note_chapter15 electromagnetic induction-0809
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PHYSICS CHAPTER 15
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CHAPTER 15: CHAPTER 15: Electromagnetic inductionElectromagnetic induction
(4 Hours)(4 Hours)
PHYSICS CHAPTER 15
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to:
Define and useDefine and use magnetic fluxmagnetic flux,,
Learning Outcome:15.1(a) Magnetic flux (1 hour)
cosBAAB
PHYSICS CHAPTER 15
15.1(a).115.1(a).1 Phenomenon of electromagnetic inductionPhenomenon of electromagnetic induction Consider some experiments were conducted by Michael Faraday that led
to the discovery of the Faraday’s law of induction as shown in Figures A, B,C, D, and E.
15.1(a) Magnetic flux
No movementNo movement
0v
Figure AFigure A
PHYSICS CHAPTER 15
Figure BFigure B
Figure CFigure C
NNSSMove towards the coilMove towards the coil
v
II
No movementNo movement
0v
PHYSICS CHAPTER 15
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Figure DFigure D
Figure EFigure E
NNMove away from the coilMove away from the coil
v
II
NN SSMove towards the coilMove towards the coil
v
II
SS
PHYSICS CHAPTER 15
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From the experiments: bar magnet is stationarybar magnet is stationary,
- galvanometer not show any deflection (no current no current flows in the coilflows in the coil).
bar magnet is moved relatively towards the coilbar magnet is moved relatively towards the coil, - the galvanometer shows a momentary deflection to the
right (Figure B). - When the bar magnet is moved relatively away from the
coil, the galvanometer is seen to deflect in the opposite direction (Figure D).
ThereforeTherefore :- when there is any relative motion between the coil any relative motion between the coil
and the bar magnetand the bar magnet , the current known as induced induced current will flow momentarilycurrent will flow momentarily through the
galvanometer. - this current due to an induced e.m.fcurrent due to an induced e.m.f across the coil.
PHYSICS CHAPTER 15
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ConclusionConclusion : When the magnetic field lines through a coil changesmagnetic field lines through a coil changes
thus the induced emf will existinduced emf will exist across the coil.
The magnitude of the induced e.m.f. depends on the induced e.m.f. depends on the speed of the relative motionspeed of the relative motion where if the
Therefore vv is proportional to the induced emf is proportional to the induced emf.
v increases induced emf increases
v decreases induced emf decreases
PHYSICS CHAPTER 15
15.1(a).2 Magnetic flux of a uniform magnetic field15.1(a).2 Magnetic flux of a uniform magnetic field
defined as the scalar product between the magnetic flux density, the scalar product between the magnetic flux density, BB with the vector of the area, with the vector of the area, AA.
flux magnetic :
cosΦ BAAB
(7.1)(7.1)
where
AB
and ofdirection ebetween th angle :densityflux magnetic theof magnitude :B
coil theof area :A
PHYSICS CHAPTER 15
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scalar quantityscalar quantity unit is weber (Wb)weber (Wb) OR tesla meter squared ( T mtesla meter squared ( T m22).). Consider a uniform magnetic field B passing through a surface area A of
a single turn coil as shown in Figures A and B.
From the Figure A, the angle is 0 thus the magnetic flux is given by
Figure AFigure A
B
A
cosΦ BA0cosBA
BA maximummaximum
area
PHYSICS CHAPTER 15
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From the Figure B, the angle is 90 thus the magnetic flux is given by
Figure BFigure B
B
A
cosΦ BA90cosBA
0Note:Note: Direction of vector A always perpendicular (normal)perpendicular (normal) to the
surface area, A. The magnetic flux is proportional to the number of field lines magnetic flux is proportional to the number of field lines
passing through the area.passing through the area.
area
90
PHYSICS CHAPTER 15
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A single turn of rectangular coil of sides 10 cm 5.0 cm is placed between north and south poles of a permanent magnet. Initially, the plane of the coil is parallel to the magnetic field as shown in Figure 15.3.
If the coil is turned by 90 about its rotation axis and the magnitude of magnetic flux density is 1.5 T, Calculate the change in the magnetic flux through the coil.
Example 1 :
SSNNP
QR
S
I I
Figure 15.3Figure 15.3
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Solution :Solution :
The area of the coil is
Initially,
Finally,
Therefore the change in magnetic flux through the coil is
T 51.B
2322 m 100510051010 ..A
From the figure, =90 thus the initial magnetic flux through the coil is
A
B
cosΦi BA90cosBA
0Φi B
A
From the figure, =0 thus the final magnetic flux through the coil is
cosΦf BA 0cos100.55.1 3
Wb105.7Φ 3f
if ΦΦΦ 0105.7 3
Wb105.7 3
PHYSICS CHAPTER 15
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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 angle 30 to the direction of the magnetic field. If the magnetic flux through the area of the coil is 1.20 mWb, calculate the magnitude of the magnetic field.
Solution :Solution :
The area of the coil is
Example 2 :
Wb1020.1 m; 100.3 32 d
4
2dA
4
100.322
A
24 m 1007.7 A
3030
coil
BA
PHYSICS CHAPTER 15
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Solution :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
Wb1020.1 m; 100.3 32 d
603090
cosΦ BA
60cos1007.71020.1 43 B
T 40.3B
PHYSICS CHAPTER 15
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The three loops of wire as shown in Figure 15.4 are all in a region of space with a uniform magnetic field. Loop 1 swings back and forth as the bob on a simple pendulum. Loop 2 rotates about a vertical axis and loop 3 oscillates vertically on the end of a spring. Which loop or loops have a magnetic flux that changes with time? Explain your answer.
Example 3 :
Figure 15.4Figure 15.4
PHYSICS CHAPTER 15
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Solution :Solution :
Only loop 2 has a changing magnetic fluxloop 2 has a changing magnetic flux.
Reason :
Loop 1 moves back and forth, and loop 3 moves up and down, but Loop 1 moves back and forth, and loop 3 moves up and down, but since the magnetic field is uniform, the flux always constant with since the magnetic field is uniform, the flux always constant with time.time.
Loop 2 on the other hand changes its orientation relative to the field Loop 2 on the other hand changes its orientation relative to the field as it rotates, hence its flux does change with time.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: At the end of this chapter, students should be able to: Explain Explain induced emfinduced emf.. StateState Faraday’s lawFaraday’s law.. ApplyApply formulaeformulae,,
Learning Outcome:15.1(b) Induced emf (2 hours)
dt
d
PHYSICS CHAPTER 15
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Derive and use Derive and use formula of induced emf for: formula of induced emf for:
a coil, a coil,
a rotating coil, a rotating coil,
Learning Outcome:15.1(b) Induced emf (2 hours)
dt
dBA
tNAB sin
ORORdt
dAB
PHYSICS CHAPTER 15
15.1(b).1 Faraday’s law of electromagnetic induction15.1(b).1 Faraday’s law of electromagnetic induction
states that the magnitude of the induced emf is proportional to the the magnitude of the induced emf is proportional to the rate of change of the magnetic fluxrate of change of the magnetic flux.
The negative signnegative sign indicates that the direction of induced emfdirection of induced emf always oppose the change of magnetic flux producing it (Lenz’s law)oppose the change of magnetic flux producing it (Lenz’s law).
15.1(b) Induced emf
dt
dΦ OR
dt
dΦ (7.2)(7.2)
where
flux magnetic theof change :Φd 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 coil of NN turns is placed in the changing magnetic field turns is placed in the changing magnetic field BB, the
induced emf is given by
dt
dN
Φ (7.3)(7.3)
, then eq. (7.3) can be written asif ΦΦ Φ d
dt
N if ΦΦ (7.4)(7.4)
flux magnetic final :Φf
flux magnetic initial :Φi
where
dt
dN
Φ cosΦ BAand
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For a coil of coil of NN turns is placed in a uniform magnetic field turns is placed in a uniform magnetic field B B but
changing in the coil’s area changing in the coil’s area AA, the induced emf is given by
dt
BAdN
cos
dt
dBNA cos (7.5)(7.5)
dt
dN
Φ cosΦ BAand
dt
BAdN
cos
dt
dANB cos (7.6)(7.6)
PHYSICS CHAPTER 15
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For a coil is connected in series to a resistor of resistance a coil is connected in series to a resistor of resistance RR and
the induced emf exist in the coil as shown in Figure 7.5,
Figure 15.5Figure 15.5
RII
dt
dN
Φ IRand
dt
dNIR
Φ (7.7)(7.7)
Note:Note: To calculate the magnitude of induced emfmagnitude of induced emf, the negative sign negative sign
can be ignoredcan be ignored. For a coil of N turns, each turn will has a magnetic flux of
BAcos through it, therefore the magnetic flux linkagemagnetic flux linkage (refer to the combined amount of flux through all the turns) is given by
the induced current I is given by
Φlinkageflux magnetic N
PHYSICS CHAPTER 15
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The magnetic flux passing through a single turn of a coil is increased quickly but steadily at a rate of 5.0102 Wb s1. If the coil have 500 turns, calculate the magnitude of the induced emf in the coil.
Solution :Solution :
By applying the Faraday’s law equation for a coil of N turns , thus
Example 15.4 :
12 s Wb100.5 turns;500
dt
dN
dt
dN
Φ
2100.5500
V 25
PHYSICS CHAPTER 15
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A coil having an area of 8.0 cm2 and 50 turns lies perpendicular to a magnetic field of 0.20 T. If the magnetic flux density is steadily reduced to zero, taking 0.50 s, determine
a. the initial magnetic flux linkage.
b. the induced emf.
Solution :Solution :
a. The initial magnetic flux linkage is given by
B
Example 15.5 :
iΦlinkageflux magnetic initial N
0;T; 0.20 turns;50 ;m 100.8 fi24 BBNA
s 0.50dt
0A
cosi ANB
PHYSICS CHAPTER 15
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Solution :Solution :
a.
b. The induced emf is given by
0;T; 0.20 turns;50 ;m 100.8 fi24 BBNA
s 0.50dt
0cos108.00.2050 4 Wb108.0linkageflux magnetic initial 3
dt
dBNA cos if BBdB and
dt
BBNA ifcos
50.0
20.000cos100.850 4
V 106.1 2
PHYSICS CHAPTER 15
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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, the diameter 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 :Solution :
Example 15.6 :
m; 103.5m; 100.4 turns;10 2f
2i
ddNs 0.25 T; 2.1 dtB
0
A
B
B
A
Initial Final
PHYSICS CHAPTER 15
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Solution :Solution :
a. The change in the area of the coil is given by
if AAdA
44
2i
2f dd
24 m 105.9 dA
m; 103.5m; 100.4 turns;10 2f
2i
ddNs 0.25 T; 2.1 dtB
2i
2f4
dd
2222 100.4103.5
4
PHYSICS CHAPTER 15
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Solution :Solution :
b. Given
The induced emf in the coil is
Therefore the induced current in the coil is given by
4.2R
dt
dANB cos
m; 103.5m; 100.4 turns;10 2f
2i
ddNs 0.25 T; 2.1 dtB
25.0
105.90cos2.110
4
V 1056.4 2
IR
A 109.1 2I 4.21056.4 2 I
PHYSICS CHAPTER 15
15.2 Induced emf (2 hours)
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: StateState Lenz’s lawLenz’s law.. ApplyApply formulaeformulae,,
Learning Outcome:
)(cos tdt
dNBA
PHYSICS CHAPTER 15
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states that an induced electric current always flows in such a an induced electric current always flows in such a direction that it opposes the change producing itdirection that it opposes the change producing it.
This law is essentially a form of the law of conservation of energyconservation of energy.
15.2(a) Lenz’s law
Direction of Direction of induced current – induced current – Right hand grip Right hand grip rule.rule.
I
INN
North pole
Experiment 1 :Experiment 1 :
From From lenz’s law experimentslenz’s law experiments : :
Figure shows the magnitude
of the magnetic field at the
solenoid increases as the bar
magnet is moved towards it.
PHYSICS CHAPTER 15
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an emf is induced in the solenoid and the galvanometer indicates that a current is flowing.
direction of the current deflection of the galvanometer needle shows the direction of the current.
the current through the solenoid is in the direction that make current through the solenoid is in the direction that make the solenoid upper end becomes a north polethe solenoid upper end becomes a north pole. This opposes the opposes the motion of the bar magnetmotion of the bar magnet and obey the lenz’s lawobey the lenz’s law.
PHYSICS CHAPTER 15
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Experiment 2 :Experiment 2 : Consider two solenoids P and Q arranged coaxially closed to each
other as shown in Figure A.
At the moment when the switch S is closedswitch S is closed, current I begins to flow in the solenoid P and producing a magnetic field inside the solenoid P. Suppose that the field points towards the solenoid Q.
Figure AFigure A
S,SwitchPP QQ
II
NNSS SSNN
indI indI--++
ind
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The magnetic flux through the solenoid Q increases with timeincreases with time. According to Faraday’s law ,an induced current due to induced emf will exist in solenoid Q.
The induced current flows in solenoid Q must produce a magnetic field that oppose the change producing it (increase in flux). Hence based on Lenz’s law, the induced current flows in circuit consists of solenoid Q is anticlockwise anticlockwise (Figure A) and the galvanometer shows a deflection.
QQSSwitch,
PP
NNSS
II
SS NN
indI indI-- ++
ind
Figure BFigure B
PHYSICS CHAPTER 15
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At the moment when the switch S is openedswitch S is opened, the current I starts to decrease in the solenoid P and magnetic flux through the solenoid Q decreases with timedecreases with time.
According to Faraday’s law ,an induced current due to induced emf will exist in solenoid Q.
The induced current flows in solenoid Q must produce a magnetic field that oppose the change producing it (decrease in flux).
Hence based on Lenz’s law, the induced current flows in circuit consists of solenoid Q is clockwise clockwise (Figure B) and the galvanometer seen to deflect in the opposite direction of Figure 15.7.
PHYSICS CHAPTER 15
A single turn of circular shaped coil has a resistance of 20 and an area of 7.0 cm2. It moves toward the north pole of a bar magnet as shown in Figure 7.10.
If the average rate of change of magnetic flux density through the coil is 0.55 T s1, a. determine the induced current in the coil b. state the direction of the induced current observed by the observer shown in figure.
Example 15.7 :
PHYSICS CHAPTER 15
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Solution :Solution :
a. By applying the Faraday’s law of induction, thus
Therefore the induced current in the coil is given by
dt
dN
BAdt
dN
A 1093.1 5I
124 s T 55.0 ;m 100.7 ; 20 turn;1 dt
dBARN
dt
dBNA
55.0100.71 4
180cosΦ BAand
V 1085.3 4
IR 201085.3 4 I
PHYSICS CHAPTER 15
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Solution :Solution :
b. Based on the lenz’s law, hence the direction of induced current is
clockwiseclockwise as shown in figure below.
NNSS indI
PHYSICS CHAPTER 15
Consider a rectangular coil of N turns, each of area A, being rotated
mechanically with a constant angular velocity in a uniform magnetic field
of flux density B about an axis as shown in figure.
When the vector of area, A is at an angle to the magnetic field B, the magnetic flux through each turn of the coil is given by
15.2(b) Induced emf in a rotating coil
side viewside view
B
NN SSA
ω
coil
cosBA t and
tBA cos
PHYSICS CHAPTER 15
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By applying the equation of Faraday’s law for a coil of N turns, thus the induced emf is given by
The induced emf is maximum when hence
dt
dN
tBAdt
dN cos
tdt
dNBA cos
tNBA sin (7.8)(7.8)
time:twhere1sin t
NBAmax (7.9)(7.9)
Tf
22 where
PHYSICS CHAPTER 15
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Eq. (7.8) also can be written as
Conclusion Conclusion : A coil rotating with constant angular velocity in a uniform magnetic field produces a sinusoidally alternating emfsinusoidally alternating emf as shown by the
induced emf against time t graph in Figure 15.11.
sinNBA (7.10)(7.10)
BA
and between angle :where
Figure 15.11Figure 15.11
ωtεε sinmax
0
max
max
V ε
tTT5.0 T5.1 T2
B
Note:Note:
This phenomenon was the important part in the development of the electric electric generator or generator or dynamodynamo.
PHYSICS CHAPTER 15
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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 uniform magnetic field of flux density 5.0 T. Calculate
a. 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 :Solution :
The area of the coil is
and the constant angular velocity in rad s1 is
Example 15.8 :
2222 m 105.110151010 A
T 0.5 turns;100 BN
s 06
min 1
rev 1
rad 2
min1
rev 200
1s rad 9.20
PHYSICS CHAPTER 15
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Solution :Solution :
a. The maximum emf produced by the coil is given by
b.
NBAmax
TBN 0.5 turns;100
9.20105.10.5100 2V 157max
B
A
38
From the figure, the angle is
Therefore the induced emf is given by
523890
sinNBA 52sin9.20105.10.5100 2
V 124
PHYSICS CHAPTER 15
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Exercise :
1. A bar magnet is held above a loop of wire in a horizontal plane, as shown in figure.
The south end of the magnet is toward the loop of the wire. The magnet is dropped toward the loop. Determine the direction of the 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(Physics for scientists and engineers,6thth edition, Serway&Jewett, Q15, p.991)edition, Serway&Jewett, Q15, p.991)
ANS. :ANS. : U thinkU think
PHYSICS CHAPTER 15
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2. A coil of area 0.100 m2 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(Physics for scientists and engineers,6thth edition,Serway&Jewett, Q35, edition,Serway&Jewett, Q35, p.996)p.996)
ANS. :ANS. : 7.547.54101033 V V
3. A circular coil has 50 turns and diameter 1.0 cm. It rotates at a constant angular velocity of 25 rev s1 in a uniform magnetic field of flux density 50 T. Determine the induced emf when the plane of the coil makes an angle 55 to the magnetic field.
ANS. :ANS. : 1.771.77101055 V V
PHYSICS CHAPTER 15
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