1 things to know (a)deduce from faraday’s experiments on electromagnetic induction or other...

1

Things to know

(a)deduce from Faraday’s experiments on electromagnetic induction or other appropriate

experiments: (i) that a changing magnetic field can induce an e.m.f. in a circuit (ii) that the direction of the induced e.m.f. opposes the change producing it (iii) the factors affecting the magnitude of the induced e.m.f. (b) describe a simple form of a.c. generator (rotating

coil or rotating magnet) and the use of slip rings

(where needed)

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• (c) sketch a graph of voltage output against time for a simple a.c. generator • (d) describe the structure and principle of operation of a simple iron-cored transformer as used for voltage transformations • (e) recall and apply the equations VP / Vs = NP / Ns and VPIP = VsIs to new situations or to solve related problems (for an ideal transformer)• (f) describe the energy loss in cables and deduce the advantages of high voltage transmission

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Electromagnetic Induction

Definition:

Electromagnetic induction is the production of electricity using magnetism.

•Need to know: Describe an experiment which shows that a changing magnetic field can induce an e.m.f. in a circuit

4

N S

In this experiment, no battery is connected to the coil. Hence no e.m.f. is found in the coil.

SensitiveGalvanometer

0-1-2 2

1

Hollow paper or plastic tube

A stationary magnet is near

the coil

5

N S

motion

When the magnet is moving towards the coil, an electric current is induced simultaneously.

G

Induced I

0-1-2 2

1


6

Faster motion

When the magnet is moving faster, the induced current is more.

N S

G

Induced I more

0-1-2 2

1


7

Not moving

When the magnet is not moving, no current is induced even though the magnetic flux is linked with the coil closely.

N S

G

No current

0-1-2 2

1


8

Faraday’s Law of Electromagnetic Induction:

The magnitude (how strong) of the induced emf (or induced current) is directly proportional to the

rate of change of the magnetic flux linked with the coil

or the rate at which the magnetic flux and wire are cutting each other.

This means that when the magnetic field is not moving in relation to the coil, there will be NO

induced emf at all.

9

Not moving

There is plenty of magnetic flux linkage with the coil, but there is no motion. Is there any induced current in the coil now? Answer: _________

N S

G

0-1-2 2

1

Self Test Question

Please draw the needle of the galvanometer.

10

What law did you apply when you answer the question in the previous slide?

11

G

0-1-2 2

1

Self Test Question

BE CD A

Moving constantly

N SBE CD A

Deflection of G or emf induced

12

G

0-1-2 2

1

Self Test Question: Sketch the graph as the magnet moves from A to E

BE CD A

Constant speed moving

N S BE CD A

Deflection of G or emf induced

`

Playing back of the graph

13

By now, you have learned that the size or strength of the induced current (or induced e.m.f.) is determined by the speed of change of the magnetic flux linkage with the coil.

There is still one more thing about electromagnetic induction you need to investigate. Look at the next slide.

14

When a current is induced in a coil, it has to flow in the certain direction.

What factor determines the direction of the induced current?

15

Lenz’s Law of electromagnetic induction: The direction of the induced current is such that its

own magnetic effect always opposes the change producing it.

This law is actually related to the Law of Conservation of Energy. The coil needs to oppose something in order to obtain energy from it. The coil itself cannot CREATE energy!

16

Beware of a different way the coil can be wound:

The paper tube can be taken away to test you

17

Can you spot the difference of winding?

Note: The dotted parts are at the back. The solid lines are at the front.

18

N S

Motion

Please mark one arrow on the left end of the coil and one arrow through the bulb to show how the induced current should flow:

Induced CURRENT

19

N S

Motion

Please mark + or – signs at the points X and Y to show the induced e.m.f. :

X Y+

20

N S

Motion

Please mark + or – signs at the points X and Y to show the presence of induced e.m.f. :

X Y

This induced emf is still there as long as the magnet is moving, even though the circuit is broken and the induced current cannot flow.

Induced emf+

21G

N

S

Coil is stationary

22G

N

S

Induced current

Coil is in motion, approaching the magnet

coil motion

N

Induced current

23

G

N

S

coil motion

N

Coil is in motion, approaching the magnet

Induced current

24

G

N

S

coil motion

N

Coil is approaching the magnet

Can you see the Right Hand Rule ?

This is also called the Dynamo Rule

Induced current

Existing flux Motion of wire

25

G

N

S

Coil is stationary again

Existing flux

No more induced current here

26

Fleming’s Right Hand Rule is also called the Dynamo Rule.

thuMb -- the Motion of the wire

seCond finger -- induced Current

First finger -- magnetic Flux (Field)

This is actually the result of Lenz’s Law

So, sometimes you use the right hand rule instead of Lenz’s Law.

27

This straight wire is moving along the

magnetic field

No current is induced

Wire moving vertically to the

flux

Current is induced towards you

Wire stops moving

No current is induced at all

Straight wire moving vertically to magnetic field.Current is induced away from you

The straight wire stops moving.No current is induced at all.

This straight wire is not moving

No current is induced

30

wire cutting flux obliquelywire cutting

flux vertically

wire moving alongside flux

wire moving alongside flux

No current is induced in the wire

No current is induced in wire

Strong induced current

Weak induced current

31

NS

Magnetic flux

32

NS

33

NS

A

BC

D

Motion

Motion

34

Motion

NS

A

B

C

D

Motion

Indu

ced

curre

nt

Magnetic Flux

35

NSA

B

C

D

motion

motion

No

indu

ced

curre

nt

No

pole

is n

eede

dNo induced current

36

NSA

B

C

D

motion

motion

37

NS

DA

BCMotion

38

NS

A

B

C

D

Motion

Indu

ced

curre

nt

Magnetic Flux

39

NS

A

B

C

D

motion

No induced current

No

indu

ced

curre

nt

Magnetic flux

40

NS

A

B

C

D

motion

motion

41

NS

A

BC

D

Motion

Motion

42

N SA

B C

D

Induced

current

Time

A D

Red arrows represents magnetic

flux

43

N SA

B

C

D

Induced

current

Time

AD

Red arrows represents magnetic

flux

44

N SA

B

C

D

motion

motion

Induced

current

Time

AD

45

N SD A

BC

Induced

current

Time

AD

Flux

motion

46

N S

A

B

C

D

motion

Induced

current

TimeA

D

47

N SA

B C

D

Induced

current

Time

A D

48

Induced current / emf

Time

49

Induced emf

Time

2.5V

-2.5V

50

The iron core

G

How would you demonstrate electromagnetic induction here?

51

The iron core

AC Source

Np turns

Ns turns

output emf

Input emf

Insulated copper wire

Primary windings

Insulated copper wire

Secondary windings

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The iron core

AC Source

Np turns

Ns turns

Secondary emf

Primary emf

emf primary

emf secondary

Np

Ns=

53

AC Source

Np turns

Ns turns

Secondary emf

Primary emf

emf

Time

54

emf

Time

Primary Alternating emf Peak 12V, 50 Hz

Secondary Alternating emf Peak 18V 50 Hz

55

The iron core

AC Source

Np turns

Ns turns

Secondary emf

Primary emf

emf primary

emf secondary

Np

Ns= 1

Step-up Transformer

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The iron core

AC Source

Np turns

Ns turns

Secondary emf

Primary emf

emf primary

emf secondary

Np

Ns= 1

Step-down Transformer

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AC Source

Np 200 turns

Ns 600 turns

output emf

Input emf 150V

Calculate (i) the output emf (ii) the induced current

60

(iii) the primary current

58

AC Source

Np 200 turns

Ns 600 turns

output emf

Input emf 150V

Calculate (i) the output emf

60

Vp

Vs Ns=

Np 150

Vs 600=

200

Vs = 450 V peak

59

AC Source

Np 200 turns

Ns 600 turns

output emf

Input emf 150V


60

Vs = 450 V peak(ii) the induced current

I = VR I = 450

60 = 7.5 A Peak

60

emf , current

Time

Secondary Alternating emf Peak 450 V, freq 50 Hz

7.5A

450V

61

AC Source

Np 200 turns

Ns 600 turns

output emf

Input emf 150V


60

Vs = 450 V peak(iii) the primary current

Power= VI, Output power = Input power

Vsx Is = Vpx Ip 450x 7.5 = 150x Ip

(assuming that the transformer is 100% effecient)

(ii) Is = 7.5 A

,

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AC Source

Np 200 turns

Ns 600 turns

output emf

Input emf 150V


60

Vs = 450 V peak(iii) the primary current

Ip

Is

Ns

=Np provided that the transformer

has an efficiency of 100%

63

emf , current

Time

Secondary Alternating emf Peak 450V, freq 50 Hz

7.5A

450V

150V

22.5A

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The iron core

AC Source

Np turns

Ns turns

Secondary emf

Primary emf

Power lost in a transformer, Textbook Page 350

Power losses are :(i) due to the electrical resistance in both the windings

Heat is generated in the wires unnecessarily

(ii) due to the production of the eddy currents in the iron core.

eddy current

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The iron core

AC Source

Np turns

Ns turns

Secondary emf

Primary emf

Such power losses are minimized:

(i) by using thicker copper wires in the windings.Less heat is generated in the wires.

(ii) by using a laminated iron core.

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Power Loss in Cables

67

P.S.

L

N

10KV R = 20

2MW

0V

Heating power in the transmitting cables = ?

R = 20

P = V I 2,000000 = 10000 x I

I = 200 A

Heating power in the transmitting cables = I2 R

Heating power in cables = 2002 x 40 = 1,600000 W = 1.6 MW A huge loss

0.4MWPower transmission

68

P.S.

L

N

20KV R = 20

2MW

0V

Heating power in the transmitting cables = ?

R = 20

P = V I 2,000000 = 20000 x I

I = 100 A

Heating power in the transmitting cables = I2 R

Heating power in cables = 1002 x 40 = 400,000 W = 0.4 MW A smaller loss

1.6MWPower transmission

1 things to know (a)deduce from faraday’s experiments on electromagnetic induction or other...

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