37

Electro MagnetismUNIT 2 ELECTRO MAGNETISM

Structure 2.1 Introduction

Objectives

2.2 Magnetic Circuit 2.3 Magnetisation Curve or B-H Curve 2.4 Composite Series Magnetic Circuit 2.5 Parallel Magnetic Circuit 2.6 Magnetic Hysteresis 2.7 Self Inductance 2.8 Mutual Inductance 2.9 Coefficient of Coupling 2.10 Dot Convention 2.11 Inductive Coupling in Series

2.11.1 Series Aiding

2.11.2 Series Opposing

2.12 Inductive Coupling in Parallel 2.12.1 Parallel Aiding

2.12.2 Parallel Opposing

2.13 Summary 2.14 Answers to SAQs

2.1 INTRODUCTION

Electromagnetism describes the relationship between electricity and magnetism. Nearly everyone, at some time or another, has had the opportunity to play with magnets. Most of us are acquainted with bar magnets or those thin magnets that usually end up on refrigerators. These magnets are known as permanent magnets. Although permanent magnets receive a lot of exposure, we use and depend on electromagnets much more in our everyday lives. Electromagnetism is essentially the foundation for all of electrical engineering. We use electromagnets to generate electricity, store memory on our computers, generate pictures on a television screen, diagnose illnesses, and in just about every other aspect of our lives that depends on electricity. Electromagnetism is the key to the operation of a large number of electrical devices used in home as well as industry. Transformers, motors, generators, circuit breakers and relays are some examples of widely used electromagnetic devices.

Electromagnetism works on the principle that an electric current through a wire generates a magnetic field. Whenever electric current flows through a conductor, a magnetic field is produced in the space surrounding the conductor. This magnetic field is the same force that makes metal objects stick to permanent magnets. In a bar magnet, the magnetic field runs from the north to the south pole. In a wire, the magnetic field forms around the wire. If we wrap that wire around a metal object, we can often magnetize that object. In this way, we can create an electromagnet.

In this unit, we will discuss about electromagnetism and study the relationship that exists between electric current and magnetic flux. Magnetic circuit provides path for magnetic flux and it is the basis of electromagnetism. We shall study about various types of magnetic circuits which may be series or parallel.

38

The way in which electric current magnetizes a magnetic material will be discussed in magnetization curve or B-H curve. The magnetization curve caused by an increasing current is not the same when the current decreases and this phenomenon is called magnetic hysteresis and this will also be discussed.

Electrical

A current carrying coil has self inductance which is a property by which it can store energy. The inductance between two coils is called mutual inductance. We shall elaborate on these with examples. When two coils are coupled together, flux in one coil is linked with the other. The coefficient of coupling between two coils gives idea about the degree of this linking. We shall study about it. When two coils are mutually coupled, their emfs may be aiding or opposing and it is determined by dot convention which will be illustrated. Two current carrying coils may be connected in series or parallel. Series and parallel connection may again be aiding or opposing. We shall study about them in details. Objectives After studying this unit, you should be able to

explain what is meant by a magnetic circuit and understand the magnetization curve. You should be able to write the equations for a series magnetic circuit made of different materials and also having air-gap in it,

differentiate between parallel and series-parallel magnetic circuit, write the emf equation for a current carrying coil and find the self-

inductance,

determine the relationship between self-inductance, mutual inductance and co-efficient of coupling, and

determine the effective inductance for inductors coupled in series and inductors coupled in parallel, each with aiding and opposing combination.

2.2 MAGNETIC CIRCUIT

You know that electric circuit provides a path for electric current. Similarly, magnetic circuit provides a path for magnetic flux.

Example of a magnetic circuit is shown in Figure 2.1. Here we are considering an iron ring having a magnetic path of l meter, cross sectional area A m2 and a coil of N turns carrying I Amperes wound on it as shown in Figure 2.1.

Figure 2.1 : Magnetic Circuit

Now we shall get acquainted with some important terms related to a magnetic circuit. These are magnetomotive force, magnetic field intensity, flux density, permeability, reluctance, etc. Magnetic field is measured in terms of flux which has unit weber (Wb). Flux density

(B) is the flux per unit area, i.e. BA= . Magnetomotive Force (m.m.f.) is the force which

39

Electro Magnetismdrives flux through a magnetic circuit. Unit of m.m.f. is Ampere Turns (AT) and it is

defined as the product of current and number of turns in a magnetic circuit. So, Magnetomotive force (m.m.f) NI= AT. Magnetomotive force per unit length of magnetic flux path is known as magnetic field intensity (H) and it is defined as

AT/mNIHl

= .

So, 200 Wb/mr

rNIB H

l = = . . . (2.1)

0 is the absolute permeability = 4 10 7 henry/m and r is the relative permeability of the medium.

Total Flux produced 0 WbrA NIB A

l = = . . . (2.2)

0

Wb

r

NIl

A

=

. . . (2.3)

Reluctance (S)

It is the property of a material which opposes the passage of magnetic flux in it.

Reluctance 0 r

lSA

= AT/Wb . . . (2.4)

From Eq. (2.3),

m.m.fFlux =reluctance

or, WbNIS

= . . . (2.5) Permeance

It is the reciprocal of reluctance and its unit is AT/Wb.

Example 2.1

A coil is wound uniformly with 300 turns over a steel ring of relative permeability 900, having a mean circumference of 40mm and cross-sectional area of 50 mm2. If a current of 25A is passed through the coil, find (a) m.m.f, (b) reluctance of the ring, and (c) flux.

Solution

(a) Coil m.m.f = NI AT = 300 25 = 7500 AT.

(b) Reluctance of the ring 3

7 60

40 104 10 900 50 10r

lA

= = = 0.707 106 AT/Wb.

(c) 6m.m.f 7500Flux 10.6 m Wb

reluctance 0.707 10= = = .

2.3 MAGNETISATION CURVE OR B-H CURVE

If you want to magnetize a magnetic material like iron, you need to apply a magnetizing force. This can be done by placing the unmagnetised iron within the field of a solenoid. In Figure 2.2 we have an electrical circuit which is used to magnetise an iron bar AX. If current I in the circuit is increased by varying the variable resistance R, the field

40

H (= NI/l) produced by the solenoid is increased. Correspondingly the flux density B in the iron bar AX is also increased. If flux density B is plotted as a function of field intensity H, we get magnetization curve or B-H curve. The approximate magnetization curves of a few magnetic materials are shown in Figure 2.3.

Electrical

Figure 2.2 : Magnetising Circuit

2000 4000 6000 8000 10000 12000 14000

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0

H in AT/m

Flux Density B in Wb/m2

Cast Iron

Wrought Iron

Steel Casting

Figure 2.3 : Magnetisation Curve or B-H Curve

2.4 COMPOSITE SERIES MAGNETIC CIRCUIT

A magnetic circuit may consist of different magnetic materials. Now let us consider a composite series magnetic circuit consisting of three different magnetic materials A, B and C of different permeabilities as well as lengths and air gap as shown in Figure 2.4. Here the total reluctance will be the sum of individual reluctances as each path has its own reluctance.

Figure 2.4 : Composite Series Magnetic Circuit

41

Electro Magnetism

Total reluctance 0 r

lA

=

1

31 210 0 2 2 0 3 3 0

a

r r r

l ll l

gA A A= + + + A . . . (2.6)

Example 2.2

A ring is composed of three sections. The cross-sectional area is 0.001 m2 for each section. The mean lengths of each section are la = 0.3 m, lb = 0.2m, lc = 0.1 m. An air-gap length of 0.1 mm is cut in the ring. Relative permeabilities for sections a, b, c are 5000, 1000 and 10000 respectively. Flux in the air gap is 7.5 10 4 Wb and the coil has 100 turns. Find (a) Total m.m.f, and (b) exciting current.

Solution

Reluctance of section a of ring

70

0.3 47746.37 AT/Wb4 10 5000 0.001

aa

ra

lSA

= = =

Reluctance of section b of ring

70

0.2 159154.57 AT/Wb4 10 1000 0.001

bb

rb

lSA

= = =

Reluctance of section c of ring

70

0.1 7957.73 AT/Wb4 10 10000 0.001

cc

rc

lSA

= = =

Reluctance of air-gap

3

70

0.1 10 79577.73 AT/Wb4 10 0.001

gg

lS

A

= = =

Total Reluctance

294436.4 AT/Wba b c gS S S S S= + + + =(a) Total m.m.f = Flux Reluctance = 7.5 10 4 Wb 294436.4 AT/Wb

= 220.83 AT.

(b) Exciting Current = m.m.f 220.83 2.21 AmpNo. of turns 100

= =

SAQ 1 (a) A ring has a diameter of 21cm and a cross-sectional area of 10cm2. The ring

is made up of semicircular sections of cast iron and cast steel, with each joint having an air-gap of 0.2 mm. Find the ampere-turns required to produce a flux of 8 10 4 Wb. The relative permeabilities of cast iron and cast steel are 166 and 800 respectively.

(b) A circular iron ring, having a cross-sectional area of 10 cm2 and a length of 4 cm in iron, has an air gap of 0.4 mm made by a saw-cut. The relative permeability of iron is 103. The ring is wound with a coil of 2000 turns and carries 2 mA current. Determine the air gap flux.

42

Electrical 2.5 PARALLEL MAGNETIC CIRCUIT

Parallel magnetic circuit consists of two parallel magnetic paths acted upon by the same m.m.f. as shown in Figure 2.5. Here ACB and ADB are the two magnetic paths acted upon by the same m.m.f. Each magnetic path has an average length of 2 (l1 + l2).

/2 /2

l1 l1

l2 l2

A

B

CD

Figure 2.5 : Parallel Magnetic Circuit

Example 2.3

For the magnetic circuit shown in Figure 2.6 the flux in the right limb is 0.48 m Wb and the number of turns wound on the central limb is 1000. Calculate (a) flux in the central limb, and (b) the current required.

The magnetiSation curve for the core is given as below :

H (AT/m) 200 400 500 600 800 1060 1400

B (AT/m) 0.4 0.8 1.0 1.1 1.2 1.3 1.4

Figure 2.6

Solution

Area of cross-section in side limbs = 4 3 =12 cm2

Area of cross-section in core = 4 3 = 12 cm2

Flux in side limbs = 0.48 m Wb

Flux density in side limbs 3

24

0.48 10 0.4 Wb/m12 10

= =

For the flux density of 0.4 Wb/m2, H = 200 AT/m

Since the coil is wound on the central limb and the magnetic circuit is symmetrical, flux in the central limb = 2 0.48 = 0.96 m Wb.

43

Electro Magnetism

Flux density in the central limb 3

24

0.96 10 0.8 Wb/m12 10

= =

For the flux density of 0.8 Wb/m2, H = 400 AT/m

Side limb has a path of length 50 cm.

Central limb has a path of length 20 cm.

Total m.m.f required = H length = 200 50 10 2 + 400 20 10 2 = 180 AT

Current required 180 0.18 Amp1000

= = .

SAQ 2 A cast steel magnetic structure made of a bar of cross-section 4 cm2 is shown in Figure 2.7. Find the current that the 500 turn magnetizing coil on the left limb should carry so that a flux of 2 mWb is produced in the right limb. Iron has r 600.

Figure 2.7

2.5.1 Series-Parallel Magnetic Circuit After knowing series and parallel magnetic circuits, we now take example of a combination of the two. Such a circuit is shown in Figure 2.8. It has two parallel magnetic circuits ACB and ADB connected across the common magnetic path AB which contains air gap of length lg. The flux in the common core is divided equally at point A between the two parallel paths which have equal reluctance. The reluctance of the path AB consists of (i) air gap reluctance, and (ii) the reluctance of the central core.

/2 /2

A

B

CD Airg

ap

Figure 2.8 : Series-Parallel Magnetic Circuit

Example 2.4

Figure 2.9 shows an electromagnet made of cast steel has a coil of 500 turns wound on the central limb. The cross-sectional area of the outer limbs and central limb are 2.5 cm2 and 6 cm2 respectively. Determine the current that the coil should carry to produce a flux of 0.6 m Wb in the air-gap. The magnetization curve for cast steel is as given below :

Flux Density B (Wb/m2)

0.2 0.5 0.7 1.0 1.2

H (AT/m) 300 540 650 900 1150

44

Electrical

Figure 2.9

Solution There are two equal parallel paths ACDE and AGE. Flux density in either parallel path is half of that in the central path as flux divides into two equal parts at point A. Total m.m.f required for the whole electromagnet = m.m.f required for path EF + m.m.f required for the air-gap + m.m.f required for either of the two parallel paths, say path ACDE.

Flux density in the central limb and air-gap 3

24

0.6 10 1 Wb/m6 10

= = .

Corresponding value of H found from the given data = 900 AT/m. So, AT for central limb = 900 25 10 2 = 225 AT

H in the air-gap 3

70

0.6 10 477.46 AT/m4 10

B

= = =

So, AT required for air-gap = 477.46 0.8 10 2 = 3.82 AT. Flux density in the path ACDE is 0.5 Wb/m2 and corresponding H is 540 AT/m. So, AT required for path ACDE = 540 100 10 2 = 540 AT Total AT required = 225 + 3.82 + 540 = 768.82

Current required 768.82 1.54 Amp500

= = . SAQ 3

A magnetic circuit made of wrought iron is shown in Figure 2.10. The central limb has a cross-sectional area of 8 cm2 and each side of the limb has a cross-sectional area of 5 cm2. Calculate the ampere turns required to produce a flux of 1 m Wb in the central limb. The magnetization of wrought iron is given by :

Flux Density (Wb/m2) 1.00 1.25

Magnetic Field (AT/m) 200 500

15cm

1mm5cm2

8cm2

AC

D E

F

G

Figure 2.10

45

Electro Magnetism2.6 MAGNETIC HYSTERESIS

You have seen in the plot of Figure 2.3 that if you want to magnetize a magnetic material, you need to apply a magnetizing force by a current flow through the coil. When you increase magnetic field intensity H from zero to a certain maximum value by increasing current I, the flux density B also increases and reaches saturation as shown by plot OA in Figure 2.11. The material becomes magnetically saturated for H = OM and has a corresponding maximum flux density of Bmax in it. If you now decrease H from its maximum value to zero by decreasing the current I, the flux density B will not decrease along AO, but will decrease less rapidly along AC. When H is zero, B is not zero but has a definite value OC. This means that even on removing the magnetizing force H, the magnetic material does not lose its magnetism completely and this is called residual magnetism.

If we want to demagnetize the electromagnet, we shall have to apply magnetizing force in the reverse direction. When we reverse the current through the coil, then B becomes zero at point D where H = OD.

If the value of H is further increased in the negative direction, the magnetic material again reaches magnetic saturation, but in negative direction. Now by taking H back from its value corresponding to negative saturation (OL) to its value for positive saturation (OM), a similar curve EFGA is obtained once again.

You must be observing from the Figure 2.11 that B always lags behind H. The magnetization curves for increasing and decreasing H are not the same and this phenomenon is called hysteresis. The loop traced out by the magnetization curve is known as hysteresis loop.

Figure 2.11 : Hysteresis Loop

2.7 SELF INDUCTANCE

You know that when current flows in a coil, flux is produced. If the current is changing with time, the flux linked with the coil is also time varying and an electromotive force (e.m.f.) is induced given by the law

diedt

or, die Ldt

= . . . (2.6)

46

In Eq. (2.6), the constant of proportionality L is known as self-inductance of the coil and its unit is Henry.

Electrical

According to Faradays law of electromagnetic induction, the induced e.m.f. in a coil having N turns is given by

de Ndt= . . . (2.7)

From Eqs. (2.6) and (2.7), we can write

di dL Ndt dt

=

or, dL Ndt= . . . (2.8)

As versus i is linear, Eq. (2.8) can be expressed as

L Ni= . . . (2.9)

Since we know from Eq. (2.5) that

0

0

r

r

N ANi Ni ilS lA

= = =

So, 0 rN A

i l = . . . (2.10)

From Eqs. (2.9) and (2.10) we now obtain 2

0 rN A N ALl l

= = . . . (2.11)

where 0 r = SAQ 4

(a) A coil of 150 turns is linked with a flux of 0.01 Wb when carrying a current of 10 A. Calculate the inductance of the coil. If the current is uniformly reversed in 0.01 sec, calculate the induced e.m.f.

(b) An iron rod 2 cm in diameter and 20 cm long is bent into a closed ring and is wound with 3000 turns of wire. It is found that when a current of 0.5 A is passed through this coil, the flux density in the coil is 0.5 Wb/m2. Assuming that all the flux is linked with every turn of the coil, find (a) the B/H ratio for iron, (b) the inductance of the coil, and (c) the voltage that would be developed across the coil if the current through the coil is interrupted and the flux in the iron falls to 10% of its former value in 0.001 sec?

2.8 MUTUAL INDUCTANCE

Two circuits may be coupled magnetically as shown in Figure 2.12. Here coil1 and coil 2 have magnetic coupling in between them. Coil1 has N1 number of turns and coil 2 has N2

47

Electro Magnetismnumber of turns. Alternating source Vs establishes an alternating current i1 which

produces an alternating flux 1 in coil1. 11 is that portion of flux 1 which completes its magnetic path around coil1 and the other part 12 is the mutual flux linked with coil 2.

r1 r2

Coil 1 N1 Turns

Coil 2 N1 Turns

Figure 2.12 : Magnetic Coupling

So, 1 = 11 + 12 . . . (2.12)

The mutual flux 12, which is alternating in nature, produces an induced e.m.f e2 in coil 2 according to Faradays law of electromagnetism,

122 2

de Ndt= . . . (2.13)

Again e2 is proportional to the rate of change of i1, i.e.

12

diedt

. . . (2.14)

or, 12 12die Mdt

= . . . (2.15)

where M12 is the constant of proportionality known as mutual inductance between two coils. Its unit is Henry (H).

From Eqs. (2.13) and (2.15) we get,

12 12 12d dN M

dt dt = i . . . (2.16)

or, 1212 21

dM Ndi= . . . (2.17)

The induced voltage e2 produced in coil 2 causes alternating current i2 to flow in coil 2. Current i2 produces an alternating flux 2 in coil 2. Out of flux 2, a portion 22 completes its path around the coil. The other portion 21 is linked with coil 1 and it produces an induced e.m.f given by

21 21 21

d dN Mdt dt

i = . . . (2.18)

So, from Eq. (2.18), we get,

2121 1

2

dM Ndi= . . . (2.19)

If the flux and current are having a linear relationship which we assume to be true, Eq. (2.17) and Eq. (2.19) can be written as

1212 2

1M N

i= . . . (2.20)

48

Electrical and 2121 1

2M N

i= . . . (2.21)

Assuming that the permeability of the mutual flux path is constant, we have

12 21M M M= = Example 2.5

A flux of 0.5 m Wb is produced by coil A of 600 turns wound on a ring with a current of 2 A in it. Calculate (a) self inductance of coil A, (b) the e.m.f induced in coil A when a current of 6 A flowing through it is switched off, assuming the current to fall to zero in 2 millisecond, and (c) the mutual inductance between the coils, if a second coil B of 400 turns is uniformly wound over the first coil A.

Solution

(a) Self inductance of coil 3

11

1

600 0.5 10 0.15 H2

A Ni

= = =

(b) 30.15 (6 0) 450 V

2 10die Ldt

= = =

(c) 3

12

1

400 0.5 10 0.1 H2

M Ni

= = = .

2.9 COEFFICIENT OF COUPLING

When two coils are coupled magnetically, it is important to know how much of the flux produced by one coil is linking with the other. Coefficient of coupling gives an idea about that.

Let us again consider the magnetically coupled coils of Figure 2.12. Following Eq. (2.9), inductance of coil1 and coil 2 can be written as :

11 1

1L N

i= . . . (2.22)

22 2

2L N

i= . . . (2.23)

Let 1 be the total flux produced by current i1 in coil 1 and 12 be the part of 1 that is linked with coil 2.

Let 12 1 1k = From Eq. (2.20),

12 1 12 2

1 1

kM N Ni i = = . . . (2.24)

Similarly, let 2 be the total flux produced by current i2 in coil 2 and 21 be the part of 2 that is linked with coil1.

Let 21 2 2k = From Eq. (2.21),

21 2 21 1

2 2

kM N Ni i = = . . . (2.25)

49

Electro MagnetismMultiplying Eqs. (2.24) and (2.25), we get,

2 1 2 1 2 1 2 1 1 2 21 2 1 2 1 2

1 2 1 2

k k N N N NM k k k k L Li i i i

= = =

1 2 1 2 1 2M k k L L k L L= = . . . (2.26) where 1 2k k k= So, From Eq. (2.26), we obtain,

1 2

MkL L

= . . . (2.27)

Constant k is called coefficient of coupling and it is defined as the ratio of mutual inductance M to the square root of the product of inductances of coil 1 and coil 2.

If k =1, you know that the flux due to one coil is fully linked with the other. If k = 0, the flux in one coil does not link with the other coil at all.

Example 2.6

The number of turns in two coupled coils A and B are 600 and 1700 respectively. When a current of 6 A flows in coil B, the total flux in this coil is 0.8 m Wb and the flux linking the first coil is 0.5 m Wb. Calculate self inductances of coils A and B, mutual inductance between the coils and coefficient of coupling.

Solution

N1 = 600, N2 = 1700, i2 = 6 A, 2 = 0.8 m Wb, 21 = 0.5 m Wb 3

22 2

2

1700 0.8 10 0.227 H6

L Ni

= = =

3

213

2

0.5 10 0.6250.8 10

k

= = =

Self inductance 2N ALl=

21

1N AL

l=

22

2N AL

l=

22 2

21 1

L NL N

=

2 21

1 2 2 22

(600)0.227 0.028 H(1700)

NL LN

= = =

Mutual Inductance

1 2 0.625 0.028 0.227 0.05 HM k L L= = = SAQ 5

The coefficient of coupling between two coils is 0.75. There are 250 turns in coil 1. The total flux of coil 1 is 0.4 m Wb when the current in this coil is 3 A. When current in coil1 is changed from 3 A to zero linearly in 3 milliseconds, the voltage induced in coil 2 is 70 V. Calculate L1, L2, M and N2.

50

Electrical 2.10 DOT CONVENTION

In a circuit, the e.m.f induced due to mutual inductance may be aiding or opposing the e.m.f induced due to self inductance. In order to determine the nature of mutually induced e.m.f, dot convention is useful. With the help of this dot convention, we can know about the sign of mutually induced e.m.f as illustrated in Figure 2.13.

M Negative

Figure 2.13 : Dot Convention

2.11 INDUCTIVE COUPLING IN SERIES

When two inductors are coupled in series, the series combination may be such that the mutual inductance existing between them is either aiding or opposing. Let us find the equivalent inductance of the series coupling for both these cases.

2.11.1 Series Aiding Figure 2.14 shows two coils coupled in series aiding. Let L1 be the self-inductance of coil 1, L2 be the self-inductance of coil 2 and M be the mutual inductance between the two coils.

Figure 2.14 : Inductive Coupling in Series (Flux Aiding)

Let us consider coil 1 first.

Coil 1

Self-induced e.m.f 1 1die Ldt

= . . . (2.28)

Mutually induced e.m.f 1die Mdt

= [due to change of current in coil 2]. . . (2.29)

Coil 2

Let us consider coil 2 now.

Self-induced e.m.f 2 2die Ldt

= . . . (2.30)

51

Electro Magnetism

Mutually induced e.m.f 2die Mdt

= [due to change of current in coil 1]. . . (2.31) Therefore the total induced e.m.f of the series coupled coils can be written as

1 2 1 2e e e e e = + + +

or, 1 2 1 22 ( 2di di di die L L M L L Mdt dt dt dt

= = + + ) . . . (2.32) If L is the equivalent inductance of the coil, then we can write

die Ldt

= . . . (2.33) From Eqs. (2.32) and (2.33), we get

or, 1 2( 2di diL L L Mdt dt

= + + ) or, . . . (2.34) 1 2 2L L L M= + +

2.11.2 Series Opposing Figure 2.15 shows two coils coupled in series opposition as their fluxes are in opposite direction as per dot convention.

Figure 2.15 : Inductive Coupling in Series (Flux Opposing)

Let us consider coil 1 and coil 2 individually.

Coil 1

Self-induced e.m.f 1 1die Ldt

= . . . (2.35)

Mutually induced e.m.f 1die Mdt

= [due to change of current in coil 2] . . . (2.36) Coil 2

Self-induced e.m.f 2 2die Ldt

= . . . (2.37)

Mutually induced e.m.f 2die Mdt

= [due to change of current in coil 1] . . . (2.38) Therefore the total induced e.m.f of the combination can be written as

1 2 1 2e e e e e = + + +

1 2 1 22 ( 2di di di die L L M L L Mdt dt dt dt

= + = + ) . . . (2.39) If L is the equivalent inductance of the combination, then we can write

die Ldt

= . . . (2.40)

52

From Eqs (.2.39) and (2.40), we get Electrical

1 2( 2di diL L L Mdt dt

= + ) or, 1 2 2L L L M= + . . . (2.41) Example 2.7

Two coils with a coefficient of coupling of 0.6 between them are connected in series so as to magnetize (i) in the same direction, and (ii) in the opposite direction. The corresponding values of equivalent inductance are obtained as 1.8 H for (i) and 0.8 H for (ii). Find the self inductance of the two coils and the mutual inductance between them.

Solution

Coefficient of coupling k = 0.6

(a) For magnetization in the same direction,

Equivalent inductance L = L1+ L2 + 2M

or, 1.8 = L1+L2 + 2M . . . (2.42)

(b) For magnetization in the opposite direction,

Equivalent inductance L = L1 + L2 2M or, 0.8 = L1 + L2 2M . . . (2.43) Subtracting Eq. (2.43) from Eq. (2.42), we get

4M = 1

or, M = 0.25 H

Adding Eq. (2.42) and Eq. (2.43), we get

2 (L1 + L2) = 2.6

or, L1 + L2 = 1.3 . . . (2.44)

We know, 1 2M k L L=

2 2

1 2 2 2(0.25) 0.1736(0.6)

ML Lk

= = = . . . (2.45)

Now (L1 L2)2 = (L1 + L2)2 4 L1 L2 = (1.3) 2 4 0.1736 = 0.9956 1 2 0.9978L L = . . . (2.46) Adding Eq. (2.44) and Eq. (2.46), we get

L1 = 1.1489 H

2 11.3 1.3 1.1489 0.1511 HL L= = = SAQ 6

The combined inductance of two coils connected in series is 0.6 H and 0.1 H depending on the relative directions of the current in the coils. If one of the coils when isolated has a self-inductance of 0.2 H, calculate (a) mutual inductance, and (b) coupling coefficient.

53

Electro Magnetism2.12 INDUCTIVE COUPLING IN PARALLEL

When two inductors are coupled in parallel, the parallel combination may be such that the mutual inductance existing between them is either aiding or opposing. Let us find the equivalent inductance of the combination in both these cases.

2.12.1 Parallel Aiding Figure 2.16 shows two coils coupled in parallel where the fluxes are additive as per dot convention.

Figure 2.16 : Inductive Coupling in Parallel (Flux Aiding)

Using Kirchoffs voltage law, we can write

1 21

di diV L Mdt dt

= + . . . (2.47)

Also, 2 12di diV L Mdt dt

= + . . . (2.48) From Eqs. (2.42) and (2.43), we get

1 2 21 2

di di di diL M L Mdt dt dt dt

+ = + 1

1

. . . (2.49)

Now 1 2i i i= + . . . (2.50) 2i i i= Substituting i2 from Eq. (2.45) in Eq. (2.44), we get

1 1 11 2

( ) ( )di d i i d i i diL M L Mdt dt dt dt

+ = + 1

or, 1 1 11 2 2di di di didi diL M M L L Mdt dt dt dt dt dt

+ = + 1

or, 1 11 2 2( 2 ) ( )di diL L M L Mdt dt

+ =

2 21 2

( )2

di L M didt L L M dt

= + 1 . . . (2.51)

Similarly, 2 11 2

( )2

di L M didt L L M dt

= + . . . (2.52)

Using Eqs. (2.46) and (2.47) in Eq. (2.42), we obtain

2 11

1 2 1 2

)2 2

L M L Mdi diV L ML L M dt L L M d

= + + + t . . . (2.53)

or, 2

1 2 1 1

1 2 2L L L M L M M diV

L L M dt + = + . . . (2.54)

54

Electrical or,

21 2

1 2 2L L M diV

L L M dt= + . . . (2.55)

Let L be the equivalent inductance of the parallel combination, then we can write

diV Ldt

= . . . (2.56) From Eqs. (2.50) and (2.51), we obtain

21 2

1 2 2L L Mdi diL

dt L L M dt = +

2

1 2

1 2 2L L ML

L L M= + . . . (2.57)

2.12.2 Parallel Opposing Figure 2.17 shows two coils coupled in parallel where the mutually induced e.m.f in a coil due to change in current in the other coil is negative.

V L1 L2M

i

i1 i2

Figure 2.17 : Inductive Coupling in Parallel (Flux Opposing)

Using Kirchoffs voltage law, we can write

1 21

di diV L Mdt dt

= . . . (2.58)

Also, 2 12di diV L Mdt dt

= . . . (2.59) From Eqs. (2.53) and (2.54), we get

1 2 21 2

di di di diL M L Mdt dt dt dt

= 1 . . . (2.60)

Now 1 2i i i= + 2i i 1i= . . . (2.61)

Substituting i2 from Eq. (2.56) in Eq. (2.55), we get

1 1 11 2

( ) ( )di d i i d i i diL M L Mdt dt dt dt

1 =

or, 1 1 11 2 2di di di didi diL M M L L Mdt dt dt dt dt dt

+ = 1

or, 1 11 2 2( 2 ) ( )di diL L M L Mdt dt

+ + = +

1 21 2

( )2

di L M didt L L M dt

+= + + . . . (2.62)

55

Electro Magnetism

Similarly, 2 11 2

( )2

di L M didt L L M dt

+= + + . . . (2.63)

Using Eqs. (2.57) and (2.58) in Eq. (2.53), we obtain

2 11

1 2 1 2

)2 2

L M L Mdi diV L ML L M dt L L M dt

+ += + + + +

. . . (2.64)

or, 2

1 2 1 1

1 2 2L L L M L M M diV

L L M dt+ = + + . . . (2.65)

or, 2

1 2

1 2 2L L M diV

L L M d= + + t . . . (2.66)

Let L be the equivalent inductance of the parallel combination, then we can write

diV Ldt

= . . . (2.67) From Eqs. (2.61) and (2.62), we obtain

21 2

1 2 2L L Mdi diL

dt L L M dt = + +

2

1 2

1 2 2L L ML

L L M= + + . . . (2.68)

Example 2.8

Two coils of inductances 4H and 6H are connected in parallel. If their mutual inductance is 3 H, calculate the equivalent inductance of the combination if (i) mutual inductance assists the self-inductance, and (ii) mutual inductance opposes the self-inductance.

Solution

L1 = 4 H L2 = 6 H M = 3 H

(a) 2 2

1 2

1 2

4 6 3 3.75 H2 4 6 2 3

L L MLL L M

= = =+ +

(b) 2 2

1 2

1 2

4 6 3 0.9375 H2 4 6 2 3

L L MLL L M

= = =+ + + +

SAQ 7 A coil of inductance 200 mH is magnetically coupled to another coil of inductance 800 mH. The coefficient of coupling between the coils is 0.5. Calculate the equivalent inductance of (a) series aiding, (b) series opposing, (c) parallel aiding, and (d) parallel opposing.

2.13 SUMMARY

In this unit we studied about electromagnetism and the relationship that exists between electric current and magnetic flux. The way in which electric current magnetizes a

56

magnetic material was discussed in magnetization curve or B-H curve. In relation to B-H curve, phenomenon of magnetic hysteresis was also discussed.

Electrical

Magnetic circuit provides path for magnetic flux and it is the basis of electromagnetism. Types of magnetic circuits which may be series or parallel were discussed.

A current carrying coil has self inductance and the inductance between two coils is called mutual inductance. We had elaborate discussion on these with examples. The coefficient of coupling between two coils determines the degree of linking between their fluxes and it was studied.

Dot convention determines whether e.m.fs between two mutually coupled coils are aiding or opposing and this was illustrated with examples.

Series or parallel connection between two current carrying coils was explained. Series and parallel connection may again be aiding or opposing. We studied about them in details.

2.14 ANSWERS TO SAQs

SAQ 1

(a) Total ampere-turns (AT) required is the sum of the AT required in the air-gap, AT required in the cast iron path and AT required in the cast steel path. Length for air-gap is 2 0.2 mm, length for cast iron path and cast steel path are both Diameter/2 = 21/2 cm. Now using Eq. (2.1), AT for air-gap is found to be 255 as AT = H l where H = B/0 (B = /A, 0 = 4 10 7). Similarly, ATs for the cast iron path and cast steel path are found to be 1265 and 263 from Eq. (2.1) as AT = H l where H = B/0 r. Hence total AT required is 1783.

(b) Using Eq. (2.1), ampere turns required for the air-gap is found as AT = Bla/0 = 1000B (la is air gap length). Length of the iron ring, li = (4 0.04 ) cm. So ampere turn required for iron ring is AT = Bli/0 r = 99B. Total AT required is 1000 B + 99 B = 1099 B. Again given that AT applied is 2000 2 10 3 = 4, we have 1099B = 4 and B is obtained as 3.64 10 3 Wb/m2. So air gap flux = BA = 3.64 10 4 Wb.

SAQ 2

In Figure 2.5, path C and path D are parallel with each other with respect to path E. So the m.m.f across the two paths is the same. If flux produced in the left limb is , flux in limb C is 1 and flux in limb D is 2 (2 = 2 m Wb), then = 1 + 2. If S1 and S2 are reluctances of path C and path D, then m.m.f across C is 1 S1= m.m.f across D 1 S1. S = l/0 r A and we get from here 1 = 10/3 mWb. Thereby we get = 10/3 + 2 = 16/3 m Wb.

Flux density in path E = /A. Flux density in path D = 2/A.

Total AT required = AT required for path E + AT required for either of two paths

C or D. Using Eq. (2.1),

0

flux density length of magnetic pathAT r

=

We get, AT required for path E = 4420. AT required for path D = 1658.

Total AT = 6078 and current needed = 6078/500 = 12.16 A.

SAQ 3

Follow exactly as in Example 4, Flux density in the central limb and air-gap is found to be 1.25 Wb/m2. Corresponding H from the given table = 500 AT/m. AT required for the central limb is 75. AT for the air-gap is 994.72.

57

Electro MagnetismFlux divides equally at A in Figure 2.7 along two parallel paths. So flux through a

parallel path ACDE is 0.5 m Wb and flux density B = 1 Wb/m2. Corresponding value of H from the given table is 200 AT/m and AT required for path ACDE is 68.

Total AT required = 75 + 994.72 + 68 = 1137.72.

SAQ 4

(a) Applying Eq. (2.9) L Ni= , we get L = 0.15 H

Induced e.m.f die Ldt

= . Change of current is 20 A in 0.01 sec. So e = 300 V.

(b) From NiHl

= , we get H = 7500 AT/m. So 56.67 10 Henry/mBH

=

Diameter d = 2 cm and cross-sectional area 2

,4dA B A= = .

L Ni= . Substituting the values, we get L = 0.942 H.

Voltage across the coil de Ndt= .

d = 90% of the final flux, dt = 0.001 sec and N = 3000. Substituting these values, we get

e = 424.11 V

SAQ 5

From 1 111

NLi= , we get L2 = 33.34 mH.

Voltage induced in coil 2 is 12die Mdt

=

e2 = 70V, di1 = 3 A and dt = 3 m sec. Using these values, we get M = 70 mH.

1 2M k L L= and so 2

2 21

0.261 HMLk L

= =

From the relation 2

22

1 1

L NL

2

N= , we get 22 1

1

LN NL

= and thus obtain N2 = 700.

SAQ 6

L = 0.6 H when flux is additive and L = 0.1 H when flux is subtractive 0.6 = L1 + L2 + 2 M and 0.1 = L1 + L2 2 M. From these equations, subtracting we get M = 0.125 H and adding we have

0.35 = 0.2 + L2Let L1 = 0.2 H (when isolated), then we get

L2 = 0.15 H

Coupling Coefficient 1 2

0.72MkL L

= =

58

SAQ 7 Electrical

From 1 2M k L L= , we get M = 200 mH In series aiding L = L1 + L2 + 2 M = 1400 mH

In series opposing L = L1 + L2 2 M = 600 mH

In parallel aiding 2

1 2

1 2200 mH

2L L ML

L L M= =+

In parallel opposing 2

1 2

1 285.71 mH

2L L ML

L L M= =+ + .

UNIT 2 ELECTRO MAGNETISM Structure 2.1 INTRODUCTION 2.2 MAGNETIC CIRCUIT 2.3 MAGNETISATION CURVE OR B-H CURVE 2.4 COMPOSITE SERIES MAGNETIC CIRCUIT SAQ 1

2.5 PARALLEL MAGNETIC CIRCUIT SAQ 2 2.5.1 Series-Parallel Magnetic Circuit SAQ 3

2.6 MAGNETIC HYSTERESIS 2.7 SELF INDUCTANCE SAQ 4

2.8 MUTUAL INDUCTANCE 2.9 COEFFICIENT OF COUPLING SAQ 5

2.10 DOT CONVENTION 2.11 INDUCTIVE COUPLING IN SERIES 2.11.1 Series Aiding 2.11.2 Series Opposing SAQ 6

2.12 INDUCTIVE COUPLING IN PARALLEL 2.12.1 Parallel Aiding 2.12.2 Parallel Opposing SAQ 7

2.13 SUMMARY 2.14 ANSWERS TO SAQs