
THEORY
For electric fields, we argued that the energy was really
stored in the potential energy of the particles positions, since
it would require that much energy to take separate charges
and form that distribution from a universe with equally
distributed charges.
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.
Energy in Electric Field

Lecture 26
THEORY
We can use the principle of virtual work to determine forces as
we did for electric forces.
Lecture 9
Be very careful using the virtual work principle
s
U
F mag
mag



Energy stored in magnetic field
Position variable
Gives correct magnitude
Lecture 10
97.315 Basic E&M and Power Engineering Topic: Virtual work
EXAMPLE
EXAMPLE
Example (Solution)
Using the principle of virtual work obtain an expression for the force
on a plate of a parallel plate capacitor. The plates are oppositely charged (+Q, -
Q) and separated by a distance S. Assume that the plates have an area A.
o
E

S
S
F

F

+Q
-Q
2
2
2
CV
QV
U 

We have shown in lecture 6 that the electrical energy stored in the electric field
between the plates of a parallel plates capacitor is given by:
where
S
A
C o

 and S
E
V o

Lecture 10
EXAMPLE
EXAMPLE
Example (Solution)
o
E

S
S
F

F

+Q
-Q
2
2
2
CV
QV
U 

S
A
C o


S
E
V o

 
A
S
Q
A
Q
AS
AS
E
U
o
o
o
o
o




2
2
2
2
2
2











An expression of the energy in terms of
plate separation S
A
Q
E
o
o


Lecture 10
EXAMPLE
EXAMPLE
Example (Solution)
o
E

S
S
F

F

+Q
-Q
A
S
Q
U
0
2
2

We can now apply the principle of virtual work to obtain the
force on the plates
2
2
2
2
o
o
o
QE
A
Q
Q
A
Q
S
U
F 







S
U
F


 With

Lecture 26
THEORY
MUTUAL INDUCTANCE

 2
v
2
i
Enclosed surface S2

 1
v
1
i
B

Loop 1
Loop 2
Enclosed surface S1
1
 2

We shall consider two current loops close together.

Lecture 26
THEORY
MUTUAL INDUCTANCE

 1
v
1
i
B

Loop 1 Loop 2
1
 2

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

 


2
2
1
12
S
a
d
B


Magnetic field of loop 1 in the region of loop 2
Integral over loop 2 surface
Flux of loop 2 produced by current in loop 1
Now some math!!!!
1
S 2
S

Lecture 26
THEORY
MUTUAL INDUCTANCE
 


2
2
1
12
S
a
d
B


 
 




2
2
1
12
S
a
d
A



Using magnetic vector potential
Using Stoke’s theorem
 


2
2
1
12



d
A
Using definition of magnetic vector potential
 











 
2 1
2
21
1
1
12
4




d
r
d
i
o


 



 
2 1
21
2
1
1
12
4 r
d
d
i o






Rearrange terms

Lecture 26
THEORY
MUTUAL INDUCTANCE
 



 
2 1
21
2
1
1
12
4 r
d
d
i o






12
1
12
M
i



Constant that depends on loop geometry
Flux in loop 2 due to current in loop 1

 1
v
1
i
B

Loop 1 Loop 2
1
 2

1
S 2
S

Lecture 26
THEORY
MUTUAL INDUCTANCE

 1
v
2
i
B

Loop 1 Loop 2
1
 2

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
1
2
21
S
a
d
B


Magnetic field of loop 2 in the region of loop 1
Integral over loop 1 surface
Flux of loop 1 produced by current in loop 2
Now some math!!!!
1
S 2
S

Lecture 26
THEORY
MUTUAL INDUCTANCE
 


1
1
2
21
S
a
d
B


 
 




1
1
2
21
S
a
d
A



Using magnetic vector potential
Using Stoke’s theorem
 


1
1
2
21



d
A
Using definition of magnetic vector potential
 











 
1 2
1
12
2
2
21
4




d
r
d
i
o


 



 
1 2
12
1
2
2
21
4 r
d
d
i o






Rearrange terms

Lecture 26
THEORY
MUTUAL INDUCTANCE
 



 
1 2
12
1
2
2
21
4 r
d
d
i o






21
2
21
M
i


Constant that depends on loop geometry
Flux in loop 1 due to current in loop 2

 1
v
2
i
B

Loop 1 Loop 2
1
 2

1
S 2
S

Lecture 26
THEORY
MUTUAL INDUCTANCE
 



 
2 1
21
2
1
1
12
4 r
d
d
i o






12
1
12
M
i



 



 
1 2
12
1
2
2
21
4 r
d
d
i o






21
2
21
M
i


Conclusion
M’s are geometrical factors
M
M
M 

 21
12
MUTUAL INDUCTANCE BETWEEN LOOPS

Lecture 26
THEORY
MUTUAL INDUCTANCE
General result
dt
di
M
dt
di
L
dt
d
dt
d
v 2
1
1
21
1
1







dt
di
L
dt
di
M
dt
d
dt
d
v 2
2
1
12
2
2







Sign convention


1
v
1
i
primary


2
v
2
i
Indicates v2 positive when v1 is positive
Lecture 26
THEORY
THEORY
MUTUAL INDUCTANCE

 2
v
2
i
Enclosed surface S2

 1
v
1
i
B

Loop 1
Loop 2
Enclosed surface S1
1
 2


Lecture 26
THEORY
Example (Question)
Find the inductance per unit length of a coaxial conductor
shown in the figure.
a
b

I

Lecture 26
THEORY
Example (Solution)
a
b

I
We can apply
Ampere’s law
for the closed
path shown in
blue.


 ˆ
2 r
I
B o

 Direction determined using
right hand rule.


ˆ
2 r
I
H 

r

Lecture 26
THEORY
Example (Solution)
The two
conductors are
linked by the
flux through the
surface of
constant angle


 ˆ
2 r
I
B o


a
b

I

 


S
A
d
B


12 with
IM

12
1
2

ˆ


drd
A
d 

Lecture 26
THEORY
Example (Solution)
a
b

I
 




0
12
ˆ
ˆ
2




d
dr
r
I
b
a
o
IM

12
1
2

ˆ


drd
A
d 








a
b
I
o ln
2
12 


I
M 12



 I
M 12









a
b
M o ln
2

 END

Lecture 26
THEORY
Example (Question)
a
b

I
Same example but with a different approach to the solution

Lecture 26
THEORY
Example (Solution)
Find the inductance per unit length of a coaxial conductor shown in the figure.
a
b

I


 ˆ
2 r
I
B o




ˆ
2 r
I
H 

r
2
2
1
LI
W 
 
 

volume
dv
H
B
W


2
1
The expression for energy stored in a
magnetic field can provide an alternate
definition for the inductance.

Lecture 26
THEORY
Example (Solution)
Find the inductance per unit length of a coaxial conductor shown in the figure.
a
b

I


 ˆ
2 r
I
B o




ˆ
2 r
I
H 

r
The expression for energy stored in a
magnetic field can provide an alternate
definition for the inductance.
 
 

volume
dv
H
B
I
M


2
1
 









0
2
0
2
2
2
2
4
dz
rdrd
r
I
I
M
b
a
o 










a
b
M o ln
2










a
b
M o ln
2


END

Lecture 26
ASSIGNMENT
These questions are straight forward. Plug in the numbers and get your answer. Being able to
solve this type of question ensures you of at least a grade of 25% on a quiz or final exam
containing questions related to this lecture.
These questions require a few manipulations of equations or numbers before the answer can be
obtained. Being able to solve this type of question ensures you of at least a grade of 50% on a
quiz or final exam containing questions related to this lecture.
These questions are the most difficult and require a thorough understanding of the topic material
and also pull in topics from other lectures and disciplines. Being able to solve this type of
question ensures you an A grade on a quiz or final exam containing questions related to this
lecture.
These question are quite involved and requires a thorough understanding of the topic material.
Being able to solve this type of question ensures you of at least a grade of 75% on a quiz or final
exam containing questions related to this lecture.
25
50
75
100
75 100 These form excellent review questions when preparing for the quiz and final exam.
25 50 75 100
SELF EVALUATION SCALE

Lecture 26
ASSIGNMENT
25 Find the mutual inductance M between two concentric
circular wire loops of radius r1 and r2 respectively where r1
<< r2.

Lecture 26
ASSIGNMENT
50 Show that the inductance of the toroid is:







a
b
h
N
L o ln
2
2


h
a
b
N turns
c

Lecture 26
ASSIGNMENT
50 A transmission line consists of two parallel conductors of
separation b and radius a as shown where b >> a. Find
the inductance per unit length of the line assuming that
the conductors are thin walled tubes.
I
I
Radius a
Radius a
b

Lecture 26
ASSIGNMENT
75 A coax transmission line has a solid metal inner
conductor of radius a and a thin outer conductor of
radius b. Estimate the inductance per unit length of the
transmission line assuming current flow is distributed
uniformly over the cross-section of the center conductor.

Lecture 26
ASSIGNMENT
50 A very long solenoid with 2 X 2 cm cross-section has an
iron core (r = 1000) and 4000 turns per meter. If it carries
a current of 500 mA, find a) its self inductance per meter
and b) the energy per meter stored in the magnetic field.
m
J
b
ans
m
H
a
ans
/
005
.
1
:
)
(
/
042
.
8
:
)
(

Lecture 26
ASSIGNMENT
75 Determine the self-inductance of a coax cable of inner
radius a and outer radius b if the inner conductor is made of
a inhomogeneous material having:





1
2 o
 Is a radial coordinate inside the conductor.























a
b
a
b
L
ans o
o
1
1
ln
ln
8
:



 

Lecture 26
ASSIGNMENT
75 Determine the inductance per unit length of a two wire
transmission line with separation distance d. Each wire has
a radius a.











 


a
a
d
L
ans ln
4
1
:



Lecture 26
REFERENCES
REFERENCES
(0) Textbook: U. S. Inan, A. S. Inan
“Engineering Electromagnetics”
(1) J.D. Kraus, K. R. Carver “Electromagnetics” 2nd
(2) Reitz, Milford, Christy “Foundations of Electromagnetic
theory” 4th
(3) M. Plonus “Applied Electromagnetics”
(4) R. P. Winch “Electricity and Magnetism”
(5) P. Lorrain, D. Corson “Electromagnetic fields and Waves”
2nd
(6) Duckworth “Electricity and Magnetism”
(7) J.D. Jackson “Classical Electrodynamics” 2nd
(8) F. Ulaby, “Fundamentals of applied Electromagnetics”
(0) Inan p. 246 - 255
(1) Kraus p. 12 - 15
(2) Reitz p. 27 - 31
(3) Plonus p. 2 - 4
(4) Winch p. 258 - 266
(5) Lorrain p. 40 - 42
(6) Duckworth p. 5 - 8
(7) Jackson p. 27 - 28
(8) Ulaby p. 7, 143 - 144

Lecture 26
SUMMARY
SELF INDUCTANCE
Li



MUTUAL INDUCTANCE
12
1
12
M
i



Lecture 26
THEORY
THEORY
Lecture 26
Lecture 26
THEORY
THEORY

2
2
B
VOLUME
W

Energy density
 

A
B
W 2
2
1

 Total magnetic energy stored in solenoid
Energy density

2
2
B

EXPRESSION
VALID
FOR
ALL
Lecture 26
THEORY
THEORY
dt
dI
L
dt
d
V 


I
V
Power
dt
dW



THEN
dt
I
dt
d
dW 



THEN
t
d
I
t
d
d
dW 



 


I
I
d
I
L
W
0
THEN
2
2
LI
W 
Energy stored
Lecture 26
THEORY
THEORY
Current out of page
AREA
A

 NI
B



May have core with
constant permeability

I
V
Power 

d t
d I
L
d t
d
V 



Lecture 26
END
END LECTURE 26

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