Note Physics 2 SP025

Substitute in (1) for d1 = 14.12 cm, d2 = 15.04 cm and
d3 = 15.90 cm

For d1 = 14.12 cm

m1  Bqr1
m1 
v 1019 )(14.122102 )
(0.56)(1.6

2.1429 105

 2.95201026 kg

For d2 = 15.04 cm

m2  Bqr2

m2  v 19 )(15.042 10 2

(0.56 )(1.6  10 )

2.1429 105

 3.1443 1026 kg

52

For d3 = 15.90 cm

m3  Bqr3

m3  v 19 )(15 .90  10 2
2
(0.56)(1.6  10 )

2.1429 105

 3.3241 1026 kg

53

Example :

A 0.13 MV is applied across the plates of a mass
spectrometer to determine the velocity of electrons which
are detected with photographic plate.

a) Find the velocity of the electrons that come out of the
plates.

b) The electrons are then entering evacuated chamber
under a centripetal force in a circular path of radius
1.50 m. What is the magnetic field used for this
circular motion?

Answer : (a) 2.14 x 108 m s-1 (b) 8.12 x 10-4 m s-1

TOPIC 5 :
ELECTROMAGNETIC INDUCTION

SP025 (2 HOURS)
PREPARED BY CHONGYL&FAUZIAHDAWAM/KMJ/2019/2020

5.1 Magnetic flux
5.2 Induced emf
5.3 Self Inductance
5.4 Energy stored in inductor
5.5 Mutual Inductance

1

TOPIC 5 :
ELECTROMAGNETIC INDUCTION

2

5.1 MAGNETIC FLUX ( )
LEARNING OUTCOME:

At the end of this chapter, students should be able to:
a) Define and use magnetic flux,

  B  A  BA cos  (Lecture :C2, CLO1, PLO1,MQF LOD1)

b ) Use magnetic flux linkage,

  N (Tutorial:C3,CTPS3,MQF LOD1)

3

5.2 INDUCED EMF
LEARNING OUTCOME:

At the end of this chapter, students should be able to:
a) Explain induced emf by using faraday’s experiment
b) State and use Faraday’s law,

c) State and use Lenz’s law to determine the direction of 4
induced current

(Lecture :C1 & C2, CLO1, PLO1,MQF LOD1)
(Tutorial:C3, CLO3, PLO4, CTPS3, MQF LOD6)

5.2 INDUCED EMF
LEARNING OUTCOME:

At the end of this chapter, students should be able to:

d) Derive and use induced emf :
i. a straight conductor,

ii. a coil, ,
iii. a rotating coil,

(Lecture :C1 & C2, CLO1, PLO1,MQF LOD1) 5
(Tutorial:C3, CLO3, CTPS3, MQF LOD6)

5.3 SELF INDUCTANCE
LEARNING OUTCOME:

At the end of this chapter, students should be able to:

a) Define self-inductance. (Lecture :C1, CLO1, PLO1,MQF LOD1)

b) Apply self-inductance,

for coil and solenoid, where iii)

i) ii)

(Tutorial:C3, CLO3, PLO4, CTPS3,MQF LOD1)

6

5.4 ENERGY STORED IN INDUCTOR
LEARNING OUTCOME:

At the end of this chapter, students should be able to:
a) Derive and use the energy stored in an inductor,

(Lecture :C2, CLO1, PLO1,MQF LOD1)
(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

7

5.5 MUTUAL INDUCTANCE
LEARNING OUTCOME:

At the end of this chapter, students should be able to:
a) Define mutual inductance.
b) Use mutual inductance ,

between two coaxial solenoids or a coaxial coil and a solenoid.

(Lecture : C1, CLO1, PLO1, MQF LOD1)
(Tutorial : C3, CLO3,PLO4, CTPS 3, MQF LOD6)

8

5.1 MAGNETIC FLUX ( )

o Magnetic flux is number of Magnetic flux is defined as the scalar product
magnetic field lines that passing between the magnetic flux density, B with the
vector of the surface area, A
through a given area.
=

=

where : magnetic flux
: magnetic field strength

: Area

: angle between and vector 9
always perpendicular to the surface area, A
CAUTION : Direction of vector

o Scalar quantity Circular coil with magnetic field
o SI unit : Wb OR Tm2 parallel to the plane of the coil

Circular coil with magnetic field = °
perpendicular to the plane of the coil

Let : 01 = 02 = 03 =
= =
= 6(5)cos60
= 15 Wb is zero
T

10

is maximum >0

MAGNETIC FLUX LINKAGE,

o Magnetic flux linkage (Φ) is a
measure of the total magnetic
flux of passing through a coil
with turns.

o Magnetic flux linkage,

Φ = Nϕ

where is the number of turns
of the coil
is the magnetic flux

passing through one tur1n1

Example 5.1.1 Given:
B = 6.0 T
A flat surface with area 3.0 cm2 is A = 3.0 cm2 = 3.0 × 10–4 m2
placed in a uniform magnetic field.
If the magnetic field strength is 6.0 Normal A θ = 90 – 30 = 60°
T, making an angle 30° with the
surface area, find the magnetic : angle between and vector
flux through this area.

12

Example 5.1.2 Given:
B = 0.6 T
A coil with 500 turns as a core with a
radius of 2 cm. It is placed in a field of : angle between and vector
0.6 T such that there is an angle of 300
between the field and the normal to the magnetic flux linkage
cross-sectional area. Calculate the Φ = Nϕ = (500)
magnetic flux and the magnetic flux
linkage.

Normal A

13

Example 5.1.3

A circular coil of radius 0.10 m is Given:
rotating in a uniform magnetic B = 0.6 T
Field of 0.20 T. Determine the
Magnetic flux through the coil : angle between and vector
when the plane of the loop and the
Magnetic field vector are parallel.

Normal A

14

5.2 INDUCED EMF
ELECTROMAGNETIC INDUCTION

Electromagnetic Induction is the production of induced emf (or induced
current) whenever the magnetic flux through a loop, coil or circuit is
changed.

The meaning of changing in magnetic flux : (2) the number of magnetic field
(1) there is a relative motion of loop & lines passing through a coil are
increased or decreased.
magnet fieldlines are ‘ cut ‘

15

Faraday’s experiment

Consider the experiment below:

1

When there is no relative motion between the magnet & the loop, G shows no 16
deflection. No induced current.

Faraday’s experiment

2

Moving the magnet toward the loop increases the number of magnetic field lines
passing through loop. The G needle is deflected indicating an induced current is 17
produced.

Faraday’s experiment

3

Moving the magnet away from the loop decreases the number of magnetic field
18

lines passing through the loop. The induced current is now in opposite direction.

From the experiments, it can seen that emf is induced only when the
magnetic flux through the coil change.

NO change in magnetic flux,
electromagnetic induction CANNOT occur.

It was further shown that induced emf increase when: 19
(a) a stronger magnet is used, i.e, magnetic flux is
increased

(b) the magnet is pushed faster into the coil, i.e the
speed of magnet is increased.

(c) the area of the coil is greater.

(d) the number of turns increased.

Faraday’s Faraday’s Lenz’s law
Experiment law (1834)

(1831)

o A steady o the magnitude of o an induced
magnetic flux the induced emf is current always
/field produces proportional to the flow in a
no current rate of change of direction that
the magnetic flux. opposes the
o A changing change in
magnetic o If the circuit / coil magnetic flux
flux/field can with turns : that causes it.
produce an
electric current 20
 induced
current

A changing magnetic flux can produce an electric current & emf
 electromagnetic induction

Faraday’s Lenz’s
law law

(1831) (1834)

o the magnitude of the induced emf is o an induced current always
proportional to the rate of change of the
flow in a direction that
magnetic flux.
opposes the change in

magnetic flux that causes it.

o If the circuit / coil with turns : 21

Example 5.2.1

(a) When the bar magnet is held (b) When the magnet is moved towards the loop, the ammeter
motionless near the loop, there is no
induced current. is detected, determine the direction of an induced current, I ?

(Ans : counterclockwise ) 22

Faraday’s Law of Induction

The induced emf is proportional to the
rate of change of magnetic flux.

Michael Faraday    d
English physicist
dt
(1791-1867)

Lenz’s Law

An induced current always flow in a
direction that opposes the change in
magnetic flux that causes it.

Heinrich F.E. Lenz 23
Russian physicist

(1804-1865)

Right hand grip rule

CW current creates
into the page

Fingers curl in the direction of , ACW current creates
your right thumb point in the out the page
direction of

24

Faraday’s experiment

The solenoid in figure is moved at constant velocity towards a fixed
bar magnet. Using Lenz’s law,
determine the direction of the induced current through the resistor.

Movement of solenoid

ab

25

1 South pole North pole 3 Lenz’s law, direction of current

When Movement of solenoid induced opposes the change.The
solenoid bring right end of solenoid must form
N north pole to oppose incoming
towards north pole of the bar magnet.
magnet bar, it b
experience an a
increasing in

flux.

2

Flux change → current induced.

USE LENZ’S LAW TO DETERMINE THE DIRECTION OF INDUCED
CURRENT

Problem solving strategy ❶As the loop being pulled in, flux
to find direction increases.
SN
① If flux decreases  replace it ❷opposes it by create opposite
type of ab
by create same type of ; if flux
increases  opposes it by create ❸ ❶ a receding magnet create
opposite pole to attract
opposite type of OR current induced flows
clockwise
For an incoming magnet create
same pole to repel ; a receding
magnet create opposite pole to
attract

② Once you know the direction of

induced magnetic field, use right

hand grip rule to find the current induced in the 27
solenoid flows from a to b
direction of the induced current.

How can we induce emf ?

DERIVE INDUCED EMF! By changing magnetic flux

In a plane coil Straight Rotate Changing
conductor coil in current,
magnetic
moving field, B
through a
magnetic field

By changing By changing Self Mutual
area, A in magnetic = Induction Induction
magnetic field
strength, B
field

28
= −


= −

INDUCED EMF IN A COIL BY CHANGING THE MAGNETIC
FIELD STRENGTH, B WITH TIME

Example 5.2.2

Initial A coil having an area of 8.0 cm2
and 50 turns lies perpendicular to

a magnetic field of 1.80 T. If the

If is perpendicular magnetic flux density is steadily
reduced to 0.6 T in 0.02 s,

to the plane of coil  determine the induced emf.

= 0, = 1 & SOLUTION

area, is constant,
Final then



29

INDUCED EMF IN A COIL BY CHANGING THE AREA, A OF THE
COIL WITH TIME

Example 5.2.3

A narrow coil of 10 turns and area of

48 m2 is placed perpendicular to a

uniform magnetic field of 1.20 T.

After 5 s, the area of the coil is

If is perpendicular increased to 83 m2. Calculate the

Initial to the plane of coil  emf induced.

= 0, = 1 & SOLUTION

is constant, then

30

Final Lenz’s law

INDUCED EMF IN A STRAIGHT CONDUCTOR MOVING IN A
MAGNETIC FIELD

02 From Faraday’s law (noting that ) emf
induced is given by

Assume rod moves a distance in time

{− : Lenz’s law}

03 In general, the magnitude of the induced
emf in the straight conductor is given by

01 Area of the loop increases as the moveable 31
rod travels to the right with velocity .
 there is a change in magnetic flux : angle between and

04 The direction of the induced current in Example 5.2.4 SOLUTION
the straight conductor can be determined
da (a)
by using the (1) Fleming right hand rule
L
or (2) Right hand palm rule   Blv sin 
cb
IR  Blv sin 
Assume that R = 12 Ω,
L = 1.2m & a uniform 2.50 v  IR
T magnetic field is
directed out the page. Bl sin 
(a) At what speed should
 0.5(12)
the bar be moved to 2.50(1.2) sin 90
produced a current of
0.5A in the resistor. v  2 m s1
(b) What is the direction of
the induced current? (b) Clockwise abcd



32



INDUCED EMF IN A ROTATING COIL

As coil rotates mechanically with a constant angular velocity, in a
uniform magnetic field, , angle between magnetic field and the
normal of coil surface area varies in turn causes magnetic flux through
the area enclosed by the coil changes with time, thus an emf induced.

The induced emf in the coil is given by:

{}

{ } where
N : number of turns

B : Magnetic field

A : Area
= : angular velocity 33

o emf induced ε V ε  NBA sin ωt Induced emf is maximum

varies  max T 1.5T when = 1 hence
sinusoidally

in time. * = 2 =

o an alternating 0 0.5T 2T t

voltage   max 
B
because has

positive value

as well as

negative value.

o when is parallel to the plane of the coil and the 34

rate of change of flux is maximum

o when is perpendicular to the plane of the coil and

the rate of change of flux is zero.

Example 5.2.5 Induced emf is maximum when = 1

A square coil of cooper wire of sides 0.075 m has hence
25 turns and is placed in a uniform magnetic field
of flux density 0.85 T as shown in figure.
The coil is then rotated at a constant speed of 2
revolution per second about the axis of rotation
shown. What is the value of the maximum emf
induced?

35

5.3 SELF INDUCTANCE



Self induction −the production of emf in a
coil due to the change of current in the coil
ifself.

+ increasing − Running a changing current (by changing
R), creates a changing magnetic flux
linkage through the solenoid.
The change in magnetic flux induces an emf
in the solenoid itself, in accord with
Faraday’s law.

Polarity of the induced emf is indicated by the dashed battery.
36

Induced emf is called back emf

Self induced emf is Self inductance, Example 5.3.1
always proportional to the ratio of self Induced emf of 5.0 V is
the time rate of change induced emf to the rate developed across a coil
of the current. of change of current in when the current flowing
through it changes at 25 A
the coil. ( s–1. Determine the self
inductance of the coil.
is a constant of where
proportionality called : number of turns SOLUTION
self inductance. : magnetic flux that pass
through one turn 37

: current
SI unit : Henry (H)

Self inductance for a coil Self inductance for a solenoid

: permeability of free space : permeability of free space
(4 T m A−1) (4 × 10 T m A−1)

: number of turns : number of turns
: area of the coil : cross sectional area of the soleno3i8d
: radius of the coil : length of the solenoid

Example 5.3.2 i) the magnetic flux through the
solenoid
A solenoid of 200 turns has s self
inductance of 50 mH. It carries a 6 A N  LI
current. Calculate 200  25103 (6)
i) the magnetic flux through the   7.5104Wb

solenoid (ii) magnetic flux linkage
ii) the magnetic flux linkage through
Φ = Nϕ = (200)
the solenoid

39

5.4 ENERGY STORED IN INDUCTOR

o Functions of an inductor:
(1) to control current

(2) to store energy in its magnetic field.

o A circuit element (coil or o Energy stored in an inductance, ,
solenoid) that has a large self- carrying a current , is:

inductance is called an

inductor.

40

o Circuit symbol :

Magnitude of induced emf produce in an inductor is given by : DERIVATION
Power received by the inductor at time t is :

Power × Time P  I dI
= Energy dt
P  IL

Pdt  I L dI

dU  I L dI

Total energy U received by the inductor when the current change from 0 to a steady value I :

I

U   dU
0
U  I I dI  1 L I2 41
L 0 2

Example 5.4.1   IR

A coil has an inductance of 30 mH 12  I (0.25 )
and a resistance of 0.25 Ω. If the I  48 A
potential difference of 12 V is
applied across the coil, calculate the U  1 LI 2
energy in the magnetic field of the 2
coil.

U  1 (30103 )(48)2
2

U  34.56J

42

Example 5.4.2 .

A 25 cm solenoid has a cross sectional area
of 15 cm2. Energy of 0.022 J is stored when
a current of 8 A flows through it.
Determine
(i) the self inductance of the solenoid.
(ii) the total flux linkage passing through

the solenoid.
(iii) the number of turns of the solenoid.

(iii)   .× ×
× ×
 

43

5.5 MUTUAL INDUCTANCE

Mutual induction − Process of producing o Magnetic flux caused by the current
an induced emf in one coil due to the in coil 1 passing through coil 2
which is nearby
change of current in another coil.
o As the current in coil 1
changing with time, magnetic
flux passing through coil 2
changing with time too.

o The change in flux in coil 2

induces an emf in coil 2, in

accord with Faraday’s law.

o If the change in current takes place in
coil 2 then emf will be induced in coil
1. 44

emf induced in one coil Mutual inductance, Mutual inductance
is proportional to the between two coaxial
is defined as ratio of solenoids or a
rate at which the coaxial coil and a
induced emf in a coil to solenoid

current in the other coil the rate of change of 45

is changing. current in another coil

( )




Equation :

=



is a constant of Magnetic flux
proportionality called N Flux lingkage
mutual inductance.
SI unit : Henry (H)

Example 5.5.1

1. What is the mutual inductance of the two coils? 2.

2. If Ns=500turns, Nc=10turns, Rs=3.10cm, 46
l1=75.0cm, and the current in the solenoid is
changing at a rate of 200 A s-1, what is the emf
induced in the surrounding coil?