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BASIC ELECTRICAL TECHNOLOGY

DET 211/3

Chapter 4 – Magnetic Circuits

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Magnetic Materials and Circuits

IntroductionMagnet contains a north pole and south pole.

Magnet flux leaves the magnet as the north pole and the place where the flux returns to the magnet as the south pole.

Two types of magnet:

• Permanent magnet

• Electromagnet

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Right Hand Rule and Ampere’s Law

When a conductor carries current a magnetic field is produced around it.

Fingers– indicate current direction

Thumb – indicate the direction of magnetic flux is wrapping around the wire

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The relationship between current and magnetic field intensity can be obtained by using Ampere’s Law.

Right Hand Rule and Ampere’s Law

Ampere’s Law states that the line integral of the magnetic field intensity, H around a closed path is equal to the total current linked by the contour.

idl.H

H: the magnetic field intensity at a point on the contour

dl: the incremental length at that point

If θ: the angle between vectors H and dl then

icosHdl

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Right Hand Rule and Ampere’s LawConsider a rectangular core with N winding

NiHlc

Nii cldl

Therefore

clNiH

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Relationship between B-H

The magnetic field intensity, H produces a magnetic flux density, B everywhere it exists.

Tesla 2 or)m/weber(HB

T 20 or)m/wb(HB r

- Permeability of the medium

0 - Permeability of free space, m.t.Awb-710 x 4

0 r - Relative permeability of the medium

For free space or electrical conductor (Al or Cu) or insulators, is unity

r

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MAGNETIC EQUIVALENT CIRCUIT

A simple magnetic circuit having a ring shaped magnetic core (toroid) and a coil that extends around the entire circumference

When current i flows through the coil of N turns, a magnetic flux is produced.

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Assumption:

•All fluxes are confined to the core

•The fluxes are uniformly distributed in the core

MAGNETIC EQUIVALENT CIRCUIT

The flux outside the toroid (called leakage flux), is so small (can be neglected)

Use Ampere’s Law,

Nidl.H

Nir.H 2NiHl

FNiHl

F = Magnetomotive force (mmf)

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MAGNETIC EQUIVALENT CIRCUIT

)m/At(l

NiH

HB

)T(lNiB

Where;

N – no of turns of coil

i – current in the coil

H – magnetic field intensity

l – mean length of the core

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Magnetic Flux Density (B) and Magnetizing Curve (B-H Curve)

Case 1: Non magnetic material core (Cu, Al, air, plastic, wood,..)

HB 0

)T(lNiB 0

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Magnetic Flux Density (B) and Magnetizing Curve (B-H Curve)

Case 2: Ferromagnetic material core (iron, steel, ferrite,..)

HHB r 0

)T(l

NiB r 0

The magnetic flux density, B increases almost linear in the region of low values of magnetic intensity, H.

At higher value of H, the change of B is nonlinear.

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MAGNETIC EQUIVALENT CIRCUIT

The flux in the coil,

)weber(BA

RF

RNi

AlNiA

lNi

Where

Ф – flux in the coil (wb)

F – magnetomotive force (mmf)

R – 1/μA = 1/P ,Reluctance

P = permeance

A – cross sectional area

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ANALOGY BETWEEN MAGNETIC CIRCUIT AND ELECTRIC CIRCUIT

a) Magnetic equivalent circuit b) Electric equivalent circuit

To solve magnetic equivalent circuit – Kirchhoff Voltage and Current Laws (KVL & KCL)

Electric circuit Magnetic circuitDriving force EMF (E) MMF (F)

Produces Current (i) Flux (Ф)

Limited by Resistance (R) Reluctance (R)

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MAGNETIC CIRCUIT WITH AIR GAP

In electric machines, the rotor is physically isolated from the stator by the air gap.

Practically the same flux is present in the poles (made by magnetic core) and the air gap.

To maintain the same flux density, the air gap will require much more mmf than the core.

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MAGNETIC CIRCUIT WITH AIR GAP

cc

cc A

lR

gg

gg A

lR

gc RRNi

ggcc IHlHNi

Where lc – mean length of the core

lg – the length of the air gap

c

cc A

B

g

gg A

B

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FRINGING EFFECT

Fringing Effect: Bulging of the flux lines in the air gap.

Effect: The effective cross section area of air gap increase so the reluctance of the air gap decrease. The flux density Bg < Bc, Bc is the flux density in the core.

If the air gaps is small, the fringing effect can be neglected. So

cg AA

ccg A

BB

In practical, large air gap will be divided into several small air gaps to reduce the fringing effect.

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INDUCTANCE

A coil wound on a magnetic core as shown in figure above, is frequently used in electric circuits. This coil may be represented by an ideal circuit element, called inductance, which is defined as the flux linkage of the coil per ampere of its current.

N

iL

Flux linkage

Inductance

RN

Al

N

NHlHAN

iHAN

iNBA

iNL

22

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15cm

30cm

15cm

15cm 10cm

l2

30cm

30cm15cm 10cm

N=200turns

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Example

l1

A ferromagnetic core is shown in Figure above. Three sides of this core are of uniform width, while the fourth side is somewhat thinner. The depth of the core

(into the page) is 10cm and the other dimensions are shown in figure. There is 200 turn coil wrapped around the left side of the core. Assuming relative permeability

r of 2500, how much flux will be produce by a 1A input current?

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Solution Example

The mean path length of region 1 is 45cm and the cross-sectional area is 10 x 10 cm = 100cm2. Therefore, the reluctance in the first region is:

WbturnsAmx

mA

lAlR

or

/.300,14)01.0)(104)(2500(

45.027

1

1

1

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The mean path length of region 2 is 130cm and the cross-sectional area is 15 x 10 cm = 150cm2. Therefore, the reluctance in the second region is:

WbturnsAmx

mA

lAlR

or

/.600,27)015.0)(104)(2500(

3.127

2

2

2

22

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Therefore, the total reluctance in the core is:

Solution Example

WbturnsAWbturnsAWbturnsA

RRReq

/.900,41/.600,27/.300,14

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turnsAAturnsNIF .200)0.1)(200(

The total magnetomotive force (MMF) is:

The total flux in the core is given by:

WbWbturnsA

turnsARF

0048.0/.900,41

.200

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Assignment 3

A ferromagnetic core is shown in Figure above. All side of this core are uniform width. The depth of the core is 10cm. Assuming relative permeability r of 2500, how much flux will be produce by a 1A input current?

15cm

30cm

15cm

15cm 15cm

lc

30cm

30cm15cm 15cm

N=200turns

i