**Unformatted text preview: **is defined as the production of an induced
the production of an induced
e.m.f.
e.m.f. in a conductor/coil whenever the
magnetic flux through the conductor/coil
changes. CHAPTER 20:
CHAPTER 20:
Electromagnetic induction
(6 Hours) 1 20.1.1 Magnetic flux of a uniform magnetic field φ : magnetic flux φ = B • A = BA cos θ (7.1)
(7.1) is defined as the scalar product between the magnetic flux
the scalar product between the magnetic flux
density, B with the vector of the area, A.
Mathematically, where B 3 θ : angle between th e direction of B and A
B : magnitude of the magnetic flux density
A : area of the coil A
area φ = 90 Figure 7.2b = BA cos 90 φ = BA cos θ φ = 0 5 From the Figure 7.2a, the angle φ is 90° thus the magnetic flux
is given by Note: Direction of vector A always perpendicular (normal) to
the surface area, A.
The magnetic flux is proportional to the number of
magnetic flux is proportional to the number of
field lines passing through the area.
field lines passing through the area. Magnetic flux (1/2 hour) LEARNING OUTCOME:
20.1
Define and use magnetic flux, At the end of this chapter, students should be able to: φ = B • A = BA cos θ
φ = B • A = BA cos θ It is a scalar quantity and its unit is weber (Wb) OR tesla
(Wb)
meter squared
meter squared ( T m2). area A B 2 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. φ = BA maximum φ = BA cos θ
= BA cos 0 Q S
S
S
S 6 4 Figure 7.2a
From the Figure 7.2a, the angle φ is 0° thus the magnetic flux is
given by Example 1 : R I P I 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. N
N
N
N
S Figure 7.3
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. Solution :
Solution : B = 1.5 T A B From the figure, θ =0° thus the final
magnetic flux through the coil is From the figure, θ =90° thus the initial
B magnetic flux through the coil is The area of the coil is
Initially, Finally, A Therefore the change in magnetic flux through the coil is
7 Example 2 : θ 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 20.2 LEARNING OUTCOME: dt I ε = −B N
N ε = NABω sin ωt dB
dt Derive and use induced emf:
emf:
I) in straight conductor, ε = lvB sin θ
ii) in coil,
ii) in coil,
OR
OR ε = −A iii) in rotating coil, S
S
I Figure 7.1b
Figure 7.1b
Figure 7.1b
Figure 7.1b Figure 7.1c
Figure 7.1c dA
dt No movement v=0 Move towards the coil
Move towards the coil v At the end of this chapter, students should be able to:
At the end of this chapter, students should be able to:
Use Faraday's experiment to explain induced emf.
emf.
State Faraday’s law and Lenz’s law to determine the
direction of induced current.
dφ
Apply formulae, ε = − d φ
formulae, Induced emf (2 hours) Solution :
area, A is given by 11 9 The angle between the direction of magnetic field, B and vector of Therefore the magnitude of the magnetic field is 20.2.1 MAGNETIC FLUX
20.1.1(a) Phenomenon of electromagnetic induction
20.1.1(a) Phenomenon of electromagnetic induction No movement v=0 Consider some experiments were conducted by Michael
Faraday that led to the discovery of the Faraday’s law of
induction as shown in Figures 7.1a, 7.1b, 7.1c, 7.1d and 7.1e. Figure 7.1a 8 10 12 N
N N
N I I S
S
I
Figure 7.1d S
S
I
Figure 7.1e
Figure 7.1e v 15 13 Move away from the coil v
Move towards the coil
Move towards the coil The magnitude of the induced e.m.f. depends on the
speed of the relative motion where if the
induced emf increases
v increases v decreases
induced emf decreases
Therefore v is proportional to the induced emf
emf. Solution :
Only loop 2 has a changing magnetic flux
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 always
but since the magnetic field is uniform, the flux always
constant with time.
Loop 2 on the other hand changes its orientation relative to the
field as it rotates, hence its flux does change with time. 17 From the experiments:
When the bar magnet is stationary the galvanometer not
bar magnet is stationary,
no current flows in the coil).
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
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.
induced emf will exist 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. 20.2.2 INDUCED EMF
20.2.2(a) Faraday’s law of electromagnetic induction dφ
dt OR ε = − dφ
dt (7.2) states that the magnitude of the induced emf is proportional
to the rate of change of the magnetic flux
flux.
Mathematically, ε ∝ −
ε ∝ −
where dφ : changeof the magneticflux
dt : change of time
ε : induced emf
The negative sign indicates that the direction of induced emf
always oppose the change of magnetic flux producing it
(Lenz’s law)
law). 14 16 18 ε = −N dφ
dt
(7.3)
(7.3) dΦ = Φ f − Φ i , then eq. (7.3) can be written as
d (φ f − φ i )
dt and φ = BA cos θ φi : initial magnetic flux
φ f : final magnetic flux ε = −N
where dφ
ε = −N
dt (7.4) 19 ε = −N d (BA cos θ )
dt dB dt ε = − NA(cosθ )
(7.5)
(7.5) For a coil of N turns is placed in a uniform magnetic field B dφ
dt but changing in the coil’s area A, the induced emf ε is given
ε
by
and φ = BA cos θ ε = −N d (BA cos θ )
ε = −N
dt dA dt ε = − NB(cosθ )
(7.6) 20 Example 4 : dφ
dt
and ε = IR
(7.7) dB
dt and
and By applying the Faraday’s law equation for a coil of N turns , thus For a coil is connected in series to a resistor of resistance R
a coil is connected in series to a resistor of resistance R ε = −N dφ
dt
dt and the induced emf ε exist in the coil as shown in Figure 7.5,
the induced current I is given by I IR = −N
IR = −N 21 Solution :
a. b. The induced emf is given by ε = − NA cosθ 24 22 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 :
Solution : I
R Figure 7.5
Figure 7.5
Note:
To calculate the magnitude of induced emf, the negative sign
magnitude of induced emf
negative sign
can be ignored.
can be ignored
For a coil of N turns, each turn will has a magnetic flux φ of
BAcosθ through it, therefore the magnetic flux linkage (refer to
the combined amount of flux through all the turns) is given by magnetic flux linkage = Nφ 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 : B
A θ =0 a. The initial magnetic flux linkage is given by
23 Example 6 : Initial A B
A B θ =0
Final 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 Ω, determine the induced
current in the coil.
Solution : Solution : R = 2.4 Ω The induced emf in the coil is b. Given X X X
X X X X Xv X X X
X X X
X X X X X FX X QX X X X X X X X X X X X X X X X X X Therefore the induced current in the coil is given by X X PX X X X X
Figure 7.7
Figure 7.7 X
X
X X I
X X To determine the direction of the current through the
galvanometer which corresponds to a deflection in a particular
current through the solenoid seen is in the
sense, then the current through the solenoid seen is in the
direction that make the solenoid upper end becomes a
direction that make the solenoid upper end becomes a
north pole This opposes the motion of the bar magnet and
pole.
obey the lenz’s law.
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 lenz’s law.
Hence galvanometer shows a
deflection. 25 27 29 Solution :
Solution :
a. The change in the area of the coil is given by Direction of
induced current –
Right hand grip
Right hand grip
rule.
rule. North pole Figure 7.6
Figure 7.6 I I 20.2.2 (B) LENZ’S LAW
states that an induced electric current always flows in such
an induced electric current always flows in such
a direction that it opposes the change producing it.
a direction that it opposes the change producing it
This law is essentially a form of the law of conservation of
energy.
energy
An illustration of lenz’s law can be explained by
the following experiments.
1st experiment:
1 experiment:
In Figure 7.6 the magnitude
of the magnetic field at the
solenoid increases as the
bar magnet is moved
towards it. ν (motion )
Note:
Thumb – direction of Motion
Thumb – direction of Motion
First finger – direction of Field
Second finger – direction of induced
Second finger – direction of induced
Second finger – direction of induced
Second finger – direction of induced Its direction is in the opposite direction of the motion
opposite direction of the motion. Since the induced current flows in the conductor PQ and is
placed in the magnetic field then this conductor will
experience magnetic force
force. current OR induced emf
Figure 7.8
Therefore the induced current flows from Q to P as shown in
Figure 7.7. induced I OR
induced I OR
induced emf B To determine the direction of the induced current (induced
emf)
emf) flows in the conductor PQ, the Fleming’s right hand
(Dynamo) rule is used as shown in Figure 7.8. An emf is induced in the
solenoid and the
galvanometer indicates that
a current is flowing. N 26 28 30 N
I N
+
+
+
+
I ind Q
Q
Q
Q ε ind 3 experiment:
3rd experiment:
Consider two solenoids P and Q arranged coaxially closed to
each other as shown in Figure 7.9a. S
P
P
P
P I Switch , S
Figure 7.9a
At the moment when the switch S is closed current I begins
switch S is closed,
to flow in the solenoid P and producing a magnetic field inside
the solenoid P. Suppose that the field points towards the
solenoid Q. and ind 33 31 S
S
-I At the moment when the switch S is opened the current I
switch S is opened,
starts to decrease in the solenoid P and magnetic flux through
the solenoid Q decreases with time According to Faraday’s
time.
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 (Figure 7.9b) and
the galvanometer seen to deflect in the opposite direction of
Figure 7.9a. Solution : dφ
dt a. By applying the Faraday’s law of induction, thus ε = −N Therefore the induced current in the coil is given by 35 N
I S
S ε ind
ind I ind Figure 7.9b Q
Q
I ind 34 32 N
+ The magnetic flux through the solenoid Q increases with
increases with
time.
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 (Figure 7.9a)
and the galvanometer shows a deflection. S
P
P I Switch, S Example 7 :
A single turn of circular shaped coil has a resistance of 20 Ω and an
area of 7.0 cm2. It moves toward the north pole of a bar magnet as
shown in Figure 7.10. 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. Solution : N
N
N
N clockwise
clockwise as shown in figure below. b. Based on the lenz’s law, hence the direction of induced current is S
S
SI
S I ind 36 X X Area, A X ε ind 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.
P
X X X X X X X X
X X X X X X X XB
X
X
X
X
X X X X X X X
X X
X ind X X X X X x X QX X
Figure 7.11
Figure 7.11 X I X
X X X X
X l
lX
X X
X vX X
X
X X X X
X
X
X X X
X X X ε = lvB sin θ
(7.9)
(7.9) In general, the magnitude of the induced emf in the straight
magnitude
conductor is given by A = lx When the conductor moves through a distance x in time t, the
area swept out by the conductor is given by Note:
Note: 37 41 39 where θ : angle between v and B
This type of induced emf is known as motional induced emf.
motional induced emf.
The direction of the induced current or induced emf in the
straight conductor can be determined by using the Fleming’s
straight conductor can be determined by using the Fleming’s
Fleming’s
Fleming’s
right hand
right hand rule (based on Lenz’s 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. (7.10) Eq. (7.9)
Eq. (7.9) also can be used for a single turn of rectangular coil
single turn of rectangular coil
moves across the uniform magnetic field
field. ε = NlvB sin θ For a rectangular coil of N turns
rectangular coil of N turns, Solution : R = 15 Ω a. By applying the equation for motional induced emf, thus
and b. Given
i. By applying the Ohm’s law, thus By using the Fleming’s right hand rule, the direction of the
induced current is from
ii. Given
The total charge passing through the resistor is given by Q = It and φ = Blx θ = 0 Since the motion of the conductor is perpendicular to the
magnetic field B hence the magnetic flux cutting by the
conductor is given by φ = BA cos θ
φ = Blx cos 0
dφ
dt ε = Blv
(7.8) d
ε = (Blx )
dt
dx
dx
=v
ε = Bl
and
dt
dt ε = According to Faraday’s law, the emf is induced in the conductor
and its magnitude is given by Example 8 : B 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.
C 25 m s −1 Figure 7.12
D
a. Calculate the motional induced emf in the rod.
b. If the rod is connected in series to the resistor of resistance
15 Ω, determine
i. the induced current and its direction.
ii. the total charge passing through the resistor in two minute. 20.2.4 INDUCED EMF IN A ROTATING COIL ω
θ A S
S
S
S Consider a rectangular coil of N turns, each of area A, being
rotated mechanically with a constant angular velocity ω in a
uniform magnetic field of flux density B about an axis as shown
in Figure 7.13.
B N
N
N
N coil
Figure 7.13: side view
Figure 7.13: side view φ = BA cos ω t When the vector of area, A is at an angle θ to the magnetic
field B, the magnetic flux φ through each turn of the coil is given
by
φ = BA cos θ and θ = ω t 38 40 42 (7.11) By applying the equation of Faraday’s law for a coil of N turns,
thus the induced emf is given by = NBAω ω = 2πf = 2π
T (7.12) sin ωt = 1 hence dφ
ε = −N
dt
d
= − N (BA cos ωt )
dt
d
= − NBA (cos ωt )
= − NBA (cos ωt )
dt ε = NBAω sin ωt ε
where max where t : time
The induced emf is maximum when Example 9 : 43 45 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 : N = 100 turns; B = 5.0 T
The area of the coil is
and the constant angular velocity in rad s−1 is loop and moves away from it.
th
(Physics for scientists and engineers,6th
edition, Serway&Jewett, Q15, p.991) ANS. : U think
47 b. after the magnet has passed through the a. while the magnet falling toward the loop, 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 bar magnet is held above a loop of wire in a horizontal
plane, as shown in Figure 7.15. Exercise 20.1 :
1. Figure 7.15
Figure 7.15 Note:
Note: Eq. (7.11) also can be written as ε = NBA ω sin θ 0.5T 1.5T ε = ε max sin ωt T From the figure, the angle θ is (7.13) Therefore the induced emf is given by 2T 46 44 t where φ : anglebetween A and B
Conclusion : A coil rotating with constant angular velocity in a
uniform magnetic field produces a sinusoidally alternating emf
as shown by the induced emf ε against time t graph in Figure
7.14.
ε (V ) ε max 0 Figure 7.14 N = 100 turns; B = 5.0 T This phenomenon
was the important − ε max
part in the
development of
the electric
generator or
dynamo.
dynamo Solution : A B a. The maximum emf produced by the coil is given by b. 38 θ 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 V
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
b. What is the orientation of the coil with respect to the
magnetic field when the maximum induced emf occurs?
th edition,Serway&Jewett,
(Physics for scientists and engineers,6th edition,Serway&Jewett, Q35,
p.996) ANS. : 7.54×10 V
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 µT. Determine the induced emf when the plane of
the coil makes an angle 55° to the magnetic field.
48
−
−
−
−
−
−
ANS. : 1.77×10−5 V
1.77× −5 B Self-inductance (1 hour) LEARNING OUTCOME:
20.3 ε
ε µ N A
µ N2A
= 0
l At the end of this chapter, students should be able to:
Define self-inductance.
Define self-inductance.
Apply formulae
Apply formulae L=−
dI dt for a loop and solenoid.
for a loop and solenoid. N
N
N
N
N
N
N+
N+ I S
S
S
S
I ind Iinduced According to the Faraday’s law, an emf has to be induced in
emf
induced in
the solenoid itself since the flux linkage changes
changes.
In accordance with Lenz’s law, the induced emf opposes the
induced emf opposes
changes that has induced it
changes that has induced it and it is known as a back emf.
back emf I εind For the current I increases : S
S
S
S
I ind
Figure 7.16b: I increases Dire...

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