ELECTROMAGNETIC INDUCTION

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UNIT 20 ELECTROMAGNETIC INDUCTION
Electromagnetic induction is the production of
an electrical potential difference (induced
emf) across a conductor situated in a changing
magnetic field.
20.1 Magnetic flux 20.2 Induced emf 20.3
Self-inductance 20.4 Mutual inductance 20.5
Energy stored in inductor
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20.1 MAGNETIC FLUX ,F

• is defined as the scalar product between
• the magnetic flux density, B and the vector
• of the surface area, A.

UnitT.m2 or Wb
? 90?
? 0?
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Example 20.1.1

• A small surface of area 10 mm2 inside a uniform
  magnetic field of strength 0.10 T is inclined at
  an angle a to the direction of the field.
  Determine the magnetic flux through the surface
  if
• a 0º,
• a 30º
• a 90º

Solution
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20.2 INDUCED EMF

• An electric current produces a magnetic field.
• (chapter 19)

If electric currents produce a magnetic field, is
it possible that a magnetic field can produce an
electric current ?

• Scientists (American Joseph Henry and the
• Englishman Michael Faraday) independently
• found that is possible.

• Henry actually made the discovery first, but
• Faraday published his results earlier and
• investigated the subject in more detail.

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20.2 INDUCED EMF

• The diagram below shows the apparatus used
• by Faraday in his attempt to produce an
• electric current from a magnetic field.

Faradays experiment to induce an emf
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20.2 INDUCED EMF

• In this experiment, Faraday hoped by using a
• strong enough battery, a steady current in X
• would produce a current in a second coil Y but
• failed.
• Faraday saw the galvanometer in circuit Y
• deflect strongly at the moment he closed the
• switch in circuit X.

• And the galvanometer deflected strongly in
• the opposite direction when he opened the
• switch.

• A steady current in X had produced no
• current in Y.

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20.2 INDUCED EMF

• Only when the current in X was starting or
• stopping was a current produced in Y.
• Faraday concluded that although a steady
• magnetic field produces no current, a
• changing magnetic field can produce an
• electric current.
• Such a current is called an induced current.
• We therefore say that an induced current is
• produced by a changing magnetic field.
• The corresponding emf required to cause
• this current is called an induced emf.

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20.2 INDUCED EMF

• Induced emf is an electromotive force
• resulting from the motion of a conductor
• through a magnetic field , or from a change in
• the magnetic flux that threads a conductor.

• Faraday did further experiments on
  electromagnetic induction, as this phenomenon is
  called.( refer diagram )

• A current is induced when a magnet is
• moved toward a coil/loop.

b) The induced current is opposite when the
magnet is moved away from the coil/loop.
c) No current is induced if the magnet does
not move relative to the coil/loop.
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20.2 INDUCED EMF
Micheal Faradays experiment
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20.2 INDUCED EMF
Micheal Faradays experiment
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20.2 INDUCED EMF

• Direction of the induced current depends on
• i ) the direction of the magnets motion and
• ii) the direction of the magnetic field.

• Magnitude of the induced current depends on
• i ) the speed of motion (v ?,Iind?)
• ii) the number of turns of the coil (N ?, Iind?)
• iii)the strength of the magnetic field (B?,Iind?)

• From the observations, Michael Faraday
• found that,

the current/emf is induced in a coil/loop or
complete circuit whenever there is a change in
the magnetic flux through the area surrounded by
the coil
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20.2 INDUCED EMF
Faradays law and Lenzs law
Faradays law
the magnitude of the induced e.m.f. is
proportional to the rate of change of the
magnetic flux
Lenzs law
an induced electric current always flows in such
a direction that it opposes the change producing
it.
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20.2 INDUCED EMF
Faradays law and Lenzs law

• These two laws are summed up in the
• relationship,

or
The (-) sign indicates that the direction of
induced e.m.f. always opposes the change of
magnetic flux producing it (Lenzs law).
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20.2 INDUCED EMF
Faradays law and Lenzs law

• The concept of Faraday's Law is that any change
• in the magnetic environment of a coil of wire
  will
• cause a voltage (emf) to be "induced" in the
  coil.
• No matter how the change is produced, the
• voltage will be generated.
• The change could be produced by
• a) changing the magnetic field strength,
• b) moving a magnet toward or away from the
• coil,
• c) moving the coil into or out of the magnetic
• field,
• d) rotating the coil relative to the magnet,
  etc.

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(A) Induced emf in coil
20.2 INDUCED EMF
Faradays law and Lenzs law
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(A) Induced emf in coil
20.2 INDUCED EMF
Faradays law and Lenzs law
Notes
i ) the magnitude of induced emf,
ii) the flux through the coil can change in any
of 3 ways,
a) B , b) A , c) ?
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(A) Induced emf in coil
20.2 INDUCED EMF
Faradays law and Lenzs law
Notes
iii)
If the coil is connected in series to a resistor
of resistance R and the induced e.m.f ? exist in
the coil as shown in figure below.
and
-

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• Lenz's Law (based on censervation of energy)
• When an emf is generated by a change in magnetic
  flux according to Faraday's Law, the polarity of
  the induced emf (next slide) is such that it
  produces a current whose magnetic field opposes
  the change which produces it.
• The induced magnetic field inside any loop of
  wire always acts to keep the magnetic flux in the
  loop constant.
• In the examples below, if the B field is
  increasing, the induced field acts in opposition
  to it.
• If it is decreasing, the induced field acts in
  the direction of the applied field to try to keep
  it constant.

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(A) Induced emf in coil
7.2 INDUCED EMF
Faradays law and Lenzs law
The polarity of the induced emf Induced current
is directed out of the positive terminal, through
the attached device (resistance) and into the
negative terminal.
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(A) Induced emf in coil
Faradays law and Lenzs law
Example 20.2.1
A coil of wire 8 cm in diameter has 50 turns and
is placed in a B field of 1.8 T. If the B field
is reduced to 0.6 T in 0.002 s , calculate the
induced emf.
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Solution
Faradays law and Lenzs law
d 8 cm, N 50 turns, B from 1.8 T to 0.6 T in
0.002 s
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(A) Induced emf in coil
Faradays law and Lenzs law
Example 20.2.2
An elastic circular loop in the plane of the
paper lies in a 0.75 T magnetic field pointing
into the paper. If the loops diamater changes
from 20.0 cm to 6.0 cm in 0.50 s,

1. What is the direction of the induced current,
2. What is the magnitude of the average induced emf,
   and
3. If the loops resistance is 2.5 O, what is the
   average induced current during the 0.50 s ?

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Faradays law and Lenzs law
Solution
B0.75 T, di 20.0 cm, df 6.0 cm, t 0.50 s

• Direction of the induced current,
• b) Magnitude of the average induced emf,

c) R 2.5 O,
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Example 20.2.3
Faradays law and Lenzs law
A circular shaped coil 3.05 cm in radius,
containing 40 turns and have a resistance of
3.55 ? is placed perpendicular to a magnetic
field of flux density of 1.25 x 10-2 T. If the
magnetic flux density is increased to 0.450 T in
time of 0.250 s, calculate the induced current
flows in the coil.
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(A) Induced emf in coil
Faradays law and Lenzs law
How to determine the direction of induced
current.- Lenzs law
Case A
Thumb induced magnetic field Fingers - induced
current
N
Direction of induced current induced-current
right hand rule.
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Faradays law and Lenzs law
How to determine the direction of induced
current.- Lenzs law
Case A

• Consider a bar magnet that is moved
• towards a solenoid.

• As the north pole of the magnet approaches
• the solenoid, the amount of magnetic field
• passing through the solenoid increases ,
• thus increasing the magnetic flux through
• the solenoid.

• The increasing flux induces an emf
• (current) in the solenoid and galvanometer
• indicates that a current is flowing.

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Faradays law and Lenzs law
How to determine the direction of induced
current.- Lenzs law
Case A

• The direction of the induced current is
• such as to generate a magnetic field in the
• direction that opposes the change in the
• magnetic flux, so the direction of the
• induced field must be in the direction that
• make the solenoid right end becomes a
• north pole.

• This opposes the motion of the bar magnet
• and obey the Lenzs law.

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Faradays law and Lenzs law
How to determine the direction of induced
current.- Lenzs law
Case B

• When the magnet is moved toward the stationary
• conducting loop, a current is induced in the
• direction shown.

(b) This induced current produces its own
magnetic field (Binduced) directed to the
left that counteracts the increasing
external flux.
Binduced
Bexternal
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Faradays law and Lenzs law
How to determine the direction of induced
current.- Lenzs law
Case B

(c) When the magnet is moved away from the
stationary conducting loop, a current is induced
in the direction shown.
(d) This induced current produces a magnetic
field (Binduced) directed to the right and
so counteracts the decreasing external flux.
Binduced
Bexternal
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(A) Induced emf in coil
Faradays law and Lenzs law
Faradays law and Lenzs law
Example 20.2.4
Calculate the current through a 37 O resistor
connected to a single turn circular loop 10 cm in
diameter, assuming that the magnetic field
through the loop is increasing at a rate of 0.050
T/s. State the direction of the current.
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Faradays law and Lenzs law
Example 20.2.4
R 37 O , d 10 cm dB/dt 0.050 T/s.
I induced
S
N
I induced
Direction of Iinduced from b to a.
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(B) Induced emf of a straight conductor

• Consider a straight conductor of length l is
  moved at a speed v to the right on a U-shaped
  conductor in a uniform magnetic field B that
  points out the paper.

• This conductor travels a distance dx vdt in a
  time dt.

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(B) Induced emf of a straight conductor

• The area of the loop increases by an amount

• According to Faradays law, the e.m.f. is
  induced in the conductor and its magnitude is
  given by

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(B) Induced emf of a straight conductor
? angle between v and B 90 o

• This induced emf is called motional induced emf.

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(B) Induced emf of a straight conductor

• As the conductor is moved to the right (Fapplied
  to the right) with speed v, the magnetic flux
  through the loop increases.

• A current is induced in the loop.

• The induced current flows in the direction that
  tends to oppose this change.

Fapplied
FB

• In order to oppose this change, the current
  through the conductor must produce a magnetic
  force (FBIL) directed to the left.

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(B)Induced emf of a straight conductor
Faradays law and Lenzs law

• The direction of the induced current due to
  induced e.m.f. flows in the linear conductor can
  be determine by using Flemings right hand rule
  (based on lenzs law).

P

• The induced current flows from P to Q.

Fapplied
FB
Fapplied
Q
Thumb direction of Motion First finger
direction of Field Second finger direction of
Induced current or
Induced e.m.f.
Only for the straight conductor.
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Polarity
(B)Induced emf of a straight conductor

• When the conductor is moved to the right
  (Fapplied to the right) with speed v, the
  electrons in the rod move with the same speed.

• Therefore, each feels a force FBqv, which acts
  upward in the figure.

• If the rod were not in contact with the U-shaped
  conductor, electrons would collect at the upper
  end of the rod, leaving the lower end positive.
  There must thus be an induced emf.

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Induced emf of a straight conductor
Example 20.2.5
Suppose the length in figure above is 0.10 m, the
velocity z is 2.5 m/s, the total resistance of
the loop is 0.030 O and B is 0.60 T. Calculate
a) the induced emf b) the induced current
c) the force acting on the rod d) the power
dissipated in the loop
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Induced emf of a straight conductor
Example 20.2.6
A 0.2-m length of wire moves at a constant
velocity of 4 m/s in a direction that is 40 o
with respect to a magnetic flux density of 0.5 T.
Calculate the induced emf.
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Induced emf of a straight conductor
Example 20.2.7
In figure above, a rod with length l 0.400 m
moves in a magnetic flux with magnitude B 1.20
T. The emf induced in the moving rod is 3.60 V.

• Calculate the speed of the rod.
• If the total resistance is 0.900 O,
• calculate the induced current.
• What force does the field exert on the
• rod as a result of this current?

7.50 m/s , 4.00 A , 1.92 N to the left
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Fig 31-CO, p.967
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(C) Induced emf in a rotating coil
An ac generator / dynamo (transforms mechanical
energy into electric energy)
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(C) Induced emf in a rotating coil
An ac generator / dynamo (transforms mechanical
energy into electric energy)
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(C) Induced emf in a rotating coil

• Consider a coil of N turns each of area A and is
  being rotated about a horizontal axis in its own
  plane at right angle to a uniform magnetic field
  of flux density B.

• As the coil rotates with the angular speed ?,
  the orientation of the loop changes with time.

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(C) Induced emf in a rotating coil

• The emf induced in the loop is given by
  Faradays law,

• The emf induced in the loop varies sinusoidally
  in time.

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(C) Induced emf in a rotating coil
The alternating emf induced in the loop plotted
as a function of time.
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Example 20.2.8
Induced emf in a rotating coil
The armature of a simple ac generator consists of
100 turns of wire, each having an area of 0.2 m2
. The armature is turned with a frequency of 60
rev/s in a constant magnetic field of flux
density 10-3 T. Calculate the maximum emf
generated.
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Example 20.2.9
Induced emf in a rotating coil

• The drawing shows a plot of the output emf of
  a generator as a function of time t. The coil of
  this device has a cross-sectional area per turn
  of 0.020 m2 and contains 150 turns. Calculate
• The frequency of the generator in hertz.
• The angular speed in rad/s
• The magnitude of the magnetic field.

2.4 Hz , 15 rad/s , 0.62 T
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Example 20.2.10
Induced emf in a rotating coil
An amarture in ac generator consists of 500
turns, each of area 60 cm2 . The amarture is
rotated at a frequency of 3600 rpm in a uniform 2
mT magnetic field. Calculate a) the frequency
of the alternating emf b) the maximum emf
generated c) the instantaneous emf at time when
the plane of the coil makes an angle of 60o
with the magnetic field ?
380 rad/s, 1.13 V, 2.26 V
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20.3 SELF-INDUCTANCE

• Self-induction is defined as the process of
  producing an induced e.m.f. in the coil due to a
  change of current flowing through the same coil.

• Consider a current is present in the circuit
  above.

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20.3 SELF-INDUCTANCE

• This current produces a magnet field in the coil
  that causes a magnetic flux through the same
  coil.

• This flux changes when the current changes.

• An emf is induced in this coil called a
  self-induced emf.

• This coil is said to have self-inductance
• (inductance).

• A coil that has inductance is called an
• inductor.

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20.3 SELF-INDUCTANCE

• The symbol for an inductor is
• if air-cored, and if it
  has
• a core of magnetic material.

• By Lenzs law, the induced current opposes
• the change that cause it.

• If the current is increasing, the direction of
• the induced field and emf are opposite to that
• of the current, to try to decrease the current.

• If the current is decreasing, the direction of
• the induced field and emf are in the same
• direction as the current, to try to increase
  the
• current.

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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.
(c) The polarity of the induced emf reverses if
the current decreases.
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20.3 SELF-INDUCTANCE

• The magnetic flux in a coil is proportional
• to the current

• From the Faradays law,

. (1)
. (2)
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20.3 SELF-INDUCTANCE
Self-inductance, L is defined as the ratio of the
self induced e.m.f. to the rate of change of
current in the coil.
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20.3 SELF-INDUCTANCE
(1) (2)
If the coil has N turns, hence
- scalar quantity - unit is henry (H).
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20.3 SELF-INDUCTANCE

• 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 (?).

• Self-inductance does not depend on current.

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20.3 SELF-INDUCTANCE
Self-inductance of a Loop and Solenoid
From
And
By substituting we get,
or
For the medium-core solenoid
or
where
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20.3 SELF-INDUCTANCE
Example 20.3.1
If the current in a 230 mH coil changes steadily
from 20.0 mA to 28.0 mA in 140 ms, what is the
induced emf ?
Example 20.3.2
(Given ?0 4? x 10-7 H m-1)
Suppose you wish to make a solenoid whose
self-inductance is 1.4 mH. The inductor is to
have a cross-sectional area of 1.2 x 10 -3 m2 and
a length of 0.052 m. How many turns of wire
needed ?
220 turns
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20.3 SELF-INDUCTANCE
Example 20.3.3
The current in a coil of wire is initially zero
but increases at a constant rate after 10.0 s it
is 50.0 A. The changing current induces an emf of
45.0 V in the coil.
a) Calculate the self inductance of the coil.
b) Calculate the total magnetic flux through
the coil when the current is 50.0 A.
a)
b)
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20.3 SELF-INDUCTANCE
Example 20.3.4
A 40.0 mA current is carried by a uniformly wound
air-core solenoid with 450 turns, a 15.0 mm
diameter and 12.0 cm length. Calculate a) the
magnetic field inside the solenoid. b) the
magnetic flux through each turn. c) the
inductance of the solenoid.
(Given ?0 4? x 10-7 H m-1)
a)
b)
c)
or
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20.4 MUTUAL INDUCTANCE
20.4 MUTUAL INDUCTANCE
Mutual Inductance for two coaxial solenoids

• Consider a long solenoid with length l and cross
  sectional area A is closely wound with N1 turns
  of wire. A second solenoid with N2 turns
  surrounds it at its centre as shown in figure
  above.

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20.4 MUTUAL INDUCTANCE
Mutual Inductance for two coaxial solenoids

• The first solenoid is the one connected to an ac
  
• generator, which sends an alternating current
  I1
• through it.

• The current I1 produces a magnetic field lines
• inside it and this field lines also pass
  through the
• solenoid 2 as shown in figure.

• If the current I1 changes with time, the
  magnetic
• flux through the solenoids 1 and 2 will change
  with
• time simultaneously.

• Due to the change of magnetic flux through the
• solenoid 2, an e.m.f. is induced in solenoid 2.

• This process is known as mutual induction.

• At the same time, the self-induction occurs in
  the
• solenoid 1 since the magnetic flux through it
  changes.

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Mutual Inductance for two coaxial solenoids
20.4 MUTUAL INDUCTANCE

• Mutual induction is defined as the process of
  producing an induced e.m.f.in one circuit/coil
  due to the change of current in another
  circuit/coil.

Mutual inductance, M

• If the current I1 in solenoid 1 is continously
  changing,
• then the flux it produces will also change
  continously.

• The changing magnetic flux from the solenoid 1
• induces an emf in the solenoid 2.

• The induced emf in the solenoid 2 is
  proportional to
• the rate of change of the current I1 in
  solenoid 1.

.. (1)
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20.4 MUTUAL INDUCTANCE
Mutual Inductance for two coaxial solenoids
Mutual inductance, M

• Also the induced emf in the solenoid 1 is
  proportional
• to the rate of change of the current I2 in
  solenoid 2.

• The mutual inductance of the two solenoids is
  the
• same if current flows in the solenoid 2 and
  flux links
• the solenoid 1, causing an induced emf when a
• change in flux linkage occurs.

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20.4 MUTUAL INDUCTANCE
Mutual Inductance for two coaxial solenoids
Mutual inductance, M

• Rearrange,

M is defined as the ratio of the induced emf in
one solenoid/coil/ to the rate of change of
current in the other solenoid/coil.

• Unit M Henry (H)

• From Faradays law,

.. (2)
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Mutual Inductance for two coaxial solenoids
20.4 MUTUAL INDUCTANCE
Mutual inductance, M
(1) (2)

• Since M12M21M, equation above can be
• written as

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Mutual Inductance for two coaxial solenoids
20.4 MUTUAL INDUCTANCE
Mutual inductance, M

• From

and

• He mutual inductance of the solenoid 2 is,

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20.4 MUTUAL INDUCTANCE
Mutual inductance, M
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20.4 MUTUAL INDUCTANCE
Example 20.5.1

• The primary coil of a solenoid of radius 2.0 cm
  has 500 turns and length of 24 cm. If the
  secondary coil with 80 turns surrounds the
  primary coil at its centre, calculate
• a. the mutual inductance of the coils
• b. the magnitude of induced e.m.f. in secondary
  coil if the current in primary coil changes at
  the rate 4.8 A s-1.

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7.5 MUTUAL INDUCTANCE
Solution 20.5.1
rp 2.0 cm , Np 500 , lp 24 cm Ns 80
dIs/dt 4.8 A s-1
a)
b)
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20.4 MUTUAL INDUCTANCE
Transformer

• A transformer is a device for increasing or
• decreasing an ac voltage.

• The operation of transformer is based on the
• principle of mutual induction and
  self-induction.

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20.4 MUTUAL INDUCTANCE
Transformer

• Two types of transformer
• a) step-up transformer (Ns gt Np)
• b) step-down transformer (Np gt Ns).

• There are three assential parts
• (1) a primary coil connected to an ac
  source
• (2) secondary coil
• (3) soft iron core

• When ac voltage is applied to the input coil
• (primary coil), the alternating current
  produces
• an alternating magnetic flux that is
  concentrated
• in the iron core, without any leakage of flux
• outside the core.

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20.5 ENERGY STORED IN INDUCTOR

• The functions of an inductor are

• to control current
• to keep energy in the form of magnetic field

• An inductor carrying current has energy
• stored in it.
• It is because a generator does work to
• establish a current in an inductor.
• Suppose an inductor is connected to a
• generator whose terminal voltage can be
• varied continously from zero to some final
• value.

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20.5 ENERGY STORED IN INDUCTOR

• As the voltage is increased, the current I in
  the
• circuit rises continously from zero to its
  final value.
• While the current is rising, an emf (back
• emf) is induced in the inductor.
• Because of this, the generator that supplies
  the
• current must maintain a potential difference
• between its terminals while the current is
  rising
• (changing), and therefore it must supply
  energy to
• the inductor.
• Thus, the generator must do work to push the
• charges through the inductor against this
  induced
• emf.

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20.5 ENERGY STORED IN INDUCTOR

• To do this, power has to be supplied by the
• generator to the inductor.

• The total work done while the current is
  changed
• from zero to its final value is given by

• This work is stored as energy in the inductor.

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20.5 ENERGY STORED IN INDUCTOR

• For a long air-core solenoid, the self-inductance
  is

• Therefore the energy stored in the solenoid is
  given by

Example 20.5.1
How much energy is stored in a 0.085-H inductor
that carries a current of 2.5 A ?
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20.5 ENERGY STORED IN INDUCTOR
Example20.5.2
A steady current of 2.5 A in a coil of 500
turns causes a flux of 1.4 x 10-4 Wb to link
(pass through) the loops of the coil. Calculate
a) the average back emf induced in the coil if
the current is stopped in 0.08 s b) the
inductance of the coil and the energy
stored in the coil (inductor).