Magnetic Circuits - Full Notes

School of Engineering Robert Gordon University EN1560-Introduction to Electrical Engineering 1

Magnetic Circuits

1.1

Magnetic field

A permanent magnet is 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 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 seeking seeking end or the north pole, N of the magnet. Similarly the other end is known as the south seeking seeking end or the south pole, S of the magnet. 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 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 Fig(1.1). Fig(1.1). These curved lines can be used to visualise the magnetic condition of the space around the magnet, which may be identified as the magnetic field. field. Also these lines lead to the idea of lines of magnetic flux flux 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 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 magnet.

Lines of magnetic flux

N

S

Fig(1.1) Lines of magnetic flux 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

EN1560-Magnetic Circuits Copyrig Copyright ht 2006 by Dr G W D M Gunawar Gunawardene, dene, © 2011 G Dunbar Dunbar

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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. When two magnets are arranged in such away that unlike poles are next to each other, as shown in Fig(1.2)(a), Fig(1.2)(a), 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 towards each other. If the magnets are arranged so that the like poles are near to each other, 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 between the two magnets are pointing in the same direction thus pushing the two magnets away from each other.

N

S

N

N

S

N

S

S

Fig(1.2)(a) Attraction between magnets

1.2

Magnetic flux

Fig(1.2)(b) Repulsion between magnets

and flux density B

The amount of magnetic field produced by a magnetic source is known as the magnetic flux and the symbol used is the Greek letter . The unit of magnetic flux is the weber or Wb. Wb. The Magnetic flux density B B 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 tesla or T. Thus we can write the equation

B

A

Wb/m 2 , Wb.m

2

or tesla

Eq(1.1)

2

2

where A is the area in m . It is seen from Eq(1.1) that Eq(1.1) that 1 T is equivalent to 1 Wb/(m ) which is another way of defining the units of flux density B. Example 1.1 The magnetic flux crossing the air gap of the magnet shown in Fig(1.3) is Fig(1.3) is 12 mWb. Determine the flux density in the air gap if the magnet has dimensions shown. Solution 1.1 Cross sectional area A is

A

8 10

2

15 10

2

120 10

4

m2

0.012 m 2

This gives EN1560-Magnetic Circuits Copyrig Copyright ht 2006 by Dr G W D M Gunawar Gunawardene, dene, © 2011 G Dunbar Dunbar

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B

A

12 mWb mWb

0.012

0.012

0.012

1 T

N

S

8 cm

Flux

15 cm Magnet

Fig(1.3) Magnet with air gap

1.3

Magnetic field due to an electric current

A fundamental law of electromagnetism is that a magnetic field is produced around a 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 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 basic sign convention to indicate the direction of the magnetic field.

1.3.1 Sign convention of electromagnetic electromagnetic field Consider a wire carrying an electric current which has a cross section as shown in Fig(1.4). Fig(1.4). In Fig(1.4)(a) the Fig(1.4)(a) 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. Another method of representing this is to place a corkscrew along the conductor, which 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 field. In Fig(1.4)(b) the Fig(1.4)(b) the current flow is reversed, ie flowing out of the paper, which is indicated by the dot. In this case it is obvious that the direction of the field is anticlockwise and again flux lines are shown by the concentric circles.

Clockwise rotation of field

Anticlockwise rotation of field

Wire

Fig(1.4)(a)

Fig(1.4)(b)

Fig(1.4) Sign convention of electromagnetic field

1.3.2 Ampere’s Law EN1560-Magnetic Circuits Copyrig Copyright ht 2006 by Dr G W D M Gunawar Gunawardene, dene, © 2011 G Dunbar Dunbar

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The Ampere’s Law is particularly useful in determining magn etic field strength near current 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 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 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. 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 Fig(1.5), Fig(1.5), enclosing a current carrying conductor, it experiences a force H, which is tangential to the path, as shown in Fig(1.5). Fig(1.5). When the unit pole is moved an infinitely small distance ℓ along the path the work done is the product of H and ℓ. The Ampere’s law states that the sum of the product of H ℓ, which is the total work 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 

H



I

( A)

Eq(1.2)

0

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

H 0



I

( A)

Eq(1.3)



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

Closed path A Closed path B

H Unit pole

Conductor

Fig(1.5) Closed path enclosing a conductor The Ampere’s law is very simple to use when the tangential force H is a constant and t his is the case for all examples considered here. Thus if H is constant then Eq(1.3) becomes Eq(1.3) becomes

H

 0

H

I

( A)



H

I 

Eq(1.4)

( A / m)

This force H is called the magnetising force or the magnetic field strength. strength.


4

As an example it is required to calculate the magnitu de 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 shown in Fig(1.4) and Fig(1.4) 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 line is constant). From Fig(1.6) and Fig(1.6) and Eq(1.4) the field strength H at a distance r is given as

H

I

I



2 r

( A / m)

Eq(1.5)

Circular path of length ℓ = 2 r

Wire

r

Fig(1.6) Circular path around a current carrying conductor

1.4 Magnetomotive force mmf Fm 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 magnetomotive force mmf or Fm, caused by a current flowing through one or more turns. Thus a coil, as 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 magnetic fields. N turns

I Amp l meters

Fig(1.7) A coil with N turns Therefore we can write an equation for F m as

mmf F m

NI

(ampere - turns or AT)

Eq(1.6)

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 circuit has a uniform cross section, the magnetomotive force per unit length of the magnetic


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circuit is the magnetising force or magnetic field strength discussed in Section 1.3.2. 1.3.2. As shown in Fig(1.7) if Fig(1.7) if ℓ is the mean length (meters) of the magnetic circuit then magnetic field strength H is given as

H

NI

( A / m)



Eq(1.7)

Example 1.2 A circular wooden ring of mean diameter 20 cm has a coil of 800 turns uniformly wound around it. If the magnetic field strength is 5000 A/m calculate the current in the coil. Solution 1.2 The mean length (circumference) of the wooden ring is ℓ = Eq(1.7) we Eq(1.7) we have

H

I

5000

20 10

N

2

800

1.5 Permeability

10 8

d =

-2

20 10 m and from

3.93 A

and B-H curves

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 permeability for free space and has the symbol 0. ie

B 0

H

4

10

7

Wb / m 2

Wb

AT / m

mAT mAT

Eq(1.8)

-7

Note that this constant has a numerical value of 4 10 Wb/(mAT) as given in Eq(1.8). Eq(1.8). Another unit for this constant is henrys/m and the interested reader can look elsewhere for details. All non-magnetic materials are considered to have the same permeability 0 as free space. Fig(1.8) space. Fig(1.8) shows shows a graph of flux density B plotted against the magnetic field strength H for free space known as a B-H curve. curve. This curve is linear and the slope of the straight line is 0. B Wb/m

Slope =

0

H AT/m

Fig(1.8) B-H curve for free space

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


6

B 0

r

H

Wb /(mAT mAT )

Eq(1.9)

where r is the relative permeability of the ferromagnetic material which has no units. From 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 three different magnetic materials are shown in Fig(1.9). Fig(1.9) . It is noted that these curves are not linear any longer and the relative permeability r of the ferromagnetic material is proportional to the slope of the B-H curve and therefore varies with the magnetic field strength H.

Flux Density Density B (Wb/m (Wb/m ) 1.8

Mild Steel

1.6

1.4

Cast Steel 1.2

1

0.8

Cast Iron

0.6

0.4

0.2

0 20

40

60

80

100

Magnetic Field Strength (AT/m)

Fig(1.9) B-H curves for different magnetic materials Example 1.3 A coil of 200 tur ns is wound uniformly over a wooden ring having a mean circumference of 60 2 cm and a uniform cross sectional area of 5 cm . If the current through the coil is 4 A calculate (a) the magnetic field strength (b) the flux density and (c) the total flux. Solution 1.3 2

-4

Here N = 200, I = 4 A, cross sectional area A = 5 cm = 5 10

2

m and the mean

-2

circumference ℓ = 60 10 m. (a) From Eq(1.7)


7

H

NI

200 4 2

60 10



2000 4

1333 A / m

6

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

B

0

r

4

H

10

7

1333

1675 10

6

r is

1 and using Eq(1.9)

1675 μT

(c) Using Eq(1.1)

BA

6

1675 10

5 10

4

0.1675 5 10

6

0.8375 μWb

Example 1.4 Calculate the magnetomotive force required to produce a flux of 0.015 Wb across an air gap 2 2.5 mm long, having an effective area of 30 cm . Solution 1.4 Area A of the air gap is

A

30 cm 2

30 10

4

m2

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

B

0.015 A

30 10

0.015 104

150

30

30

4

5T

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

H

B 0

5 4

10

0.398 107 A / m

7

Therefore from Eq(1.7) mmf is

mmf NI

H 

0.398 10 7

2.5 10

3

0.9947 10 4

9947 AT

1.6 Reluctance S and the magnetic circuit In an electric circuit an electromotive electromotive force or an emf E E will force a current I to flow in the circuit and the opposition to the flow of current is the resistance resistance R. In a similar manner a magnetomotive force or mmf Fm will force a magnetic flux to flow in a magnetic circuit circuit and the opposition to the flow of flux is the reluctance S. reluctance S. This is illustrated in Fig(1.10). Fig(1.10). I

E

V

I R

N

Magnetic circuit

Electric circuit

Fig(1.10) Analogy between magnetic and electric circuits EN1560-Magnetic Circuits Copyrig Copyright ht 2006 by Dr G W D M Gunawar Gunawardene, dene, © 2011 G Dunbar Dunbar

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

E

V

IR

volts

Eq(1.10)

The analogy for the magnetic circuit is that the mmf F m is equivalent to the product of the flux and the reluctance S. Ie

F m

AT

S

Eq(1.11)

By comparing Eq(1.10) and Eq(1.11) Eq(1.11) it seen that F m (mmf), (flux) and S (reluctance) are analogous to E (emf), I (current) and R (resistance) respectively. Now we can develop an equation for S. From Eq(1.11)

S

F m

(AT/Wb)

Eq(1.12)

Substituting for F m from Eq(1.6) and for

S

NI BA

from Eq(1.1) we Eq(1.1) we have

(AT/Wb)

Eq(1.13)

Now we can replace NI by H ℓ (see Eq(1.7)) Eq(1.7)) and Eq(1.13) becomes Eq(1.13) becomes

S

H  BA

(AT/Wb)

Eq(1.14)

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

S

m



B H

(Wb / m 2 )

A

( At / m)

m

2

m Wb m 2

m AT

At

At / Wb

Wb

m2

Eq(1.15)

Now the ratio B/H is the permeability of the material considered and therefore the equation for reluctance S becomes

S



A

m

 0

A r

(Wb / mAt )m 2

m Wb

or At / Wb m

Eq(1.16)

2

mAT 2

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

1.6.1 Comparison of the electric and magnetic circuits It is helpful to present various electric and magnetic quantities and their relationship in tabular 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 denote the electromotive force emf and the electric filed strength, which may be confusing at times. Normally bold letter E is used to represent the electric filed strength and care must be taken in using this symbol.


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Electric circuit Quantity Units Emf (E) Volt(V) Electric field strength Volts per meter (V/m) (E) Current (I) Ampere (A) Equivalent to emf per resistance Current density Ampere per squared 2 meter (A/m ) Resistance (R)

Magnetic Quantity Mmf (Fm) Magnetic filed strength (H) Magn Magnet etic ic flux flux ( ) Equivalent to mmf per reluctance Flux density (B)

circuit

Reluctance (S)

Ohm ( )

Units Ampere turns(AT) Ampere turns per meter (AT/m) Weber (Wb)

Tesla or Weber per squared meter 2 (Wb/m ) Ampere turns per Weber (AT/Wb)

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

Example 1.5 2

The radius and the cross sectional area of a mild steel ring are 5 cm and 400 mm 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 the relative permeability is 200. Solution 1.5 Here Length of the ring ℓ Cross Cross sectio sectional nal area A Current I in the coil Flux Flux Relative permeability r

-2

= 2 r = 2 (5 10 ) m, -6 2 = 400 10 m = 0.5 A -3 = 0.1 0.1 10 Wb = 200

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

2



S

A

0

4

r

5 10 2 10

7

10

5 10

7

2

200 400 10

2

5 10 7

2 4 10

2

16

6

3.125 10 6

AT / Wb

(b) From Eq(1.11) From Eq(1.11)

F m

3.125 106

S

0.1 10

3

0.3125 103

312.5 ( AT )

and from Eq(1.7)

F m

NI N

F m

312.5

I

0.5

625 (turns)


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Example 1.6 A coil of 300 turns is wound uniform ly over an iron ring having a uniform cross sectional area 2 of 500 mm and a mean circumference of 400 mm. If the coil coil has a resistance of 8 and is is connected across a 20 V dc supply, calculate the required mmf, H, and S. Assume that r is 950. Solution 1.6 Here Curren Currentt I in the coil coil

= 20 8 = 2.5 A

Length of the ring ℓ Cross Cross sectio sectional nal area A Number of turns Relative permeability r

= 0.4 m, -6 2 = 500 10 m = 300 = 950

(i)

F m

NI

300 2.5

NI

750



0.4

750 AT

(ii)

H

1875 ( AT / m)

(iii) From Eq(1.9)

B 0

r

H

B

0

This gives

H

4

r

10

7

900 1875

4

0.9 0.1875

2.12 Wb / m 2

as

BA

2.12 500 10

6

1060 10

6

1.06 mWb

(iv) From Eq(1.12)

S

F m

750 1.06 10

3

707.355 10 3 AT / Wb

1.7 Composite magnetic circuit Consider a magnetic circuit which consists of two specimens of iron arranged as shown in Fig(1.11). Fig(1.11). Let ℓ1 and ℓ2 be the mean lengths of specimen 1 and specimen2 in meters, A 1 and A2 be their respective cross sectional areas in square meters, and 1 and 2 be their respective relative permeabilities. The reluctance of specimen 1 is given as 1

S 1 0

1 A1

( AT / Wb)

Eq(1.17)

and that for specimen 2 is


11

2

S 2 0

2 A2

( AT / Wb)

Eq(1.18)

Specimen 1 ℓ1

A1

A2 Specimen 2 ℓ1

Fig(1.11) Composite magnetic circuit

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 individual reluctances S 1 and S2. This is equivalent to adding the resistances of a series circuit. Thus the total reluctance is given by

S

S 1

1

S 2

A1

0

and the total flux

2

1

0

A2

AT / Wb

Eq(1.19)

2

is

mmf mmf

NI

AT

1

S 0

1

A1

2 0

2

( AT / Wb)

Wb

Eq(1.20)

A2

Example 1.7 A closed magnetic circuit made out of mild steel consists of two parts. The m ean length of first 2 part is 6 cm and its cross sectional area is 1 cm . The second part is 2 cm long having a cross 2 sectional area is 0.5 cm . 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 permeability of mild steel is 750. Solution 1.7 Here Length of part 1 ℓ1

= 6 cm = 0.06 m

Length of part 2 ℓ2 Cross sectional area of part 1 A 1 Cross sectional area of part 2 A 2 Number of turns Current I in the coil Relative permeability of both parts

= 2 cm = 0.02 m 2 -4 2 = 1 cm = 1 10 m 2 -4 2 = 0.5 cm = 0.5 10 m = 200 = 0.4 A = 750

r

Reluctance of part 1 is EN1560-Magnetic Circuits Copyrig Copyright ht 2006 by Dr G W D M Gunawar Gunawardene, dene, © 2011 G Dunbar Dunbar

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1

S 1 0

6 10 A1

1

4

7

10

2

6 10 7

750 1 10

4

4

6.366 105 AT / Wb

7.5

and that for specimen 2 is 2

S 2

0

2 10

A2

4

7

10

2

2 10 7

750 0.5 10

4

4

4.244 105 AT / Wb

3.75

Therefore the total reluctance is

S

S 1

S 2

4.244) 10 5

(6.366

10.61 10 5

AT / Wb

Therefore the total flux is

mmf mmf S

200 0.4 10.61 10

7.54 10

5

5

Wb

Therefore the total flux density in part 2 is

B

7.54 10 A2

0.5 10

5 4

1.508 Wb / m 2

Example 1.8 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 2

ℓ1 of the iron is 40 cm and its cross sectional area is 0.001 m and its relative permeability is

850. The length ℓ2 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 iron passes through the air gap calculate (a) the reluctance of the iron path (b) the reluctance of the air gap (c) the total reluctance of the magnetic circuit (d) the mmf needed to produce the flux (e) the flux density in the air gap (f) the flux density in the iron (g) the magnetic field strength in the air gap (h) the magnetic field strength in the iron (i) the current in the coil ℓ1

Air gap

I N

ℓ2

Fig(1.12) Magnetic circuit with an air gap


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Solution 1.8 Length of iron ℓ1

= 40 cm = 0.4 m

Length of air gap ℓ2 Area of iron and air gap Number of turns Flux in the iron and air gap Relative permeability of iron

= 1 mm = 0.001 m 2 = 0.001 m = 2000 = 1.5 mWb = 850

r

(a)

S 1

A

0

(b) As

4

1

r for

10

7

4

850 0.001

8.5

6

0.374 10 AT / Wb

air is 1

2

S 2

4 107

0.4

1

A

4

0

0.001

1 10

7

4

10

0.001

7

0.796 106 AT / Wb

(c) Therefore the total reluctance is

S

S 1

0.796) 10 6

(0.374

S 2

1.17 10 6

AT / Wb

(d) Therefore the total mmf is

mmf

1.17 10 6 1.5 10

S

3

1755 AT

(e) Flux density of air gap is

B2

1.5 10

3

1.5 Wb / m 2 or T (tesla)

0.001

A

(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 2 of air gap is

H 2

1.5

B 4

0

10

1.194 10 6 AT / m

7

(h) Magnetic field strength H 1 of iron is

H 1

0

1.194 106

1.5

B r

4

10

7

850

850

1404 AT / m

(i) Current in the coil is

I

mmf mmf

1755

N

2000

0.8775 A

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Magnetic Circuits - Full Notes

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