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Page 1: Ch20 Electromagnetic Induction

Electromagnetic Induction

Page 2: Ch20 Electromagnetic Induction

2

Define and use magnetic flux,

20.1 Magnetic Flux

cosBAAB

Page 3: Ch20 Electromagnetic Induction

• is defined as the scalar product between the magnetic flux density, B and the vector of the surface area, A

where,: magnetic flux

: angle between B and AB: magnetic flux density

A: area of the coil• Scalar quantity• Unit: T m2 or Wb

Magnetic Flux, : a measure of the number of magnetic field lines that cross a given area

A

B = 90

BA = 0

3

Direction of vector A always perpendicular to the surface area, A.

cosBAAB

Page 4: Ch20 Electromagnetic Induction

4

A flat surface with area 3.0 cm2 is placed in a uniform magnetic field. If the magnetic field strength is 6.0 T, making an angle 30° with the surface area, find the magnetic flux through this area.

Example 20.1Solution:

3030

coil

B

Normal A

Using: cosAB23 m T 109.0

θ = 90 – 30 = 60°

Page 5: Ch20 Electromagnetic Induction

5

REVISION: Magnetic Flux cosBAAB

Page 6: Ch20 Electromagnetic Induction

6

• A solenoid 4.00 cm in diameter and 20.0 cm long has 250 turns and carries a current of 15.0 A. Calculate the magnetic flux through the circular cross sectional area of the solenoid

• A long, straight wire carrying a current of 2.0 A is placed along the axis of a cylinder of radius 0.50 m and a length of 3.0 m. Determine the total magnetic flux through the cylinder.

REVISION: Magnetic Flux

A

area90

II I

SN

= 2.96×10–5 Tm2

Page 7: Ch20 Electromagnetic Induction

7

The three loops of wire 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 20.2

Page 8: Ch20 Electromagnetic Induction

8

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

Solution:

Page 9: Ch20 Electromagnetic Induction

9

1A long, straight wire carrying a current of 2.0 A is placed along the axis of a cylinder of radius 0.50 m and a length of 3.0 m. Determine the total magnetic flux through the cylinder.[Zero]

2A solenoid 4.00 cm in diameter and 20.0 cm long has 250 turns and carries a current of 15.0 A. Calculate the magnetic flux through the circular cross sectional area of the solenoid.[2.96×10–5 T m2]

Exercise 20.1

Page 10: Ch20 Electromagnetic Induction

10

(d)Derive and use induced emf:- in straight conductor,

- in coil,

- in rotating coil,

(a)Use Faraday’s experiment to explain induced emf

(b)State Faraday’s law and use Lenz’s law to determine the direction of induced current

(c) Use induced emf,

20.2 Induced emf

dtd

sinBlv

dtdBNA

dtdANB

tNAB sin

Page 11: Ch20 Electromagnetic Induction

11

History – Faraday’s experiment to induced emf

• In this experiment, Faraday hoped by using a strong enough battery, a steady current in X would produce a current in a second coil Y but failed

• Current carrying conductor magnetic field

• Magnetic field electric current????

• The diagram below shows the apparatus used by Faraday in his attempt to produce an electric current from a magnetic field

Page 12: Ch20 Electromagnetic Induction

12

• This is Faraday’s apparatus for demonstrating that a magnetic field can produce a current

• A change in the field produced by the top coil induces an EMF and, hence, a current in the bottom coil

• When the switch is opened and closed, the galvanometer registers currents in opposite directions

• No current flows through the galvanometer when the switch remains closed or open

REVISION: Faraday’s experiment to induced emf

OBSERVATION

GALVANOMETER DEFLECTION

Switch on YESSwitch of YESSteady current

NO

Faraday’s experiment used a magnetic field that was changing because the current producing it was changing

Page 13: Ch20 Electromagnetic Induction

13

• Faraday concluded that although a steady magnetic field produces no current, a changing magnetic field can produce an electric current

• Such a current is called an induced current

• We therefore say that an induced current is produced by a changing magnetic field

• The corresponding emf required to cause this current is called an induced emf

Faraday’s experiment to induced emf

OBSERVATION

GALVANOMETER DEFLECTION

Switch on YESSwitch of YESSteady current

NO

Page 14: Ch20 Electromagnetic Induction

14

• a magnetic field that is changing because the magnet is moving

A changing magnetic field induces an emf

Page 15: Ch20 Electromagnetic Induction

15

Electromagnetic Induction is the production of induced e.m.f.s or induced currents whenever the

magnetic flux through a loop, coil or circuit is changed

meaning of changing in magnetic flux

a relative motion of loop &

magnet field lines are ‘cut’

number of magnetic field lines passing through a coil are

increased or decreased

To increase induced e.m.fUse a

stronger

magnet can

increase magneti

c flux

Push the

magnet faster into the coil to increase speed

The area of the coil is

greater

The number of turns increa

sed

Electromagnetic Induction

Page 16: Ch20 Electromagnetic Induction

16

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

Lenz’s Law

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

Faraday’s Law

• These two laws are summed up in the relationship

• For a coil of N turns

• Since d = final - initial

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

dtdΦ

dtdN Φ

dt

N if ΦΦ

Page 17: Ch20 Electromagnetic Induction

17

Solution:A coil of wire 8 cm in diameter has 50 turns and is placed in a B field of 1.8 T. If the B field is reduced to 0.6 T in 0.002 s , calculate the induced emf.

Example 20.3

tN

dtdΦN initialfinalB

tBB

NA initialfinal

tBBdN initialfinal

2

2

V 151

dtdN Φ

Note:

To calculate the magnitude of induced emf, the negative sign can be ignored

Page 18: Ch20 Electromagnetic Induction

18

Solution:

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

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.

Example 20.412 s Wb1005

turns;500

.dtdN

dtdN Φ

2100.5500

V 25

Page 19: Ch20 Electromagnetic Induction

19

• When an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it

• The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant

• In the examples below, if the B field is increasing, the induced field acts in opposition to it

• If it is decreasing, the induced field acts in the direction of the applied field to try to keep it constant

Lenz’s Law

Page 20: Ch20 Electromagnetic Induction

20

Lenz’s Law

Page 21: Ch20 Electromagnetic Induction

21

Lenz’s Law

N

I

Induced current flows in counterclockwise

N

S Induced current flows in clockwise

Page 22: Ch20 Electromagnetic Induction

22

Lenz’s Law

(1834)

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

Faraday’s Law

(1831)

• the magnitude of the induced emf is proportional to the rate of change of the magnetic flux

Faraday’s Experime

nt

• a steady magnetic flux/ field produces no current

• a changing magnetic flux/ field can produce an electric current induce current

Magnetic flux,

dtdΦ

REVISION

Page 23: Ch20 Electromagnetic Induction

23

• Faraday's Law : change in magnetic flux INDUCED CURRENT/ EMF, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it

• The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant

Lenz’s Law • In the examples below, if the B field is increasing, the induced

field acts in opposition to it• If it is decreasing, the induced field acts in the direction (SAME

DIRECTION) of the applied field to try to keep it constant

Page 24: Ch20 Electromagnetic Induction

24

Lenz’s Law

N

I

Induced current flows in counterclockwise

Page 25: Ch20 Electromagnetic Induction

25

Lenz’s Law

Induced current flows in clockwise

N

S

Page 26: Ch20 Electromagnetic Induction

26

Another great example of Lenz's law is to take a copper tube (it's conductive but non-magnetic) and drop a piece of steel

down through the tube. The piece of steel will fall through, as you might expect.  It accelerates very close to the acceleration

due to gravity.

Now take the same copper tube and drop a magnet through it. You will notice that the magnet falls very slowly. This is because the copper tube "sees" a changing magnetic field from the falling magnet. This changing magnetic field induces a current in the copper tube. The induced current in the copper tube creates its

own magnetic field that opposes the magnetic field that created it.

Page 27: Ch20 Electromagnetic Induction

27

Faraday’s Law dtdΦ

How can we induce emf?By changing magnetic flux/

field

In a plane coil

By changing area in

magnetic field

By changing magnetic

field strength,

B

Straight conductor

moving through a magnetic

field

Rotating Coil

Page 28: Ch20 Electromagnetic Induction

28

• From

Induced emf in a plane coil by changing area in B

dtdN Φ

cosBAΦand

dtcosBAdN

dtdAcosNB

Stretching the coil reduces the area of the coil magnetic flux through coil is decreased and, current is induced in the coil

Flux through coil is decreased

Page 29: Ch20 Electromagnetic Induction

29

The flexible loop has a radius of 12 cm and is in a magnetic field of strength 0.15 T. The loop is grasped at point A and B and stretched until its area is nearly zero. If it takes 0.20 s to close the loop, find the magnitude of the average induced emf in it during this time.

Example 20.5

Page 30: Ch20 Electromagnetic Induction

30

Solution:

dtd

dANBdt

final initial( )A ANBt

2(0 )rNB

t

2(0 (0.12) )(1)(0.15)0.20

V 104.3 2

Page 31: Ch20 Electromagnetic Induction

31

• From

Induced emf in a plane coil by changing magnetic field, B strength

dtdN Φ

cosBAΦand

dtcosBAdN

As the magnetic field strength, B is increasing or decreasing with time, the magnetic flux through the area changes, therefore induces an emf in the coil

dtdBcosNA

Page 32: Ch20 Electromagnetic Induction

32

A circular coil has 200 turns and diameter 36 cm. the resistance of the coil is 2.0 Ω. A uniform magnetic field is applied perpendicularly to the plane of the coil. If the field changes uniformly from 0.5 T to 0 T in 0.8s.(a) Find the induced e.m.f. &

current in the coil while the field is changed.

(b) Determine the direction of the current induced.

Example 20.6Solution:

1. Calculate area, A = r2

2. Determine emf,

3. Determine I from dtdBNA

IRA36.6

Page 33: Ch20 Electromagnetic Induction

33

Solution: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 20.7

0

A

B

B

A

Initial Final

dtdAcosNB

Page 34: Ch20 Electromagnetic Induction

34

Solution: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

Example 20.8

B

0A

Wb108.0 3initial

V 106.1 2

dtdBcosNA

Page 35: Ch20 Electromagnetic Induction

35

Calculate the current through a 37 Ω resistor connected to a single turn circular loop 10 cm in diameter, assuming that the magnetic field through the loop is increasing at a rate of 0.050 T/s. State the direction of the current.

Example 20.9

Page 36: Ch20 Electromagnetic Induction

36

• Calculate, Solution:

dtdΦ

2

2

dA,

dtdBA

RI

I induced

I induced

V. 410933

A 10 x 1.06 5-R

I

Page 37: Ch20 Electromagnetic Induction

37

• Induced emf in a plane coil:– by changing area in

magnetic field

– by changing magnetic field strength

dtdAcosNB

dtdBcosNA

How can we induce emf?By changing magnetic flux/ field

In a plane coil

By changing area in

magnetic field

By changing magnetic

field strength,

B

Straight conductor

moving through a magnetic

field

Rotating Coil

REVISION

dtdΦ

Page 38: Ch20 Electromagnetic Induction

38

Induced emf in a straight conductor moving through a magnetic field

• As a conductor is moved through a magnetic field, current is induced and the bulb is lightened up

Page 39: Ch20 Electromagnetic Induction

39

• Consider a straight conductor of length l is moved at a speed v to the right on a U-shaped conductor in a uniform magnetic field B that points out the paper

• This conductor travels a distance dx =vdt in a time dt

• The area of the loop increases by an amount:

Induced emf in a straight conductor moving through a magnetic field

ldxdA

Page 40: Ch20 Electromagnetic Induction

40

• This induced emf is called motional induced emf

• The direction of the induced current or induced emf in the straight conductor can be determined by using the Lenz’s Law

• If B field is increasing, the induced field acts in opposition

to it• If B is decreasing, the induced field acts in the direction (SAME DIRECTION) of the applied field

to try to keep it constant

• According to Faraday’s law, the e.m.f. is induced in the conductor and its magnitude is given by

angle between v and B

Induced emf in a straight conductor moving through a magnetic field

dtd

dtdAB

dtdxBl v

dtdx

and

sinBlvBlv

Page 41: Ch20 Electromagnetic Induction

41

Page 42: Ch20 Electromagnetic Induction

42

Consider the arrangement shown below. Assume that R = 6 Ω, L = 1.2 m & a uniform 2.50 T magnetic field is directed into the page.(a)At what speed should the

bar be moved to produced a current of 0.5A in the resistor

(b)what is the direction of the induced current?

Example 20.10

Page 43: Ch20 Electromagnetic Induction

43

(a) = Blv sin , = IR

(b)From Lenz’s Law

Solution:

1s m1 v

Page 44: Ch20 Electromagnetic Induction

44

(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

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.

Example 20.11

C

D

B

1s m 52

Page 45: Ch20 Electromagnetic Induction

45

Induced emf in a rotating coil

An ac generator / dynamo: transforms mechanical energy into electric energy

Page 46: Ch20 Electromagnetic Induction

46

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

• Consider a coil of N turns each of area A and is being rotated about a horizontal axis in its own plane at right angle to a uniform magnetic field of flux density B.

• As the coil rotates with the angular speed ω, the orientation of the loop changes with time.

Induced emf in a rotating coil

cosBA t andtBA cos

dtdN

tBAdtdN cos

tdtdNBA cos

tNBA sin

Page 47: Ch20 Electromagnetic Induction

47

• The emf induced in the loop varies sinusoidally in time

• The induced emf is maximum when hence

where

Induced emf in a rotating coil

NBAmax

Tf 22

Note:

This phenomenon was the important part in the

development of the electric generator or

dynamo.

Page 48: Ch20 Electromagnetic Induction

48

iii) The emf induced in a coil varies sinusoidally with time.

iv) Maximum voltage (ξ max = NBA ω) is produced when the coil is parallel to the magnetic field (sin ωt = 1).

v) No voltage exists when the coil is perpendicular to the magnetic field

• The graph show that :i) The magnitude of

induced emf is depends on the angle between the field and the coil.

ii) The induced emf is an alternating voltage because has positive value as well as negative value.

Induced emf in a rotating coil

Page 49: Ch20 Electromagnetic Induction

(a)Define self-inductance(b)Apply self-inductance,

for coil and solenoid

20.3 Self Inductance

dtdI

L

Page 50: Ch20 Electromagnetic Induction

50

• Running a changing current (by changing R), creates a changing magnetic field, which creates an induced emf that fights the change

• Unit: Henry (V s A-1)

Self Induction – the production of e.m.f. in a circuit due to the change of current in the circuit itself

Page 51: Ch20 Electromagnetic Induction

51

When the switch S is closed, a current I begins to flow in the solenoid

The current produces a magnetic field whose field lines through the solenoid and generate the magnetic flux linkage

If the resistance of the variable resistor changes, thus the current flows in the solenoid also changed, then so too does magnetic flux linkage

Self Induction

S RII

NS

Page 52: Ch20 Electromagnetic Induction

52

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

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

• For an increasing current, the direction of the induced field and emf are opposite to that of the current, to try to decrease the current

• For the current I increases:Self Induction

indε- +

NSI

ISN

Direction of the induced emf is in the opposite direction of the current I.

Page 53: Ch20 Electromagnetic Induction

53

• If the current is decreasing, the direction of the induced field and emf are in the same direction as the current, to try to increase the current

• This coil is said to have self-inductance (inductance)

• A coil that has inductance is called an inductor use to store energy in the form of magnetic field

• For the current I decreases:Self Induction

+ -indε

NSI IindI

indI

NS

Direction of the induced emf is in the same direction of the current I.

Page 54: Ch20 Electromagnetic Induction

54

(a) A current in the coil produces a magnetic field directed to the left

Self Induction

(b) If the current increases, the increasing magnetic flux creates an induced emf having the polarity shown by the dashed battery

(a)The polarity of the induced emf reverses if the current decreases

Page 55: Ch20 Electromagnetic Induction

55

• From the Faraday’s law, thus

• From the self-induction phenomenon, we get

whereL: self inductance of the coil,

unit: Henry (H) or Wb A-1

I: current

Self Inductance, L: the ratio of the self induced e.m.f. to the rate of change of current in the coil

ILΦ

LILΦdt

d L

LIdtd

dtdIL

dtdIL

/

• The value of the self-inductance depends on(a) the size and shape of the coil(b) the number of turn (N)(c) the permeability of the medium in the

coil ()

Page 56: Ch20 Electromagnetic Induction

56

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

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

• The magnetic flux passing through each turn of the solenoid always maximum and is given by

Self Inductance of a solenoid

lNI

B 0

0cosBA

AlNI

0

lNIA0

INL

lNIA

INL 0

lAN

L2

0

Page 57: Ch20 Electromagnetic Induction

57

Suppose you wish to make a solenoid whose self-inductance is 1.4 mH. The inductor is to have a cross-sectional area of 1.2 x 10 -3 m2 and a length of 0.052 m. How many turns of wire needed?

Example 20.14Induced emf of 5.0 V is developed across a coil when the current flowing through it changes at 25 A s-1. Determine the self inductance of the coil.

Example 20.12

H.20

dtdIL

Example 20.13If the current in a 230 mH coil changes steadily from 20.0 mA to 28.0 mA in 140 ms, what is the induced emf?

220 turnsl

ANL

20

Page 58: Ch20 Electromagnetic Induction

58

Solution

a. The change in the current is

Therefore the inductance of the solenoid is given by

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.

Example 20.15;A 5.2;0 m;10 0.8 turns;500 fi

2 IIlNV 012.0 s; 35.0 dt

if IIdI 05.2 dIA 5.2dI

dtdIL

35.05.2012.0 L

H 1068.1 3L

Page 59: Ch20 Electromagnetic Induction

59

c. The final magnetic flux linkage is given by

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

lAN

L2

0

2

273

100.85001041068.1

A

24 m 1028.4 A

ffL LI

5.21068.1 3 Wb102.4 3

fL

Page 60: Ch20 Electromagnetic Induction

• An inductor is a circuit component (coil or solenoid) which produced an self induced emf

• Function of an inductor:(1)to control current(2)store energy in form of

magnetic field• Back emf produce in an

inductor is given by:

Derive and use the energy stored in an inductor,

20.4 Energy Stored In Inductor

2

21 LIU

Inductor

dtdIL

Page 61: Ch20 Electromagnetic Induction

61

• The total work done while the current is changed from zero to its final value is given by

and analogous toin capacitor

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

• Therefore the energy stored in the solenoid is given by

• The electrical power P in overcoming the back emf in the circuit is given by

Energy Stored In Inductor

IP

dtdILIP

dILIPdt dILIdU

IUIdILdU

00

2

21 LIU

2

21 CVU

lAN

L2

0

2

21 LIU

lAIN

U22

0

21

Page 62: Ch20 Electromagnetic Induction

62

A 400 turns solenoid has a cross sectional area 1.81×10-3 m2 and length 20 cm carrying a current of 3.4 A.(i) Calculate the inductance of

the solenoid(ii) Calculate the energy stored

in the solenoid.(iii)Calculate the induced emf in

the solenoid if the current drops uniformly to zero in 55 ms.

How much energy is stored in a 0.085-H inductor that carries a current of 2.5 A?

Example 20.16

2

21 LIU

Example 20.17 A steady current of 2.5 A in a coil of 500 turns causes a flux of 1.4 x 10-4 Wb to link (pass through) the loops of the coil. Calculate(a)the average back emf induced

in the coil if the current is stopped in 0.08 s

(b)the inductance of the coil and the energy stored in the coil (inductor).

Example 20.18

HL 31082.1

J21005.1

V1125.0

Page 63: Ch20 Electromagnetic Induction

63

Solution:(a)1. Calculate A2. Use

(b) Use

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)

Example 20.19

lANL

20

2

21 LIU

H 1088.2 5LJ 1069.3 5U

Page 64: Ch20 Electromagnetic Induction

• Mutual Induction: emf induced in a circuit by a changing current in another nearby circuit

(a)Define mutual inductance

(b)Use mutual inductance

between two coaxial solenoids or a coaxial coil and a solenoid

20.5 Mutual Inductance

lANNM 210

Page 65: Ch20 Electromagnetic Induction

65

The induced current in this loop that caused by the change of current in neighbouring loop Mutual Induction

NS N S

Increasing current in loop 1 produces a change in magnetic flux. This changes is experienced by loop 2 that placed nearby

According to Faraday’s law, emf is induced in loop 2 to oppose the changes.

12

Page 66: Ch20 Electromagnetic Induction

66

• In mutual induction, the e.m.f. induced in one coil is proportional to the rate at which the current in the other coil is changing

• If we assume that the current in coil 1 changes at a rate of dI1/dt, the magnetic flux will change by dΦ1/dt and this changes is experienced by coil 2

• The induced e.m.f. in coil 2 is

• If vice versa, the induced emf in coil 1, 1 is given by where

Mutual Inductance, M: the ratio of induced emf in a coil to the rate of change of current in another coil

dtdI1

2 dtdIM 1

122

dtdIM 2

211 MMM 2112

Keep In mind emf induced in coil 2 is due to the current change in coil 1

emf induced in coil 1 by changing current in coil 2

Page 67: Ch20 Electromagnetic Induction

67

• From Faraday’s Law

and

• Since M12 = M21 = M

• Rearrange,

Mutual Inductance, M

dtdIM 1

122 dtdIM 1

2

dtdIM 2

211 dtdIM 2

2

dtdI

dtdIM

2

1

1

2

dtdN 2

22

dtdN

dtdIM 2

21

12

22112 dNdIM

1

2212 I

NM

22112 NIM

2

1121 I

NM

2

11

1

22

IN

INM

Page 68: Ch20 Electromagnetic Induction

68

Solution:(a)Using

(b)From

Two coils, X & Y are magnetically coupled. The emf induced in coil Y is 2.5 V when the current flowing through coil X changes at the rate of 5 A s-1. Determine:(a)the mutual inductance of the

coils(b)the emf induced in coil X if

there is a current flowing through coil Y which changes at the rate of 1.5 A s-1.

Example 20.201552 As

dtdI,V. x

Y

dtdI

MX

Y

H5.0M

dtdIM Y

X

V75.0

Page 69: Ch20 Electromagnetic Induction

69

• 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

• When a current I1 flows in the primary coil (N1), it produces a magnetic field B1,

• Magnetic flux,

Mutual inductance, M between two coaxial solenoids

lIN

B 1101

l

I1 I1

N1N2

A

N1: primary coil

N2: secondary coil

0cos11 ABl

AIN 1101

Page 70: Ch20 Electromagnetic Induction

70

• If no 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

Mutual inductance, M between two coaxial solenoids

21

lAIN

INM 110

1

212

1

2212 I

NM

lANN

MM 21012

Page 71: Ch20 Electromagnetic Induction

71

Solution:N1= 1000; l = 50×10-2 m;d1 = 3×10-2 m, N2 = 50;

(a)Using

(b)In secondary coil,

Primary coil of a cylindrical former with the length of 50 cm and diameter 3 cm has 1000 turns. If the secondary coil has 50 turns, calculate:(a)its mutual inductance(b)the induced emf in the

secondary coil if the current flowing in the primary coil is changing at the rate of 4.8 A s-1.

Example 20.2111 84 As.

dtdI

lANNM 210

HM 51088.8

dtdIM 1

2

V41025.4

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72

Solution:(a)Using

(b)Using

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

e.m.f. in secondary coil if the current in primary coil changes at the rate 4.8 A s-1.

Example 20.22

P

PS

lANNM 0

dtdIM P

S

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73

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.

Example 20.23

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74

(b)The mutual inductance of the coils is given by

(a)The self-inductance of coil C is given by

and the energy stored in C is

Solution Wb;25.0 Wb;75.0 A; 0.3 DCC I

turns5000 turns;1000 DC NN

C

CCC I

NL

0.375.01000

C L

H 250C L

2CCC 2

1 ILU

20.325021

J 1125C U

C

DD

INM

0.3

25.05000

H 417M

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75

Given The induced emf in coil D is given by

Solution Wb;25.0 Wb;75.0 A; 0.3 DCC I

turns5000 turns;1000 DC NN A 0.30.30 s; 25.0 C dIdt

dtdI

M CD

25.0

0.3417

V 5004D

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Next Chapter…CHAPTER 21 :Alternating current