chapter20 student

8/10/2019 Chapter20 Student

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PHYSICS CHAPTER 20

20.0 ELECTROMAGNETIC INDUCTION

20.1 Magnetic flux

20.2 Induced emf

20.3 Self-inductance

20.4 Mutual Inductance

20.5 Energy stored in inductor 2


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

Define and use magnetic flux,

Learning Outcome:

20.1 Magnetic flux ( 1/2 hour)

w w w

. k m s

. m a t r i k

. e d u

. m y / p h y s i c s


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20.1.1 Phenomenon of electromagnetic inductionConsider some experiments were conducted by MichaelFaraday that led to the discovery of the Faradays law ofinduction as shown in Figures 7.1a, 7.1b, 7.1c, 7.1d and 7.1e.

20.1 Magnetic flux

No movement

0v

Figure 7.1a


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Figure 7.1b

Figure 7.1c

NS Move towards the coil

v

I I

No movement

0v


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Figure 7.1d

Figure 7.1e

N Move away from the coil

v

I I

N SMove towards the coil

v

I I

S

Simulation 7.1
http://localhost/var/www/apps/conversion/tmp/scratch_10/AF_3101.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_10/AF_3101.html


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From the experiments:When the bar magnet is stationary , the galvanometer notshow 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 relativelyaway from the coil, the galvanometer is seen to deflect in the

opposite direction (Figure 7.1d).Therefore when there is any relative motion between thecoil and the bar magnet , the current known as inducedcurrent 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 changesthus the induced emf will exist across the coil.


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The magnitude of the induced e.m.f. depends on thespeed of the relative motion where if the

Therefore v is proportional to the induced emf .

20.1.2 Magnetic flux of a uniform magnetic field

is defined as the scalar product between the magnetic fluxdensity, B with the vector of the area, A .Mathematically,

v increases induced emf increasesv decreases induced emf decreases

A B andof directione between thangle: where fluxmagnetic:

(7.1)


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It is a scalar quantity and its unit is weber (Wb) OR teslameter squared ( T m 2).

Consider a uniform magnetic field B passing through a surfacearea 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 isgiven by

Figure 7.2a

B

A

cos BA0cos BA

BA maximum

area


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From the Figure 7.2a, the angle is 90 thus the magnetic fluxis given by

Figure 7.2b

B A

cos BA90cos BA

0Note:

Direction of vector A always perpendicular (normal) tothe surface area, A .The magnetic flux is proportional to the number offield lines passing through the area.

area90


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A single turn of rectangular coil of sides 10 cm 5.0 cm is placedbetween north and south poles of a permanent magnet. Initially, theplane of the coil is parallel to the magnetic field as shown in Figure7.3.

If the coil is turned by 90 about its rotation axis and the magnitudeof magnetic flux density is 1.5 T, Calculate the change in themagnetic flux through the coil.

Example 20.1 :

SNP

QR

S

I I

Figure 7.3


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Solution :The area of the coil is

Initially,

Finally,

Therefore the change in magnetic flux through the coil is

T51. B

2322 m100510051010 .. AFrom the figure, =90 thus the initialmagnetic flux through the coil is

A B

cos i BA90cos BA

B

A

From the figure, =0 thus the finalmagnetic flux through the coil is

cos

f BA 0cos100.55.1 3

if 0105.7 3


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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 angle30 to the direction of the magnetic field. If the magnetic fluxthrough the area of the coil is 1.20 mWb, calculate the magnitude ofthe magnetic field.Solution :

The area of the coil is

Example 20.2 :

Wb1020.1m;100.3 32 d

4

2d A

4100.3

22 A

24

m1007.7 A

3030

coil

B A


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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

Wb1020.1m;100.3 32 d

603090

cos BA

60cos1007.71020.1 43 B


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The three loops of wire as shown in Figure 7.4 are all in a region ofspace with a uniform magnetic field. Loop 1 swings back and forthas the bob on a simple pendulum. Loop 2 rotates about a verticalaxis and loop 3 oscillates vertically on the end of a spring. Whichloop or loops have a magnetic flux that changes with time? Explain

your answer.

Example 20.3 :

Figure 7.4


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Solution :Only 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 alwaysconstant with time.

Loop 2 on the other hand changes its orientation relative to thefield as it rotates, hence its flux does change with time.


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

Use Faradays experiment to explain induced emf.

State Faradays law and use Lenzs law to determine thedirection of induced current.

Apply induced emf ,

Learning Outcome:

20.2 Induced emf (2 hours)

w w w

. k m s

. m a t r i k

. e d u



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At the end of this chapter, students should be able to:Derive and use induced emf :

in straight conductor,

in coil,

in rotating coil,

Learning Outcome:

20.2 Induced emf (2 hours)

w w w

. k m s

. m a t r i k

. e d u


OR


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20.2.1 Faradays law of electromagnetic inductionstates that the magnitude of the induced emf is proportionalto 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) .

20.2 Induced emf

dt d OR (7.2)

where fluxmagnetictheof change:d timeof change:dt

emf induced:


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For a coil of N turns, eq. (7.2) can be written as

Since

For a coil of N turns is placed in the changing magneticfield B , the induced emf is given by

(7.3)

, then eq. (7.3) can be written asif

d

(7.4)

fluxmagneticfinal: f fluxmagneticinitial: i

where

dt

d N

cos BAand


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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 givenby

dt

BAd N

cos

(7.5)

dt

d N

cos BAand

dt

BAd N cos

(7.6)


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For a coil is connected in series to a resistor of resistanceR and the induced emf exist in the coil as shown in Figure 7.5,

Figure 7.5

R

I I dt

d N

IR and

dt

d N IR

(7.7)

Note:To calculate the magnitude of induced emf , the negative signcan 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 tothe combined amount of flux through all the turns) is given by

the induced current I is given by


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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 coilhave 500 turns, calculate the magnitude of the induced emf in thecoil.Solution :

By applying the Faradays law equation for a coil of N turns , thus

Example 20.4 :

12 sWb100.5turns;500 dt

d N

dt

d N

2

100.5500


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A coil having an area of 8.0 cm 2 and 50 turns lies perpendicular to a

magnetic field of 0.20 T. If the magnetic flux density is steadilyreduced to zero, taking 0.50 s, determinea. the initial magnetic flux linkage.b. the induced emf.Solution :

a. The initial magnetic flux linkage is given by

B

Example 20.5 :

ilinkagefluxmagneticinitial N

0;T;0.20turns;50;m100.8f i

24 B B N A

s0.50dt

0 A

cosi A NB


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Solution :

a.

b. The induced emf is given by

0;T;0.20turns;50;m100.8 f i24 B B N A

s0.50dt

0cos108.00.2050 4

dt dB NA cos if B BdB and

dt

B B NA if cos

50.0

20.000cos100.850 4


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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, thediameter 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 :

Example 20.6 :

m;103.5m;100.4turns;10 2f 2

i d d N

s0.25T;2.1 dt B

0

A

B B

A

Initial Final


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Solution :

a. The change in the area of the coil is given by

if A AdA

44

2i

2f

d d

m;103.5m;100.4turns;10 2f 2

i d d N

s0.25T;2.1 dt B

2i2f 4

d d

2222

100.4103.54


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Solution :

b. GivenThe induced emf in the coil is

Therefore the induced current in the coil is given by

4.2 R

dt

dA NB cos

m;103.5m;100.4turns;10 2f 2

i d d N

s0.25T;2.1 dt B

25.0105.90cos2.110

4

V1056.4 2

IR 4.21056.4 2 I


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Direction ofinduced current Right hand griprule.

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 ofenergy .

20.2.2 Lenzs law

I

I NNorth pole

An illustration of lenzs law can be explained bythe following experiments.

1st

experiment:In Figure 7.6 the magnitudeof the magnetic field at thesolenoid increases as thebar magnet is moved

towards it. An emf is induced in thesolenoid and thegalvanometer indicates thata current is flowing.

Figure 7.6


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To determine the direction of the current through thegalvanometer which corresponds to a deflection in a particular

sense, then the current through the solenoid seen is in thedirection that make the solenoid upper end becomes anorth pole . This opposes the motion of the bar magnet andobey the lenzs law .

v

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X XP

Q

F

2nd experiment:

Consider a straight conductor PQis placed perpendicular to themagnetic field and move theconductor to the left with constantvelocity v as shown in Figure 7.7.

When the conductor move to theleft thus the induced currentneeds to flow in such a way tooppose the change which hasinduced it based on lenzs law.Hence galvanometer shows adeflection.

Figure 7.7

I

Simulation 7.2


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To determine the direction of the induced current (inducedemf) 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 isplaced in the magnetic field then this conductor willexperience magnetic force .

Its direction is in the opposite direction of the motion .

Thumb direction of Motion

First finger direction of FieldSecond finger direction of induced

current OR induced emf

Note: B

)motion(

OR induced I emf induced

Figure 7.8


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3 rd 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 beginsto flow in the solenoid P and producing a magnetic field insidethe solenoid P. Suppose that the field points towards thesolenoid Q.

Figure 7.9a

S,SwitchP Q

I I

NS SN

ind I ind I -+

ind


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The magnetic flux through the solenoid Q increases withtime . 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 amagnetic field that oppose the change producing it (increasein flux). Hence based on Lenzs law, the induced current flowsin circuit consists of solenoid Q is anticlockwise (Figure 7.9a)and the galvanometer shows a deflection.

QSSwitch,

P

NS

I I

S N

ind I ind I - +

ind

Figure 7.9b


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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 Faradayslaw ,an induced current due to induced emf will exist insolenoid Q.The induced current flows in solenoid Q must produce amagnetic field that oppose the change producing it (decreasein flux). Hence based on Lenzs law, the induced current flowsin circuit consists of solenoid Q is clockwise (Figure 7.9b) andthe galvanometer seen to deflect in the opposite direction ofFigure 7.9a.

Simulation 7.3


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A single turn of circular shaped coil has a resistance of 20 and an

area of 7.0 cm 2. It moves toward the north pole of a bar magnet asshown in Figure 7.10.

If the average rate of change of magnetic flux density through thecoil is 0.55 T s 1,a. determine the induced current in the coilb. state the direction of the induced current observed by the

observer shown in Figure 7.10.

Example 20.7 :

Figure 7.10


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Solution :a. By applying the Faradays law of induction, thus

Therefore the induced current in the coil is given by

dt

d N

BAdt

d N

124 sT55.0;m100.7;20turn;1dt

dB A R N

dt

dB NA

55.0100.71 4

180cos BAand

V1085.34

IR

201085.3 4 I


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Solution :b. Based on the lenzs law, hence the direction of induced current is

clockwise as shown in figure below.

NS ind I


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Consider a straight conductor PQ of length l is moved

perpendicular with velocity v across a uniform magnetic field Bas shown in Figure 7.11.

When the conductor moves through a distance x in time t , thearea swept out by the conductor is given by

20.2.3 Induced emf in a straight conductor

Figure 7.11

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

X X X X

P

Q

l

x

B

v

lx A

Area, A

ind I

ind


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Since the motion of the conductor is perpendicular to themagnetic field B hence the magnetic flux cutting by the

conductor is given by

According to Faradays law, the emf is induced in the conductorand its magnitude is given by

0 cos BA and0cos Blx Blx

dt

d

Blxdt

d

Blv

dt

dx Bl v

dt

dxand

(7.8)


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In general, the magnitude of the induced emf in the straightconductor is given by

This type of induced emf is known as motional induced emf.The direction of the induced current or induced emf in thestraight conductor can be determined by using the Flemingsright hand rule (based on Lenzs law).In the case of Figure 7.11, the direction of the induced current orinduced emf is from Q to P. Therefore P is higher potential thanQ.

(7.9)

Bv and betweenangle:where

Note:

Eq. (7.9) also can be used for a single turn of rectangular coilmoves across the uniform magnetic field .

For a rectangular coil of N turns ,

(7.10)


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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 rodis 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.

Example 20.8 :

Figure 7.12

C

D

B

1sm52


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Solution :a. By applying the equation for motional induced emf, thus

b. Giveni. By applying the Ohms law, thus

By using the Flemings right hand rule, the direction of the

induced current is from D to C .ii. Given

The total charge passing through the resistor is given by

15 R

sinlvB

T;10502;sm25m;1020 312 Bvl

90sin10250251020 32

IR 1525.1 I

90 and

s120602 t

It Q 1201033.8 2Q

PHYSICS CHAPTER 20


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Consider a rectangular coil of N turns, each of area A , being

rotated mechanically with a constant angular velocity in auniform magnetic field of flux density B about an axis as shownin Figure 7.13.

When the vector of area, A is at an angle to the magneticfield B , the magnetic flux through each turn of the coil is givenby

20.2.4 Induced emf in a rotating coil

Figure 7.13: side view

B

N S A

coil

cos BA t andt BA cos

PHYSICS CHAPTER 20


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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

dt d N

t BAdt

d N cos

t dt d NBA cos

(7.11)

time:t where 1sin t (7.12)

T f

22 where

PHYSICS CHAPTER 20


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Eq. (7.11) also can be written as

Conclusion : A coil rotating with constant angular velocity in auniform magnetic field produces a sinusoidally alternating emf as shown by the induced emf against time t graph in Figure7.14.

(7.13)

B A and betweenangle: where

Figure 7.14

t sinmax

0

max

max

V

t T T 5.0 T 5.1 T 2

B

Note:

This phenomenonwas the importantpart in thedevelopment ofthe electricgenerator ordynamo .

Simulation 7.4

PHYSICS CHAPTER 20


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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 uniformmagnetic field of flux density 5.0 T. Calculatea. 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

Example 20.9 :

2222 m105.110151010 AT0.5turns;100 B N

s06min1

rev1rad2

min1rev200

1srad9.20

PHYSICS CHAPTER 20


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Solution :a. The maximum emf produced by the coil is given by

b.

NBAmax

T B N 0.5turns;100

9.20105.10.5100 2

B

A

38

From the figure, the angle is

Therefore the induced emf is given by523890

sin NBA

52sin9.20105.10.5100 2

PHYSICS CHAPTER 20


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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 theloop of the wire. The magnet is droppedtoward the loop. Determine the direction ofthe 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,6 thedition, Serway&Jewett, Q15, p.991)ANS. : U think

Figure 7.15

PHYSICS CHAPTER 20


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2. A straight conductor of length 20 cm moves in a uniform magneticfield of flux density 20 mT at a constant speed of 10 m s -1 . Thevelocity makes an angle 30 to the field but the conductor isperpendicular to the field. Determine the induced emf.

ANS. : 2.0 10 2 V3. A coil of area 0.100 m 2 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,6 th edition,Serway&Jewett, Q35,p.996)

ANS. : 7.54 10 3 V4. 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 offlux density 50 T. Determine the induced emf when the plane ofthe coil makes an angle 55 to the magnetic field.

ANS. : 1.77 10 5 V

PHYSICS CHAPTER 20


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At the end of this chapter, students should be able to:Define self-inductance.Apply self-inductance,

for a loop and solenoid.

Learning Outcome:

20.3 Self-inductance (1 hour)

w w w

. k m s

. m a t r i k

. e d u

. m y / p h y s i c

s

PHYSICS CHAPTER 20


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20.3.1 Self-inductionConsider a solenoid which is connected to a battery , a switch Sand variable resistor R , forming an open circuit as shown inFigure 7.16a.

20.3 Self-inductance

Figure 7.16a: initial

S R I I

When the switch S is closed, a current

I begins to flow in the solenoid.The current produces a magneticfield whose field lines through thesolenoid and generate the magneticflux linkage .

If the resistance of the variableresistor changes , thus the currentflows in the solenoid also changed ,then so too does magnetic fluxlinkage .

NS


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Self-induction experimentThe 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 lampshave the same brightness in their respective circuits with steadycurrent flowing.

When the switch S is closed, the lamp A 2 with variable resistor Ris seen to become bright almost immediately but the lamp A 1

with iron-core coil L increases slowly to full brightness.

Figure 7.17a

switch, S

coil, L

iron-core

R

lamp A 1

lamp A 2

PHYSICS CHAPTER 20


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Reason:The coil L undergoes the self-induction and induced emf

in it. The induced or back emf opposes the growth ofcurrent so the glow in the lamp A 1 increases slowly .

The resistor R , however has no back emf, hence the lampA2 glow fully bright as soon as switch S is closed .

This effect can be shown by the graph of current I againsttime t through both lamps in Figure 7.17b.

lamp A 2 with resistor R

t

I

0Figure 7.17b

0 I

lamp A 1 with coil L

PHYSICS CHAPTER 20


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coil, L

switch, S

R

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.

Example 20.10 :

Figure 7.18

The switch S is closed for a longtime and is suddenly opened.Explain why a spark jump across theswitch contacts S .

Solution :

When the switch S is suddenly opened, the current in thecircuit starts to fall very rapidly and induced a maximumemf in the coil L which tends to maintain the current.This back emf is high enough to break down the insulation ofthe air between the switch contacts S and a spark can easily

appear at the switch.

PHYSICS CHAPTER 20


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From the self-induction phenomenon, we get

From the Faradays law, thus

20.3.2 Self-inductance, L

I L

wherecurrent: I

(7.14)

linkagefluxmagnetic:L

dt

d L

LI dt d

(7.15)

PHYSICS CHAPTER 20


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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

anddt dI L

dt

d N

dt

dI L

d N dI L N LI

(7.16)

magnetic flux linkage


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The magnetic flux density at the centre of the air-core

solenoid is given by

The magnetic flux passing through each turn of the solenoidalways maximum and is given by

Therefore the self-inductance of the solenoid is given by

20.3.3 Self-inductance of a solenoid

l NI B 0

0cos BA

Al

NI

0

l

NIA0

I N L

l NIA

I N L 0

(7.17)

PHYSICS CHAPTER 20


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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 theinduced emf is 0.012 V. Calculatea. 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

Example 20.11 :

if I I dI

;A5.2;0m;100.8turns;500 f i2 I I l N

V012.0s;35.0 dt

05.2 dI A5.2dI

dt

dI L

35.0

5.2012.0 L


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At the end of this chapter, students should be able to:Define mutual inductance.Use mutual inductance of two coaxial coils,

Explain the working principle of transformer.

Learning Outcome:

20.4 Mutual inductance ( 2 hours)

w w w

. k m s

. m a t r i k

. e d u


PHYSICS CHAPTER 20


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1 I

1 B

Coil 1 Coil 2

1 B20.4.1 Mutual induction

Consider two circular close-packed coils near eachother and sharing a commoncentral axis as shown inFigure 7.20.

A current I 1 flows in coil 1,produced by the battery inthe external circuit.

The current I 1 produces amagnetic field lines insideit and this field lines alsopass through coil 2 asshown in Figure 7.20.

20.4 Mutual inductance

Figure 7.20

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If the current I 1 changes with time , the magnetic flux throughcoils 1 and 2 will change with time simultaneously.

Due to the change of magnetic flux through coil 2, an emf isinduced in coil 2 . This is in accordance to the Faradays lawof induction .

In other words, a change of current in one coil leads to theproduction of an induced emf in a second coil which ismagnetically linked to the first coil .This process is known as mutual induction.Mutual induction is defined as the process of producing aninduced emf in one coil due to the change of current inanother coil.

At the same time, the self-induction occurs in coil 1 since themagnetic flux through it changes .

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From the Figure 7.20, consider the coils 1 and 2 have N 1 and

N 2 turns respectively.If the current I 1 in coil 1 changes, the magnetic flux through coil2 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) .

20.4.2 Mutual inductance, M

dt

dI 12

(7.21)

(7.22)

M M M 2112where : Mutual inductance

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Mutual inductance is defined as the ratio of induced emf in acoil to the rate of change of current in another coil .

From the Faradays law for the coil 2, thus

dt

d N 222

dt

d N

dt

dI M 22

112

22112 d N dI M

1

2212

I N M

22112 N I M

and

2

11

21 I

N M

(7.23)

magnetic flux linkagethrough coil 2

magnetic flux linkagethrough coil 1

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Consider a long solenoid with length l and cross sectional area

A is closely wound with N 1 turns of wire. A coil with N 2 turnssurrounds it at its centre as shown in Figure 7.21.

When a current I 1 flows in the primary coil ( N 1), it produces amagnetic field B 1,

20.4.3 Mutual inductance for two solenoids

l

I 1

I 1

N 1 N 2

A

N 1: primary coil

N 2: secondary coil

Figure 7.21

l

I N B 110

1

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and then the magnetic flux 1,

If no magnetic flux leakage , thus

If the current I 1 changes, an emf is induced in the secondarycoils, therefore the mutual inductance occurs and is given by

21

0cos11

A Bl

A I N 1101

l

A I N

I

N M 110

1

2

12

1

2212

I

N M

(7.24)

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S CS C 0

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A current of 3.0 A flows in coil C and is produced a magnetic fluxof 0.75 Wb in it. When a coil D is moved near to coil C coaxially, aflux of 0.25 Wb is produced in coil D. If coil C has 1000 turns andcoil 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

Example 20.12 :

Wb;25.0Wb;75.0A;0.3 DCC I

turns5000turns;1000 DC N N

C

CCC

I

N L

0.3

75.01000C L

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Solution :

a. and the energy stored in C is

b. The mutual inductance of the coils is given by

2CCC

2

1 I LU

20.32502

1

Wb;25.0Wb;75.0A;0.3 DCC I turns5000turns;1000 DC N N

C

DD

I

N M

0.3

25.05000

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Solution :

c. GivenThe induced emf in coil D is given by

A0.30.30s;25.0 C dI dt

Wb;25.0Wb;75.0A;0.3 DCC I turns5000turns;1000 DC N N

dt

dI M CD

25.0

0.3417

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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 N P > N S the transformer is a step-down transformer .

If N P < N S the transformer is a step-up transformer .

20.4.4 Transformer

Figure 7.22

laminated iron core

primary coil secondary coil

N P

turns N

Sturns

alternatingvoltage source

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The symbol of transformer in the electrical circuit is shown inFigure 7.23.

Working principle of transformer When an alternating voltage source is applied to the primarycoil, the alternating current produces an alternating magneticflux concentrated in the iron core.Without no magnetic flux leakage from the iron core, the samechanging magnetic flux passes through the secondary coil andinducing 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 .

Formula of transformer According to the mutual inductance, the induced emf in theprimary and secondary coils are given by

For an ideal transformer, there is no flux leakage thus

dt d N PPP (7.25)

(7.26)dt

d N SSS and

dt

d

dt

d SP

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By dividing eqs. (7.25) and (7.26), hence

There is no dissipation of power for the ideal transformer ,

therefore

In general,

dt

d N

dt

d N

SS

P

P

S

P

S

P

S

P

N N

SP P P SSPP I I

P

S

S

P

I

I

(7.27)

primaryof power:P P wheresecondaryof power:S P

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Energy losses in transformer Although transformers are very efficient devices, small energylosses do occur in them owing to four main causes:

Resistance of coilsThe wire used for the primary and secondary coils hasresistance and so ordinary ( I 2 Rt ) heat losses occur.Overcome : The transformer coils are made of thickcopper wire .

HysteresisThe magnetization of the core is repeatedly reversed bythe alternating magnetic field . The resulting expenditureof energy in the core appears as heat .Overcome : By using a magnetic material (such asMumetal) which has a low hysteresis loss .Flux leakageThe 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 iswound directly on top of the other rather than having two

separate coils.

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In an alternating current (ac) transformer in which the primary andsecondary windings are perfectly coupled, there is no current flowsin the primary when there is no load in the secondary. When thesecondary is connected to resistors, a current of 5 A is observed toflow in the primary under an applied voltage of 100 V. If the primarycontains 100 turns and the secondary 25000 turns, calculatea. the voltage,b. the current in the secondary.Solution :

a. By applying the formula of transformer,

Example 20.13 :

turns;100V;100A;5 PPP N V I turns25000S N

S

P

S

P

N

N

V

V

25000100100

SV


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Exercise 20.2 :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 turnssurrounds the primary coil at its centre, calculatea. the mutual inductance of the coilsb. the magnitude of induced emf in the secondary coil if the

current in primary coil changes at the rate 4.8 A s 1.

ANS. : 2.63 10 2 H; 0.126 V2. A transformer, assumed to be 100% efficient, is used with a

supply voltage of 120 V. The primary winding has 50 turns.The required output voltage is 3000 V. The output power is200 W.a. Name this type of transformer.b. Calculate the number of turns in the secondary winding.c. Calculate the current supplied to the primary winding

ANS. : 1250 turns; 1.67 A

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3. A transformer with a 100 turns primary coil and a 500 turnssecondary coil is connected to a supply voltage of 2.0 V.Calculate the output voltage and the maximum current insecondary coil if the current in primary coil is to be limited to0.10 A.

ANS. : 10 V; 0.020 A

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

Learning Outcome:

20.5 Energy stored in an inductor ( hour)

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Consider an inductor of inductance L . Suppose that at time t ,the current in the inductor is in the process of building up to itssteady 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 circuitis given by

20.5 Energy stored in an inductor

dt dI L

I P

dt

dI LI P LIdI Pdt and dU Pdt LIdI dU (7.18)

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The total energy stored in the inductor , U as the currentincreases from 0 to I can be found by integrating the eq. (7.18).

Thus

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

Therefore the energy stored in the solenoid is given by

I U IdI LdU 00

(7.19)

and analogous to 2

21

CV U in capacitor

l

A N L

20

2

2

1 LI U (7.20)

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A solenoid of length 25 cm with an air-core consists of 100 turnsand diameter of 2.7 cm. Calculatea. the self-inductance of the solenoid, andb. 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

Example 20.14 :

m107.2m;1025turns;100 22 d l N

24222 m1073.54

107.2

4

d A

l

A N L

20

2

427

10251073.5100104

L

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Solution :b. Given

By applying the equation of energy stored in the inductor, thus2

21

LI U

25

6.11088.221

m107.2m;1025turns;100 22 d l N A6.1 I

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Exercise 20.3 :Given 0 = 4 10 7 H m 1

1. An emf of 24.0 mV is induced in a 500 turns coil at an instantwhen the current is 4.00 A and is changing at the rate of10.0 A s -1 . Determine the magnetic flux through each turn ofthe coil.(Physics for scientists and engineers,6 th edition,Serway&Jewett,

Q6, p.1025)ANS. : 1.92 10 5 Wb2. 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 cmlength. 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

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3. A current of 1.5 A flows in an air-core solenoid of 1 cm radiusand 100 turns per cm. Calculatea. 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 ata rate of 0.0640 A s 1, the magnitude of the back emf is0.016 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, determinei. 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 J5. At a particular instant the electrical power supplied to a

300 mH inductor is 20 W and the current is 3.5 A. Determinethe rate at which the current is changing at that instant.

ANS. : 19 A s

1

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Next ChapterCHAPTER 21 :

Alternating current

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