Magnetic Circuits
Introduction
• The region around a magnet where its poles
exhibit a force of attraction or repulsion is
called magnetic field.
• Although magnetic flux lines have no real
existence, yet their concept is very useful to
understand various magnetic effects.
• The number of magnetic flux lines is called
magnetic flux φ, with the unit Webber (Wb).
• The direction of magnetic flux lines is from
N-pole to the S-pole outside the magnet. But
inside the magnet their direction is from S-
pole to N-pole forming a closed path.
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Ohm’s Law for Magnetic Circuits
• The closed path followed by magnetic flux
is called a magnetic circuit.
• The Figure shows a solenoid having N
(turns) wound on an iron core. When
current I (A) is passed through the solenoid,
magnetic flux φ (Wb) is set-up in the core.
Ohm’s law for magnetic circuits:
𝑁𝐼 = φ𝑅 Reluctance
Magnetomotive force (mmf) Flux direction is determined by right hand rule
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Ohm’s Law for Magnetic Circuits
mmf is the driving force that produces the flux in
a magnetic circuit (A.turn).
Reluctance is the opposition of a magnetic path to
the flow of flux lines (A.turn/Wb).
Let 𝑙 = mean length of magnetic circuit in (𝑚);
𝑎 = cross-sectional area of core in (𝑚2 );
μ= Permeability of core material (Wb/A.turn.m).
Reluctance is defined by:
𝑙
𝑅=
μ𝑎
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Ohm’s Law for Magnetic Circuits
• Permeability (μ) is the measure of the ease of the flow of magnetic flux lines in a
material. The higher the value the easier the flow of flux in the material.
• Air permeability is the poorest and is given by: 𝝁𝟎 = 𝟒𝝅𝟏𝟎−𝟕 (Wb/A.turn.m).
• The ratio of the material permeability (μ) to the permeability of air or vacuum (μ0 ) is
called the relative permeability (μ𝑟 ) of the material (unitless).
• Materials with high relative permeability are termed ‘ferromagnetic materials’.
• On the other hand, Non-magnetic materials have unity relative permeability (𝜇𝑟 = 1)
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Example (1)
An iron ring of 400 cm mean circumference is made from round iron of cross-section 20𝑐𝑚2 . Its
relative permeability is 500. If it is wound with 400 turns, what current would be required to produce a
flux of 0.001 Wb?
Mean length of magnetic path, 𝑙= 400 cm = 4 m
Area of iron ring, a = 20𝑐𝑚2 = 20 . 10−4 𝑚2
Absolute permeability, μ0 = 4π . 10−7
Ohm’s law for magnetic circuits:
𝑁𝐼 = φ𝑅
Reluctance is defined by:
𝑙
𝑅=
μ𝑎
4
Therefore, 400𝐼 = 0.001 I=7.96A
(4π . 10−7 )(500)(20 . 10−4 )
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Example (2)
A coil of insulated wire of 500 turns and of resistance 4Ω is closely wound on iron ring. The ring has a
mean diameter of 0.25 m and a uniform cross-sectional area of 700 mm2. Calculate the total flux in the
ring when a DC supply of 6V is applied to the ends of the winding. Assume a relative permeability of 550.
Mean length of magnetic path is the ring circumference, 𝑙 = π𝐷 = 0.25 π m
Area of iron ring, a = 700𝑚𝑚2 = 700 . 10−6 𝑚2
Absolute permeability, μ0 = 4π . 10−7
𝑉 6
Current flowing in the coil 𝐼 = = = 1.5𝐴
𝑅 4
Ohm’s law for magnetic circuits:
𝑁𝐼 = φ𝑅
Reluctance is defined by:
𝑙
𝑅=
μ𝑎
0.25 π
Therefore, 500 . (1.5) = φ φ = 0.46 𝑚𝑊𝑏
(4π . 10−7 )(550)(700 .10−6 ) 7
Magnetic and Electric Circuits
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Series Magnetic Circuits
• A magnetic circuit that has a number of parts of different dimensions and/or materials
carrying the same magnetic field is called a series magnetic circuit.
R2
R1
R3
φ
mmf
Rg 9
Example (3)
An iron ring is made up of three parts; 𝑙1 = 10 cm, 𝑎1 = 5 𝑐𝑚2 ; 𝑙2 = 8 cm, 𝑎2 = 3 𝑐𝑚2 ; 𝑙3 = 6 cm, 𝑎3 =
2.5 𝑐𝑚2 . It is wound with a 250 turns coil. Calculate current required to produce flux of 0.4 mWb,
knowing that μ1 = 2670, μ2 = 1050, and μ3 = 600.
R2
R1
R3
φ
mmf
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Example (3)
Ohm’s law for magnetic circuits:
𝑁𝐼 = φ𝑅𝑒𝑞 R2
where, 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3
𝑙1 10 . 10−2 R1
𝑅1 = =
μ1 𝑎1 (4π . 10−7 )(2670)(5 . 10−4 ) R3
−2 φ
𝑙2 8 . 10
𝑅2 = =
μ2 𝑎2 (4π . 10−7 )(1050)(3 . 10−4 ) mmf
𝑙3 6 . 10−2
𝑅3 = =
μ3 𝑎3 (4π . 10−7 )(600)(2.5 . 10−4 )
Therefore, 𝐼 = 0.93𝐴 11
Parallel Magnetic Circuits
• A magnetic circuit which has two or more than two paths for the magnetic flux is called a
parallel magnetic circuit.
φ3 φ2
R1
R3 R2
φ
1
mmf
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Example (4)
The magnetic structure of a synchronous machine is shown in Figure. Assuming the stator and rotor iron have
infinite permeability. Find the air gap flux given that N=1000 turns, I=10A, 𝑔= 1 cm, and 𝐴𝑔 = 2000 𝑐𝑚2
φ3 φ2
R
g2
Rrt
Rst Rst
φ
1
mmf
Rg1
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Example (4)
Ohm’s law for magnetic circuits: Parallel branches
𝑁𝐼 = φ𝑅𝑒𝑞
𝑅𝑠𝑡 𝑅𝑠𝑡
where, 𝑅𝑒𝑞 = 𝑅𝑔1 + 𝑅𝑔2 + 𝑅𝑟𝑡 + R g2
𝑅𝑠𝑡 +𝑅𝑠𝑡
Note that the permeability of the iron core is infinity, hence φ
𝑅𝑠𝑡 = 𝑅𝑟𝑡 = 0
mmf
The two air gaps are identical, hence
𝑔 1 . 10−2
𝑅𝑔1 = 𝑅𝑔2 = =
μ0 𝐴𝑔 (4π . 10−7 )(1)(2000 . 10−4 ) Rg1
Therefore, φ = 0.13𝑊𝑏
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