chapter20 student
TRANSCRIPT
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PHYSICS CHAPTER 20
20.0 ELECTROMAGNETIC INDUCTION
20.1 Magnetic flux
20.2 Induced emf
20.3 Self-inductance
20.4 Mutual Inductance
20.5 Energy stored in inductor 2
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At the end of this chapter, students should be able to:
Define and use magnetic flux,
Learning Outcome:
20.1 Magnetic flux ( 1/2 hour)
w w w
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. e d u
. m y / p h y s i c s
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20.1.1 Phenomenon of electromagnetic inductionConsider some experiments were conducted by MichaelFaraday that led to the discovery of the Faradays law ofinduction as shown in Figures 7.1a, 7.1b, 7.1c, 7.1d and 7.1e.
20.1 Magnetic flux
No movement
0v
Figure 7.1a
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Figure 7.1b
Figure 7.1c
NS Move towards the coil
v
I I
No movement
0v
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Figure 7.1d
Figure 7.1e
N Move away from the coil
v
I I
N SMove towards the coil
v
I I
S
Simulation 7.1
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From the experiments:When the bar magnet is stationary , the galvanometer notshow any deflection ( no current flows in the coil ).When the bar magnet is moved relatively towards the coil,the galvanometer shows a momentary deflection to the right(Figure 7.1b). When the bar magnet is moved relativelyaway from the coil, the galvanometer is seen to deflect in the
opposite direction (Figure 7.1d).Therefore when there is any relative motion between thecoil and the bar magnet , the current known as inducedcurrent will flow momentarily through the galvanometer.This current due to an induced e.m.f across the coil.
Conclusion :When the magnetic field lines through a coil changesthus the induced emf will exist across the coil.
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The magnitude of the induced e.m.f. depends on thespeed of the relative motion where if the
Therefore v is proportional to the induced emf .
20.1.2 Magnetic flux of a uniform magnetic field
is defined as the scalar product between the magnetic fluxdensity, B with the vector of the area, A .Mathematically,
v increases induced emf increasesv decreases induced emf decreases
A B andof directione between thangle: where fluxmagnetic:
(7.1)
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It is a scalar quantity and its unit is weber (Wb) OR teslameter squared ( T m 2).
Consider a uniform magnetic field B passing through a surfacearea A of a single turn coil as shown in Figures 7.2a and 7.2b.
From the Figure 7.2a, the angle is 0 thus the magnetic flux isgiven by
Figure 7.2a
B
A
cos BA0cos BA
BA maximum
area
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From the Figure 7.2a, the angle is 90 thus the magnetic fluxis given by
Figure 7.2b
B A
cos BA90cos BA
0Note:
Direction of vector A always perpendicular (normal) tothe surface area, A .The magnetic flux is proportional to the number offield lines passing through the area.
area90
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A single turn of rectangular coil of sides 10 cm 5.0 cm is placedbetween north and south poles of a permanent magnet. Initially, theplane of the coil is parallel to the magnetic field as shown in Figure7.3.
If the coil is turned by 90 about its rotation axis and the magnitudeof magnetic flux density is 1.5 T, Calculate the change in themagnetic flux through the coil.
Example 20.1 :
SNP
QR
S
I I
Figure 7.3
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Solution :The area of the coil is
Initially,
Finally,
Therefore the change in magnetic flux through the coil is
T51. B
2322 m100510051010 .. AFrom the figure, =90 thus the initialmagnetic flux through the coil is
A B
cos i BA90cos BA
B
A
From the figure, =0 thus the finalmagnetic flux through the coil is
cos
f BA 0cos100.55.1 3
if 0105.7 3
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A single turn of circular coil with a diameter of 3.0 cm is placed in
the uniform magnetic field. The plane of the coil makes an angle30 to the direction of the magnetic field. If the magnetic fluxthrough the area of the coil is 1.20 mWb, calculate the magnitude ofthe magnetic field.Solution :
The area of the coil is
Example 20.2 :
Wb1020.1m;100.3 32 d
4
2d A
4100.3
22 A
24
m1007.7 A
3030
coil
B A
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Solution :
The angle between the direction of magnetic field, B and vector of
area, A is given by
Therefore the magnitude of the magnetic field is
Wb1020.1m;100.3 32 d
603090
cos BA
60cos1007.71020.1 43 B
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The three loops of wire as shown in Figure 7.4 are all in a region ofspace with a uniform magnetic field. Loop 1 swings back and forthas the bob on a simple pendulum. Loop 2 rotates about a verticalaxis and loop 3 oscillates vertically on the end of a spring. Whichloop or loops have a magnetic flux that changes with time? Explain
your answer.
Example 20.3 :
Figure 7.4
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Solution :Only loop 2 has a changing magnetic flux .
Reason :Loop 1 moves back and forth, and loop 3 moves up and down,but since the magnetic field is uniform, the flux alwaysconstant with time.
Loop 2 on the other hand changes its orientation relative to thefield as it rotates, hence its flux does change with time.
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At the end of this chapter, students should be able to:
Use Faradays experiment to explain induced emf.
State Faradays law and use Lenzs law to determine thedirection of induced current.
Apply induced emf ,
Learning Outcome:
20.2 Induced emf (2 hours)
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At the end of this chapter, students should be able to:Derive and use induced emf :
in straight conductor,
in coil,
in rotating coil,
Learning Outcome:
20.2 Induced emf (2 hours)
w w w
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. e d u
. m y / p h y s i c s
OR
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20.2.1 Faradays law of electromagnetic inductionstates that the magnitude of the induced emf is proportionalto the rate of change of the magnetic flux .Mathematically,
The negative sign indicates that the direction of induced emf always oppose the change of magnetic flux producing it(Lenzs law) .
20.2 Induced emf
dt d OR (7.2)
where fluxmagnetictheof change:d timeof change:dt
emf induced:
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For a coil of N turns, eq. (7.2) can be written as
Since
For a coil of N turns is placed in the changing magneticfield B , the induced emf is given by
(7.3)
, then eq. (7.3) can be written asif
d
(7.4)
fluxmagneticfinal: f fluxmagneticinitial: i
where
dt
d N
cos BAand
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For a coil of N turns is placed in a uniform magnetic field B
but changing in the coils area A , the induced emf is givenby
dt
BAd N
cos
(7.5)
dt
d N
cos BAand
dt
BAd N cos
(7.6)
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For a coil is connected in series to a resistor of resistanceR and the induced emf exist in the coil as shown in Figure 7.5,
Figure 7.5
R
I I dt
d N
IR and
dt
d N IR
(7.7)
Note:To calculate the magnitude of induced emf , the negative signcan be ignored .
For a coil of N turns, each turn will has a magnetic flux of BAcos through it, therefore the magnetic flux linkage (refer tothe combined amount of flux through all the turns) is given by
the induced current I is given by
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The magnetic flux passing through a single turn of a coil is
increased quickly but steadily at a rate of 5.0 10 2 Wb s 1. If the coilhave 500 turns, calculate the magnitude of the induced emf in thecoil.Solution :
By applying the Faradays law equation for a coil of N turns , thus
Example 20.4 :
12 sWb100.5turns;500 dt
d N
dt
d N
2
100.5500
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A coil having an area of 8.0 cm 2 and 50 turns lies perpendicular to a
magnetic field of 0.20 T. If the magnetic flux density is steadilyreduced to zero, taking 0.50 s, determinea. the initial magnetic flux linkage.b. the induced emf.Solution :
a. The initial magnetic flux linkage is given by
B
Example 20.5 :
ilinkagefluxmagneticinitial N
0;T;0.20turns;50;m100.8f i
24 B B N A
s0.50dt
0 A
cosi A NB
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Solution :
a.
b. The induced emf is given by
0;T;0.20turns;50;m100.8 f i24 B B N A
s0.50dt
0cos108.00.2050 4
dt dB NA cos if B BdB and
dt
B B NA if cos
50.0
20.000cos100.850 4
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A narrow coil of 10 turns and diameter of 4.0 cm is placed
perpendicular to a uniform magnetic field of 1.20 T. After 0.25 s, thediameter of the coil is increased to 5.3 cm.a. Calculate the change in the area of the coil.b. If the coil has a resistance of 2.4 , determine the induced
current in the coil.Solution :
Example 20.6 :
m;103.5m;100.4turns;10 2f 2
i d d N
s0.25T;2.1 dt B
0
A
B B
A
Initial Final
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Solution :
a. The change in the area of the coil is given by
if A AdA
44
2i
2f
d d
m;103.5m;100.4turns;10 2f 2
i d d N
s0.25T;2.1 dt B
2i2f 4
d d
2222
100.4103.54
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Solution :
b. GivenThe induced emf in the coil is
Therefore the induced current in the coil is given by
4.2 R
dt
dA NB cos
m;103.5m;100.4turns;10 2f 2
i d d N
s0.25T;2.1 dt B
25.0105.90cos2.110
4
V1056.4 2
IR 4.21056.4 2 I
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Direction ofinduced current Right hand griprule.
states that an induced electric current always flows in such
a direction that it opposes the change producing it .This law is essentially a form of the law of conservation ofenergy .
20.2.2 Lenzs law
I
I NNorth pole
An illustration of lenzs law can be explained bythe following experiments.
1st
experiment:In Figure 7.6 the magnitudeof the magnetic field at thesolenoid increases as thebar magnet is moved
towards it. An emf is induced in thesolenoid and thegalvanometer indicates thata current is flowing.
Figure 7.6
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To determine the direction of the current through thegalvanometer which corresponds to a deflection in a particular
sense, then the current through the solenoid seen is in thedirection that make the solenoid upper end becomes anorth pole . This opposes the motion of the bar magnet andobey the lenzs law .
v
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X XP
Q
F
2nd experiment:
Consider a straight conductor PQis placed perpendicular to themagnetic field and move theconductor to the left with constantvelocity v as shown in Figure 7.7.
When the conductor move to theleft thus the induced currentneeds to flow in such a way tooppose the change which hasinduced it based on lenzs law.Hence galvanometer shows adeflection.
Figure 7.7
I
Simulation 7.2
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To determine the direction of the induced current (inducedemf) flows in the conductor PQ, the Flemings right hand
(Dynamo) rule is used as shown in Figure 7.8.
Therefore the induced current flows from Q to P as shown in
Figure 7.7.Since the induced current flows in the conductor PQ and isplaced in the magnetic field then this conductor willexperience magnetic force .
Its direction is in the opposite direction of the motion .
Thumb direction of Motion
First finger direction of FieldSecond finger direction of induced
current OR induced emf
Note: B
)motion(
OR induced I emf induced
Figure 7.8
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3 rd experiment:Consider two solenoids P and Q arranged coaxially closed to
each other as shown in Figure 7.9a.
At the moment when the switch S is closed , current I beginsto flow in the solenoid P and producing a magnetic field insidethe solenoid P. Suppose that the field points towards thesolenoid Q.
Figure 7.9a
S,SwitchP Q
I I
NS SN
ind I ind I -+
ind
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The magnetic flux through the solenoid Q increases withtime . According to Faradays law ,an induced current due to
induced emf will exist in solenoid Q.The induced current flows in solenoid Q must produce amagnetic field that oppose the change producing it (increasein flux). Hence based on Lenzs law, the induced current flowsin circuit consists of solenoid Q is anticlockwise (Figure 7.9a)and the galvanometer shows a deflection.
QSSwitch,
P
NS
I I
S N
ind I ind I - +
ind
Figure 7.9b
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At the moment when the switch S is opened , the current I starts to decrease in the solenoid P and magnetic flux through
the solenoid Q decreases with time . According to Faradayslaw ,an induced current due to induced emf will exist insolenoid Q.The induced current flows in solenoid Q must produce amagnetic field that oppose the change producing it (decreasein flux). Hence based on Lenzs law, the induced current flowsin circuit consists of solenoid Q is clockwise (Figure 7.9b) andthe galvanometer seen to deflect in the opposite direction ofFigure 7.9a.
Simulation 7.3
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A single turn of circular shaped coil has a resistance of 20 and an
area of 7.0 cm 2. It moves toward the north pole of a bar magnet asshown in Figure 7.10.
If the average rate of change of magnetic flux density through thecoil is 0.55 T s 1,a. determine the induced current in the coilb. state the direction of the induced current observed by the
observer shown in Figure 7.10.
Example 20.7 :
Figure 7.10
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Solution :a. By applying the Faradays law of induction, thus
Therefore the induced current in the coil is given by
dt
d N
BAdt
d N
124 sT55.0;m100.7;20turn;1dt
dB A R N
dt
dB NA
55.0100.71 4
180cos BAand
V1085.34
IR
201085.3 4 I
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Solution :b. Based on the lenzs law, hence the direction of induced current is
clockwise as shown in figure below.
NS ind I
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Consider a straight conductor PQ of length l is moved
perpendicular with velocity v across a uniform magnetic field Bas shown in Figure 7.11.
When the conductor moves through a distance x in time t , thearea swept out by the conductor is given by
20.2.3 Induced emf in a straight conductor
Figure 7.11
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
X X X X
P
Q
l
x
B
v
lx A
Area, A
ind I
ind
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Since the motion of the conductor is perpendicular to themagnetic field B hence the magnetic flux cutting by the
conductor is given by
According to Faradays law, the emf is induced in the conductorand its magnitude is given by
0 cos BA and0cos Blx Blx
dt
d
Blxdt
d
Blv
dt
dx Bl v
dt
dxand
(7.8)
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In general, the magnitude of the induced emf in the straightconductor is given by
This type of induced emf is known as motional induced emf.The direction of the induced current or induced emf in thestraight conductor can be determined by using the Flemingsright hand rule (based on Lenzs law).In the case of Figure 7.11, the direction of the induced current orinduced emf is from Q to P. Therefore P is higher potential thanQ.
(7.9)
Bv and betweenangle:where
Note:
Eq. (7.9) also can be used for a single turn of rectangular coilmoves across the uniform magnetic field .
For a rectangular coil of N turns ,
(7.10)
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A 20 cm long metal rod CD is moved at speed of 25 m s 1 across a
uniform magnetic field of flux density 250 mT. The motion of the rodis perpendicular to the magnetic field as shown in Figure 7.12.
a. Calculate the motional induced emf in the rod.b. If the rod is connected in series to the resistor of resistance
15 , determine
i. the induced current and its direction.
ii. the total charge passing through the resistor in two minute.
Example 20.8 :
Figure 7.12
C
D
B
1sm52
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Solution :a. By applying the equation for motional induced emf, thus
b. Giveni. By applying the Ohms law, thus
By using the Flemings right hand rule, the direction of the
induced current is from D to C .ii. Given
The total charge passing through the resistor is given by
15 R
sinlvB
T;10502;sm25m;1020 312 Bvl
90sin10250251020 32
IR 1525.1 I
90 and
s120602 t
It Q 1201033.8 2Q
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Consider a rectangular coil of N turns, each of area A , being
rotated mechanically with a constant angular velocity in auniform magnetic field of flux density B about an axis as shownin Figure 7.13.
When the vector of area, A is at an angle to the magneticfield B , the magnetic flux through each turn of the coil is givenby
20.2.4 Induced emf in a rotating coil
Figure 7.13: side view
B
N S A
coil
cos BA t andt BA cos
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By applying the equation of Faradays law for a coil of N turns,thus the induced emf is given by
The induced emf is maximum when hence
dt d N
t BAdt
d N cos
t dt d NBA cos
(7.11)
time:t where 1sin t (7.12)
T f
22 where
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Eq. (7.11) also can be written as
Conclusion : A coil rotating with constant angular velocity in auniform magnetic field produces a sinusoidally alternating emf as shown by the induced emf against time t graph in Figure7.14.
(7.13)
B A and betweenangle: where
Figure 7.14
t sinmax
0
max
max
V
t T T 5.0 T 5.1 T 2
B
Note:
This phenomenonwas the importantpart in thedevelopment ofthe electricgenerator ordynamo .
Simulation 7.4
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A rectangular coil of 100 turns has a dimension of 10 cm 15 cm. It
rotates at a constant angular velocity of 200 rpm in a uniformmagnetic field of flux density 5.0 T. Calculatea. the maximum emf produced by the coil,b. the induced emf at the instant when the plane of the coil makes
an angle of 38 to the magnetic field.
Solution :The area of the coil is
and the constant angular velocity in rad s 1 is
Example 20.9 :
2222 m105.110151010 AT0.5turns;100 B N
s06min1
rev1rad2
min1rev200
1srad9.20
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Solution :a. The maximum emf produced by the coil is given by
b.
NBAmax
T B N 0.5turns;100
9.20105.10.5100 2
B
A
38
From the figure, the angle is
Therefore the induced emf is given by523890
sin NBA
52sin9.20105.10.5100 2
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Exercise 20.1 :1. A bar magnet is held above a loop of wire in a horizontal
plane, as shown in Figure 7.15.The south end of the magnet is toward theloop of the wire. The magnet is droppedtoward the loop. Determine the direction ofthe current through the resistor
a. while the magnet falling toward the loop,
b. after the magnet has passed through the
loop and moves away from it.(Physics for scientists and engineers,6 thedition, Serway&Jewett, Q15, p.991)ANS. : U think
Figure 7.15
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2. A straight conductor of length 20 cm moves in a uniform magneticfield of flux density 20 mT at a constant speed of 10 m s -1 . Thevelocity makes an angle 30 to the field but the conductor isperpendicular to the field. Determine the induced emf.
ANS. : 2.0 10 2 V3. A coil of area 0.100 m 2 is rotating at 60.0 rev s -1 with the axis of
rotation perpendicular to a 0.200 T magnetic field.a. If the coil has 1000 turns, determine the maximum emf
generated in it.b. What is the orientation of the coil with respect to the
magnetic field when the maximum induced emf occurs?(Physics for scientists and engineers,6 th edition,Serway&Jewett, Q35,p.996)
ANS. : 7.54 10 3 V4. A circular coil has 50 turns and diameter 1.0 cm. It rotates at a
constant angular velocity of 25 rev s 1 in a uniform magnetic field offlux density 50 T. Determine the induced emf when the plane ofthe coil makes an angle 55 to the magnetic field.
ANS. : 1.77 10 5 V
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At the end of this chapter, students should be able to:Define self-inductance.Apply self-inductance,
for a loop and solenoid.
Learning Outcome:
20.3 Self-inductance (1 hour)
w w w
. k m s
. m a t r i k
. e d u
. m y / p h y s i c
s
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20.3.1 Self-inductionConsider a solenoid which is connected to a battery , a switch Sand variable resistor R , forming an open circuit as shown inFigure 7.16a.
20.3 Self-inductance
Figure 7.16a: initial
S R I I
When the switch S is closed, a current
I begins to flow in the solenoid.The current produces a magneticfield whose field lines through thesolenoid and generate the magneticflux linkage .
If the resistance of the variableresistor changes , thus the currentflows in the solenoid also changed ,then so too does magnetic fluxlinkage .
NS
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Self-induction experimentThe effect of the self-induction can be demonstrated by the
circuit shown in Figure 7.17a.
Initially variable resistor R
is adjusted so that the two lampshave the same brightness in their respective circuits with steadycurrent flowing.
When the switch S is closed, the lamp A 2 with variable resistor Ris seen to become bright almost immediately but the lamp A 1
with iron-core coil L increases slowly to full brightness.
Figure 7.17a
switch, S
coil, L
iron-core
R
lamp A 1
lamp A 2
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Reason:The coil L undergoes the self-induction and induced emf
in it. The induced or back emf opposes the growth ofcurrent so the glow in the lamp A 1 increases slowly .
The resistor R , however has no back emf, hence the lampA2 glow fully bright as soon as switch S is closed .
This effect can be shown by the graph of current I againsttime t through both lamps in Figure 7.17b.
lamp A 2 with resistor R
t
I
0Figure 7.17b
0 I
lamp A 1 with coil L
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coil, L
switch, S
R
A circuit contains an iron-cored coil L, a switch S, a resistor R and
a dc source arranged in series as shown in Figure 7.18.
Example 20.10 :
Figure 7.18
The switch S is closed for a longtime and is suddenly opened.Explain why a spark jump across theswitch contacts S .
Solution :
When the switch S is suddenly opened, the current in thecircuit starts to fall very rapidly and induced a maximumemf in the coil L which tends to maintain the current.This back emf is high enough to break down the insulation ofthe air between the switch contacts S and a spark can easily
appear at the switch.
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From the self-induction phenomenon, we get
From the Faradays law, thus
20.3.2 Self-inductance, L
I L
wherecurrent: I
(7.14)
linkagefluxmagnetic:L
dt
d L
LI dt d
(7.15)
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Self-inductance is defined as the ratio of the self induced(back) emf to the rate of change of current in the coil .
OR
For the coil of N turns, thus
anddt dI L
dt
d N
dt
dI L
d N dI L N LI
(7.16)
magnetic flux linkage
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The magnetic flux density at the centre of the air-core
solenoid is given by
The magnetic flux passing through each turn of the solenoidalways maximum and is given by
Therefore the self-inductance of the solenoid is given by
20.3.3 Self-inductance of a solenoid
l NI B 0
0cos BA
Al
NI
0
l
NIA0
I N L
l NIA
I N L 0
(7.17)
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A 500 turns of solenoid is 8.0 cm long. When the current in the
solenoid is increased from 0 to 2.5 A in 0.35 s, the magnitude of theinduced emf is 0.012 V. Calculatea. the inductance of the solenoid,b. the cross-sectional area of the solenoid,c. the final magnetic flux linkage through the solenoid.
(Given 0 = 4 10 7 H m 1)Solution :
a. The change in the current is
Therefore the inductance of the solenoid is given by
Example 20.11 :
if I I dI
;A5.2;0m;100.8turns;500 f i2 I I l N
V012.0s;35.0 dt
05.2 dI A5.2dI
dt
dI L
35.0
5.2012.0 L
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At the end of this chapter, students should be able to:Define mutual inductance.Use mutual inductance of two coaxial coils,
Explain the working principle of transformer.
Learning Outcome:
20.4 Mutual inductance ( 2 hours)
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1 I
1 B
Coil 1 Coil 2
1 B20.4.1 Mutual induction
Consider two circular close-packed coils near eachother and sharing a commoncentral axis as shown inFigure 7.20.
A current I 1 flows in coil 1,produced by the battery inthe external circuit.
The current I 1 produces amagnetic field lines insideit and this field lines alsopass through coil 2 asshown in Figure 7.20.
20.4 Mutual inductance
Figure 7.20
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If the current I 1 changes with time , the magnetic flux throughcoils 1 and 2 will change with time simultaneously.
Due to the change of magnetic flux through coil 2, an emf isinduced in coil 2 . This is in accordance to the Faradays lawof induction .
In other words, a change of current in one coil leads to theproduction of an induced emf in a second coil which ismagnetically linked to the first coil .This process is known as mutual induction.Mutual induction is defined as the process of producing aninduced emf in one coil due to the change of current inanother coil.
At the same time, the self-induction occurs in coil 1 since themagnetic flux through it changes .
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From the Figure 7.20, consider the coils 1 and 2 have N 1 and
N 2 turns respectively.If the current I 1 in coil 1 changes, the magnetic flux through coil2 will change with time and an induced emf will occur in coil 2,
2 where
If vice versa, the induced emf in coil 1, 1
is given by
It is a scalar quantity and its unit is henry (H) .
20.4.2 Mutual inductance, M
dt
dI 12
(7.21)
(7.22)
M M M 2112where : Mutual inductance
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Mutual inductance is defined as the ratio of induced emf in acoil to the rate of change of current in another coil .
From the Faradays law for the coil 2, thus
dt
d N 222
dt
d N
dt
dI M 22
112
22112 d N dI M
1
2212
I N M
22112 N I M
and
2
11
21 I
N M
(7.23)
magnetic flux linkagethrough coil 2
magnetic flux linkagethrough coil 1
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Consider a long solenoid with length l and cross sectional area
A is closely wound with N 1 turns of wire. A coil with N 2 turnssurrounds it at its centre as shown in Figure 7.21.
When a current I 1 flows in the primary coil ( N 1), it produces amagnetic field B 1,
20.4.3 Mutual inductance for two solenoids
l
I 1
I 1
N 1 N 2
A
N 1: primary coil
N 2: secondary coil
Figure 7.21
l
I N B 110
1
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and then the magnetic flux 1,
If no magnetic flux leakage , thus
If the current I 1 changes, an emf is induced in the secondarycoils, therefore the mutual inductance occurs and is given by
21
0cos11
A Bl
A I N 1101
l
A I N
I
N M 110
1
2
12
1
2212
I
N M
(7.24)
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A current of 3.0 A flows in coil C and is produced a magnetic fluxof 0.75 Wb in it. When a coil D is moved near to coil C coaxially, aflux of 0.25 Wb is produced in coil D. If coil C has 1000 turns andcoil D has 5000 turns.a. Calculate self-inductance of coil C and the energy stored in C
before D is moved near to it.
b. Calculate the mutual inductance of the coils.c. If the current in C decreasing uniformly from 3.0 A to zero in
0.25 s, calculate the induced emf in coil D.
Solution :
a. The self-inductance of coil C is given by
Example 20.12 :
Wb;25.0Wb;75.0A;0.3 DCC I
turns5000turns;1000 DC N N
C
CCC
I
N L
0.3
75.01000C L
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Solution :
a. and the energy stored in C is
b. The mutual inductance of the coils is given by
2CCC
2
1 I LU
20.32502
1
Wb;25.0Wb;75.0A;0.3 DCC I turns5000turns;1000 DC N N
C
DD
I
N M
0.3
25.05000
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Solution :
c. GivenThe induced emf in coil D is given by
A0.30.30s;25.0 C dI dt
Wb;25.0Wb;75.0A;0.3 DCC I turns5000turns;1000 DC N N
dt
dI M CD
25.0
0.3417
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is an electrical instrument to increase or decrease the emf(voltage) of an alternating current.Consider a structure of the transformer as shown in Figure 7.22.
If N P > N S the transformer is a step-down transformer .
If N P < N S the transformer is a step-up transformer .
20.4.4 Transformer
Figure 7.22
laminated iron core
primary coil secondary coil
N P
turns N
Sturns
alternatingvoltage source
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The symbol of transformer in the electrical circuit is shown inFigure 7.23.
Working principle of transformer When an alternating voltage source is applied to the primarycoil, the alternating current produces an alternating magneticflux concentrated in the iron core.Without no magnetic flux leakage from the iron core, the samechanging magnetic flux passes through the secondary coil andinducing an alternating emf.
After that the induced current is produced in the secondary coil.
Figure 7.23
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The characteristics of an ideal transformer are:Zero resistance of primary coil .
No magnetic flux leakage from the iron core .No dissipation of energy and power .
Formula of transformer According to the mutual inductance, the induced emf in theprimary and secondary coils are given by
For an ideal transformer, there is no flux leakage thus
dt d N PPP (7.25)
(7.26)dt
d N SSS and
dt
d
dt
d SP
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By dividing eqs. (7.25) and (7.26), hence
There is no dissipation of power for the ideal transformer ,
therefore
In general,
dt
d N
dt
d N
SS
P
P
S
P
S
P
S
P
N N
SP P P SSPP I I
P
S
S
P
I
I
(7.27)
primaryof power:P P wheresecondaryof power:S P
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Energy losses in transformer Although transformers are very efficient devices, small energylosses do occur in them owing to four main causes:
Resistance of coilsThe wire used for the primary and secondary coils hasresistance and so ordinary ( I 2 Rt ) heat losses occur.Overcome : The transformer coils are made of thickcopper wire .
HysteresisThe magnetization of the core is repeatedly reversed bythe alternating magnetic field . The resulting expenditureof energy in the core appears as heat .Overcome : By using a magnetic material (such asMumetal) which has a low hysteresis loss .Flux leakageThe flux due to the primary may not all link the secondary.Some of the flux loss in the air .Overcome : By designing one of the insulated coils iswound directly on top of the other rather than having two
separate coils.
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In an alternating current (ac) transformer in which the primary andsecondary windings are perfectly coupled, there is no current flowsin the primary when there is no load in the secondary. When thesecondary is connected to resistors, a current of 5 A is observed toflow in the primary under an applied voltage of 100 V. If the primarycontains 100 turns and the secondary 25000 turns, calculatea. the voltage,b. the current in the secondary.Solution :
a. By applying the formula of transformer,
Example 20.13 :
turns;100V;100A;5 PPP N V I turns25000S N
S
P
S
P
N
N
V
V
25000100100
SV
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Exercise 20.2 :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 turnssurrounds the primary coil at its centre, calculatea. the mutual inductance of the coilsb. the magnitude of induced emf in the secondary coil if the
current in primary coil changes at the rate 4.8 A s 1.
ANS. : 2.63 10 2 H; 0.126 V2. A transformer, assumed to be 100% efficient, is used with a
supply voltage of 120 V. The primary winding has 50 turns.The required output voltage is 3000 V. The output power is200 W.a. Name this type of transformer.b. Calculate the number of turns in the secondary winding.c. Calculate the current supplied to the primary winding
ANS. : 1250 turns; 1.67 A
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3. A transformer with a 100 turns primary coil and a 500 turnssecondary coil is connected to a supply voltage of 2.0 V.Calculate the output voltage and the maximum current insecondary coil if the current in primary coil is to be limited to0.10 A.
ANS. : 10 V; 0.020 A
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At the end of this chapter, students should be able to:Derive and use the energy stored in an inductor,
Learning Outcome:
20.5 Energy stored in an inductor ( hour)
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Consider an inductor of inductance L . Suppose that at time t ,the current in the inductor is in the process of building up to itssteady value I at a rate dI/dt .
The magnitude of the back emf is given by
The electrical power P in overcoming the back emf in the circuitis given by
20.5 Energy stored in an inductor
dt dI L
I P
dt
dI LI P LIdI Pdt and dU Pdt LIdI dU (7.18)
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The total energy stored in the inductor , U as the currentincreases from 0 to I can be found by integrating the eq. (7.18).
Thus
For a long air-core solenoid, the self-inductance is
Therefore the energy stored in the solenoid is given by
I U IdI LdU 00
(7.19)
and analogous to 2
21
CV U in capacitor
l
A N L
20
2
2
1 LI U (7.20)
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A solenoid of length 25 cm with an air-core consists of 100 turnsand diameter of 2.7 cm. Calculatea. the self-inductance of the solenoid, andb. the energy stored in the solenoid,if the current flows in it is 1.6 A.(Given 0 = 4 10 7 H m 1)
Solution :a. The cross-sectional area of the solenoid is given by
Hence the self-inductance of the solenoid is
Example 20.14 :
m107.2m;1025turns;100 22 d l N
24222 m1073.54
107.2
4
d A
l
A N L
20
2
427
10251073.5100104
L
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Solution :b. Given
By applying the equation of energy stored in the inductor, thus2
21
LI U
25
6.11088.221
m107.2m;1025turns;100 22 d l N A6.1 I
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Exercise 20.3 :Given 0 = 4 10 7 H m 1
1. An emf of 24.0 mV is induced in a 500 turns coil at an instantwhen the current is 4.00 A and is changing at the rate of10.0 A s -1 . Determine the magnetic flux through each turn ofthe coil.(Physics for scientists and engineers,6 th edition,Serway&Jewett,
Q6, p.1025)ANS. : 1.92 10 5 Wb2. 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 cmlength. Calculate
a. the magnetic field inside the solenoid,b. the magnetic flux through each turn,c. the inductance of the solenoid.
ANS. : 1.88 10 4 T; 3.33 10 8 Wb; 3.75 10 4 H
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3. A current of 1.5 A flows in an air-core solenoid of 1 cm radiusand 100 turns per cm. Calculatea. the self-inductance per unit length of the solenoid.b. the energy stored per unit length of the solenoid.
ANS. : 0.039 H m 1; 4.4 10 2 J m 1
4. At the instant when the current in an inductor is increasing ata rate of 0.0640 A s 1, the magnitude of the back emf is0.016 V.a. Calculate the inductance of the inductor.b. If the inductor is a solenoid with 400 turns and the current
flows in it is 0.720 A, determinei. the magnetic flux through each turn,
ii. the energy stored in the solenoid.ANS. : 0.250 H; 4.5 10 4 Wb; 6.48 10 2 J5. At a particular instant the electrical power supplied to a
300 mH inductor is 20 W and the current is 3.5 A. Determinethe rate at which the current is changing at that instant.
ANS. : 19 A s
1
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Next ChapterCHAPTER 21 :
Alternating current
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