Magnetic circuits

By
Dr. Krishna Roy
Assistant Professor
Electrical Engineering Department
NITR
 Syllabus

Magnetic circuits:
Review of fundamental laws of electromagnetic induction
Transformer and rotational emfs
Solution of magnetic circuits
 Syllabus

Magnetic circuits:
Review of fundamental laws of electromagnetic induction
Transformer and rotational emfs
Solution of magnetic circuits
Review of fundamental laws
of electromagnetic induction

4
 Review of fundamental laws of
electromagnetic induction
Faraday’s first law:
Whenever there is a change in magnetic flux linking
with a conductor, an EMF is induced in the conductor.

• If the circuit is closed, then the induced EMF will

circulate current through the closed circuit.
• This phenomenon is known as electromagnetic
induction.

The first law of electromagnetic induction gives a

qualitative description of the EMI process.
5
 Review of fundamental laws of
electromagnetic induction
Faraday’s second law:
The amount of EMF induced in the conductor is
proportional to the rate of change of flux linking with
the conductor.

If, e = induced EMF d

e
 = magnetic flux dt
N = Number of turns in the coil d  N 
e
 = Flux linkage = N dt
t = Time d
e N
dt
6
 Review of fundamental laws of
electromagnetic induction
Faraday’s second law:
d
e N
dt

• Direction (polarity) of the induced EMF is given by

the third law of electromagnetic induction or to be
more precise, by the Lenz’s law.

7
 Review of fundamental laws of
electromagnetic induction
Lenz’s law:
d
e N
dt
• The direction of induced EMF is such that the current
set up by it produces a magnetic field that opposes the
change in original flux.
• In general, Lenz’s law states that the effect always opposes the cause.
• The phenomena explained by Lenz’s law can be incorporated by making
(-1) as the constant of proportionality.
• Thus final expression of EMF induced due to electromagnetic induction

d
can be written as:
eN
dt
8
 Review of fundamental laws of
electromagnetic induction
d
eN
dt

• Thus, whenever there is a relative change between

magnetic flux and a coil, EMF is induced in the coil.

• There are several ways in which such relative change

can be achieved:
• Dynamically induced EMF
• Statically induced EMF

9
 Syllabus

Magnetic circuits:
Review of fundamental laws of electromagnetic induction
Transformer and rotational emfs
Solution of magnetic circuits
 Syllabus

Magnetic circuits:
Review of fundamental laws of electromagnetic induction
Transformer and rotational emfs
Solution of magnetic circuits
Transformer and Rotational
emfs

12
 Transformer and Rotational emfs
• There are several ways in which relative change
between magnetic flux and a coil can be achieved:
• Dynamically induced EMF or Rotational EMF
• Statically induced EMF or Transformer EMF

13
 Dynamically induced EMF or
Rotational emfs
There are two ways in which an
EMF can be induced in a
conductor dynamically:
•When the magnetic field is
stationary and constant
•But the conductor or coil
physically moves in the
magnetic field

14
 Dynamically induced EMF or
Rotational emfs
•When the conductor or coil is stationary
•But the magnet or the magnetic field moves
physically

15
 Dynamically induced EMF or
Rotational emfs
•In both cases, there will be a change in the flux
linking with the conductor
•Hence EMF will be induced in the conductor
•This EMF is called dynamically induced EMF or
rotational EMF.

16
 Dynamically induced EMF or
Rotational emfs
• Flux density = B
• Length of the conductor = L
• Velocity of the conductor = V
•If the conductor moves •If the conductor moves at an
perpendicular to the flux lines, angle to the flux lines, then
then dynamically induced EMF: dynamically induced EMF:

e  BLV e  BLV sin 

V


V
 Dynamically induced EMF or
Rotational emfs
•Direction (polarity) of the induced EMF can be found out
using Fleming’s right hand rule.
Dynamically induced EMF or
Rotational emfs
B
F

I
 Statically induced EMF or
Transformer EMF
•When both magnet as well as the conductor is stationary,
but the magnetic field itself is time-varying
•There will be a change in the flux linking with the
conductor
•Hence EMF will be induced in the conductor
•This EMF is called statically induced EMF or
transformer EMF.
 Statically induced EMF or
Transformer EMF
N
S NS

In general, if the magnetic flux through a N-turn coil

increases by dΦ webers in dt seconds e.m.f. induced in
coil, d
eN
dt
 Syllabus

Magnetic circuits:
Review of fundamental laws of electromagnetic induction
Transformer and rotational emfs
Solution of magnetic circuits
 Syllabus

Magnetic circuits:
Review of fundamental laws of electromagnetic induction
Transformer and rotational emfs
Solution of magnetic circuits
Solution of magnetic circuits

24
 Magnetic Circuits
A magnetic circuit is defined as the path followed by magnetic
flux.

I I
Iron
Ring (Toroid )

Magnetic
flux lines

25
 Magnetic Circuits

l = Total length of magnetic path along the toroid ring (i.e. average
perimeter of the ring)
A = Area of cross section of the ring
N = number of turns in the solenoid
I = Current in the solenoid
26
 Magnetic Circuits
Magnetic field intensity developed by the solenoid is:
NI
H AT/m
l
Flux density within the core of the iron toroid is :
B  0  r H Wb/m2

The amount of flux flowing through the toroid :

  BA
  0 r HA
NI
  0 r A
l
NI

l
0 r A
27
 Magnetic Circuits
NI
 Magneto Motive Force (MMF)
l
0 r A
• MMF is the source that produces
magnetism in a magnetic circuit.

• MMF is analogous to EMF in an

electric circuit.

• MMF can be defined as the magnetic potential difference that

forces magnetic flux to flow through a magnetic circuit.

• MMF is expressed in the unit of Ampere-turns or in short AT.

28
 Magnetic Circuits
NI
 Magnetic reluctance

NI
l
l S
0  r A S
0  r A

• Reluctance of the magnetic flux path is a measure of how much

opposition the flux faces while flowing through the magnetic circuit.

• It is analogous to resistance in electric circuits.

• Property of a magnetic material that opposes the creation of flux in it.

• Reluctance is inversely proportional to relative permeability (r).

• It is obvious that a good magnetic material (high value of r) will

offer less opposition (low reluctance) to the flux flowing through it.
MMF

Reluctance 29
 Comparison between magnetic and
electric circuits
Similarities
Magnetic circuit Electric circuit
Driving force MMF (AT) = NI EMF (Volts)
MMF EMF
Effect  
Flux, Reluctance Current, I
Resistance
Opposition Reluctance, S
1 l
0  r A
Resistance, R
l
A

Reluctivity,
1
Resistivity, 
0 r
1
Permeability, 0  r Conductivity,  


30
 Comparison between magnetic and
electric circuits
Differences
Magnetic circuit Electric circuit

Magnetic flux can pass through Electric current cannot pass

almost anything through insulators
Magnetic flux can pass through air Electric current cannot pass
also through air (unless sparking)
Permeability of a material depends Resistivity does not depend on
on flux and flux density current (with constant temperature)

Some magnetic materials can Current is immediately off in a

retain magnetic property even after conductor when the source is
the source is removed removed
31
 Composite Magnetic Circuits

• Like electrical circuits can be made of series or

parallel connection of different resistances, so are
magnetic circuits.

• Magnetic circuits are often composed of more than

one type of materials having different values of
relative permeabilities.

• Thus, each part of the magnetic circuit will have

different reluctance values.

32
 Series Magnetic Circuit
• The composite magnetic circuit shown below consists of three
different magnetic materials of different permeabilitis, different
lengths and may be different cross sections also.
• Each part will have their own reluctance values.

33
 Series Magnetic Circuit
• The source of MMF (coil or solenoid) will drive flux through the
composite circuit whose value will be decided by the overall
reluctance.
• Same flux flows through all the three different parts (series circuit).

Flux lines

34
 Series Magnetic Circuit

Total reluctance S  
li l1 l2 l3
  
i  0  r i Ai  0  r 1 A1  0  r 2 A2  0  r 3 A3
MMF NI NI
Resultant flux through the circuit   Reluctance  S 
 
li
i 0 ri Ai
35
 Series Magnetic Circuit

• In a series electric circuit the same current flows through all parts of
the circuit.
• In a series magnetic circuit also, same flux flows through all parts of
the circuit.
• Moreover, the total MMF required to drive flux through the entire
circuit is summation of the MMFs required for individual parts of the
series circuit.
• This is like in electric circuit (KVL). 36
 Parallel Magnetic Circuit

• In a parallel magnetic circuit, the flux gets divided among the

branches
• Like current gets divided in parallel electric circuit
• The MMF required is same for all the branches in parallel
• Like voltage is equal across each of the parallel branches in an
electric circuit
37
 Flux leakage
• Leakage fluxes are those which do not flow through the entire length of
the magnetic circuit, but complete their path through intermediate
portions of the circuit.

38
 Flux Fringing
• While crossing an air gap, flux lines have a tendency to bulge out as
they cross the air gap.
• This effect is called fringing.

• Fringing effect is more severe when length of the air gap is more.

39
 Magnetization characteristics of a magnetic
material
• Relation between the magnetizing force (H) and the magnetic
property developed [flux density (B)]
B  H   0  r H

NI
H
l H
• Initially linear increase in B, but saturated afterwards
• When all the magnetic domains get fully oriented then saturated
• B-H characteristics is thus non-linear
 Magnetization with AC supply
• Input voltage is AC, i.e. varies sinusoidally from + to -
• Thus, input current I is also AC, i.e.
varies sinusoidally from + to -
NI
H
l
• Thus, magnetizing force H is also AC,
i.e. varies sinusoidally from + to -

• Thus, the magnetizing force starts from zero, rises up to a positive

maximum value, then comes down back to zero, drops down to a
negative minimum value and goes back again to zero. This cycle
continues.
 Magnetization with AC supply

NI
H
l H
Magnetization with AC supply
• The loop CDEFQC is
called hysteresis loop

• C = Positive saturation
• OD = Retentivity
(residual flux density)
• OE = Coercivity
• F = Negative saturation
Loss of energy during magnetization with AC
supply

• Hysteresis loss

• Eddy current loss

44
 Loss of energy during magnetization with AC
supply • Energy spent during
magnetization is more than
the energy released during de-
magnetization.
• This is evident from the
residual magnetism.
• The amount of energy thus
lost during each cycle of
magnetization is called
Hysteresis Loss.
• This heats up the material
during magnetization.

Total hysteresis loss in a magnetic material is proportional to the

area of its hysteresis loop.
The loop area (EDCQFE) is equal to hysteresis loss (in W) per
cycle of the input AC signal.
 Loss of energy during magnetization with AC
supply
• The second type of loss that takes place in a magnetic
material during magnetization by an alternating field is
called Eddy current loss.
• Most magnetic materials are either metallic or are alloys of
metallic base.
• This makes them conducting to some extent in addition to
being magnetic.
• The flux that passes through the material is time varying.
 Loss of energy during magnetization with AC
supply
• The flux that passes through the material is time varying.
• This will induce voltage inside the body of the material.
• This causes flow of currents in small loops within body of the
material.
Eddy current

• This causes I2R power loss in the material.

• This loss is called eddy current loss.
 Loss of energy during magnetization with AC
supply

• Hysteresis loss and eddy current loss are

collectively called iron loss or core loss
because they take place in the core material
(iron) of all electrical machines including
transformers, motors, and generators.

48
Example1: An iron ring of mean length 30 cm has an air gap of 2
mm and a coil with 200 turns on it. If relative permeability of iron is
300 when a current of 1 A flows through the coil; find the flux
density.

Length of flux path = mean length of the ring l = 30 cm = 0.3 m

Relative permeability r = 300

Length of air gap l1 = 2 mm = 0.002 m

Number of turns in coil N = 200

Current through coil I=1A

 MMF = NI = 200 AT

49
Example1: An iron ring of mean length 30 cm has an air gap of 2
mm and a coil with 200 turns on it. If relative permeability of iron is
300 when a current of 1 A flows through the coil; find the flux
density.

Length of flux path = mean length of the ring l = 30 cm = 0.3 m

Relative permeability r = 300
Length of air gap l1 = 2 mm = 0.002 m

Let, cross sectional area of the ring is A m2.

MMF NI NI
Resultant flux    
Reluctance S l1 l2

 0  r 1 A1  0  r 2 A2

Putting l1 = 0.002 m, l2 = 0.298 m, r1 = 1, r2 = 300, and A1 = A2 = A.

50
Example1: An iron ring of mean length 30 cm has an air gap of 2
mm and a coil with 200 turns on it. If relative permeability of iron is
300 when a current of 1 A flows through the coil; find the flux
density.

MMF NI NI
  
Reluctance S l1 l2

 0  r 1 A1  0  r 2 A2
l1 = 0.002 m, l2 = 0.298 m, r1 = 1, r2 = 300, and A1 = A2 = A
200
  0.085 A Wb
0.002 0.298

4  10  1  A 4  10 7  300  A
7

 0.085 A
Flux density B    0.085 Wb/m2
A A

51
Example2:A magnetic circuit comprises three parts in series, each of
uniform cross-sectional area (c.s.a.). They are:
(a) a length of 80 mm and c.s.a. 50 mm2,
(b) a length of 60 mm and c.s.a. 90 mm2,
(c) an airgap of length 0.5 mm and c.s.a. 150 mm2.
A coil of 4000 turns is wound on part (b), and the flux density in the airgap is
0.30 T. Assuming that all the flux passes through the given circuit, and that
the relative permeability μr is 1300, estimate the coil current to produce such
a flux density.

Flux passing through the air-gap,

  Bc Ac  0.3 150 106 Wb  0.45  104 Wb
MMF required by the section a,
la 0.45  104  80  103
Fa  Sa    AT  44.1 AT
 0 r Aa 410  1300  50  10
7 6

52
Example2:A magnetic circuit comprises three parts in series, each of
uniform cross-sectional area (c.s.a.). They are:
(a) a length of 80 mm and c.s.a. 50 mm2,
(b) a length of 60 mm and c.s.a. 90 mm2,
(c) an airgap of length 0.5 mm and c.s.a. 150 mm2.
A coil of 4000 turns is wound on part (b), and the flux density in the airgap is
0.30 T. Assuming that all the flux passes through the given circuit, and that
the relative permeability μr is 1300, estimate the coil current to produce such
a flux density.

MMF required by the section b,

lb 0.45  104  60  103

Fb  Sb    AT  18.4 AT
 0 r Ab 4 10  1300  90  10
7 6

MMF required by the section c,

lc 0.45  104  0.5  103

Fc  Sc    AT  119.3 AT
0 r Ac 4 10  1 150  10
7 6

53
Example2:A magnetic circuit comprises three parts in series, each of
uniform cross-sectional area (c.s.a.). They are:
(a) a length of 80 mm and c.s.a. 50 mm2,
(b) a length of 60 mm and c.s.a. 90 mm2,
(c) an airgap of length 0.5 mm and c.s.a. 150 mm2.
A coil of 4000 turns is wound on part (b), and the flux density in the airgap is
0.30 T. Assuming that all the flux passes through the given circuit, and that
the relative permeability μr is 1300, estimate the coil current to produce such
a flux density.

Total MMF required,

F  Fa  Fb  Fc   44.1  18.4  119.3 AT  181.8 AT

Now, F  IN
F 181.8
Or, I   A  45.4 mA
N 4000

54
Example3:In the magnetic circuit detailed in Fig. (a), with a uniform cross-
sectional area of 10cm2, the airgap length is 1 mm, the total mean length on
flux Φ path is 450mm and the same mean circumference of Φ1 path (or, Φ2
path) is 780 mm. Calculate the required number of turns to be placed on the
core in order to establish a flux of 0.8 mWb in the air gap. Moreover, the B-
H curve of the core material is given in Fig. (b). Neglect the fringing effect
and leakage flux.

55
Example3:In the magnetic circuit detailed in Fig. (a), with a uniform cross-
sectional area of 10cm2, the airgap length is 1 mm, the total mean length on
flux Φ path is 450mm and the same mean circumference of Φ1 path (or, Φ2
path) is 780 mm. Calculate the required number of turns to be placed on the
core in order to establish a flux of 0.8 mWb in the air gap. Moreover, the B-
H curve of the core material is given in Fig. (b). Neglect the fringing effect
and leakage flux.

Given,
Cross-sectional area a  10 cm 2 10  104 m 2
Air-gap length lg  1 mm  1 103 m
Length of core on flux Φ path
lc   450  1 mm  449 103 m
Length of core on flux Φ1 path

ls  780 mm  780  103 m

Flux in the air-gap  g  0.8 mWb  0.8  103 Wb
g 0.8 103
 Bg   Wb/m 2
 0.8 Wb/m 2

a 10 104 56
Example3:In the magnetic circuit detailed in Fig. (a), with a uniform cross-
sectional area of 10cm2, the airgap length is 1 mm, the total mean length on
flux Φ path is 440mm and the same mean circumference of Φ1 path (or, Φ2
path) is 780 mm. Calculate the required number of turns to be placed on the
core in order to establish a flux of 0.8 mWb in the air gap. Moreover, the B-
H curve of the core material is given in Fig. (b). Neglect the fringing effect
and leakage flux.

Flux in the central limb 0.8 mWb gets divided

into two equal paths.
Therefore, flux in side limbs
0.8
1   2  mWb  0.4 103 Wb
2
Flux density in central limb
g 0.8 103
Bc   Wb/m 2
 0.8 Wb/m 2

a 10 104
Flux density in side limbs
1 0.4 103
Bs   Wb/m 2
 0.4 Wb/m 2

a 10 104 57
Example3:In the magnetic circuit detailed in Fig. (a), with a uniform cross-
sectional area of 10cm2, the airgap length is 1 mm, the total mean length on
flux Φ path is 440mm and the same mean circumference of Φ1 path (or, Φ2
path) is 780 mm. Calculate the required number of turns to be placed on the
core in order to establish a flux of 0.8 mWb in the air gap. Moreover, the B-
H curve of the core material is given in Fig. (b). Neglect the fringing effect
and leakage flux.

Now, from Fig. (b),

Hc Bc  0.8Wb/m 2
 400 AT/m
Hs Bs  0.4Wb/m 2
 200 AT/m
Bg 0.8
Hg   AT/m
0 410 7

 636619.77 AT/m

58
Example3:In the magnetic circuit detailed in Fig. (a), with a uniform cross-
sectional area of 10cm2, the airgap length is 1 mm, the total mean length on
flux Φ path is 440mm and the same mean circumference of Φ1 path (or, Φ2
path) is 780 mm. Calculate the required number of turns to be placed on the
core in order to establish a flux of 0.8 mWb in the air gap. Moreover, the B-
H curve of the core material is given in Fig. (b). Neglect the fringing effect
and leakage flux.
Total MMF required,
F  H c lc  H g lg  H s ls
  400  449  636619.77  1  200  780   103 AT
 179.6  636.6  156  AT
 972.2 AT
Now, current through the coil, I  10 A
Hence, F  NI
F 972.2
Or, N    98 turns
I 10 59
Thank You

60