Lecture 3 Magnetic Circuits

Lecture 3
Magnetic Circuits
Electromagnetic fields play an important role in the conversion of electromechanical
energy in the rotating machines and transformers. In this lecture some basic concepts
of electromagnetic theory are reviewed, typical magnetic circuits are analyzed and
other parameter, including flux linkages and inductances, are defined.
Production of Magnetic Field
When a current i flows through a conductor, magnetic fluxes are generated and wraps
around in a similar manner as that of direction of travel of cork-screw when turned in
clock-wise direction.
The study and the design of electrical machinery include the analysis of the magnetic
circuits involved in these machines. A magnetic circuit consists of a magnetic
structure built mainly of high permeability magnetic material. Thus, magnetic flux is
confined to the paths presented by the high-permeability material, just as the electric
current is confined to the paths presented by the high-conductivity conductors of the
electric circuit.
Consider the magnetic core shown in the diagram of Fig 2. It has a current (I)
carrying winding of N turns that generates a magnetomotive fource (mmf) F of NI.
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This mmf creates a magnetic field in the core having an intensity of H c ampereturn/meter
along the length of the magnetic path. According to Ampere’s law
∮ (1)
If H c is constant, and l c is the mean path length of the core, then (1) gives,
H c l c =NI=F= mmf (2)
The core is usually made of ferromagnetic material. The magnetic flux density B
(weber/m 2 ) in the core is related to the magnetic field H according to the saturation
curve, or B-H curve (Fig 3). The slope of this is designated as µ, the permeability of
the material, thus giving relationship between B and H as
B = µH (3)
As seen from Fig 3, the slope of the curve that µ is not constant and depends of the
operating value of magnetic flux density ( ). The permeability of a magnetic material
is usually given relative to the permeability of the free-space, µ 0 . Thus
µ= µ 0 µ r . (4)
where, µ r is the relative permeability. In SI units, the permeability of the free-space is
µ 0 = 410 -7 H/m.
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Fig 3: Typical B-H curves of different materials.
The magnetic flux (expressed in webers) through a given surface is found as
follows:
∫ (5)
Since the flux density in the core B c has the same direction as
the cross-sectional area, A c , Equation (5) reduces to
and is uniform over
F NI Hclc
NI
Bc Ac Hc Ac Ac
NI
l
H
c
c
l
l
Ac
F
c
NI
c
(6)
lc
= reluctance of the magnetic circuit in AT/wb. Eq (6) is analogous to Ohm’s
A
c
law for resistive circuits.
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Table : Analogy between Magnetic and Electric circuits
Electric circuit
Magnetic circuit
I=current (A)
= flux (wb)
V= emf (V)
F = mmf (AT)
R= resistance ()
R = reluctance (AT/wb)
= conductivity (S/m) µ = permeability (H/m)
Self study: Example 4.1, 4.2, 4.4
Magnetic Circuit with Air-gap and Flux Fringing
At an air-gap in a magnetic core, the flux fringes out into neighboring air paths as
shown in Figure. The result is non-uniform flux density in the air-gap (decreasing
outward), enlargement of the effective air-gap area and a decrease in the average gap
flux density. The fringing effect also disturbs the core flux pattern to some depth near
the gap. The effect of fringing increases with the air-gap length.
Fig: Flux fringing at the air-gap
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If fringing is neglected, then A g =A c ; B g =B c , g = c ,
but still H g H c and also R g R c due to change in permeability in core and air-gap.
Series Parallel Magnetic Circuits
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Similar to Kirchhoff’s voltage law, we may write, for any closed path in a magnetic
circuit,
Around a closed path in a magnetic circuit the algebraic sum of ampere-turns is equal
to the algebraic sum of the products of the reluctances and fluxes.
Again, similar to Kirchhoff’s current law for a junction, for any closed path in a
magnetic circuit,
Which states that the algebraic sum of all the magnetic fluxes flowing out of a
junction in a magnetic circuit is zero.
Problems
D4.1 The magnetic circuit shown in Fig. below has an air gap cut in the right leg of
the core. The air gap is 0.1mm long. The coil is connected to a voltage source, and the
current drawn is adjusted so that the magnetic flux density in the air gap is 1.5T.
Assume that flux fringing in the air gap is negligible. The magnetic circuit has the
following dimensions: A c =16cm 2 , l c =40cm, and N=350 turns. The relative
permeability of the core is µ r =50,000.
(i)
(ii)
(iii)
Find the value of the current.
Calculate the magnetic flux
Determine the flux linkage of the coil
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Solution:
Here, l c = 40cm=0.4m; A c =16cm 2 = m 2 , B g =B c = 1.5T
Reluctance of the core,
Reluctance of the air gap,
Total Reluctance, R t =R c +R g = 3,979+49,736=53,715 AT/wb
(i) Flux, =BA=1.5 =2.4 mWb
(ii)
(iii)
Current
Flux,
(iv) Flux linkage, =N =350 =0.84 wb-t
Problem 4.2 is similar to D4.1.
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D4.2 A magnetic core is built in the form of a circular ring having a mean radius of 10
cm. A coil containing 150 turns is wound uniformly throughout the length of the core.
The coil is connected to a voltage source, and it draws a current of 15A.
a. Determine the of the coil
b. Calculate the magnetic field intensity in the core
Solution:
Given, N=150 turns; I =15A; r = 10cm = 0.1m
a.
b. l c = 2r = 20.1=0.628 m
c.
D4.3 The circular ring of has a mean cross-sectional area of
25cm 2 . The relative permeability of the material of the ring is 1500. Calculate
a. The magnetic flux in the core
b. The magnetic flux density in the core
c. The flux linkage of the coil
d. The reluctance of the core
Solution:
Here, A c = 25cm 2 =2510 -4 m 2 ; µ r =1500
a. B=µH =µ 0 µ r H =410 -7 15003581=6.75 T
b. =BA=6.752510 -4 wb=16.875 mwb
c. =N=150 =2.53 wb-t
d.
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D4.4 A magnetic core has a circular cross-sectional area of 2.0 inch 2 , a mean path
length of 10 inch, and an air-gap length of 0.125 inch. A 350-turn coil is wound
around the magnetic core, and a current of 5A is supplied to the coil. Assume that the
relative permeability of the core is and fringing of the flux in the air gap is
negligible.
a. Calculate the reluctance of the magnetic circuit
b. Find the magnetic flux density in the air gap
Solution:
A = 2.0 inch 2 = ( )
g = 0.125 inch =
N = 350 turn; I = 5A
a.
b. =1.96
c.
D4.5 Repeat assuming that the core has a relative permeability
µ r =5000.
Solution:
a. µ r =5000
AT/wb
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. =1.96 AT/wb
c. Total Reluctance, R t =R c +R g = +1.96 =1.99 AT/wb
Flux,
=0.68 T
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