matriculation physics ( electromagnetic induction )

1

PHYSICS CHAPTER 7

is defined as the production of an induced the production of an induced e.m.f. in a conductor/coil whenever the e.m.f. in a conductor/coil whenever the magnetic flux through the conductor/coil magnetic flux through the conductor/coil changes.changes.

CHAPTER 7: CHAPTER 7: Electromagnetic inductionElectromagnetic induction

(7 Hours)(7 Hours)

PHYSICS CHAPTER 7

2

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

Define and useDefine and use magnetic flux, magnetic flux,

Learning Outcome:

7.1 Magnetic flux (1 hour)

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cosBAAB

PHYSICS CHAPTER 7

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7.1.17.1.1 Phenomenon of electromagnetic inductionPhenomenon of electromagnetic induction 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.

7.1 Magnetic flux

No movementNo movement

0v

Figure 7.1aFigure 7.1a

PHYSICS CHAPTER 7

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Figure 7.1bFigure 7.1b

Figure 7.1cFigure 7.1c

NNSSMove towards the coilMove towards the coil

v

II

No movementNo movement

0v

PHYSICS CHAPTER 7

5

Figure 7.1dFigure 7.1d

Figure 7.1eFigure 7.1e

NNMove away from the coilMove away from the coil

v

II

NN SSMove towards the coilMove towards the coil

v

II

SS

Stimulation 7.1

PHYSICS CHAPTER 7

6

From the experiments: When the bar magnet is stationarybar magnet is stationary, the galvanometer not

show any deflection (no current flows in the coilno 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 any relative motion between the coil and the bar magnetcoil and the bar magnet , the current known as induced induced current will flow momentarilycurrent will flow momentarily through the galvanometer. This current due to an induced e.m.fcurrent due to an induced e.m.f across the coil.

Conclusion : When the magnetic field lines through a coil changesmagnetic field lines through a coil changes

thus the induced emf will existinduced emf will exist across the coil.

PHYSICS CHAPTER 7

7

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

Therefore vv is proportional to the induced emf is proportional to the induced emf.

7.1.27.1.2 Magnetic flux of a uniform magnetic fieldMagnetic flux of a uniform magnetic field is defined as the scalar product between the magnetic flux the scalar product between the magnetic flux

density, density, BB with the vector of the area, with the vector of the area, AA.

Mathematically,

v increases induced emf increases

v decreases induced emf decreases

AB

and ofdirection ebetween th angle :where flux magnetic :

cosΦ BAAB

densityflux magnetic theof magnitude :Bcoil theof area :A

(7.1)(7.1)

PHYSICS CHAPTER 7

8

It is a scalar quantityscalar quantity and its unit is weber (Wb)weber (Wb) OR tesla tesla meter squared meter squared ( T m2).

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.

From the Figure 7.2a, the angle is 0 thus the magnetic flux is given by


B

A

cosΦ BA0cosBA

BA maximummaximum

area

PHYSICS CHAPTER 7

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


B

A

cosΦ BA90cosBA

0Note:Note: Direction of vector A always perpendicular (normal)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.

area

90

PHYSICS CHAPTER 7

10

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.

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.

Example 1 :

SSNNP

QR

S

I I

Figure 7.3Figure 7.3

PHYSICS CHAPTER 7

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Solution :Solution :

The area of the coil is

Initially,

Finally,

Therefore the change in magnetic flux through the coil is

T 51.B

2322 m 100510051010 ..A

From the figure, =90 thus the initial magnetic flux through the coil is

A

B

cosΦi BA90cosBA

0Φi B

A

From the figure, =0 thus the final magnetic flux through the coil is

cosΦf BA 0cos100.55.1 3

Wb105.7Φ 3f

if ΦΦΦ 0105.7 3

Wb105.7 3

PHYSICS CHAPTER 7

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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 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.



Example 2 :

Wb1020.1 m; 100.3 32 d

4

2dA

3030

coil

BA

4

100.322

A

24 m 1007.7 A

PHYSICS CHAPTER 7

13


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

Wb1020.1 m; 100.3 32 d

603090

cosΦ BA 60cos1007.71020.1 43 B

T 40.3B

PHYSICS CHAPTER 7

14

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.

Example 3 :


PHYSICS CHAPTER 7

15


Only loop 2 has a changing magnetic fluxloop 2 has a changing magnetic flux.

Reason :

Loop 1 moves back and forth, and loop 3 moves up and down, 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.constant with time.

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

PHYSICS CHAPTER 7

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Explain Explain induced emf.induced emf. StateState Faraday’s law and Lenz’s law. Faraday’s law and Lenz’s law. ApplyApply formulae formulae,,

Derive Derive induced emf of a straight conductor and a coil in induced emf of a straight conductor and a coil in changing magnetic flux.changing magnetic flux.

Learning Outcome:

7.2 Induced emf (2 hours)

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dt

d

PHYSICS CHAPTER 7

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

a straight conductor, a straight conductor,

a coil, a coil,

a rotating coil, a rotating coil,

Learning Outcome:

7.2 Induced emf (2 hours)

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sinlvB

dt

dBA

tNAB sin

ORORdt

dAB

PHYSICS CHAPTER 7

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7.2.17.2.1 Faraday’s law of electromagneticFaraday’s law of electromagnetic inductioninduction states that the magnitude of the induced emf is proportional the magnitude of the induced emf is proportional

to the rate of change of the magnetic fluxto the rate of change of the magnetic flux.Mathematically,

The negative signnegative sign indicates that the direction of induced emfdirection of induced emf always oppose the change of magnetic flux producing it oppose the change of magnetic flux producing it (Lenz’s law)(Lenz’s law).

7.2 Induced emf

dt

dΦ OR

dt

dΦ (7.2)(7.2)

where flux magnetic theof change :Φd timeof change :dt

emf induced :

PHYSICS CHAPTER 7

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

Since

For a coil of coil of NN turns is placed in the changing magnetic turns is placed in the changing magnetic

field field BB, the induced emf is given by

dt

dN

Φ (7.3)(7.3)

, then eq. (7.3) can be written asif ΦΦ Φ d

dt

N if ΦΦ (7.4)(7.4)

flux magnetic final :Φf

flux magnetic initial :Φi

where

dt

dN

Φ cosΦ BAand

PHYSICS CHAPTER 7

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For a coil of coil of NN turns is placed in a uniform magnetic field turns is placed in a uniform magnetic field B B but changing in the coil’s area changing in the coil’s area AA, the induced emf is given by

dt

BAdN

cos

dt

dBNA cos (7.5)(7.5)

dt

dN

Φ cosΦ BAand

dt

BAdN

cos

dt

dANB cos (7.6)(7.6)

PHYSICS CHAPTER 7

21

For a coil is connected in series to a resistor of resistance a coil is connected in series to a resistor of resistance

RR and the induced emf exist in the coil as shown in Figure 7.5,


RII

dt

dN

Φ IRand

dt

dNIR

Φ (7.7)(7.7)

Note:Note: To calculate the magnitude of induced emfmagnitude of induced emf, the negative sign negative sign

can be ignoredcan be ignored. For a coil of N turns, each turn will has a magnetic flux of

BAcos through it, therefore the magnetic flux linkagemagnetic flux linkage (refer to the combined amount of flux through all the turns) is given by

the induced current I is given by

Φlinkageflux magnetic N

PHYSICS CHAPTER 7

22

The magnetic flux passing through a single turn of a coil is increased quickly but steadily at a rate of 5.0102 Wb s1. If the coil have 500 turns, calculate the magnitude of the induced emf in the coil.


By applying the Faraday’s law equation for a coil of N turns , thus

Example 4 :

12 s Wb100.5 turns;500

dt

dN

dt

dN

Φ

2100.5500

V 25

PHYSICS CHAPTER 7

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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.


a. The initial magnetic flux linkage is given by

B

Example 5 :

iΦlinkageflux magnetic initial N

0;T; 0.20 turns;50 ;m 100.8 fi24 BBNA

s 0.50dt

0A

cosi ANB

PHYSICS CHAPTER 7

24


a.

b. The induced emf is given by

0;T; 0.20 turns;50 ;m 100.8 fi24 BBNA

s 0.50dt

0cos108.00.2050 4 Wb108.0linkageflux magnetic initial 3

dt

dBNA cos if BBdB and

dt

BBNA ifcos

50.0

20.000cos100.850 4

V 106.1 2

PHYSICS CHAPTER 7

25

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 , determine the induced

current in the coil.


Example 6 :

m; 103.5m; 100.4 turns;10 2f

2i

ddNs 0.25 T; 2.1 dtB

0

A

B

B

A

Initial Final

PHYSICS CHAPTER 7

26


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

if AAdA

44

2i

2f dd

24 m 105.9 dA

m; 103.5m; 100.4 turns;10 2f

2i


2i

2f4

dd

2222 100.4103.5

4

PHYSICS CHAPTER 7

27


b. Given

The induced emf in the coil is

Therefore the induced current in the coil is given by

4.2R

dt

dANB cos

m; 103.5m; 100.4 turns;10 2f

2i


25.0

105.90cos2.110

4

V 1056.4 2

IR

A 109.1 2I 4.21056.4 2 I

PHYSICS CHAPTER 7

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Direction of Direction of induced current – induced current – Right hand grip Right hand grip rule.rule.

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 ita direction that it opposes the change producing it.

This law is essentially a form of the law of conservation of conservation of energyenergy.

7.2.2 Lenz’s law

I

INN

North pole

An illustration of lenz’s law can be explained by the following experiments.

11stst experiment: experiment:

In Figure 7.6 the magnitude of the magnetic field at the solenoid increases as the bar magnet is moved towards it.

An emf is induced in the solenoid and the galvanometer indicates that a current is flowing.


PHYSICS CHAPTER 7

29

To determine the direction of the current through the galvanometer which corresponds to a deflection in a particular sense, then the current through the solenoid seen is in 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 polenorth pole. This opposes the motion of the bar magnetopposes the motion of the bar magnet and obey the lenz’s lawobey the lenz’s law.

v

XX XX XX XX

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XX XX XX XXP

Q

F

22ndnd experiment: 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 induced current needs to flow in such a way to needs to flow in such a way to oppose the change which has oppose the change which has induced itinduced it based on lenz’s law. Hence galvanometer shows a deflection.


I

Stimulation 7.2

PHYSICS CHAPTER 7

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To determine the direction of the induced current (induced direction of the induced current (induced emf)emf) flows in the conductor PQ, the Fleming’s right hand Fleming’s right hand (Dynamo) rule(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 is placed in the magnetic field then this conductor will conductor will experience magnetic forceexperience magnetic force.

Its direction is in the opposite direction of the motionopposite direction of the motion.

Thumb Thumb – direction of – direction of MotionMotion

First finger First finger – direction of – direction of FieldField

Second fingerSecond finger – direction of – direction of induced induced

current OR induced emfcurrent OR induced emf

Note:Note:B )motion(

OR induced Iemf induced


PHYSICS CHAPTER 7

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33rdrd experiment: 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 closedswitch S is closed, current I begins to flow in the solenoid P and producing a magnetic field inside the solenoid P. Suppose that the field points towards the solenoid Q.


S,SwitchPP QQ

II

NNSS SSNN

indI indI--++

ind

PHYSICS CHAPTER 7

32

The magnetic flux through the solenoid Q increases with increases with timetime. 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 anticlockwise (Figure 7.9a) and the galvanometer shows a deflection.

QQSSwitch,

PP

NNSS

II

SS NN

indI indI-- ++

ind


PHYSICS CHAPTER 7

33

At the moment when the switch S is openedswitch S is opened, the current I starts to decrease in the solenoid P and magnetic flux through the solenoid Q decreases with timedecreases with 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 (decrease in flux). Hence based on Lenz’s law, the induced current flows in circuit consists of solenoid Q is clockwise clockwise (Figure 7.9b) and the galvanometer seen to deflect in the opposite direction of Figure 7.9a.

Stimulation 7.3

PHYSICS CHAPTER 7

34

A single turn of circular shaped coil has a resistance of 20 and an area of 7.0 cm2. It moves toward the north pole of a bar magnet as shown in Figure 7.10.

If the average rate of change of magnetic flux density through the coil is 0.55 T s1, 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.

Example 7 :


PHYSICS CHAPTER 7

35


a. By applying the Faraday’s law of induction, thus

Therefore the induced current in the coil is given by

dt

dN

BAdt

dN

A 1093.1 5I

124 s T 55.0 ;m 100.7 ; 20 turn;1 dt

dBARN

dt

dBNA

55.0100.71 4

180cosΦ BAand

V 1085.3 4

IR 201085.3 4 I

PHYSICS CHAPTER 7

36


b. Based on the lenz’s law, hence the direction of induced current is

clockwiseclockwise as shown in figure below.

NNSS indI

PHYSICS CHAPTER 7

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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.

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

7.2.3 Induced emf in a straight conductor


XX XX XX XX

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P

Q

l

x

B

v

lxA

Area, A

indI

ind

PHYSICS CHAPTER 7

38

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

According to Faraday’s law, the emf is induced in the conductor and its magnitude is given by

0cosΦ BA and0cosΦ Blx BlxΦ

dt

d

Blxdt

d

Blvdt

dxBl v

dt

dxand

(7.8)(7.8)

PHYSICS CHAPTER 7

39

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

This type of induced emf is known as motional induced emf.motional induced emf. The direction direction of the induced current induced current or induced emfinduced emf in the

straight conductor can be determined by using the Fleming’s Fleming’s right handright 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.

sinlvB (7.9)(7.9)

Bvθ

and between angle :where

Note:Note:

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 fieldmoves across the uniform magnetic field.

For a rectangular coil of rectangular coil of NN turns turns,

sinNlvB (7.10)(7.10)

PHYSICS CHAPTER 7

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A 20 cm long metal rod CD is moved at speed of 25 m s1 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.

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 8 :


C

D

B

1s m 52

PHYSICS CHAPTER 7

41


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

b. Given

i. By applying the Ohm’s law, thus

By using the Fleming’s right hand rule, the direction of the the direction of the

induced current is from D to Cinduced current is from D to C.

ii. Given

The total charge passing through the resistor is given by

15R

sinlvB

T; 10502;s m 25 m; 1020 312 Bvl

90sin10250251020 32 V 25.1

IRA 1033.8 2I

1525.1 I

90and

s 120602 t

ItQ C 10Q

1201033.8 2Q

PHYSICS CHAPTER 7

42

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

rotated mechanically with a constant angular velocity in a

uniform magnetic field of flux density B about an axis as shown in Figure 7.13.

When the vector of area, A is at an angle to the magnetic field B, the magnetic flux through each turn of the coil is given by

7.2.4 Induced emf in a rotating coil

Figure 7.13: side viewFigure 7.13: side view

B

NN SSA

ω

coil

cosBA t and

tBA cos

PHYSICS CHAPTER 7

43

By applying the equation of Faraday’s law for a coil of N turns, thus the induced emf is given by

The induced emf is maximum when hence

dt

dN

tBAdt

dN cos

tdt

dNBA cos

tNBA sin (7.11)(7.11)

time:twhere1sin t

NBAmax (7.12)(7.12)

Tf

22 where

PHYSICS CHAPTER 7

44

Eq. (7.11) also can be written as

Conclusion Conclusion : A coil rotating with constant angular velocity in a uniform magnetic field produces a sinusoidally alternating sinusoidally alternating

emfemf as shown by the induced emf against time t graph in Figure 7.14.

sinNBA (7.13)(7.13)

BA

and between angle :where


ωtεε sinmax

0

max

max

V ε

tTT5.0 T5.1 T2

B

Note:Note:

This phenomenon was the important part in the development of the electric electric generator or generator or dynamodynamo.

Stimulation 7.4

PHYSICS CHAPTER 7

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.



and the constant angular velocity in rad s1 is

Example 9 :

2222 m 105.110151010 A

T 0.5 turns;100 BN

s 06

min 1

rev 1

rad 2

min1

rev 200

1s rad 9.20

PHYSICS CHAPTER 7

46


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

b.

NBAmax

TBN 0.5 turns;100

9.20105.10.5100 2V 157max

B

A

38

From the figure, the angle is

Therefore the induced emf is given by

523890

sinNBA 52sin9.20105.10.5100 2

V 124

PHYSICS CHAPTER 7

47

Exercise 7.1 :

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 the loop of the wire. The magnet is dropped toward the loop. Determine the direction of the 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(Physics for scientists and engineers,6thth edition, Serway&Jewett, Q15, p.991)edition, Serway&Jewett, Q15, p.991)

ANS. :ANS. : U thinkU think


PHYSICS CHAPTER 7

48

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. :ANS. : 2.02.0101022 V V3. 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 magnetic field when the maximum induced emf occurs?

(Physics for scientists and engineers,6(Physics for scientists and engineers,6thth edition,Serway&Jewett, Q35, edition,Serway&Jewett, Q35, p.996)p.996)

ANS. :ANS. : 7.547.54101033 V V4. A circular coil has 50 turns and diameter 1.0 cm. It rotates at a

constant angular velocity of 25 rev s1 in a uniform magnetic field of flux density 50 T. Determine the induced emf when the plane of the coil makes an angle 55 to the magnetic field.

ANS. :ANS. : 1.771.77101055 V V

PHYSICS CHAPTER 7

49

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine self-inductance. self-inductance. ApplyApply formulae formulae

for a loop and solenoid.for a loop and solenoid.

Learning Outcome:

7.3 Self-inductance (1 hour)

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l

AN

dtdIL

20

PHYSICS CHAPTER 7

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7.3.17.3.1 Self-inductionSelf-induction Consider a solenoid which is connected to a battery , a switch S

and variable resistor R, forming an open circuit as shown in Figure 7.16a.

7.3 Self-inductance

Figure 7.16a: initialFigure 7.16a: initial

S RII

When the switch S is closed, a current

I begins to flow in the solenoid.

The current produces a magnetic current produces a magnetic field whose field lines through the field whose field lines through the solenoidsolenoid and generate the magnetic magnetic flux linkageflux linkage.

If the resistanceresistance of the variable resistor changeschanges, thus the current current flows in the solenoid also changedchanged, then so too does magnetic flux so too does magnetic flux linkagelinkage.

NNSS

PHYSICS CHAPTER 7

51

According to the Faraday’s law, an emf emf has to be induced in induced in the solenoid itself since the flux linkage changesthe solenoid itself since the flux linkage changes.

In accordance with Lenz’s law, the induced emf opposes induced emf opposes the changes that has induced it changes that has induced it and it is known as a back emfback emf.

For the current I increases :

indε-- ++

NNSSI

I

Figure 7.16b: Figure 7.16b: II increases increases

SSNN

indIindI

Direction of the induced emf is in the

opposite directionopposite direction of the current I.

PHYSICS CHAPTER 7

52

For the current I decreases :

This process is known as self-induction. Self-inductionSelf-induction is defined as the process of producing an the process of producing an

induced emf in the coil due to a change of current flowing induced emf in the coil due to a change of current flowing through the same coilthrough the same coil.

Figure 7.16c: Figure 7.16c: II decreases decreases

Direction of the induced emf is in the

same directionsame direction of the current I.

++ --indε

NNSSI IindI

indI

NNSS

PHYSICS CHAPTER 7

53

Self-induction experimentSelf-induction experiment The 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 lamps have the same brightness in their respective circuits with steady current flowing.

When the switch S is closed, the lamp A2 with variable resistor R is seen to become bright almost immediately but the lamp A1 with iron-core coil L increases slowly to full brightness.


switch, S

coil, L

iron-core

R

lamp A1

lamp A2

PHYSICS CHAPTER 7

54

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

in it. The induced or back emf opposes the growth of opposes the growth of current so the glow in the lamp Acurrent so the glow in the lamp A11 increases slowly increases slowly.

The resistor RR, however has no back emf, hence the lamp , however has no back emf, hence the lamp AA22 glow fully bright as soon as switch S is closed glow fully bright as soon as switch S is closed.

This effect can be shown by the graph of current I against

time t through both lamps in Figure 7.17b.

lamp A2 with resistor R

t

I

0Figure 7.17bFigure 7.17b

0I

lamp A1 with coil L

PHYSICS CHAPTER 7

55

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 10 :


The switch S is closed for a long time and is suddenly opened. Explain why a spark jump across the switch contacts S .

Solution :Solution : When the switch S is suddenly opened, the current in the current in the

circuit starts to fall very rapidlycircuit starts to fall very rapidly and induced a maximum induced a maximum emf in the coil Lemf in the coil L which tends to maintain the current.

This back emf is high enough to break down the insulation of break down the insulation of the air between the switch contacts Sthe air between the switch contacts S and a spark can easily spark can easily appearappear at the switch.

PHYSICS CHAPTER 7

56

From the self-induction phenomenon, we get

From the Faraday’s law, thus

7.3.2 Self-inductance, L

ILΦ

LILΦcoil theof inductance-self :Lwhere

current :I

(7.14)(7.14)

linkageflux magnetic :L

dt

d L

LIdt

d

dt

dIL (7.15)(7.15)

PHYSICS CHAPTER 7

57

Self-inductance Self-inductance is defined as the ratio of the self induced the ratio of the self induced (back) emf to the rate of change of current in the coil(back) emf to the rate of change of current in the coil.OR

For the coil of N turns, thus

dtdIL

/

dt

dN

and

dt

dIL

dt

dN

dt

dIL

dNdIL

NLI

II

NL L

(7.16)(7.16)

magnetic flux linkage

PHYSICS CHAPTER 7

58

It is a scalar quantityscalar quantity and its unit is henry (H)henry (H). Unit conversion :

The value of the self-inductance dependsself-inductance depends on the size and shape of the coilsize and shape of the coil, the number of turn (number of turn (NN)), the permeability of the medium in the coil (permeability of the medium in the coil ()).

A circuit element which possesses mainly self-inductance is known as an inductorinductor. It is used to store energy in the form store energy in the form of magnetic fieldof magnetic field.

The symbol of inductor in the electrical circuit is shown in Figure 7.19.

121 Am T1A Wb1H1


PHYSICS CHAPTER 7

59

The magnetic flux density at the centre of the air-core centre of the air-core solenoidsolenoid is given by

The magnetic fluxmagnetic flux passing through each turn of the solenoid alwaysalways maximum maximum and is given by

Therefore the self-inductance of the solenoidself-inductance of the solenoid is given by

7.3.3 Self-inductance of a solenoid

l

NIB 0

0cosBA

Al

NI

0

l

NIA0

I

NL

l

NIA

I

NL 0

l

ANL

20 (7.17)(7.17)

PHYSICS CHAPTER 7

60

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 the induced emf is 0.012 V. Calculate

a. the inductance of the solenoid,

b. the cross-sectional area of the solenoid,

c. the final magnetic flux linkage through the solenoid.

(Given 0 = 4 107 H m1)


a. The change in the current is

Therefore the inductance of the solenoid is given by

Example 11 :

if IIdI

;A 5.2;0 m;10 0.8 turns;500 fi2 IIlN

V 012.0 s; 35.0 dt

05.2 dIA 5.2dI

dt

dIL

35.0

5.2012.0 L

H 1068.1 3L

PHYSICS CHAPTER 7

61


b. By using the equation of self-inductance for the solenoid, thus

c. The final magnetic flux linkage is given by

;A 5.2;0 m;10 0.8 turns;500 fi2 IIlN

V 012.0 s; 35.0 dt

l

ANL

20

2

273

100.8

5001041068.1

A

24 m 1028.4 A

ffL LI

5.21068.1 3 Wb102.4 3

fL

PHYSICS CHAPTER 7

62

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

inductor, inductor,

Learning Outcome:

7.4 Energy stored in an inductor (½ hour)

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2

2

1LIU

PHYSICS CHAPTER 7

63

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

steady 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 circuit is given by

7.4 Energy stored in an inductor

dt

dIL

IP

dt

dILIP

LIdIPdt and dUPdt LIdIdU (7.18)(7.18)

PHYSICS CHAPTER 7

64

The total energy stored in the inductortotal energy stored in the inductor, UU as the current

increases 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 solenoidenergy stored in the solenoid is given by

IUIdILdU

00

2

2

1LIU (7.19)(7.19)

and analogous to 2

2

1CVU in capacitorin capacitor

l

ANL

20

2

2

1LIU

l

AINU

220

2

1 (7.20)(7.20)

PHYSICS CHAPTER 7

65

A solenoid of length 25 cm with an air-core consists of 100 turns and diameter of 2.7 cm. Calculate

a. the self-inductance of the solenoid, and

b. the energy stored in the solenoid,

if the current flows in it is 1.6 A.

(Given 0 = 4 107 H m1)


a. The cross-sectional area of the solenoid is given by

Hence the self-inductance of the solenoid is

Example 12 :

m10 7.2 m;10 25 turns;100 22 dlN

24

222

m 1073.54

107.2

4

d

A

l

ANL

20

2

427

1025

1073.5100104

L

H 1088.2 5L

PHYSICS CHAPTER 7

66


b. Given

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

2

2

1LIU

25 6.11088.22

1

J 1069.3 5U

m10 7.2 m;10 25 turns;100 22 dlNA 6.1I

PHYSICS CHAPTER 7

67

Exercise 7.2 :

Given 0 = 4 107 H m1

1. An emf of 24.0 mV is induced in a 500 turns coil at an instant when the current is 4.00 A and is changing at the rate of 10.0 A s-1. Determine the magnetic flux through each turn of the coil. (Physics for scientists and engineers,6(Physics for scientists and engineers,6thth edition,Serway&Jewett, edition,Serway&Jewett, Q6, p.1025)Q6, p.1025)

ANS. :ANS. : 1.921.92101055 Wb Wb

2. 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.

ANS. :ANS. : 1.881.88101044 T; 3.33 T; 3.33101088 Wb; 3.75 Wb; 3.75101044 H H

PHYSICS CHAPTER 7

68

3. A current of 1.5 A flows in an air-core solenoid of 1 cm radius and 100 turns per cm. Calculate a. the self-inductance per unit length of the solenoid.b. the energy stored per unit length of the solenoid.

ANS. :ANS. : 0.039 H m 0.039 H m11; 4.4; 4.4101022 J m J m11

4. At the instant when the current in an inductor is increasing at a rate of 0.0640 A s1, the magnitude of the back emf is 0.016 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, determine

i. the magnetic flux through each turn,ii. the energy stored in the solenoid.

ANS. :ANS. : 0.250 H; 4.50.250 H; 4.5101044 Wb; 6.48 Wb; 6.48101022 J J5. At a particular instant the electrical power supplied to a

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

ANS. :ANS. : 19 A s19 A s11

PHYSICS CHAPTER 7

69

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine mutual inductance. mutual inductance. Derive and useDerive and use formulae for mutual inductance of two formulae for mutual inductance of two

coaxial coils, coaxial coils,

ExplainExplain the working principle of transformer and the the working principle of transformer and the effect of eddy current in transformer.effect of eddy current in transformer.

Learning Outcome:

7.5 Mutual inductance (2 hours)

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l

ANN

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NM 210

1

12212

PHYSICS CHAPTER 7

70

1I

1B

Coil 1 Coil 2

1B7.5.17.5.1 Mutual Mutual

inductioninduction Consider two circular close-

packed coils near each other and sharing a common central axis as shown in Figure 7.20.

A current I1 flows in coil 1,

produced by the battery in the external circuit.

The current II11 produces a produces a

magnetic field linesmagnetic field lines inside it and this field lines also also pass through coil 2pass through coil 2 as shown in Figure 7.20.

7.5 Mutual inductance


PHYSICS CHAPTER 7

71

If the current II11 changes with time changes with time, the magnetic fluxmagnetic flux through

coils 1 and 2 will change with timechange with time simultaneously.

Due to the change of magnetic flux through coil 2, an emf is emf is induced in coil 2induced in coil 2. This is in accordance to the Faraday’s law Faraday’s law of inductionof induction.

In other words, a change of current in one coil leads to the change of current in one coil leads to the production of an induced emf in a second coilproduction of an induced emf in a second coil which is magnetically linked to the first coilmagnetically linked to the first coil.

This process is known as mutual induction. Mutual inductionMutual induction is defined as the process of producing an the process of producing an

induced emf in one coil due to the change of current in induced emf in one coil due to the change of current in another coil.another coil.

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

PHYSICS CHAPTER 7

72

From the Figure 7.20, consider the coils 1 and 2 have N1 and

N2 turns respectively.

If the current I1 in coil 1 changes, the magnetic flux through coil 2 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 quantityscalar quantity and its unit is henry (H)henry (H).

7.5.2 Mutual inductance, M

dt

dI12

dt

dIM 1

122 (7.21)(7.21)

dt

dIM 2

211 (7.22)(7.22)

MMM 2112where :: Mutual inductance Mutual inductance

PHYSICS CHAPTER 7

73

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

From the Faraday’s law for the coil 2, thus

dt

dN 2

22

dt

dN

dt

dIM 2

21

12

22112 dNdIM

1

2212 I

NM

22112 NIM

and

2

1121 I

NM

2

11

1

22

I

N

I

NM

(7.23)(7.23)

magnetic flux linkage through coil 2

magnetic flux linkage through coil 1

PHYSICS CHAPTER 7

74

Consider a long solenoid with length l and cross sectional area

A is closely wound with N1 turns of wire. A coil with N2 turns

surrounds it at its centre as shown in Figure 7.21.

When a current I1 flows in the primary coil (N1), it produces a

magnetic field B1,

7.5.3 Mutual inductance for two solenoids

lI1

I1

N1N2

A

N1: primary coil

N2: secondary coil


l

INB 110

1

PHYSICS CHAPTER 7

75

and then the magnetic flux Ф1,

If no magnetic flux leakageno magnetic flux leakage, thus

If the current I1 changes, an emf is induced in the secondary

coils, therefore the mutual inductance occurs and is given by

21

0cos11 ABl

AIN 1101

l

AIN

I

NM 110

1

212

1

2212 I

NM

l

ANNMM 210

12

(7.24)(7.24)

PHYSICS CHAPTER 7

76

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

a. The self-inductance of coil C is given by

Example 13 :

Wb;25.0 Wb;75.0 A; 0.3 DCC I turns5000 turns;1000 DC NN

C

CCC I

NL

0.3

75.01000C L

H 250C L

PHYSICS CHAPTER 7

77


a. and the energy stored in C is

b. The mutual inductance of the coils is given by

2CCC 2

1ILU

20.32502

1

J 1125C U


C

DD

I

NM

0.3

25.05000

H 417M

PHYSICS CHAPTER 7

78


c. Given

The induced emf in coil D is given by

A 0.30.30 s; 25.0 C dIdt


dt

dIM C

D

25.0

0.3417

V 5004D

PHYSICS CHAPTER 7

79

is an electrical instrument to increase or decrease the emf increase or decrease the emf (voltage) of an alternating current.(voltage) of an alternating current.

Consider a structure of the transformer as shown in Figure 7.22.

If NP > NS the transformer is a step-down transformerstep-down transformer.

If NP < NS the transformer is a step-up transformerstep-up transformer.

7.5.4 Transformer


laminated iron core

primary coil secondary coil

NP

turnsNS

turns

alternating voltage source

PHYSICS CHAPTER 7

80

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

Working principle of transformerWorking principle of transformer When an alternating voltage source is applied to the primary

coil, the alternating current produces an alternating magnetic flux concentrated in the iron core.

Without no magnetic flux leakage from the iron core, the same changing magnetic flux passes through the secondary coil and inducing an alternating emf.

After that the induced current is produced in the secondary coil.


PHYSICS CHAPTER 7

81

The characteristics of an ideal transformer are: Zero resistance of primary coilZero resistance of primary coil. No magnetic flux leakage from the iron coreNo magnetic flux leakage from the iron core. No dissipation of energy and powerNo dissipation of energy and power.

Formula of transformerFormula of transformer According to the mutual inductance, the induced emf in the

primary and secondary coils are given by

For an ideal transformer, there is no flux leakageideal transformer, there is no flux leakage thus

dt

dN P

PP

(7.25)(7.25)

(7.26)(7.26)dt

dN S

SS

and

dt

d

dt

d SP

PHYSICS CHAPTER 7

82

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

There is no dissipation of power for the ideal transformerno dissipation of power for the ideal transformer, therefore

In general,

dt

dN

dt

dN

SS

PP

S

P

S

P

S

P

N

N

SP PP SSPP II

P

S

S

P

I

I

P

S

S

P

S

P

S

P

I

I

N

N

V

V

(7.27)(7.27)

primary ofpower :PPwhere

secondary ofpower :SP

PHYSICS CHAPTER 7

83

Energy losses in transformerEnergy losses in transformer Although transformers are very efficient devices, small energy

losses do occur in them owing to four main causes: Resistance of coilsResistance of coils

The wire used for the primary and secondary coils has resistance and so ordinary (resistance and so ordinary (II22RR) heat losses) heat losses occur.Overcome :Overcome : The transformer coils are made of thick thick copper wirecopper wire.

HysteresisHysteresisThe magnetization of the coremagnetization of the core is repeatedly reversed by repeatedly reversed by the alternating magnetic fieldthe alternating magnetic field. The resulting expenditure resulting expenditure of energy in the core appears as heatof energy in the core appears as heat.Overcome :Overcome : By using a magnetic materialmagnetic material (such as Mumetal) which has a low hysteresis losslow hysteresis loss.

Flux leakageFlux leakageThe flux due to the primary may not all link the secondary. Some of the flux loss in the airflux loss in the air.Overcome :Overcome : By designing one of the insulated coils is wound directly on top of the other rather than having two separate coils.

PHYSICS CHAPTER 7

84

Eddy currentEddy currentThe alternating magnetic flux induces eddy currents in alternating magnetic flux induces eddy currents in the iron corethe iron core. By Lenz’s law, the eddy currents are induced in such a direction to oppose the magnetic flux changesdirection to oppose the magnetic flux changes.This current causes heating and dissipation of power in heating and dissipation of power in the corethe core.

Overcome Overcome : : The effect is reduced by using laminated iron laminated iron corecore as shown in Figures 7.24a and 7.24b.

Figure 7.24aFigure 7.24a Figure 7.24bFigure 7.24bStimulation 7.5

PHYSICS CHAPTER 7

85

In an alternating current (ac) transformer in which the primary and secondary windings are perfectly coupled, there is no current flows in the primary when there is no load in the secondary. When the secondary is connected to resistors, a current of 5 A is observed to flow in the primary under an applied voltage of 100 V. If the primary contains 100 turns and the secondary 25000 turns, calculate

a. the voltage,

b. the current in the secondary.


a. By applying the formula of transformer,

Example 14 :

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

S

P

S

P

N

N

V

V

25000

100100

S

V

V 25000S V

PHYSICS CHAPTER 7

86


b. The current in the secondary coil is given by

P

S

S

P

I

I

N

N

525000

100 SI

A 02.0S I

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

PHYSICS CHAPTER 7

87

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Explain Explain back emf and its effect on DC motor.back emf and its effect on DC motor.

Learning Outcome:

7.6 Back emf in DC motor (½ hour)

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PHYSICS CHAPTER 7

88

FF

Figure 7.25 shows a simple direct-current (dc) motor.

7.6 Back emf in DC motor


PHYSICS CHAPTER 7

89

When a current, I flows in the coil of the armature which is placed in a uniform magnetic field, a magnetic force is produced and causes the coil to rotate as shown in Figure 7.25.

As the coil rotates, its magnetic flux changes and so an emf is induced across the coil. (Faraday’s law)

By Lenz’s law this induced emf opposes the current which is

making the coil to turn and this emf is called back emf (back emf (εεBB)).

The back emf is given by

As the motor speeds upspeeds up, the back emf, back emf, εεBB increases increases because

it is proportional to the frequencyproportional to the frequency, f.

NBAB

fBfinal

initial

final

initial

f

f

so (7.28)(7.28)

PHYSICS CHAPTER 7

90

I

εB

V

RMotorMotor

When the motor is first switched onfirst switched on, the back emf is zeroback emf is zero: it risesrises as the motor speeds upspeeds up.

When the motor is running freelyrunning freely, the back emf is nearly back emf is nearly equal to the supply voltageequal to the supply voltage and so there will not be much will not be much current drawncurrent drawn.

When a load is appliedload is applied to the motor, the motor slows downmotor slows down, the back emf fallsback emf falls, and so the current in the coil increasescurrent in the coil increases.

Figure 7.25 also can be simplified into the circuit shown in Figure 7.26.

LoopLoopFigure 7.26Figure 7.26

PHYSICS CHAPTER 7

91

By applying the Kirchhoff’s 2nd law to the Figure 7.26,

Eq. (7.29) I

IRIRV B (7.29)(7.29)

RIIIV 2B

RIIIV 2B (7.30)(7.30)

where IV : power supplied

I2R : power lost as heat in coil

IεB : mechanical power

PHYSICS CHAPTER 7

92

A motor rotates at a rate of 1000 revolutions per minute. The supply voltage is 240 V and the resistance of the armature is 2.5 .a. Calculate the back emf if the current in the armature is 7.5 A.A load is applied to the motor and the speed of the rotation is found to decrease to 500 revolutions per minute. Calculateb. the back emf now.c. the new current in the armature.d. the mechanical power produced by the motor.


a. Given

The back emf in the armature is given by

Example 15 :

5.2 V; 240 rpm; 10000 RV

IRεV B

5257240 B ..ε V 221B

A 5.7I

PHYSICS CHAPTER 7

93


b. Given

The new back emf in the armature is given by

c. The new current in the armature is

d. The mechanical power produced by the motor is given by

0

f

i ε

ε

rpm 500

rpm 1000221

f

V 111f

5.2 V; 240 rpm; 10000 RVrpm 500

IRεV B

52111240 .IA 6.51I

Bpower mechanical I 1116.51

W1073.5power mechanical 3

PHYSICS CHAPTER 7

94

Exercise 7.3 :

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 emf in the secondary coil if the current in primary coil changes at the rate 4.8 A s1.

ANS. :ANS. : 2.63 2.63101022 H; 0.126 V H; 0.126 V

2. 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 is 200 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. :ANS. : 1250 turns; 1.67 A 1250 turns; 1.67 A

PHYSICS CHAPTER 7

95

3. A transformer with a 100 turns primary coil and a 500 turns secondary coil is connected to a supply voltage of 2.0 V. Calculate the output voltage and the maximum current in secondary coil if the current in primary coil is to be limited to 0.10 A.

ANS. :ANS. : 10 V; 0.020 A 10 V; 0.020 A

4. The resistance of the armature of a dc motor is 0.75 . A supply of 240 V is connected to this motor. When the motor rotates freely without load, the current in the armature is 4.0 A and the rate of rotation is 400 rpm.Calculate

a. the back emf produced.

b. the mechanical power generated.

If a load is applied, the current increases to 60 A. Calculate

c. the back emf now.

d. the mechanical power.

e. the rotation speed of the armature.

ANS. :ANS. : 237 V; 948 W; 195 V; 11.7 kW; 329 rpm 237 V; 948 W; 195 V; 11.7 kW; 329 rpm

96

PHYSICS CHAPTER 7

Next Chapter…CHAPTER 8 :

Alternating current

matriculation physics ( electromagnetic induction )

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