PHYSICS CHAPTER 15

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CHAPTER 15: CHAPTER 15: Electromagnetic inductionElectromagnetic induction

(4 Hours)(4 Hours)

PHYSICS CHAPTER 15

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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:

Define and useDefine and use magnetic fluxmagnetic flux,,

Learning Outcome:15.1(a) Magnetic flux (1 hour)

cosBAAB

PHYSICS CHAPTER 15

15.1(a).115.1(a).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 A, B,C, D, and E.

15.1(a) Magnetic flux

No movementNo movement

0v

Figure AFigure A

PHYSICS CHAPTER 15

Figure BFigure B

Figure CFigure C

NNSSMove towards the coilMove towards the coil

v

II

No movementNo movement

0v

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Figure DFigure D

Figure EFigure E

NNMove away from the coilMove away from the coil

v

II

NN SSMove towards the coilMove towards the coil

v

II

SS

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From the experiments: bar magnet is stationarybar magnet is stationary,

- galvanometer not show any deflection (no current no current flows in the coilflows in the coil).

bar magnet is moved relatively towards the coilbar magnet is moved relatively towards the coil, - the galvanometer shows a momentary deflection to the

right (Figure B). - When the bar magnet is moved relatively away from the

coil, the galvanometer is seen to deflect in the opposite direction (Figure D).

ThereforeTherefore :- when there is any relative motion between the coil any relative motion between the coil

and the bar magnetand 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.

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ConclusionConclusion : 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.

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.

v increases induced emf increases

v decreases induced emf decreases

PHYSICS CHAPTER 15

15.1(a).2 Magnetic flux of a uniform magnetic field15.1(a).2 Magnetic flux of a uniform magnetic field

defined as the scalar product between the magnetic flux density, the scalar product between the magnetic flux density, BB with the vector of the area, with the vector of the area, AA.

flux magnetic :

cosΦ BAAB

(7.1)(7.1)

where

AB

and ofdirection ebetween th angle :densityflux magnetic theof magnitude :B

coil theof area :A

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scalar quantityscalar quantity unit is weber (Wb)weber (Wb) OR tesla meter squared ( T mtesla meter squared ( T m22).). Consider a uniform magnetic field B passing through a surface area A of

a single turn coil as shown in Figures A and B.

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

Figure AFigure A

B

A

cosΦ BA0cosBA

BA maximummaximum

area

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

Figure BFigure B

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 field lines magnetic flux is proportional to the number of field lines

passing through the area.passing through the area.

area

90

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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 15.3.

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 15.3Figure 15.3

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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 15

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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.

Solution :Solution :

The area of the coil is

Example 2 :

Wb1020.1 m; 100.3 32 d

4

2dA

4

100.322

A

24 m 1007.7 A

3030

coil

BA

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Solution :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

Wb1020.1 m; 100.3 32 d

603090

cosΦ BA

60cos1007.71020.1 43 B

T 40.3B

PHYSICS CHAPTER 15

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The three loops of wire as shown in Figure 15.4 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 :

Figure 15.4Figure 15.4

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Solution :Solution :

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, but Loop 1 moves back and forth, and loop 3 moves up and down, but since the magnetic field is uniform, the flux always constant with since the magnetic field is uniform, the flux always constant with time.time.

Loop 2 on the other hand changes its orientation relative to the field Loop 2 on the other hand changes its orientation relative to the field as it rotates, hence its flux does change with time.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: At the end of this chapter, students should be able to: Explain Explain induced emfinduced emf.. StateState Faraday’s lawFaraday’s law.. ApplyApply formulaeformulae,,

Learning Outcome:15.1(b) Induced emf (2 hours)

dt

d

PHYSICS CHAPTER 15

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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: Derive and use Derive and use formula of induced emf for: formula of induced emf for:

a coil, a coil,

a rotating coil, a rotating coil,

Learning Outcome:15.1(b) Induced emf (2 hours)

dt

dBA

tNAB sin

ORORdt

dAB

PHYSICS CHAPTER 15

15.1(b).1 Faraday’s law of electromagnetic induction15.1(b).1 Faraday’s law of electromagnetic induction

states that the magnitude of the induced emf is proportional to the the magnitude of the induced emf is proportional to the rate of change of the magnetic fluxrate of change of the magnetic flux.

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

15.1(b) Induced emf

dt

dΦ OR

dt

dΦ (7.2)(7.2)

where

flux magnetic theof change :Φd 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 coil of NN turns is placed in the changing magnetic field turns is placed in the changing magnetic 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

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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)

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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,

Figure 15.5Figure 15.5

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

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

Solution :Solution :

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

Example 15.4 :

12 s Wb100.5 turns;500

dt

dN

dt

dN

Φ

2100.5500

V 25

PHYSICS CHAPTER 15

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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.

Solution :Solution :

a. The initial magnetic flux linkage is given by

B

Example 15.5 :

iΦlinkageflux magnetic initial N

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

s 0.50dt

0A

cosi ANB

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Solution :Solution :

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

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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, 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.

Solution :Solution :

Example 15.6 :

m; 103.5m; 100.4 turns;10 2f

2i

ddNs 0.25 T; 2.1 dtB

0

A

B

B

A

Initial Final

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Solution :Solution :

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

ddNs 0.25 T; 2.1 dtB

2i

2f4

dd

2222 100.4103.5

4

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Solution :Solution :

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

ddNs 0.25 T; 2.1 dtB

25.0

105.90cos2.110

4

V 1056.4 2

IR

A 109.1 2I 4.21056.4 2 I

PHYSICS CHAPTER 15

15.2 Induced emf (2 hours)

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: StateState Lenz’s lawLenz’s law.. ApplyApply formulaeformulae,,

Learning Outcome:

)(cos tdt

dNBA

PHYSICS CHAPTER 15

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states that an induced electric current always flows in such a an induced electric current always flows in such a direction that it opposes the change producing itdirection that it opposes the change producing it.

This law is essentially a form of the law of conservation of energyconservation of energy.

15.2(a) Lenz’s law

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

I

INN

North pole

Experiment 1 :Experiment 1 :

From From lenz’s law experimentslenz’s law experiments : :

Figure shows the magnitude

of the magnetic field at the

solenoid increases as the bar

magnet is moved towards it.

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

direction of the current deflection of the galvanometer needle shows the direction of the current.

the current through the solenoid is in the direction that make current through the solenoid is in the direction that make the solenoid upper end becomes a north polethe solenoid upper end becomes a north pole. This opposes the opposes the motion of the bar magnetmotion of the bar magnet and obey the lenz’s lawobey the lenz’s law.

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Experiment 2 :Experiment 2 : Consider two solenoids P and Q arranged coaxially closed to each

other as shown in Figure A.

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.

Figure AFigure A

S,SwitchPP QQ

II

NNSS SSNN

indI indI--++

ind

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The magnetic flux through the solenoid Q increases with timeincreases 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 (increase in flux). Hence based on Lenz’s law, the induced current flows in circuit consists of solenoid Q is anticlockwise anticlockwise (Figure A) and the galvanometer shows a deflection.

QQSSwitch,

PP

NNSS

II

SS NN

indI indI-- ++

ind

Figure BFigure B

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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 B) and the galvanometer seen to deflect in the opposite direction of Figure 15.7.

PHYSICS CHAPTER 15

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.

Example 15.7 :

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Solution :Solution :

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

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Solution :Solution :

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

clockwiseclockwise as shown in figure below.

NNSS indI

PHYSICS CHAPTER 15

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.

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

15.2(b) Induced emf in a rotating coil

side viewside view

B

NN SSA

ω

coil

cosBA t and

tBA cos

PHYSICS CHAPTER 15

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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 (7.8)(7.8)

time:twhere1sin t

NBAmax (7.9)(7.9)

Tf

22 where

PHYSICS CHAPTER 15

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Eq. (7.8) also can be written as

Conclusion Conclusion : A coil rotating with constant angular velocity in a uniform magnetic field produces a sinusoidally alternating emfsinusoidally alternating emf as shown by the

induced emf against time t graph in Figure 15.11.

sinNBA (7.10)(7.10)

BA

and between angle :where

Figure 15.11Figure 15.11

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.

PHYSICS CHAPTER 15

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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 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.

Solution :Solution :

The area of the coil is

and the constant angular velocity in rad s1 is

Example 15.8 :

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 15

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Solution :Solution :

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

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Exercise :

1. A bar magnet is held above a loop of wire in a horizontal plane, as shown in figure.

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

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2. 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 V

3. 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

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