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Chapter 7:Electromagnetic induction

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Title: Chapter 7:Electromagnetic induction


1
is defined as the production of an induced e.m.f.
in a conductor/coil whenever the magnetic flux
through the conductor/coil changes.
CHAPTER 20 Electromagnetic induction(6 Hours)
2
Learning Outcome
20.1 Magnetic flux (1/2 hour)
  • At the end of this chapter, students should be
    able to
  • Define and use magnetic flux,

3
  • 20.1.1 Magnetic flux of a uniform magnetic field
  • is defined as the scalar product between the
    magnetic flux density, B with the vector of the
    area, A.
  • Mathematically,

where
4
  • It is a scalar quantity and its unit is weber
    (Wb) OR tesla meter squared ( T m2).
  • Consider a uniform magnetic field B passing
    through a surface area 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 is given by

Figure 7.2a
maximum
5
  • From the Figure 7.2a, the angle ? is 90? thus
    the magnetic flux is given by

Figure 7.2b
  • Direction of vector A always perpendicular
    (normal) to the surface area, A.
  • The magnetic flux is proportional to the number
    of field lines passing through the area.

6
Example 1
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 7.3. 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.
Figure 7.3
7
Solution The area of the coil is Initially,
Finally, Therefore the change in
magnetic flux through the coil is
From the figure, ? thus the initial
magnetic flux through the coil is
From the figure, ? thus the final magnetic
flux through the coil is
8
Example 2
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 The area of the coil is
9
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
10
Learning Outcome
20.2 Induced emf (2 hours)
  • At the end of this chapter, students should be
    able to
  • Use Faraday's experiment to explain induced emf.
  • State Faradays law and Lenzs law to determine
    the direction of induced current.
  • Apply formulae,
  • Derive and use induced emf
  • I) in straight conductor,
  • ii) in coil, OR
  • iii) in rotating coil,

11
20.2.1 Magnetic flux
  • 20.1.1(a) Phenomenon of electromagnetic induction
  • Consider some experiments were conducted by
    Michael Faraday that led to the discovery of the
    Faradays law of induction as shown in Figures
    7.1a, 7.1b, 7.1c, 7.1d and 7.1e.

Figure 7.1a
12
S
N
Figure 7.1b
Figure 7.1c
13
N
S
Figure 7.1d
S
N
Figure 7.1e
14
  • From the experiments
  • When the bar magnet is stationary, the
    galvanometer not show 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 relatively away from the
    coil, the galvanometer is seen to deflect in the
    opposite direction (Figure 7.1d).
  • Therefore when there is any relative motion
    between the coil and the bar magnet , the current
    known as induced current 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
    changes thus the induced emf will exist across
    the coil.

15
  • The magnitude of the induced e.m.f. depends on
    the speed of the relative motion where if the
  • Therefore v is proportional to the induced emf.

v increases
induced emf increases
v decreases
induced emf decreases
16
Example 3
Figure 7.4
The three loops of wire as shown in Figure 7.4
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.
17
Solution
18
20.2.2 Induced emf
  • 20.2.2(a) Faradays law of electromagnetic
    induction
  • states that the magnitude of the induced emf is
    proportional to 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).

OR
where
19
, then eq. (7.3) can be written as
where
and
20
  • 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 given by

and
21
  • For a coil is connected in series to a resistor
    of resistance R and the induced emf ? exist in
    the coil as shown in Figure 7.5,

the induced current I is given by
and
Figure 7.5
  • To calculate the magnitude of induced emf, the
    negative sign can 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 to the combined
    amount of flux through all the turns) is given by

22
Example 4
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 coil have 500
turns, calculate the magnitude of the induced emf
in the coil. Solution By applying the
Faradays law equation for a coil of N turns ,
thus
23
Example 5
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 a. The initial magnetic flux
linkage is given by
24
Solution a. b. The induced emf is given by
25
Example 6
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
Initial
Final
26
Solution a. The change in the area of the coil
is given by
27
Solution b. Given The induced emf in the
coil is Therefore the induced current
in the coil is given by
28
20.2.2 (b) Lenzs law
  • 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 of energy.
  • An illustration of lenzs law can be explained by
    the following experiments.

Direction of induced current Right hand grip
rule.
  • 1st experiment
  • In Figure 7.6 the magnitude of the magnetic field
    at the solenoid increases as the bar magnet is
    moved towards it.
  • An emf is induced in the solenoid and the
    galvanometer indicates that a current is flowing.

N
Figure 7.6
29
  • To determine the direction of the current through
    the galvanometer which corresponds to a
    deflection in a particular sense, then the
    current through the solenoid seen is in the
    direction that make the solenoid upper end
    becomes a north pole. This opposes the motion of
    the bar magnet and obey the lenzs law.
  • 2nd experiment
  • Consider a straight conductor PQ is placed
    perpendicular to the magnetic field and move the
    conductor to the left with constant velocity v as
    shown in Figure 7.7.
  • When the conductor move to the left thus the
    induced current needs to flow in such a way to
    oppose the change which has induced it based on
    lenzs law. Hence galvanometer shows a deflection.

Figure 7.7
30
  • To determine the direction of the induced current
    (induced emf) 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 is placed in the magnetic field then this
    conductor will experience magnetic force.
  • Its direction is in the opposite direction of the
    motion.

Thumb direction of Motion First finger
direction of Field Second finger direction of
induced
current OR induced emf
Figure 7.8
31
  • 3rd 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 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 7.9a
32
  • The magnetic flux through the solenoid Q
    increases with time. 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 a magnetic field that oppose the change
    producing it (increase in flux). Hence based on
    Lenzs law, the induced current flows in circuit
    consists of solenoid Q is anticlockwise (Figure
    7.9a) and the galvanometer shows a deflection.

Figure 7.9b
33
  • 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 Faradays 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
    Lenzs law, the induced current flows in circuit
    consists of solenoid Q is clockwise (Figure 7.9b)
    and the galvanometer seen to deflect in the
    opposite direction of Figure 7.9a.

34
Example 7
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 s?1, a. determine the induced
current in the coil b. state the direction of
the induced current observed by the
observer shown in Figure 7.10.
Figure 7.10
35
Solution a. By applying the Faradays law of
induction, thus Therefore the
induced current in the coil is given by
36
Solution b. Based on the lenzs law, hence the
direction of induced current is clockwise as
shown in figure below.
37
20.2.3 Induced emf in a straight conductor
  • Consider a straight conductor PQ of length l is
    moved perpendicular with velocity v across a
    uniform magnetic field B as shown in Figure 7.11.
  • When the conductor moves through a distance x in
    time t, the area swept out by the conductor is
    given by

Area, A
Figure 7.11
38
  • Since the motion of the conductor is
    perpendicular to the magnetic field B hence the
    magnetic flux cutting by the conductor is given
    by
  • According to Faradays law, the emf is induced in
    the conductor and its magnitude is given by

and
and
39
  • In general, the magnitude of the induced emf in
    the straight conductor is given by
  • This type of induced emf is known as motional
    induced emf.
  • The direction of the induced current or induced
    emf in the straight conductor can be determined
    by using the Flemings right hand rule (based on
    Lenzs law).
  • In the case of Figure 7.11, the direction of the
    induced current or induced emf is from Q to P.
    Therefore P is higher potential than Q.

where
  • Eq. (7.9) also can be used for a single turn of
    rectangular coil moves across the uniform
    magnetic field.
  • For a rectangular coil of N turns,

40
Example 8
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 rod is
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.
Figure 7.12
41
Solution a. By applying the equation for
motional induced emf, thus b. Given i. By
applying the Ohms law, thus By using
the Flemings right hand rule, ii. Given
The total charge passing through the resistor is
given by
42
20.2.4 Induced emf in a rotating coil
  • 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 7.13.
  • 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

Figure 7.13 side view
and
43
  • 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

where
where
44
  • Eq. (7.11) also can be written as
  • Conclusion A coil rotating with constant
    angular velocity in a uniform magnetic field
    produces a sinusoidally alternating emf as shown
    by the induced emf ? against time t graph in
    Figure 7.14.

where
This phenomenon was the important part in the
development of the electric generator or dynamo.
Figure 7.14
45
Example 9
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 The area of the coil is and
the constant angular velocity in rad s?1 is
46
Solution a. The maximum emf produced by the
coil is given by b.
From the figure, the angle ? is Therefore the
induced emf is given by
47
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 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,6th
edition, SerwayJewett, Q15, p.991) ANS. U think
Figure 7.15
48
2. A straight conductor of length 20 cm moves in
a uniform magnetic field of flux density 20 mT at
a constant speed of 10 m s-1. The velocity makes
an angle 30? to the field but the conductor is
perpendicular to the field. Determine the induced
emf. ANS. 2.0?10?2 V 3. 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,6th edition,SerwayJewett, Q35,
p.996) ANS. 7.54?103 V 4. 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 of flux density 50 ?T.
Determine the induced emf when the plane of the
coil makes an angle 55? to the magnetic
field. ANS. 1.77?10?5 V
49
Learning Outcome
20.3 Self-inductance (1 hour)
  • At the end of this chapter, students should be
    able to
  • Define self-inductance.
  • Apply formulae
  • for a loop and solenoid.

50
20.3 Self-inductance
  • 20.3.1 Self-induction
  • Consider a solenoid which is connected to a
    battery , a switch S and variable resistor R,
    forming an open circuit as shown in Figure 7.16a.
  • When the switch S is closed, a current I begins
    to flow in the solenoid.
  • The current produces a magnetic field whose field
    lines through the solenoid and generate the
    magnetic flux linkage.
  • If the resistance of the variable resistor
    changes, thus the current flows in the solenoid
    also changed, then so too does magnetic flux
    linkage.

Figure 7.16a initial
51
  • According to the Faradays law, an emf has to be
    induced in the solenoid itself since the flux
    linkage changes.
  • In accordance with Lenzs law, the induced emf
    opposes the changes that has induced it and it is
    known as a back emf.
  • For the current I increases

Figure 7.16b I increases
52
  • For the current I decreases
  • This process is known as self-induction.
  • Self-induction is defined as the process of
    producing an induced emf in the coil due to a
    change of current flowing through the same coil.

Figure 7.16c I decreases
53
20.3 SELF-INDUCTANCE
Iinduced
Iinduced
  • 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.
(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
the current decreases.
(c) The polarity of the induced emf reverses if
the current decreases.
54
  • Self-induction experiment
  • The 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 lamps have the same brightness in their
    respective circuits with steady current flowing.
  • When the switch S is closed, the lamp A2 with
    variable resistor R is seen to become bright
    almost immediately but the lamp A1 with iron-core
    coil L increases slowly to full brightness.

Figure 7.17a
55
  • Reason
  • The coil L undergoes the self-induction and
    induced emf in it. The induced or back emf
    opposes the growth of current so the glow in the
    lamp A1 increases slowly.
  • The resistor R, however has no back emf, hence
    the lamp A2 glow fully bright as soon as switch S
    is closed.
  • This effect can be shown by the graph of current
    I against time t through both lamps in Figure
    7.17b.

Figure 7.17b
56
Example 10
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.
The switch S is closed for a long time and is
suddenly opened. Explain why a spark jump across
the switch contacts S .
Figure 7.18
  • Solution
  • When the switch S is suddenly opened, the
    and ...................
    ............... ........ which tends to
    maintain the current.
  • This back emf is high enough to
    ..and a
    ..

57
20.3.2 Self-inductance, L
  • From the self-induction phenomenon, we get
  • From the Faradays law, thus

where
58
  • 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

and
59
  • It is a scalar quantity and its unit is henry
    (H).
  • Unit conversion
  • The value of the self-inductance depends on
  • the size and shape of the coil,
  • the number of turn (N),
  • the permeability of the medium in the coil (?).
  • A circuit element which possesses mainly
    self-inductance is known as an inductor. It is
    used to store energy in the form of magnetic
    field.
  • The symbol of inductor in the electrical circuit
    is shown in Figure 7.19.

Figure 7.19
60
20.3.3 Self-inductance of a solenoid
  • The magnetic flux density at the centre of the
    air-core solenoid is given by
  • The magnetic flux passing through each turn of
    the solenoid always maximum and is given by
  • Therefore the self-inductance of the solenoid is
    given by

61
Example 11
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 the induced emf
is 0.012 V. Calculate a. 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
62
Solution b. By using the equation of
self-inductance for the solenoid, thus c.
The final magnetic flux linkage is given by
63
Exercise 20.2
Given ?0 4? ? 10?7 H m?1 1. An emf of 24.0 mV
is induced in a 500 turns coil at an instant when
the current is 4.00 A and is changing at the
rate of 10.0 A s-1. Determine the magnetic flux
through each turn of the coil. (Physics for
scientists and engineers,6th edition,SerwayJewett
, Q6, p.1025) ANS. 1.92?10?5 Wb 2. 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. ANS. 1.88?10?4 T 3.33?10?8 Wb
3.75?10?4 H
64
3. A current of 1.5 A flows in an air-core
solenoid of 1 cm radius and 100 turns per cm.
Calculate a. 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 at a rate of 0.0640
A s?1, the magnitude of the back emf is 0.016
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, determine i. 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
J 5. At a particular instant the electrical
power supplied to a 300 mH inductor is 20 W and
the current is 3.5 A. Determine the rate at which
the current is changing at that instant. ANS.
19 A s?1
65
Learning Outcome
20.4 Mutual inductance (2 hours)
  • At the end of this chapter, students should be
    able to
  • Define mutual inductance.
  • Derive and use formulae for mutual inductance of
    two coaxial coils,
  • Explain the working principle of transformer and
    the effect of eddy current in transformer.

66
20.4 Mutual inductance
  • 20.4.1 Mutual induction
  • Consider two circular close-packed coils near
    each other and sharing a common central axis as
    shown in Figure 7.20.
  • A current I1 flows in coil 1, produced by the
    battery in the external circuit.
  • The current I1 produces a magnetic field lines
    inside it and this field lines also pass through
    coil 2 as shown in Figure 7.20.

Figure 7.20
67
  • If the current I1 changes with time, the magnetic
    flux through coils 1 and 2 will change with time
    simultaneously.
  • Due to the change of magnetic flux through coil
    2, an emf is induced in coil 2. This is in
    accordance to the Faradays law of induction.
  • In other words, a change of current in one coil
    leads to the production of an induced emf in a
    second coil which is magnetically linked to the
    first coil.
  • This process is known as mutual induction.
  • Mutual induction is defined as the process of
    producing an induced emf in one coil due to the
    change of current in another coil.
  • At the same time, the self-induction occurs in
    coil 1 since the magnetic flux through it changes.

68
20.4.2 Mutual inductance, M
  • From the Figure 7.20, consider the coils 1 and 2
    have N1 and N2 turns respectively.
  • If the current I1 in coil 1 changes, the magnetic
    flux through coil 2 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).

where
Mutual inductance
69
  • Mutual inductance is defined as the ratio of
    induced emf in a coil to the rate of change of
    current in another coil.
  • From the Faradays law for the coil 2, thus

magnetic flux linkage through coil 1
magnetic flux linkage through coil 2
and
70
20.4.3 Mutual inductance for two solenoids
  • Consider a long solenoid with length l and cross
    sectional area A is closely wound with N1 turns
    of wire. A coil with N2 turns surrounds it at its
    centre as shown in Figure 7.21.
  • When a current I1 flows in the primary coil (N1),
    it produces a magnetic field B1,

N1 primary coil N2 secondary coil
Figure 7.21
71
  • and then the magnetic flux ?1,
  • If no magnetic flux leakage, thus
  • If the current I1 changes, an emf is induced in
    the secondary coils, therefore the mutual
    inductance occurs and is given by

72
Mutual inductance, M
73
Example 13
A current of 3.0 A flows in coil C and is
produced a magnetic flux of 0.75 Wb in it. When a
coil D is moved near to coil C coaxially, a flux
of 0.25 Wb is produced in coil D. If coil C has
1000 turns and coil 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
74
Solution a. and the energy stored in C
is b. The mutual inductance of the coils is
given by
75
Solution c. Given The induced emf in
coil D is given by
76
20.4.4 Transformer
  • 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 NP gt NS the transformer is a step-down
    transformer.
  • If NP lt NS the transformer is a step-up
    transformer.

Figure 7.22
77
  • The symbol of transformer in the electrical
    circuit is shown in Figure 7.23.
  • Working principle of transformer
  • When an alternating voltage source is applied to
    the primary coil, the alternating current
    produces an alternating magnetic flux
    concentrated in the iron core.
  • Without no magnetic flux leakage from the iron
    core, the same changing magnetic flux passes
    through the secondary coil and inducing 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.

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  • Energy losses in transformer
  • Although transformers are very efficient devices,
    small energy losses do occur in them owing to
    four main causes
  • Resistance of coils
  • The wire used for the primary and secondary
    coils has resistance and so ordinary (I2R) heat
    losses occur.
  • Overcome The transformer coils are made of
    thick copper wire.
  • Hysteresis
  • The magnetization of the core is repeatedly
    reversed by the alternating magnetic field. The
    resulting expenditure of energy in the core
    appears as heat.
  • Overcome By using a magnetic material (such as
    Mumetal) which has a low hysteresis loss.
  • Flux leakage
  • The 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 is wound directly on top of the other
    rather than having two separate coils.

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Learning Outcome
20.5 Energy stored in an inductor (½ hour)
  • At the end of this chapter, students should be
    able to
  • Derive and use formulae for energy stored in an
    inductor,

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23.4 Energy stored in an inductor
  • Consider an inductor of inductance L. Suppose
    that at time t, the current in the inductor is in
    the process of building up to its steady 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 circuit is given by

and
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and analogous to
in capacitor
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Example 12
A solenoid of length 25 cm with an air-core
consists of 100 turns and diameter of 2.7 cm.
Calculate a. the self-inductance of the solenoid,
and b. 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
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Solution b. Given By applying the equation
of energy stored in the inductor, thus
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