CHAPTER 21
ELECTROMAGNETIC
INDUCTION
BASIC CONCEPTS
Faraday’s Law
Lenz’s Law
Motional emf
1
Electric Charge Produces an
Electric Field
Moving Electric Charge Produces a
Magnetic Field
Now Changing Magnetic Field Produces an
εmf
2
a potential difference (voltage)
What is changing and causing
the emf is the magnetic flux.
3
We defined Electric Flux
= Magnetic Flux is similar
= Flux through an area
4
We will define a vector to
represent area. The direction
of the vector will be
perpendicular to the surface.
The length of the vector will be
the area of the surface. The
vector will be .
The flux will be
5
First, flat area perpendicular to
= 0 = 6
For a surface parallel to = 90 = 0
And for in between
= Δ
= 7
If this flux changes an emf will
be induced in the area.
Consider a copper coil with one
loop
B
A
r
8
= For N loops
N loops
B
A
r
= 9
FARADAY’S LAW
The emf induced in a circuit is
directly proportional to the
time rate of change of the
magnetic flux through the
circuit.
Or
=
10
If changes with time an emf,
, will be produced.
Coil in magnetic field.
11
Flux through coil
= Flux can change:
1. Magnitude of changes.
2. Area of loop changes.
12
3. The angle changes.
4. Any combination of the
three.
13
Example:
Consider a coil with 200 turns
on a rectangular frame, 20 cm
by 30 cm. The resistance of the
coil is 2 Ω.
14
If increases uniformly from
zero to 0.5"#/% in 0.8 s
what is the current in the coil?
= &0.2%(&0.3%( = 0.060%
At = 0
= 0
At = 0.8
= =
&0.5"#/% (0.060% = 0.03"#
15
& = 0( − & = 0.8(
| | =
0.8
200&0.03"#(
| | =
0.8
| | = 7.5/
The current will be
7.5/
0= =
= 3.75
1 2Ω
16
Lenz’s Law
A changing magnetic flux will
cause an emf and if the emf is
in a conductor there will be a
current.
What will be the direction of
the current?
Lenz’s Law will answer that
question.
17
The polarity of the induced emf
is such that it tends to produce
a current that will create a
magnetic flux to oppose the
change in magnetic flux
through the loop.
18
a. Magnet moves toward loop.
Magnet field is down and
increasing. Induced emf and
thus current will be ccw to
oppose the increase.
19
b. Magnet moves away from
loop. Magnetic field is up and
but decreasing.
Induced current gives magnetic
field to oppose the decrease.
20
Motional emf
Δx
21
In time Δt the rod will move Δx.
The area covered in Δt will be
Δ
= 3Δ4
Δ4 = 5Δ
Δ
= 35Δ
Faraday’s Law
22
Δ
=
Δ
Δ = Δ
= 35Δ
35Δ
=
= 35
Δ
A conducting rod moving in a
magnetic field will have an emf
Induced across it given by this
equation.
23
Example: My airplane
Luscombe 8E
Built 1947
Wingspan 22 feet = 6.7m
Speed 100 mph = 44.7 m/s
24
If I am flying it at a point on
earth where the vertical
78
magnetic field is 5410 9 what
is the voltage drop across the
wings?
= :5
= &541078 9(&6.7%(&44.7%/(
= 1.54107 /
25
Consider a loop around a long solenoid.
26
The current in the solenoid is increasing
so the flux through the loop is
increasing.
An emf is induced in the loop by
.
27
INDUCTANCE
BASIC CONCEPTS
Mutual Inductance
Self Inductance
Magnetic Field Energy
Circuits with Inductors
28
Mutual Inductance
I1
#1
∙ =
>
If I1 changes with time t
Δ0> ?@A Δ>
BCCCD E=
Δ
Δ
29
Put second coil at P
I1
ΔF
Δ
#1
#2
?@A
BCCCD GHG:#2
30
?@A
BCCCD 0
If I1 changes with time
For example 0> = 0J GHK
Then
> = J GHK
And
= −
ΔLM
Δ
31
= −
J KK
Or
And
−
J K
=
K
0 =
1
1
I1
#1
#2
I2
32
Φ>
N=
0>
>
PJ 0> :
=
0>
>
PJ > N = PJ =
:
:
Only depends on geometry, number of
turns, area and length.
33
Self Inductance
If the current is changing then there will be
a changing magnetic flux through the coil.
34
A changing magnetic flux will induce an
emf.
= −
@L
@
Therefore we define in analogy to the
mutual inductance
Φ
3=
G
This is Self-Inductance
Solve for Φ
35
3G
Φ =
Faraday’s Law
= −
QLR
Q
= −3
Q
Q
Self-Induced emf
ΔG
= −3
Δ
36
What is the inductance, 3, of the black coil?
Area
A
ΦS
3=
0
>
TUU@ = PJ 0>
:
>
Φ = PJ 0> :
> PJ > 0> >
:
3=
= PJ
0>
:
37
Magnetic Field Energy
Consider the coil
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Current starts at zero and increases to 0.
38
/
=VW = 0 1 =
= 0/
1
Or
= = 0
And
Δ0
= −3
Δ
Δ0
Δ0
|=| = 0 X3 Y = 03
Δ
Δ
Power is work per unit time.
Δ"
==
Δ
39
Δ0
Δ" = =Δ = 03 Δ = 30Δ0
Δ
How much work to bring current through
inductor from zero to 0?
To calculate the work integrate from zero
current to current I
We won’t do it but you may see how on the
next page.
40
"=Z
\]\
\]J
\
30[0 = 3 Z 0[0
J
1 " = 30
2
_UA U
"^ BCCCD `W[ HWab
41
The energy stored in the magnetic field is
1 c = 30
2
Apply to solenoid
42
/:d%W = :
>
= PJ 0
:
So
:
0=
PJ >
> 3 = PJ
:
Put into equation for c
1 1 > :
c = 30 = PJ
X
Y
2
2
:
PJ >
43
1 :
c=
2 PJ
1 &
: (
c=
2 PJ
And
/:d%W = :
So
1 HWab 2 PJ &
: (
=
&
: (
/:d%W
1 d=
2 PJ
Energy stored in magnetic field of coil.
44
R-L Circuit
Consider this circuit
45
First consider that switch `> is closed and
calculate how the current will increase with
time.
Second, then while there is a current in the
circuit consider opening `> and closing `
and calculate how the current will decay.
Close `>
46
At time = 0 close switch.
Kirchkoff’s loop starting at the switch and
going ccw.
47
ΔG
− G1 − 3 = 0
Δ
ΔG − G1
=
Δ
3
ΔG 1
= − G
Δ 3 3
At t = 0 i=0
Then
Δ0
X Y
=
Δ ]J
3
After long time
Δ0
=0
Δ
48
And
0=
1
49
50
The L-C Circuit
The capacitor has been charged to an initial
charge e and then placed across the
inductor.
51
The L-C Circuit
Remember energy in
Capacitor
cf =
> gh
i
And
Inductor
>
c = 3G Then study
52
53
Transfromers
54