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Transcript
```UNIT 20 :
ELECTROMAGNETIC INDUCTION
Electromagnetic induction is the production
of an electrical potential difference (induced
emf) across a conductor situated in a
changing magnetic field.
20.1
20.2
20.3
20.4
20.5
Magnetic flux
Induced emf
Self-inductance
Mutual inductance
Energy stored in inductor
1
20.1 MAGNETIC FLUX ,Φ
• is defined as the scalar product between
the magnetic flux density, B and the vector
of the surface area, A.
Unit:T.m2 or Wb
 
  B  A  BA cos θ
area, A

A

B
B
 = 90
=0
A
  BA  = 0
2
Example 20.1.1
A small surface of area 10 mm2 inside a uniform magnetic field of strength
0.10 T is inclined at an angle α to the direction of the field. Determine the
magnetic flux through the surface if
i) α = 0º,
ii) α = 30º
iii) α = 90º
Solution :
 
  B  A  BA cos θ
20.2 INDUCED EMF
• An electric current produces a magnetic field.
(chapter 19)
If electric currents produce a magnetic field,
is it possible that a magnetic field can
produce an electric current ?
• Scientists (American Joseph Henry and the
found that is possible.
• Henry actually made the discovery first, but
Faraday published his results earlier and
4
investigated the subject in more detail.
20.2 INDUCED EMF
• The diagram below shows the apparatus used
by Faraday in his attempt to produce an
electric current from a magnetic field.
Faraday’s experiment to induce an emf
5
20.2 INDUCED EMF
• In this experiment, Faraday hoped by using a
strong enough battery, a steady current in X
would produce a current in a second coil Y but
failed.
• Faraday saw the galvanometer in circuit Y
deflect strongly at the moment he closed the
switch in circuit X.
• And the galvanometer deflected strongly in
the opposite direction when he opened the
switch.
current in Y.
6
20.2 INDUCED EMF
• Only when the current in X was starting or
stopping was a current produced in Y.
magnetic field produces no current, a
changing magnetic field can produce an
electric current.
• Such a current is called an induced current.
• We therefore say that an induced current is
produced by a changing magnetic field.
• The corresponding emf required to cause
this current is called an induced emf.
7
20.2 INDUCED EMF
• Induced emf is an electromotive force
resulting from the motion of a conductor
through a magnetic field , or from a change in
the magnetic flux that threads a conductor.
• Faraday did further experiments on
electromagnetic induction, as this
phenomenon is called.( refer diagram )
a) A current is induced when a magnet is
moved toward a coil/loop.
b) The induced current is opposite when the
magnet is moved away from the coil/loop.
c) No current is induced if the magnet does
not move relative to the coil/loop.
8
9
20.2 INDUCED EMF
10
20.2 INDUCED EMF
11
20.2 INDUCED EMF
• Direction of the induced current depends on :
i ) the direction of the magnet’s motion and
ii) the direction of the magnetic field.
• Magnitude of the induced current depends on :
i ) the speed of motion (v ↑,Iind↑)
ii) the number of turns of the coil (N ↑, Iind↑)
iii)the strength of the magnetic field (B↑,Iind↑)
• From the observations, Michael Faraday
found that,
“ the current/emf is induced in a coil/loop or complete
circuit whenever there is a change in the magnetic flux
12
through the area surrounded by the coil”
20.2 INDUCED EMF
“the magnitude of the induced e.m.f.
is proportional to the rate of change
of the magnetic flux”
Lenz’s law
“an induced electric current
always flows in such a direction
that it opposes the change
producing it.”
13
20.2 INDUCED EMF
• These two laws are summed up in the
relationship,
dΦB
dΦB
or   
 
dt
dt
dt : change of time
 : induced e.m.f.
dΦB : change of magnetic flux
induced e.m.f. always opposes the change of
magnetic flux producing it (Lenz’s law).
14
20.2 INDUCED EMF
• The concept of Faraday's Law is that any change
in the magnetic environment of a coil of wire will
cause a voltage (emf) to be "induced" in the coil.
• No matter how the change is produced, the
voltage will be generated.
• The change could be produced by
a) changing the magnetic field strength,
b) moving a magnet toward or away from the
coil,
c) moving the coil into or out of the magnetic
field,
d) rotating the coil relative to the magnet, etc. 15
(A) Induced emf in coil
20.2 INDUCED EMF
16
(A) Induced emf in coil
20.2 INDUCED EMF
dΦB
Notes
 
i ) the magnitude of induced emf,
dt
d
 
dt
d
or   N
dt
 N
 final  initial
t final  tinitial
dB
   NA
dt
dA
   NB
dt
ii) the flux through the coil can change in any
of 3 ways,
 
a) B , b) A , c) θ
 = B • A = BA cos θ
17
(A) Induced emf in coil
20.2 INDUCED EMF
Notes
iii) If the coil is connected in series to a
resistor of resistance R and the induced
e.m.f  exist in the coil as shown in figure
below.
dΦ
|  |
B
dt
and
-
I induced R
+ I induced
  IR  I induced 

R
18
Lenz's Law (based on censervation of energy)
• When an emf is generated by a change in magnetic
flux according to Faraday's Law, the polarity of the
induced emf (next slide) is such that it produces a
current whose magnetic field opposes the change
which produces it.
• The induced magnetic field inside any loop of wire
always acts to keep the magnetic flux in the loop
constant.
• In the examples below, if the B field is increasing, the
induced field acts in opposition to it.
• If it is decreasing, the induced field acts in the
19
direction of the applied field to try to keep it constant.
(A) Induced emf in coil
7.2 INDUCED EMF
The polarity of the induced emf
Induced current is directed out of the positive
terminal, through the attached device (resistance)
and into the negative terminal.
20
(A) Induced emf in coil
Example 20.2.1
A coil of wire 8 cm in diameter has 50 turns
and is placed in a B field of 1.8 T. If the B
field is reduced to 0.6 T in 0.002 s , calculate
the induced emf.
21
Solution
d = 8 cm, N = 50 turns, B from 1.8 T to 0.6 T
in 0.002 s
dB
   NA
dt
  final  initial 
dΦB

  N
  N 
dt
t


 BA  final  BA initial 
 B final  Binitial  
   NA

   N 
t
t




  d  2  B final  Binitial  
  151 V
   N    
t
  2  

22
(A) Induced emf in coil
Example 20.2.2
An elastic circular loop in the plane of the
paper lies in a 0.75 T magnetic field pointing
into the paper. If the loop’s diamater changes
from 20.0 cm to 6.0 cm in 0.50 s,
a)What is the direction of the induced current,
b)What is the magnitude of the average
induced emf, and
c)If the loop’s resistance is 2.5 Ω, what is the
average induced current during the 0.50 s ?
23
Solution:
B=0.75 T, di= 20.0 cm, df= 6.0 cm, t = 0.50 s
a) Direction of the induced current,
b) Magnitude of the average induced emf,
dA
  NB
dt
c) R = 2.5 Ω,
24
Example 20.2.3
A circular shaped coil 3.05 cm in radius,
containing 40 turns and have a resistance of
3.55  is placed perpendicular to a magnetic
field of flux density of 1.25 x 10-2 T. If the
magnetic flux density is increased to 0.450 T
in time of 0.250 s, calculate the induced
current flows in the coil.
25
(A) Induced emf in coil
How to determine the direction of induced
current.- Lenz’s law
Case A
Thumb – induced magnetic field
Fingers - induced current
N
+
I induced
Direction of induced current
– induced-current right hand
rule.
-
I induced
26
How to determine the direction of induced current.- Lenz’s law
Case A
• Consider a bar magnet that is moved
towards a solenoid.
• As the north pole of the magnet approaches
the solenoid, the amount of magnetic field
passing through the solenoid increases ,
thus increasing the magnetic flux through
the solenoid.
• The increasing flux induces an emf
(current) in the solenoid and galvanometer
indicates that a current is flowing.
27
How to determine the direction of induced current.- Lenz’s law
Case A
• The direction of the induced current is
such as to generate a magnetic field in the
direction that opposes the change in the
magnetic flux, so the direction of the
induced field must be in the direction that
make the solenoid right end becomes a
north pole.
• This opposes the motion of the bar magnet
and obey the Lenz’s law.
28
How to determine the direction of induced current.- Lenz’s law
Case B
(a) When the magnet is moved toward the stationary
conducting loop, a current is induced in the
direction shown.
(b) This induced current produces its own magnetic
field (Binduced) directed to the left that counteracts
the increasing external flux.
Bexternal
Binduced
29
How to determine the direction of induced current.- Lenz’s law
Case B
(c) When the magnet is moved away from the
stationary conducting loop, a current is induced
in the direction shown.
(d) This induced current produces a magnetic field
(Binduced) directed to the right and so counteracts
the decreasing external flux.
Bexternal
Binduced
30
(A) Induced emf in coil
Example 20.2.4
Calculate the current
through a 37 Ω resistor
connected to a single
turn circular loop 10 cm
in diameter, assuming
that the magnetic field
through the loop is
increasing at a rate of
0.050 T/s. State the
direction of the current.
31
Example 20.2.4
R = 37 Ω , d = 10 cm
dB/dt = 0.050 T/s.
S
N
2
dB
 d  dB
|  | A
 
dt
 2  dt
I induced
I induced

3.93 x 10-4
I 
 1.06 x 10-5 A
R
37
Direction of Iinduced : from b to a.
32
(B) Induced emf of a straight conductor
• Consider a
straight conductor
of length l is
moved at a speed
v to the right on a
U-shaped
conductor in a
uniform magnetic
field B that points
out the paper.
dA
dx  vdt
• This conductor travels a distance dx =vdt in a time
dt.
33
(B) Induced emf of a straight conductor
• The area of the
loop increases by
an amount
dA  ldx
dA  lvdt
• According to
the e.m.f. is
induced in the
conductor and
its magnitude is
given by
dA
dx  vdt
34
(B) Induced emf of a straight conductor
d
dA
 
B
dt
dt

lvdt 
 B
dt
  Blv
  Blv sin 
θ = angle between v and B
= 90 o
 
 
ε  L vB
• This induced emf is called motional induced emf.
35
(B) Induced emf of a straight conductor
• As the conductor is moved to the right (Fapplied to
the right) with speed v, the magnetic flux through
the loop increases.
• A current is induced
in the loop.
• The induced current
flows in the direction
that tends to oppose
this change.
FB
Fapplied
dA
dx  vdt
• In order to oppose this change, the current through
the conductor must produce a magnetic force (F=BIL)
directed to the left.
36
(B)Induced emf of a straight conductor
•The direction of the induced current due to
induced e.m.f. flows in the linear conductor can be
determine by using Fleming’s right hand rule
(based on lenz’s law).
• The induced current
flows from P to Q.


 (motion)
B
Fapplied
P
FB
Q
I induced or  induced
Only for the
straight
conductor.
Fapplied
dA
dx  vdt
Thumb – direction of Motion
First finger – direction of Field
Second finger – direction of Induced current
37
or Induced e.m.f.
(B)Induced emf of a straight conductor
Polarity
• When the conductor is moved to the right (Fapplied
to the right) with speed v, the electrons in the rod
move with the same speed.
• Therefore, each
feels a force F=Bqv,
which acts upward in
the figure.
dA
dx  vdt
• If the rod were not in contact with the U-shaped
conductor, electrons would collect at the upper end
of the rod, leaving the lower end positive. There
38
must thus be an induced emf.
Induced emf of a straight conductor
Example 20.2.5
Suppose the length in figure above is 0.10 m,
the velocity z is 2.5 m/s, the total resistance of
the loop is 0.030 Ω and B is 0.60 T. Calculate
a) the induced emf
b) the induced current
c) the force acting on the rod
d) the power dissipated in the loop
a) |  | Blv
b) I 

R
c) F  BIL
d) Pdissipated  I 2 R
39
Induced emf of a straight conductor
Example 20.2.6
A 0.2-m length of wire moves at a constant
velocity of 4 m/s in a direction that is 40 o with
respect to a magnetic flux density of 0.5 T.
Calculate the induced emf.
  Blv sin 
40
Induced emf of a straight conductor
Example 20.2.7
In figure above, a rod with length l = 0.400 m
moves in a magnetic flux with magnitude B =
1.20 T. The emf induced in the moving rod is
3.60 V.
a) Calculate the speed of the rod.
b) If the total resistance is 0.900 Ω,
calculate the induced current.
c) What force does the field exert on the
rod as a result of this current?
7.50 m/s , 4.00 A , 1.92 N to the left
41
42
Fig 31-CO, p.967
(C) Induced emf in a rotating coil
An ac generator / dynamo
(transforms mechanical energy into electric energy)
43
(C) Induced emf in a rotating coil
An ac generator / dynamo
(transforms mechanical energy into electric energy)
44
(C) Induced emf in a rotating coil
• Consider a coil of N
turns each of area A and
horizontal axis in its own
plane at right angle to a
uniform magnetic field of
flux density B.
• As the coil rotates with
the angular speed ω, the
orientation of the loop
changes with time.
A

45
(C) Induced emf in a rotating coil
• The emf induced in the loop is given by
d
  N
,   AB cos   AB cos t




dt
d  AB cos t 
 N
, A and B are constant
dt
  NAB  sin t 
 NAB sin t
ε  NABω sin θ
  o sin t ,  o  max  NAB
• The emf induced in the loop varies
sinusoidally in time.
46
(C) Induced emf in a rotating coil
 0  0  0
 0  0
   max
   max
   max
   max
A

The alternating emf induced in the loop plotted
47
as a function of time.
Induced emf in a rotating coil
Example 20.2.8
The armature of a simple ac generator
consists of 100 turns of wire, each having an
area of 0.2 m2 . The armature is turned with
a frequency of 60 rev/s in a constant
magnetic field of flux density 10-3 T.
Calculate the maximum emf generated.
 max  NAB
48
Induced emf in a rotating coil
Example 20.2.9
ε (V)
28 V
0.42 s
0
0.21 s
t
0.63 s
0.84 s
-28 V
The drawing shows a plot of the output emf of a
generator as a function of time t. The coil of this
device has a cross-sectional area per turn of
0.020 m2 and contains 150 turns. Calculate
a)The frequency of the generator in hertz.
c) The magnitude of the magnetic field.
2.4 Hz , 15 rad/s , 0.62 T
49
Induced emf in a rotating coil
Example 20.2.10
An amarture in ac generator consists of 500
turns, each of area 60 cm2 . The amarture is
rotated at a frequency of 3600 rpm in a
uniform 2 mT magnetic field. Calculate
a) the frequency of the alternating emf
b) the maximum emf generated
c) the instantaneous emf at time when the
plane of the coil makes an angle of 60o
with the magnetic field ?
380 rad/s, 1.13 V, 2.26 V
50
20.3 SELF-INDUCTANCE
• Self-induction is defined as the process of
producing an induced e.m.f. in the coil due
to a change of current flowing through the
same coil.
S
N
I
S
R
I
• Consider a current is present in the circuit
above.
51
20.3 SELF-INDUCTANCE
• This current produces a magnet field in the
coil that causes a magnetic flux through the
same coil.
• This flux changes when
the current changes.
• An emf is induced in
this coil called a selfinduced emf.
S
N
I
S
R
• This coil is said to have self-inductance
(inductance).
• A coil that has inductance is called an
inductor.
I
52
20.3 SELF-INDUCTANCE
• The symbol for an inductor is
if air-cored, and
a core of magnetic material.
if it has
• By Lenz’s law, the induced current opposes
the change that cause it.
• If the current is increasing, the direction of
the induced field and emf are opposite to that
of the current, to try to decrease the current.
• If the current is decreasing, the direction of
the induced field and emf are in the same
direction as the current, to try to increase the
53
current.
20.3 SELF-INDUCTANCE
Iinduced
Iinduced
(a) A current in the coil produces a magnetic field
directed to the left.
(b) If the current increases, the increasing magnetic
flux creates an induced emf having the polarity
shown by the dashed battery.
(c) The polarity of the induced emf reverses if
54
the current decreases.
20.3 SELF-INDUCTANCE
• The magnetic flux in a coil is proportional
to the current
ΦB  I
ΦB  LI L : self - inductance of the coil
dΦB …. (1)
ε
dt
dI
  L
dt
d LI 
ε
dt
…. (2)
55
dI
  L
dt
L

dI
20.3 SELF-INDUCTANCE
dt
Self-inductance, L is defined as the ratio
of the self induced e.m.f. to the rate of
change of current in the coil.
56
(1) = (2)
20.3 SELF-INDUCTANCE
dΦB
dI

 L
dt
dt
If the coil has N turns, hence
dI
dΦB
L N
dt
dt
L  dI  N  dΦB
LI  NΦB
NΦB
L
I
- scalar quantity
- unit is henry (H).
1 H  1 Wb A-1  1 T m 2 A-1
57
20.3 SELF-INDUCTANCE
NΦB
L
I
• The value of the self-inductance depends on
a) the size and shape of the coil
b) the number of turn (N)
c) the permeability of the medium in the
coil ().
• Self-inductance does not depend on current.
58
20.3 SELF-INDUCTANCE
Self-inductance of a Loop and Solenoid
NΦB
μ
NI
0
From L 
And B 
ΦB  BA cos 0 
I
l
By substituting we get,
μ0 N 2 A
2
L

μ
n
Al
or
L
0
l
N
n
l
For the medium-core solenoid :
μr μ0 N 2 A
μN 2 A or
L
L
l
l
where
μr : relative permeabili ty
μ : permeabili ty of medium
A : area of the solenoid
μ  μr μ0
59
20.3 SELF-INDUCTANCE
Example 20.3.1
If the current in a 230 mH coil changes
steadily from 20.0 mA to 28.0 mA in 140 ms,
what is the induced emf ?
dI
  L
dt
(Given 0 = 4 x 10-7 H m-1)
Example 20.3.2
μN
L

Suppose you wish to make a solenoid
l
whose self-inductance is 1.4 mH. The
inductor is to have a cross-sectional area of
1.2 x 10 -3 m2 and a length of 0.052 m. How
60
many turns of wire needed ? 220 turns
0
2
A
20.3 SELF-INDUCTANCE
Example 20.3.3
The current in a coil of wire is initially zero
but increases at a constant rate; after 10.0 s
it is 50.0 A. The changing current induces
an emf of 45.0 V in the coil.
a) Calculate the self – inductance of the coil.
b) Calculate the total magnetic flux through
the coil when the current is 50.0 A.
a)    L dI
dt
b)
NΦB
L
I
61
20.3 SELF-INDUCTANCE
Example 20.3.4
A 40.0 mA current is carried by a uniformly
wound air-core solenoid with 450 turns, a 15.0
mm diameter and 12.0 cm length. Calculate
a) the magnetic field inside the solenoid.
b) the magnetic flux through each turn.
c) the inductance of the solenoid.
(Given 0 = 4 x 10-7 H m-1)
a) B  μ0 NI
l
b) ΦB  BA cos 0 
c)
NΦB
L
I
or
μ0 N A
L
l
2
62
20.4 MUTUAL INDUCTANCE
20.4 MUTUAL INDUCTANCE
Mutual Inductance for two coaxial solenoids
A
N1
N2
I1
l
Ac generator
I1
• Consider a long solenoid with length l and cross
sectional area A is closely wound with N1 turns of wire.
A second solenoid with N2 turns surrounds it at its
centre as shown in figure above.
Mutual Inductance for two coaxial solenoids
20.4 MUTUAL INDUCTANCE
• The first solenoid is the one connected to an ac
generator, which sends an alternating current I1
through it.
• The current I1 produces a magnetic field lines
inside it and this field lines also pass through the
solenoid 2 as shown in figure.
• If the current I1 changes with time, the magnetic
flux through the solenoids 1 and 2 will change with
time simultaneously.
• Due to the change of magnetic flux through the
solenoid 2, an e.m.f. is induced in solenoid 2.
• This process is known as mutual induction.
• At the same time, the self-induction occurs in the
64
solenoid 1 since the magnetic flux through it changes.
Mutual Inductance for two coaxial solenoids
20.4 MUTUAL INDUCTANCE
• Mutual induction is defined as the process of
producing an induced e.m.f.in one circuit/coil due to
the change of current in another circuit/coil.
Mutual inductance, M
• If the current I1 in solenoid 1 is continously changing,
then the flux it produces will also change continously.
• The changing magnetic flux from the solenoid 1
induces an emf in the solenoid 2.
• The induced emf in the solenoid 2 is proportional to
the rate of change of the current I1 in solenoid 1.
dI 1
2  
dt
dI1 ….. (1)
 2   M 21
dt
65
Mutual Inductance for two coaxial solenoids
Mutual inductance, M
20.4 MUTUAL INDUCTANCE
• Also the induced emf in the solenoid 1 is proportional
to the rate of change of the current I2 in solenoid 2.
dI 2
1  
dt
dI 2
 1   M 12
dt
M 21  M12  M  a constant of proportionality
 mutual inductance
• The mutual inductance of the two solenoids is the
same if current flows in the solenoid 2 and flux links
the solenoid 1, causing an induced emf when a
66
Mutual Inductance for two coaxial solenoids
Mutual inductance, M
• Rearrange,
M 
2
dI1
dt

20.4 MUTUAL INDUCTANCE
1
dI 2
dt
M is defined as “ the ratio of the induced emf in one
solenoid/coil/ to the rate of change of current in the
other solenoid/coil.”
• Unit M : Henry (H)
d
  N
dt
d2
 2  N2
dt
….. (2)
67
Mutual Inductance for two coaxial solenoids
Mutual inductance, M
(1) = (2)
20.4 MUTUAL INDUCTANCE
dI1
d2
 M 21
 N2
dt
dt
M 21  dI1  N 2  dΦ2
M 21I1  N 2Φ2
N 2Φ2
M 21 
I1
• Since M12=M21=M, equation above can be
written as
N 2Φ2 N 1Φ1
M

I1
I2
68
Mutual Inductance for two coaxial solenoids
Mutual inductance, M
• From
N 2Φ2
M 21 
I1
20.4 MUTUAL INDUCTANCE
and 2  B1 A 
 o N1 I 1 A
l
• He mutual inductance of the solenoid 2 is,
N 2Φ2 N 2 o N1 I1 A
M 21 

I1
I1l
M 21  M 12  M 
N 2Φ2 N 1Φ1
M

I1
I2
o N 2 N1 A
l
M
o N 2 N1 A
l
69
20.4 MUTUAL INDUCTANCE
Mutual inductance, M
N 2Φ2 N 1Φ1
M

I1
I2
M
o N 2 N1 A
l
dI 2
 1   M 12
dt
70
20.4 MUTUAL INDUCTANCE
Example 20.5.1
The primary coil of a solenoid of radius 2.0 cm
has 500 turns and length of 24 cm. If the
secondary coil with 80 turns surrounds the
primary coil at its centre, calculate
a. the mutual inductance of the coils
b. the magnitude of induced e.m.f. in
secondary coil if the current in primary coil
changes at the rate 4.8 A s-1.
71
7.5 MUTUAL INDUCTANCE
Solution 20.5.1
rp = 2.0 cm , Np =500 , lp 24 cm Ns = 80
dIs/dt = 4.8 A s-1
a)
M
M
o N 2 N1 A
l
o N s N p A
lp
dI 2
b)  1   M 12
dt
 s  M
dI p
dt
72
73
20.4 MUTUAL INDUCTANCE
Transformer
Vp
(input)
primary coil
NP
turns
laminated iron
core
NS
turns
Vs
(output)
secondary coil
Symbol
in circuit
• A transformer is a device for increasing or
decreasing an ac voltage.
• The operation of transformer is based on the
74
principle of mutual induction and self-induction.
Transformer
20.4 MUTUAL INDUCTANCE
• Two types of transformer
a) step-up transformer (Ns > Np)
b) step-down transformer (Np > Ns).
• There are three assential parts;
(1) a primary coil connected to an ac source
(2) secondary coil
(3) soft iron core
• When ac voltage is applied to the input coil
(primary coil), the alternating current produces
an alternating magnetic flux that is concentrated
in the iron core, without any leakage of flux
outside the core.
75
20.5 ENERGY STORED IN INDUCTOR
• The functions of an inductor are ;
- to control current
- to keep energy in the form of magnetic field
• An inductor carrying current has energy
stored in it.
• It is because a generator does work to
establish a current in an inductor.
• Suppose an inductor is connected to a
generator whose terminal voltage can be
varied continously from zero to some final
value.
76
20.5 ENERGY STORED IN INDUCTOR
• As the voltage is increased, the current I in the
circuit rises continously from zero to its final value.
• While the current is rising, an emf (back
emf) is induced in the inductor.
• Because of this, the generator that supplies the
current must maintain a potential difference
between its terminals while the current is rising
(changing), and therefore it must supply energy to
the inductor.
• Thus, the generator must do work to push the
charges through the inductor against this induced
emf.
77
20.5 ENERGY STORED IN INDUCTOR
• To do this, power has to be supplied by the
generator to the inductor.
W U
W
P  I
P
t
dI


dW dU
P  I L 
P

 dt 
dt
dt
dU  Pdt
Pdt  LIdI
• The total work done while the current is changed
from zero to its final value is given by
U
I
0
0
 dU  L  dI
1 2
U  LI
2
• This work is stored as energy in the inductor.
78
20.5 ENERGY STORED IN INDUCTOR
• For a long air-core solenoid, the self-inductance is
μ0 N 2 A
L
l
• Therefore the energy stored in the solenoid
2
2
is given by U  1 LI 2


μ
N
AI
1
2
U  
2
0
l


Example 20.5.1
How much energy is stored in a 0.085-H
inductor that carries a current of 2.5 A ?
1 2
U  LI
2
79
20.5 ENERGY STORED IN INDUCTOR
Example20.5.2
A steady current of 2.5 A in a coil of 500 turns
causes a flux of 1.4 x 10-4 Wb to link (pass
through) the loops of the coil. Calculate
a) the average back emf induced in the coil if
the current is stopped in 0.08 s
b) the inductance of the coil and the energy
stored in the coil (inductor).
dΦB
|  |
dt
dI
L
dt
1 2
U  LI
2
80
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