Magnetic Circuits

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School of Engineering

Robert Gordon University

EN1560-Introduction to Electrical Engineering

1

1 Magnetic

Circuits

1.1 1.1 Magnetic

Magnetic field

field

A

A permanent magnetpermanent magnet is a piece of ferromagnetic material, such as iron, which attracts otheris a piece of ferromagnetic material, such as iron, which attracts other pieces of the same material. If a permanent magnet is suspended in the air so that it is free to pieces of the same material. If a permanent magnet is suspended in the air so that it is free to swing in a horizontal plane, one end of the magnet will take up a position towards the earth’s swing in a horizontal plane, one end of the magnet will take up a position towards the earth’s North Pole. This end is called the

North Pole. This end is called the north seekingnorth seeking end or the end or the north pole,north pole, NN of the magnet. of the magnet. Similarly the other end is known as the

Similarly the other end is known as the south seekingsouth seeking end or the end or the south pole,south pole, SS of the of the magnet.

magnet.

The distribution of a magnetic field can be demonstrated by the following experiment. A The distribution of a magnetic field can be demonstrated by the following experiment. A permanent magnet is placed on a table, covered it by a sheet of cardboard and some iron permanent magnet is placed on a table, covered it by a sheet of cardboard and some iron filings are sprinkled uniformly over the sheet. A slight tapping of the cardboard will cause the filings are sprinkled uniformly over the sheet. A slight tapping of the cardboard will cause the filings to position themselves in curved lines between the poles as shown in

filings to position themselves in curved lines between the poles as shown in Fig(1.1)Fig(1.1). These. These curved lines can be used to visualise the magnetic condition of the space around the magnet, curved lines can be used to visualise the magnetic condition of the space around the magnet, which may be identified as the

which may be identified as the magnetic fieldmagnetic field. Also these lines lead to the idea of. Also these lines lead to the idea of lines oflines of magnetic flux

magnetic flux which were introduced by Michael Faraday to visualise the distribution and which were introduced by Michael Faraday to visualise the distribution and density of the magnetic field. They can also be used as a vehicle to explain various effects of density of the magnetic field. They can also be used as a vehicle to explain various effects of magnetism. It should be realised that the magnetic flux occupies the whole three-dimensional magnetism. It should be realised that the magnetic flux occupies the whole three-dimensional space in the vicinity of the magnet and decreases in strength as moved away from the space in the vicinity of the magnet and decreases in strength as moved away from the magnet.

magnet.

Each line of magnetic flux is a closed loop with no beginning and no end as shown in Each line of magnetic flux is a closed loop with no beginning and no end as shown in Fig(1.1)

Fig(1.1). In fact a flux line which starts at a point on the north pole of a magnet passes. In fact a flux line which starts at a point on the north pole of a magnet passes N

N SS

Fig(1.1) Lines of magnetic flux Fig(1.1) Lines of magnetic flux

Lines of magnetic flux Lines of magnetic flux

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through the space surrounding it, enters the south pole and continues through the magnet to through the space surrounding it, enters the south pole and continues through the magnet to the starting point thus forming a closed loop. This follows that these lines never intersect. the starting point thus forming a closed loop. This follows that these lines never intersect. When two magnets are arranged in such away that unlike poles are next to each other, as When two magnets are arranged in such away that unlike poles are next to each other, as shown in

shown in Fig(1.2)(a)Fig(1.2)(a), attraction takes place. The lines of flux passing between the two, attraction takes place. The lines of flux passing between the two magnets behave as if they were trying to shorten themselves causing the magnets to attract magnets behave as if they were trying to shorten themselves causing the magnets to attract towards each other. If the magnets are arranged so that the like poles are near to each other, towards each other. If the magnets are arranged so that the like poles are near to each other, as shown in

as shown in Fig(1.2)(b)Fig(1.2)(b), then repulsion takes place. It is seen that the flux lines in the space, then repulsion takes place. It is seen that the flux lines in the space between the two magnets are pointing in the same direction thus pushing the two magnets between the two magnets are pointing in the same direction thus pushing the two magnets away from each other.

away from each other.

1.2 1.2 Magnetic

Magnetic flux

flux

and

and flux

flux density

density B

B

The amount of magnetic field produced by a magnetic source is known as the

The amount of magnetic field produced by a magnetic source is known as the magnetic fluxmagnetic flux and the

and the symbol used symbol used is the is the Greek letter Greek letter . The . The unit of unit of magnetic flux magnetic flux is theis the weberweber oror WbWb.. The

The Magnetic flux density BMagnetic flux density B is defined as the amount of flux per unit area, which is is defined as the amount of flux per unit area, which is perpendicular to the direction of the flux. The unit of flux density is

perpendicular to the direction of the flux. The unit of flux density is teslatesla oror TT. Thus we can. Thus we can write the equation

write the equation

tesla

or

Wb.m

,,

Wb/m

22 22

A

B

Eq(1.1)Eq(1.1)

where A is the area in m

where A is the area in m22. It is seen from Eq(1.1). It is seen from Eq(1.1) that 1 T is equivalent to 1 Wb/(m that 1 T is equivalent to 1 Wb/(m22) which is) which is another way of defining the units of flux density B.

another way of defining the units of flux density B. Example 1.1

Example 1.1

The magnetic flux crossing the air gap of the magnet shown in

The magnetic flux crossing the air gap of the magnet shown in Fig(1.3)Fig(1.3) is 12 mWb. Determine is 12 mWb. Determine the flux density in the air gap if the magnet has dimensions shown.

the flux density in the air gap if the magnet has dimensions shown. Solution 1.1

Solution 1.1

Cross sectional area A is Cross sectional area A is

2 2 2 2 4 4 2 2 2 2

m

012

012 ..

0

0 m

m

10

120

10

15

10

8

8 A

A

This gives This gives N N SS N N S S

Fig(1.2)(a) Attraction between Fig(1.2)(a) Attraction between

magnets magnets N N NN S S S S

Fig(1.2)(b) Repulsion between Fig(1.2)(b) Repulsion between

magnets magnets

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N N SS Flux Flux Magnet Magnet 15 cm 15 cm 8 cm 8 cm

Fig(1.3) Magnet with air gap Fig(1.3) Magnet with air gap

Fig(1.4) Sign convention of Fig(1.4) Sign convention of

electromagnetic field electromagnetic field Wire Wire Clockwise Clockwise rotation of field rotation of field Anticlockwise Anticlockwise rotation of field rotation of field Fig(1.4)(a) Fig(1.4)(b) Fig(1.4)(a) Fig(1.4)(b)

T

mW

mWb

b

A

B

1

012

012 ..

0

012

012 ..

0

012

012 ..

0

12

1.3 1.3 Magnetic

Magnetic field

field due

due to

to an

an electric

electric current

current

A

A fundamental fundamental law law of of electromagnetism electromagnetism is is that that a a magnetic magnetic field field is is produced produced around around aa conductor when that conductor carries an electric current. This phenomenon was conductor when that conductor carries an electric current. This phenomenon was demonstrated by Oersted in 1820. He noticed that when a wire carrying an electric current is demonstrated by Oersted in 1820. He noticed that when a wire carrying an electric current is placed above a magnetic needle, the needle was deflected clockwise or anticlockwise placed above a magnetic needle, the needle was deflected clockwise or anticlockwise depending on the direction of the current flow. Using his observations it is possible to form a depending on the direction of the current flow. Using his observations it is possible to form a basic sign convention to indicate the direction of the magnetic field.

basic sign convention to indicate the direction of the magnetic field.

1.3.1 Sign

1.3.1 Sign convention of electromagne

convention of electromagnetic field

tic field

Consider a wire carrying an electric current which has a cross section as shown in

Consider a wire carrying an electric current which has a cross section as shown in Fig(1.4)Fig(1.4). In. In Fig(1.4)(a)

Fig(1.4)(a) the current is flowing into the paper as indicated by the cross. The magnetic field the current is flowing into the paper as indicated by the cross. The magnetic field has a clockwise direction and the concentric circles around the wire show the flux lines. has a clockwise direction and the concentric circles around the wire show the flux lines. Another

Another method method of of representing representing this this is is to to place place a a corkscrew corkscrew along along the the conductor, conductor, whichwhich travels into the paper when rotated clockwise. The movement of the corkscrew into the paper travels into the paper when rotated clockwise. The movement of the corkscrew into the paper represents the current flow and the clockwise rotation indicates the direction of the magnetic represents the current flow and the clockwise rotation indicates the direction of the magnetic field.

field. In

In Fig(1.4)(b)Fig(1.4)(b) the current flow is reversed, ie flowing out of the paper, which is indicated by the current flow is reversed, ie flowing out of the paper, which is indicated by the dot. In this case it is

the dot. In this case it is obvious that the direction of obvious that the direction of the field is anticlockwise and again fluxthe field is anticlockwise and again flux lines are shown by the concentric circles.

lines are shown by the concentric circles.

1.3.2

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Fig(1.5) Closed path enclosing a Fig(1.5) Closed path enclosing a

conductor conductor Closed path B Closed path B Conductor Conductor Closed path A Closed path A Unit pole Unit pole H H

The Ampere’s Law is particularly useful in determining magn

The Ampere’s Law is particularly useful in determining magn etic field strength near currentetic field strength near current carrying conductors in certain geometrical arrangements. Knowing the field strength the carrying conductors in certain geometrical arrangements. Knowing the field strength the magnetic flux density at a point and the magnetic flux around a circuit can easily be magnetic flux density at a point and the magnetic flux around a circuit can easily be determined. In electrical engineering problems such as electrical machines, transformers etc, determined. In electrical engineering problems such as electrical machines, transformers etc, we are often asked to design a magnetic circuit to produce a given flux. The application of the we are often asked to design a magnetic circuit to produce a given flux. The application of the law is straight forward provided that we know the direction of the flux and the law is most law is straight forward provided that we know the direction of the flux and the law is most suitable in situations where the field patterns are predictable.

suitable in situations where the field patterns are predictable.

The Ampere’s law is a statement of fact based on experiments. If a unit pole is placed on any The Ampere’s law is a statement of fact based on experiments. If a unit pole is placed on any irregular closed path, such as path A in

irregular closed path, such as path A in Fig(1.5)Fig(1.5), enclosing a current carrying conductor, it, enclosing a current carrying conductor, it experiences a force H, which is tangential to the path, as shown in

experiences a force H, which is tangential to the path, as shown in Fig(1.5)Fig(1.5).. When the unitWhen the unit pole is moved an infinitely small distance

pole is moved an infinitely small distance ℓℓ along the path the work done is the product of H along the path the work done is the product of H and

and ℓℓ. The Ampere’s law states that the sum of the product of H. The Ampere’s law states that the sum of the product of H ℓℓ, which is the total work, which is the total work done by the unit pole in moving once around the closed path A enclosing the conductor is done by the unit pole in moving once around the closed path A enclosing the conductor is numerically equal to the current flow in the conductor. This is written as

numerically equal to the current flow in the conductor. This is written as

))

A

((

II

H

0 0     _Eq(1.2)_Eq(1.2) In the limit as

In the limit as ℓℓ tends to zero this summation becomes and integral and is written as tends to zero this summation becomes and integral and is written as

))

A

((

II

H

0 0    _Eq(1.3)_Eq(1.3)

The circle around the integral sign indicates that the integration is done around a closed path. The circle around the integral sign indicates that the integration is done around a closed path. If the unit pole is moved around any path, regular or irregular, which encloses the conductor If the unit pole is moved around any path, regular or irregular, which encloses the conductor will produce the same result. However path B in

will produce the same result. However path B in Fig(1.5)Fig(1.5) fails to link the conductor and fails to link the conductor and therefore no work is done in moving a unit pole round such a path.

therefore no work is done in moving a unit pole round such a path.

The Ampere’s law is very simple to use when the tangential force H is a constant and t The Ampere’s law is very simple to use when the tangential force H is a constant and t his ishis is the case for all examples considered here. Thus if H is constant then

the case for all examples considered here. Thus if H is constant then Eq(1.3)Eq(1.3) becomes becomes

))

m

//

A

((

II

H

))

A

((

II

H

0 0         Eq(1.4) Eq(1.4)

This force H is called the

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Fig(1.6) Circular path around a Fig(1.6) Circular path around a

current carrying conductor current carrying conductor

Wire Wire Circular path of Circular path of length length ℓℓ = = 22 rr rr

As an example it is required to calculate the magnitu

As an example it is required to calculate the magnitu de of the field strength at a point distancede of the field strength at a point distance r from the axis of a long conductor carrying a current I. The field pattern for such a case is r from the axis of a long conductor carrying a current I. The field pattern for such a case is shown in

shown in Fig(1.4)Fig(1.4) and if we consider a circular path at radius r, the field strength along this and if we consider a circular path at radius r, the field strength along this path is tangential to the path and will be constant (as the field strength at any point on a flux path is tangential to the path and will be constant (as the field strength at any point on a flux line is constant). From

line is constant). From Fig(1.6)Fig(1.6) and and Eq(1.4)Eq(1.4) the field strength H at a distance r is given asthe field strength H at a distance r is given as

))

m

//

A

((

r

2

2 II

II

H

  Eq(1.5)Eq(1.5)

1.4 Magnetomotive force mmf F

mm

In an electric circuit the current is due to the existence of an electromotive force. In a similar In an electric circuit the current is due to the existence of an electromotive force. In a similar manner the magnetic flux in a magnetic circuit is due to the existence of a

manner the magnetic flux in a magnetic circuit is due to the existence of a magnetomotive magnetomotive force mmf

force mmf oror FFmm, caused by a current flowing through one or more turns. Thus a coil, as, caused by a current flowing through one or more turns. Thus a coil, as

shown in

shown in Fig(1.7)Fig(1.7), of N turns carrying a current of I Amps is the basic force for the creation of, of N turns carrying a current of I Amps is the basic force for the creation of magnetic fields.

magnetic fields.

Therefore we can write an equation for F Therefore we can write an equation for Fmm as as

A

AT

T))

or

turns

--(ampere

(ampere

NI

F

mmf

_m_m Eq(1.6)Eq(1.6)

and has the units ampere-turns. Since N has no units sometimes it is expressed in amperes. and has the units ampere-turns. Since N has no units sometimes it is expressed in amperes. The magnetomotive force is the total current linked with the magnetic circuit. If the magnetic The magnetomotive force is the total current linked with the magnetic circuit. If the magnetic circuit has a uniform cross section, the magnetomotive force per unit length of the magnetic circuit has a uniform cross section, the magnetomotive force per unit length of the magnetic

I Amp I Amp

N turns N turns

ll meters meters

Fig(1.7) A coil with N turns Fig(1.7) A coil with N turns

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Fig(1.8) B-H curve for free space Fig(1.8) B-H curve for free space B Wb/m B Wb/m H AT/m H AT/m Slope = Slope = 00 circuit is the magnetising force

circuit is the magnetising force or magnetic field strength discussed inor magnetic field strength discussed in Section 1.3.2Section 1.3.2.. As As shown in

shown in Fig(1.7)Fig(1.7) if if ℓℓ_{is the mean length (meters) of the magnetic circuit then magnetic field}_{is the mean length (meters) of the magnetic circuit then magnetic field} strength H is given as strength H is given as

))

m

//

A

((

NI

H

  Eq(1.7)Eq(1.7) Example 1.2 Example 1.2 A

A circular circular wooden wooden ring ring of of mean mean diameter diameter 20 20 cm cm has has a a coil coil of of 800 800 turns turns uniformly uniformly woundwound around it. If the magnetic field strength is 5000 A/m calculate the current in the coil.

around it. If the magnetic field strength is 5000 A/m calculate the current in the coil. Solution 1.2

Solution 1.2

The mean length (circumference) of the wooden ring is

The mean length (circumference) of the wooden ring is ℓℓ == d d = = 2200 1100-2-2 m and from m and from Eq(1.7)

Eq(1.7) we have we have

A

93

93 ..

3

8

10

800

10

20

20 5000

5000

N

H

II

2 2  

1.5 Permeability

and B-H curves

For free space or a non-magnetic material the ratio of magnetic flux density B to magnetic For free space or a non-magnetic material the ratio of magnetic flux density B to magnetic filed strength or magnetising force H is a constant. This constant is known as the filed strength or magnetising force H is a constant. This constant is known as the permeability for free space

permeability for free space and has the symboland has the symbol 00. ie. ie

mA

mAT

T

Wb

m

AT

m

Wb

H

B

//

10

4

2 2 7 7 0 0 Eq(1.8)Eq(1.8)

Note that this constant has a numerical value of 4

Note that this constant has a numerical value of 4 1010-7-7 Wb/(mAT) as given in Wb/(mAT) as given in Eq(1.8)Eq(1.8).. Another

Another unit unit for for this this constant constant is is henrys/m henrys/m and and the the interested interested reader reader can can look look elsewhere elsewhere forfor details. All non-magnetic materials are considered to have the same permeability

details. All non-magnetic materials are considered to have the same permeability 00 as free as free

space.

space. Fig(1.8) Fig(1.8) shows a graph of flux density B plotted against the magnetic field strength H shows a graph of flux density B plotted against the magnetic field strength H for free space known as a

for free space known as a B-H curveB-H curve. This curve is linear and the slope of the straight line is. This curve is linear and the slope of the straight line is

0 0..

The

The absolute permeabilityabsolute permeability of of a a ferromagnetic ferromagnetic material material is is expressed expressed in in relation relation to to thethe permeability of free space and is given

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20 20 4040 6060 8080 100100 0 0 0.2 0.2 0.4 0.4 0.6 0.6 0.8 0.8 1 1 1.2 1.2 1.4 1.4 1.6 1.6 1.8 1.8 Flux

Flux DensiDensity B (ty B (Wb/mWb/m ))

Magnetic Field Strength (AT/m) Magnetic Field Strength (AT/m) Cast Iron Cast Iron Cast Steel Cast Steel Mild Steel Mild Steel

Fig(1.9) B-H curves for different Fig(1.9) B-H curves for different

magnetic materials magnetic materials

))

/(

0 0

Wb

mA

mAT

T

H

B

r r Eq(1.9)Eq(1.9) where

where r r is theis the relative permeabilityrelative permeability of the ferromagnetic material which has no units. Fromof the ferromagnetic material which has no units. From

this definition it is seen that the relative permeability of free space is 1. this definition it is seen that the relative permeability of free space is 1.

By plotting measured values of B against H, a B-H curve is obtained and typical curves for By plotting measured values of B against H, a B-H curve is obtained and typical curves for three different magnetic materials are shown in

three different magnetic materials are shown in Fig(1.9)Fig(1.9). It is noted that these curves are not. It is noted that these curves are not linear any longer and the relative permeability

linear any longer and the relative permeability r r of the ferromagnetic material is proportional of the ferromagnetic material is proportional

to the slope of the B-H curve and therefore varies with the magnetic field strength H. to the slope of the B-H curve and therefore varies with the magnetic field strength H.

Example 1.3 Example 1.3 A coil of 200 tur

A coil of 200 tur ns is wound uniformly over ns is wound uniformly over a wooden ring having a wooden ring having a mean circumference of a mean circumference of 6060 cm and a uniform cross sectional area of 5 cm

cm and a uniform cross sectional area of 5 cm22. If the current through the coil is 4 A calculate. If the current through the coil is 4 A calculate (a) the magnetic field strength (b) the flux density and (c) the total flux.

(a) the magnetic field strength (b) the flux density and (c) the total flux. Solution 1.3

Solution 1.3

Here N = 200, I = 4 A, cross sectional area A = 5 cm

Here N = 200, I = 4 A, cross sectional area A = 5 cm22 = = 55 1100-4-4 mm22 and the mean and the mean circumference

circumference ℓℓ = = 6600 1100-2-2 m. m.

(a) From

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Fig(1.10) Analogy between magnetic and Fig(1.10) Analogy between magnetic and

N N II R R II E E _V_V Magnetic

Magnetic circuit circuit Electric Electric circuitcircuit

m

//

A

1333

6

4

4 2000

2000

10

60

4

200

200 NI

NI

H

₂₂  

(b) As the wooden ring is made of a non-magnetic material

(b) As the wooden ring is made of a non-magnetic material r r is 1 and using is 1 and using Eq(1.9)Eq(1.9)

μT

1675

10

10 1675

1675

1333

10

4

77 66 0 0

H

B

_r_r (c) Using

(c) Using Eq(1.1)Eq(1.1)

μWb

0.8375

10

5

5 1675

1675

..

0

10

5

10

10 1675

1675

66 44 66

BA

Example 1.4 Example 1.4

Calculate the magnetomotive force required to produce a flux of 0.015 Wb across an air gap Calculate the magnetomotive force required to produce a flux of 0.015 Wb across an air gap 2.5 mm long, having an effective area of 30 cm

2.5 mm long, having an effective area of 30 cm22..

Solution 1.4 Solution 1.4

Area A of the air gap is Area A of the air gap is

2 2 4 4 2 2

m

10

30

30 cm

cm

30

30 A

A

From

From Eq(1.1)Eq(1.1) flux density B is flux density B is

T

5

30

150

30

10

015

015 ..

0

10

30

015

015 ..

0

0 A

A

B

4 4 4 4 From

From Eq(1.8)Eq(1.8) magnetic field strength H is magnetic field strength H is

m

//

A

10

398

398 ..

0

10

4

5

5 B

B

H

₇₇ 77 0 0 Therefore from

Therefore from Eq(1.7)Eq(1.7) mmf ismmf is

A

AT

T

9947

10

10 9947

9947

..

0

10

5

5 ..

2

10

398

398 ..

0

77 33 44  

H

NI

mmf

1.6 Reluctance S and the magnetic circuit

In an electric circuit an

In an electric circuit an electromotiveelectromotive force or an force or an emf emf E will force a current I to flow in the E will force a current I to flow in the circuit and the opposition to the flow of current is the

circuit and the opposition to the flow of current is the resistanceresistance R. In a similar manner a R. In a similar manner a magnetomotive force or mmf F

magnetomotive force or mmf Fmm will will force a force a magnetic magnetic flux flux to to flow flow in in a a magnetic circuimagnetic circuit t andand

the opposition to the flow of flux is the

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For the electric circuit shown in

For the electric circuit shown in Fig(1.10)Fig(1.10) emf E is equivalent to the volt drop across the emf E is equivalent to the volt drop across the resistor R and we can write the basic equation as

resistor R and we can write the basic equation as

volts

IIR

R

V

E

Eq(1.10)Eq(1.10)

The analogy for the magnetic circuit is that the mmf F

The analogy for the magnetic circuit is that the mmf Fmm is equivalent to the product of the flux is equivalent to the product of the flux

and the reluctance S. Ie and the reluctance S. Ie

A

AT

T

S

F

_m_m Eq(1.11)Eq(1.11) By comparing

By comparing Eq(1.10)Eq(1.10) andand Eq(1.11)Eq(1.11) it seen that F it seen that Fmm (mmf), (mmf), (flux) and S (reluctance) are(flux) and S (reluctance) are

analogous to E (emf), I (current) and R (resistance) respectively. Now we can develop an analogous to E (emf), I (current) and R (resistance) respectively. Now we can develop an equation for S. From

equation for S. From Eq(1.11)Eq(1.11)

(AT/Wb)

m m

F

S

Eq(1.12)Eq(1.12) Substituting for F

Substituting for Fmm from from Eq(1.6)Eq(1.6) and and for for fromfrom Eq(1.1)Eq(1.1) we have we have

(AT/Wb)

BA

NI

S

Eq(1.13)Eq(1.13)

Now we can replace NI by H

Now we can replace NI by Hℓℓ_(see_{(see Eq(1.7)}_{Eq(1.7)) and}_{) and Eq(1.13)}_{Eq(1.13) becomes}_becomes

(AT/Wb)

BA

H

S

 Eq(1.14)Eq(1.14)

If the denominator and numerator of

If the denominator and numerator of Eq(1.14)Eq(1.14) are both divided by H we haveare both divided by H we have

Wb

At

Wb

At

m

AT

m

Wb

m

At

m

Wb

m

A

H

B

S

//

))

//

((

))

//

((

22 2 2 2 2 2 2   Eq(1.15) Eq(1.15)

Now the ratio B/H is the permeability of the material considered and therefore the equation for Now the ratio B/H is the permeability of the material considered and therefore the equation for reluctance S becomes reluctance S becomes Wb Wb At At or or m m mAT mAT Wb Wb m m m m mAt mAt Wb Wb m m A A A A S S r r

//

))

//

((

22 22 0 0     Eq(1.16) Eq(1.16)

Here it is noted that

Here it is noted that ℓℓ is the length in meters and that A is the cross sectional area in m is the length in meters and that A is the cross sectional area in m22..

1.6.1 1.6.1 Comparison of the electric and magnetic circuits

Comparison of the electric and magnetic circuits

It is helpful to present various electric and magnetic quantities and their relationship in tabular It is helpful to present various electric and magnetic quantities and their relationship in tabular form and such a table is given in

form and such a table is given in Table(1.1)Table(1.1). It is noted that the same symbol E is used to. It is noted that the same symbol E is used to denote the electromotive force emf and the electric filed strength, which may be confusing at denote the electromotive force emf and the electric filed strength, which may be confusing at times. Normally bold letter

times. Normally bold letter EE is used to represent the electric filed strength and care must be is used to represent the electric filed strength and care must be taken in using this symbol.

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Electric

Electric circuit circuit Magnetic Magnetic circuitcircuit

Quantity Units Quantity Units

Emf

Emf (E) (E) Volt(V) Volt(V) Mmf Mmf (F(Fmm) ) Ampere Ampere turns(AT)turns(AT)

Electric field strength Electric field strength

((EE))

Volts

Volts per per meter meter (V/m) (V/m) Magnetic Magnetic filedfiled strength (H) strength (H)

Ampere turns per Ampere turns per

meter (AT/m) meter (AT/m) Current (I)

Current (I) Equivalent to emf per Equivalent to emf per

resistance resistance

Ampere (A)

Ampere (A) MaMagngnetetic ic flflux ux (( )) Equivalent to mmf Equivalent to mmf per reluctance per reluctance Weber (Wb) Weber (Wb) Current

Current density density Ampere Ampere per per squaredsquared meter (A/m meter (A/m22))

Flux

Flux density density (B) (B) Tesla Tesla or or Weber Weber perper squared meter squared meter

(Wb/m (Wb/m22)) Resistance (R)

Resistance (R) OOhhm m (( )) Reluctance Reluctance (S) (S) Ampere Ampere turns turns perper Weber (AT/Wb) Weber (AT/Wb) Table(1.1)

Table(1.1)..Electric and magnetic circuit parametersElectric and magnetic circuit parameters

The radius and the cross sectional area of a mild steel ring are 5 cm and 400 mm The radius and the cross sectional area of a mild steel ring are 5 cm and 400 mm 22 respectively. A current of 0.5 A flows in a coil wound around the ring and the flux produced is respectively. A current of 0.5 A flows in a coil wound around the ring and the flux produced is 0.1 mWb. Calculate (a) the reluctance of mild steel and (b) the number of turns of the coil if 0.1 mWb. Calculate (a) the reluctance of mild steel and (b) the number of turns of the coil if the relative permeability is 200.

the relative permeability is 200. Solution 1.5

Solution 1.5 Here

Here

Length of the ring

Length of the ring ℓℓ ₌_{= 2}_{2 r}_{r =}_{= 2}_{2 ((5}_{5 1}₁₀₀-2-2

) m, ) m, Cro

Cross ss secsectiotional nal area area A A = 400= 400 1010-6-6 m m22 Current

Current I I in in the the coil coil = = 0.5 0.5 AA Fl

Flux ux = = 0.0.11 1010-3-3 Wb Wb Relative permeability

Relative permeability r r = = 200200

(a) From

(a) From Eq(1.16) Eq(1.16)

Wb

AT

A

S

r r

//

10

125

125 ..

3

16

10

5

10

4

2

10

2

10

5

10

400

200

10

4

10

5

2

6 6 7 7 2 2 7 7 2 2 6 6 7 7 2 2 0 0   (b) From (b) From Eq(1.11) Eq(1.11)

))

((

5

5 ..

312

10

10 3125

3125

..

0

10

1

1 ..

0

10

125

125 ..

3

66 33 33

AT

S

F

_m_m and from

and from Eq(1.7)Eq(1.7)

))

((

625

5

5 ..

0

5

5 ..

312

312 turns

turns

I

F

N

NI

F

m m m m

(11)

Example 1.6 Example 1.6 A coil of

A coil of 300 turns 300 turns is wound uniformis wound uniform ly over an ly over an iron ring iron ring having a having a uniform cross uniform cross sectional areasectional area of 500 mm

of 500 mm22 and and a a mean circumference mean circumference of 400 of 400 mm. If mm. If the coithe coil has l has a resistance a resistance of 8 of 8 and iand iss connected

connected across a across a 20 V 20 V dc supply, dc supply, calculate the calculate the required mmf, required mmf, H, H, and S. and S. Assume thatAssume that r r is is

950. 950. Solution 1.6 Solution 1.6 Here Here Cur

Currenrent t I I in in the the coicoil l = = 2020 8 8 = = 2.5 2.5 AA Length of the ring

Length of the ring ℓℓ = 0.4 m,= 0.4 m, Cro

Cross ss secsectiotional nal area area A A = 500= 500 1010-6-6 m m22 Number

Number of of turns turns = = 300300 Relative permeability Relative permeability r r = = 950950 (i) (i)

AT

NI

F

_m_m

300

2

2 ..

5

750

(ii) (ii)

))

//

((

1875

4

4 ..

0

750

750 m

m

AT

NI

H

  (iii) From

(iii) From Eq(1.9)Eq(1.9)

2 2 7 7 r r 0 0 r r 0 0

m

//

Wb

12

12 ..

2

2 1875

1875

..

0

9

9 ..

0

4

4 1875

1875

900

10

4

4 H

H

B

H

B

This gives This gives asas

mWb

06

06 ..

1

10

10 1060

1060

10

500

12

12 ..

2

2 BA

BA

66 66 (iv) From

(iv) From Eq(1.12)Eq(1.12)

Wb

AT

F

S

mm

₇₀₇

_..

₃₅₅

₁₀

_//

10

06

06 ..

1

750

33 3 3

1.7 Composite magnetic circuit

Consider a magnetic circuit which consists of two specimens of iron arranged as shown in Consider a magnetic circuit which consists of two specimens of iron arranged as shown in Fig(1.11)

Fig(1.11). Let. Let ℓℓ11 and andℓℓ22 be the mean lengths of specimen 1 and specimen2 in meters, A be the mean lengths of specimen 1 and specimen2 in meters, A11 and and

A

A22 be their respective cross sectional areas in square meters, and be their respective cross sectional areas in square meters, and 11 and and 22 be their be their

respective relative permeabilities. respective relative permeabilities.

The reluctance of specimen 1 is given as The reluctance of specimen 1 is given as

))

//

((

1 1 1 1 0 0 1 1 1 1

AT

Wb

A

S

 Eq(1.17)Eq(1.17)

and that for specimen 2 is and that for specimen 2 is

(12)

Specimen 1 Specimen 1 Specimen 2 Specimen 2 ℓ ℓ₁₁ ℓ ℓ₁₁ A A22 A A11

Fig(1.11) Composite magnetic circuit Fig(1.11) Composite magnetic circuit

))

//

((

2 2 2 2 0 0 2 2 2 2

AT

Wb

A

S

 Eq(1.18)Eq(1.18)

If a coil of N turns carrying a current I is wound on the specimen 1 and if the magnetic flux is If a coil of N turns carrying a current I is wound on the specimen 1 and if the magnetic flux is assumed to be confined to iron core then the total reluctance is given by the sum of the assumed to be confined to iron core then the total reluctance is given by the sum of the individual reluctances S

individual reluctances S11 and S and S22. This is equivalent to adding the resistances of a series. This is equivalent to adding the resistances of a series

circuit. Thus the total reluctance is given by circuit. Thus the total reluctance is given by

Wb

AT

A

S

//

2 2 2 2 0 0 2 2 1 1 1 1 0 0 1 1 2 2 1 1     Eq(1.19) Eq(1.19) and the total flux

and the total flux isis

Wb

))

Wb

//

A

AT

T

((

A

AT

T

A

NI

S

mm

mmf

f

2 2 2 2 0 0 2 2 1 1 1 1 0 0 1 1    Eq(1.20)Eq(1.20) Example 1.7 Example 1.7

A closed magnetic circuit made out of mild steel consists of two parts. The m

A closed magnetic circuit made out of mild steel consists of two parts. The m ean length of firstean length of first part is 6 cm and its cross sectional area is 1 cm

part is 6 cm and its cross sectional area is 1 cm22. The second part is 2 cm long having a cross. The second part is 2 cm long having a cross sectional area is 0.5 cm

sectional area is 0.5 cm22. A coil of 200 turns carrying a current of 0.4 A is wound uniformly. A coil of 200 turns carrying a current of 0.4 A is wound uniformly over the first part of the circuit. Calculate the flux density in the second path if the relative over the first part of the circuit. Calculate the flux density in the second path if the relative permeability of mild steel is 750.

permeability of mild steel is 750. Solution 1.7 Solution 1.7 Here Here Length of part 1 Length of part 1ℓℓ11 = 6 cm = 0.06 m= 6 cm = 0.06 m Length of part 2 Length of part 2ℓℓ₂₂ _{= 2 cm = 0.02 m}_{= 2 cm = 0.02 m} Cross sectional area of part 1 A

Cross sectional area of part 1 A11 = = 1 1 cmcm 2 2

=

= 11 1100-4-4 m m22 Cross sectional area of part 2 A

Cross sectional area of part 2 A22 = = 0.5 0.5 cmcm 2 2

=

= 00..55 1100-4-4 m m22 Number

Number of of turns turns = = 200200 Current

Current I I in in the the coil coil = = 0.4 0.4 AA Relative permeability of both parts

Relative permeability of both parts r r = = 750750

Reluctance of part 1 is Reluctance of part 1 is

(13)

Fig(1.12) Magnetic circuit with an air gap Fig(1.12) Magnetic circuit with an air gap

ℓ ℓ₁₁ N N II ℓ ℓ₂₂ Air gap Air gap

Wb

AT

A

S

6

6 ..

366

10

10 //

5

5 ..

7

4

10

6

10

1

750

10

4

10

6

77 ₅₅ 4 4 7 7 2 2 1 1 1 1 0 0 1 1 1 1  

and that for specimen 2 is and that for specimen 2 is

Wb

AT

A

S

4

4 ..

244

10

10 //

75

75 ..

3

4

10

2

10

5

5 ..

0

750

10

4

10

2

77 ₅₅ 4 4 7 7 2 2 2 2 0 0 2 2 2 2  

Therefore the total reluctance is Therefore the total reluctance is

Wb

AT

S

₁₁ ₂₂

((

6

6 ..

366

4

4 ..

244

244 ))

10

55

10

10 ..

61

10

55

//

Therefore the total flux is Therefore the total flux is

Wb

S

mm

mmf

f

₅₅ 5 5

7

7 ..

54

10

61

61 ..

10

4

4 ..

0

200

Therefore the total flux density in part 2 is Therefore the total flux density in part 2 is

2 2 4 4 5 5 2 2

//

508

508 ..

1

10

5

5 ..

0

10

54

54 ..

7

7 m

m

Wb

A

B

The iron core of a magnetic circuit which has an air gap in it is shown in

The iron core of a magnetic circuit which has an air gap in it is shown in Fig(1.12)Fig(1.12). The length. The length ℓ

ℓ₁₁_{of the iron is 40 cm and its cross sectional area is 0.001 m}_{of the iron is 40 cm and its cross sectional area is 0.001 m}22

and its relative permeability is and its relative permeability is 850. The length

850. The length ℓℓ₂₂_{of the air gap is 1 mm. The iron core is wound with a coil of 2000 turns and}_{of the air gap is 1 mm. The iron core is wound with a coil of 2000 turns and} the current flow in the coil produces a flux in the air gap of 1.5 mWb. Assuming all the flux in the current flow in the coil produces a flux in the air gap of 1.5 mWb. Assuming all the flux in the iron passes through the air gap calculate

the iron passes through the air gap calculate (a) the reluctance of the iron path

(a) the reluctance of the iron path (b) the reluctance of the air gap (b) the reluctance of the air gap

(c) the total reluctance of the magnetic circuit (c) the total reluctance of the magnetic circuit (d) the mmf needed to produce the flux (d) the mmf needed to produce the flux (e) the flux density in the air gap

(e) the flux density in the air gap (f) the flux density in the iron (f) the flux density in the iron

(g) the magnetic field strength in the air gap (g) the magnetic field strength in the air gap (h) the magnetic field strength in the iron (h) the magnetic field strength in the iron (i) the current in the coil

(14)

Solution 1.8 Solution 1.8 Length of iron

Length of ironℓℓ11 = 40 cm = 0.4 m= 40 cm = 0.4 m

Length of air gap

Length of air gap ℓℓ22 = 1 mm = 0.001 m= 1 mm = 0.001 m

Area of iron and air gap

Area of iron and air gap = 0.001 m= 0.001 m22 Number

Number of of turns turns = = 20002000 Flux

Flux in in the the iron iron and and air air gap gap = = 1.5 1.5 mWbmWb Relative permeability of iron

Relative permeability of iron r r = = 850850

(a) (a)

Wb

AT

A

S

0

0 ..

374

10

10 //

5

5 ..

8

4

10

4

001

001 ..

0

850

10

4

4 ..

0

66 7 7 7 7 1 1 0 0 1 1 1 1   (b) As

(b) As r r for air is 1 for air is 1

Wb

AT

A

S

0

0 ..

796

10

10 //

4

10

1

001

001 ..

0

10

4

001

001 ..

0

66 7 7 7 7 0 0 2 2 2 2  

(c) Therefore the total reluctance is (c) Therefore the total reluctance is

Wb

AT

S

₁₁ ₂₂

((

0

0 ..

374

0

0 ..

796

796 ))

10

66

1

1 ..

17

10

66

//

(d) Therefore the total mmf is (d) Therefore the total mmf is

AT

S

mmf

1

1 ..

17

10

66

1

1 ..

5

10

33

1755

(e) Flux density of air gap is (e) Flux density of air gap is

))

((

//

5

5 ..

1

001

001 ..

0

10

5

5 ..

1

33 ₂₂ 2

2

Wb

m

or

T

tesla

A

B

(f) As same flux pass through iron core flux density of iron is the same as that of air gap. (f) As same flux pass through iron core flux density of iron is the same as that of air gap. (g) Magnetic field strength H

(g) Magnetic field strength H22 of air gap is of air gap is

m

AT

B

H

1

1 ..

194

10

10 //

10

4

5

5 ..

1

66 7 7 0 0 2 2

(h) Magnetic field strength H

(h) Magnetic field strength H11 of iron is of iron is

m

AT

B

H

r r

//

1404

850

10

194

194 ..

1

850

10

4

5

5 ..

1

66 7 7 0 0 1 1

(i) Current in the coil is (i) Current in the coil is

A

8775

..

0

0 2000

2000

1755

N

mm

mmf

f

II

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