chapter 3_magnetic cct (1)

MAGNETIC CIRCUITS

LNI @ Network Analysis 1 2

INTRODUCTION Magnetic Circuit

A closed path containing magnetic flux. It also contain air gaps and other materials.


INTRODUCTION (Cont…)


PROPERTIES OF MAGNETIC CIRCUIT Flux Density, B

Occur when magnetomotive force (mmf) is applied to magnetic material.

It will induced flux, Unit in Webers(wb) The number of flux line

per area is called flux density, B.

Unit in Tesla

AB


PROPERTIES OF MAGNETIC CIRCUIT (cont..) Permeability,µ

Different magnetic material has it own strength of magnet.

This strength is vary depends: Type of core being used Number of flux line through the core.

Permeability of free space (vacuum) is denoted by µo.

0

70 /104

r

Amwb

Relative permeability, µr


AMPERE’S CIRCUITAL LAW Use the same concept as electric circuit.

Flux = current. It produced in a core Magnetomotive force (mmf)= electromotive

force (emf).


AMPERE’S CIRCUITAL LAW (cont…)

REI

ICurrentEemf

,

RNI

FluxNIFmmf

,

R = Reluctance

AlR


AMPERE’S CIRCUITAL LAW (cont…) Electric Circuit

Apply KVL ΣV = 0

Magnetic Circuit Apply Ampere’S

circuital Law ΣF = 0 The mmf drop is

HlF

H : magnetizing forcel:length of the section

(At)


Determine

BAA

B

LNIH

H can be obtain from B-H curve


cba

HlNIHlHlHlNI

F

ironcobaltsteel

)()()(

0


B-H CURVE (MAGNETIZING CURVE)


SERIES MAGNETIC CIRCUITS Reluctance is seldom determined. Flux can be determined using ohm’s law

analogy. The value of B and H is determined using

B-H curve and µ is calculated if asked for.


Example 2: For magnetic circuit below, find the value of I to develop a magnetic flux of 4 x 10-4wb and µ and µr for the material under these conditions.

Cast steelL = 0.16mA = 2 x 10-3 m2 NI

I

RN=400


TBA

B

2.0102104

3

4

Flux Density, B From B-H curve : H = 170 At/m

mAI

I

NHlI

HlNIF

68400

16.0170

83.9351041018.1

/1018.11702.0

7

3

3

or

AmwbHB

Permeability, µ

Solution:


Example 3:The electromagnetic has picked up a section of cast iron. Determine the current I required to establish the indicated flux of the core.

Sheet steel

Cast steel

GivenLab = Lcd = Lef = Lfa = 10cm

Lbc = Lde = 1cm

A = 4 x 10-4 m2

Flux = 3.5 x 10-4 wb

e

f

b

a

e c

N=50t


AIR GAPS Exist when there is an

opening at the electromagnetic device.

Air gaps will cause the flux to bends outwards. The flux density is not

uniformly equal to the flux density in the surrounding magnetic material (fringing).


AIR GAPS (cont…) Flux density, Bg Magnetizing force, Hg

g

gg AB

ggg

og

g

gg

lHF

BH

The permeability of air is equal to permeability of free space

The mmf, Fg across the air gapcoreg

coreg

AA

where


Example 4: Find the value of I required to establish a magnetic flux of 0.75 x 10-4 wb in a series magnetic circuit.

Given:Air gap = 2 x 10-3 mArea = 1.5 x 10-4 m2

Lcdfeab = 100 x 10-3 m

Lbc = 2 x 10-3 m

ba

e

c

f

d

I

200t

Cast steel


Solution

TA

B 5.0105.11075.0

4

4

Flux Density, B

From B-H curve : HCS = 280 At/m

mAtB

HO

gg /1098.3

1045.0 5

7

For air gap, Hg


AI

I

NHl

I

AtHl

Hl

HlHlHl

HlNIF

gCS

12.4200824

824

1021098.310100280 353

Ampere’s Circuital Law


SERIES PARALLEL MAGNETIC CIRCUITS The close analogies between electric and magnetic circuits will lead to series-parallel

magnetic circuits. The analogy will be helpful in solving the problem. Example: Find the value of I required to establish a magnetic flux of 1.5 x 10-4wb in cd. Given

that, lbcde = lefab = 0.2m, lbe = 0.05m, cross sectional area, A = 6 x 10-4 m2

Φ1

Φ2bI

50t

f

c

de

aSheet steel

Exercise 1


Exercise 2


Exercise 3


Exercise 4


Exercise 5


chapter 3_magnetic cct (1)

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