ELECTROMAGNETISM
E2063/Unit 1/ 1
UNIT 1 ELECTROMAGNETISM
OBJECTIVES
General Objective
: To understand the basic principles of electromagnetism.
Specific Objectives
: At the end of the unit unit you will be able to :
Explain the relation between current-carrying conductor and magnetism.
Appreciate factors that influence magnetic field strength.
State magnetic quantity characteristics and use their relevant formulae.
Describe electromagnetic induction and factors that influence the value of induced current.
ELECTROMAGNETISM
E2063/Unit 1/ 2
INPUT
1.0
INTRODUCTION
The word magnet comes from THE word ‘magnetic’ (lodestrone/black stone) found in Magnesia, Turkey. There are two types of magnet i.e. permanent magnets and temporary magnets (electromagnet). A permanent magnet is a piece of ferromagnetic material (such as iron and metal) which has properties of attracting other pieces of these materials. A permanent magnet will position itself in a north direction (north pole) and south direction (south pole) when freely suspended. When we magnetize magnetic materials (such as iron), it becomes a temporary magnet (electromagnet). For instance, an electromagnet can be produced by flowing a current through a solenoid. The solenoid is very important in electr electroni onicc theory theory.. An electr electromag omagnet net provid provides es the basic basic of many many items items of electr electrica icall equipment, examples of which include electric bells, relay and telephone receiver. The distribution of a magnetic field can be investigated. investigated. The pattern of magnetic field of bar a magnet shown shown in Figure 1.1. The direction of a line of flux is from from the north north pole to the south pole on the outside of the magnet. The laws of magnetic attraction and repulsion can be demonstrated by using two bar magnets. In Fig. 1.2 (a), with unlike poles adjacent, attraction takes place. In Fig 1.2(b), with similar poles adjacent, repulsion occurs.
Source: Electrical And Electrical Principles and Technology by John Bird
ba r a magnet. Figure 1.1 The pattern of magnetic field of bar
ELECTROMAGNETISM
E2063/Unit 1/ 3
Source: Electrical And Electrical Principles and Technology by John Bird
Figure 1.2 The magnetic attraction and repulsion by using two bar magnets.
Most of the electrical apparatus that we use everyday use the electromagnetic principle. Magnetic effect is produced when this electric component is connected to a power supply. For example, magnetic effect that is produced by electric motor can causes fan to rotate.
1.1
CURRE CURRENTNT-CA CARR RRYI YING NG CONDU CONDUCT CTOR OR AND AND EL ELEC ECTR TROM OMAG AGNE NETI TISM SM
A flow of current through a wire produces a magnetic field in a circular path around the wire.The direction of magnetic line of flux around the wire is best remembered by the screw rule or the grip rule.
The screw rule states that if a normal right and thread screw is screwed along the conductor in the direction of the current, the direction of rotation of the screw is in the direction of the magnetic field.
ELECTROMAGNETISM
E2063/Unit 1/ 4
The field pattern of a current flowing to a conductor is illustrated in Figures 1.3(a) and 1.3(b)
Source: Fundamental Electrical & Electronic Principles by Christopher R Robertson
Figure 1.3 The field pattern of current flowing.
If two closed current-carrying conductors flow in the same direction, magnetic flux around that conductor will combine to create create attraction between them. If closed currentcarrying conductors flow in opposite direction, these two conductors will repulse each other. This effect is shown in Figure 1.4.
(a)
(b)
Source: Pengajian Kejuruteraan Elektrik dan Elektronik,Cetakan Pertama 2000 oleh Abd. Samad Hanif
Figure 1.4
Two closed current-carrying conductors flow in the same direction (a) flow and in opposite direction (b).
ELECTROMAGNETISM
1.2.1
E2063/Unit 1/ 5
MAGNET MAGNETIC IC FIEL FIELD D STRENG STRENGTH, TH, H (MAGNE (MAGNETISI TISING NG FORC FORCE) E)
Magnetic field strength is defined as magnetomotive force, Fm per metre length of measurement being ampere-turn per metre. H =
F m l
=
NI ampere turn / metre l
where Fm N I l
-
magnetomotive force number of turns current average le length of of magnetic ci circuit
Example 1.1
A current of 500mA is passed through a 600 turn coil wound of a toroid of mean diameter 10cm. Calculate the magnetic field strength.
Solution to Example 1.1
I N d
H =
= = =
NI l
0 .5 A 600 π x 10 x 10-2m
ampereturn / metre
=
600 600 × 0.5 0.3142
=
954 954 .81 AT/m
Example 1.2
An iron ring has a cross-sectional area of 400 mm2 and a mean diameter of 25 cm. it is wound with 500 turns. If the value of relative permeability is 250, find the total flux set up in the ring. The coil resistance is 474 Ω and the supply voltage is 20 V.
Solution to Example 1.2
ELECTROMAGNETISM
E2063/Unit 1/ 6
The condition of the problem are represented in Fig. 1.5
Fig. 1.5
I = V/ R = 240 / 474 = 0.506 A -2 l = π D = π × (25 × 10 ) = 0.7854 m H=
NI l
=
500
×
0.506
0.7854
= 322.13 AT/m
ELECTROMAGNETISM
E2063/Unit 1/ 7
Activity 1A
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!
1.1
Find Find the magnet magnetic ic field field strengt strength h appl applied ied to a magn magneti eticc circu circuit it of mean mean lengt length h 50 cm when a coil of 400 turns is applied to it carrying a current of 1.2 A.
1.2
A curr current ent of 2.5A 2.5A when when flowin flowing g thro through ugh a coil coil produc produces es an m.m. m.m.f. f. of 675 At. Calculate the number of turns on the coil.
1.3
A magne magnetiz tizing ing force force 8000 8000 A/m is applied applied to a circu circular lar magnet magnetic ic cir circui cuitt of mean mean diameter 30 cm by passing a current through a coil wound on the circuit. If the coil is uniformly wound around the circuit and has 750 turns, find the current in the coil.
ELECTROMAGNETISM
E2063/Unit 1/ 8
Feedback to the Activity 1A
1.
960A/m
2.
270 turns
3.
10.05 A
4.
2547 AT
ELECTROMAGNETISM
E2063/Unit 1/ 9
INPUT
1.3 MAGNETIC QUANTITY CHARACTERISTICS AND THEIR RELEVANT
FORMULAE 1.3.1 1.3.1
Magne Magnetic tic Flux Flux and and Flux Flux den densit sity y
Magnetic flux is the amount of magnetic filed produced by a magnetic source. The symbol for magnetic flux is Φ . The unit for magnetic flux is the weber, Wb. Magnetic flux density is the amount of flux passing through a defined area that is perpendicular to the direction of flux: Magnetic flux density =
magnetic flux
B=
area Φ A
Tesla
The symbol for magnetic flux density is B. The unit of magnetic flux density is the tesla, T, and the unit for area A is m2 where 1 T = 1 Wb/m. Example 1.3
A magnetic pole face has rectangular section having dimensions 200mm by 100mm. If the total flux emerging from the the pole is 150µ Wb, calculate the flux density. Solution to Example 1.3
Magnetic flux, Φ = 150 µ Wb = 150 x 10-6 Wb Cross sectional area, A = 200mm x 100mm = 20 000 x 10-6m2 Flux density, B =
Φ A
=
150 150 × 10 −6 20000 ×10 −6
= 7.5 mT
ELECTROMAGNETISM
1.3.2
E2063/Unit 1/ 10
Permeability
For air, or any other non-magnetic medium, the ratio of magnetic flux density to magnetic field strength is constant , i.e.
B H
= a constant. This constant is called the
permeability of free space and is equal to 4π x 10-7 H/m. For air and any other o ther nonmagnetic medium, the ratio B H
= µ 0
For all media other than free space B H
= µ 0 µ r
where µ r is the relative permeability and is defined as µ r
=
flux density in material flux density in vacuum
varies with the type of magnetic material. From its definition, µ r for a vacuum is 1. µ µ r is called the absolute absolute permeability. permeability. The approximat approximatee range of values of relative relative permeability µ r for some common magnetic materials are : µ r 0
Cast iron Mild steel Cast steel
µ
r
µ
r
µ
r
= 100 – 250 = 200 – 800 = 300 – 900
Example 1.4
A flux density of 1.2 T is produced in a piece of cast steel by a magnetizing force of 1250 A/m. Find the relative permeability of the steel under these conditions.
Solution to Example 1.4 B
= µ 0 µ r H
µ r
=
B µ 0 H
=
1.2 ( 4π ×10
2
−
)(1250 )
= 764
ELECTROMAGNETISM
1.3.4
E2063/Unit 1/ 11
Reluctance
Reluctance,S is the magnetic resistance of a magnetic circuit to presence of magnetic flux. Reluctance, S =
F m Φ
=
Hl BA
=
l ( B / H ) A
=
1 µ 0 µ r A
The unit for reluctance is 1/H or H-1 or A/Wb Ferromagnetic materials have a low reluctance and can be used as magnetic screens to prevent magnetic fields affecting materials within the screen.
Example 1.5
Determine the reluctance of a piece of mumetal of length 150mm when the relative permeability is 4 000. Find also the absolute permeability of the mumetal.
Solution to Example 1.5
Reluctance,
S
=
=
1 µ 0 µ r A
150 ( 4π ×10
7
−
10
×
3
−
)( 4000 )(1800
= 16 580 H-1 Absolute permeability, µ = µ 0 µ 7 = ( 4π ×10 )( 4000 ) = 5.027 x 10-3 H/m r
−
1.4 1.4
ELEC EL ECTR TROM OMAG AGNE NETI TIC C INDU INDUCT CTIO ION N
10
×
6
−
)
ELECTROMAGNETISM
E2063/Unit 1/ 12
When a conductor is moved across a magnetic field so as to cut through the flux, an electromagnetic force (e.m.f.) is produced in the conductor. This effect is known as electromagnetic induction. The effect of electromagnetic induction will cause induced current. Two laws of electromagnetic induction i. Faraday’s law It is a relative movement of the magnetic flux and the conductor then causes an e.m.f. and thus the current to be induced in the conductor. Induced e.m.f. on the conductor could be produced by two methods i.e. flux cuts conductor or conductor cuts flux. a.
Flux cuts conductor
When the magnet is moved towards the coil ( Fig 1.5 ), a deflection is noted on the galvanometer showing that a current has been produced in the coil.
Source: Pengajian Kejuruteraan Kejuruteraan Elektrik dan Elektronik, Cetakan Pertama 2000 oleh Mohd. Isa bin Idris
Figure 1.5 Flux cuts conductor
b.
Conductor cuts flux
When the conductor is moved through a magnetic field (Fig. 1.6 ). An e.m.f. is induced in the conductor and thus a source of e.m.f. is created between the ends of the conductor. (This is the simple concept of AC generator) g enerator) This induced electromagnetic field field is given by E = Bl v volts where B l v
= = =
flux density, T length of the conductor in the magnetic field, m conductor velocity, m/s
ELECTROMAGNETISM
E2063/Unit 1/ 13
If the conductor moves at the angle θ ° to the magnetic field, then E = Bl v sinθ ° volts
Source: Pengajian Kejuruteraan Kejuruteraan Elektrik dan Elektronik, Cetakan Pertama 2000 oleh Mohd. Isa bin Idris
Figure 1.6 Conductor cuts flux
Example 1.6
A conductor 300mm long moves at a uniform speed of 4m/s at right-angles to a uniform magnetic field of flux density 1.25T. Determine the current flowing in the conductor when (a) its ends are open-ci open-circuite rcuited d (b) its ends are connected to a load of 20 Ω resistance.
Solution To Example 1.6
When a conductor moves in a magnetic field it will have an e.m.f. induced in it but this e.m.f. can only produce a current if there is a closed circuit. Induced e.m.f. E = Blv =(1.25)(300/1000)(4) (a) If the the ends of the conductor are open circuited no current will will flow flow even though 1.5 V has been induced.
(b) From From Ohm’s Ohm’s law law I
E =
R
=
1.5 20
75 mA
ELECTROMAGNETISM
ii.
E2063/Unit 1/ 14
Lenz’z Law
The direction of an induced e.m.f. is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f.. This effect is shown in Fig. 1.7
Source: Pengajian Kejuruteraan Kejuruteraan Elektrik dan Elektronik, Cetakan Pertama 2000 oleh Mohd. Isa bin Idris
Figure 1.7 Bar magnet move in and move out from a solenoid
ELECTROMAGNETISM
E2063/Unit 1/ 15
Activity 1B
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…! 1.4
The The maxim maximum um work workin ing g flux flux dens densit ity y of a lift liftin ing g elect electro roma magne gnett is 1.8 1.8 T and the the effective area of a pole face is circular in cross-section. If the total magnetic flux produced is 353 mWb, determine the radius of the pole.
1.5
Determine Determine the magnetic magnetic field field strength strength and the m.m.f. m.m.f. require required d to produce produce a flux density of 0.25T in an air gap of length 12mm.
1.6
A coil coil of 300 300 turns turns is is wound wound unif unifor orml mly y on a ring ring of non non-m -magn agnet etic ic mate materi rial al.. The ring has a mean circumferen circumference ce of 40 cm and a uniform cross-sec cross-sectional tional area area of 2 4cm . If the current in the coil is 5 A, calculate (a) the magnetic field strength, strength, (b) the flux density and (c) the total magnetic flux in the ring.
1.7
A mild mild stee steell ring ring has has a radi radius us of 50m 50mm m and and a cross cross-s -sec ecti tion onal al area area of 400m 400mm2 m2.. A current of 0.5 A flows in a coil wound uniformly around the ring and the flux produced is 0.1 mWb. If the relative permeability at this value of current is 200, find (a) the reluctanc reluctancee of the mild mild steel steel (b) the number number of turns turns on the coil coil
1.8 At what velocity must a conductor 75mm long cut a magnetic flux of density 0.6T if
an e.mf. of 9V is to be included in it? Assume the conductor, the field and the direction of motion are mutually perpendicular. 1.9 A conductor moves with a velocity velocity of 15m/s at an angle (a) 90° (b) 60°and (c)
30° to a magnetic field produced between two square-faced poles of side length 2cm. If the flux leaving a pole face is 5 µ Wb, find the magnitude of the induced e.m.f. in each case.
1.10 1.10
A fl flux dens densiity of of 1.2 1.2wb wb/m /m2 is required in the 2 mm air gap of an electromagnet having an iron path 1 m long. Calculate the m.m.f. required, assuming a relative permeability of iron as 1500. Neglect leakage.
ELECTROMAGNETISM
E2063/Unit 1/ 16
Feedback to the activity 1B
1.4 0.1961 m 1.5 1.5 2378 2378 A 1.6 1.6 (a) 3750 3750 A/m A/m (b) (b) 4.71 4.712m 2mT T
(c) 1.88 1.885 5 µ Wb
1.7 (a) S= 3.125 x 106/H (b) N = 625 turns 1.8 200 m/s 1.9 (a) 3.75 mV (b) 3.25 mV (c) 1.875 mV 1.10 2547 AT
ELECTROMAGNETISM
E2063/Unit 1/ 17
SELF-ASSESSMENT 1
You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 1 given on the next page. If you face any problems, discuss it with your lecturer. Good luck. Question 1-1 (a)
What is the flux density in a magnetic field of cross-sectional area 20 cm2 having a flux of 3mWb.
(b) (b)
Deter Determi mine ne the the tot total al flu flux x emer emergi ging ng fro from m a magne magneti ticc pole pole face face hav havin ing g dimensions 5 cm by 6 cm, if the the flux density is 0.9 T.
(c) (c)
A sole soleno noid id 200 200 cm cm long long is is woun wound d with with 500 500 tur turns ns of of wire wire.. Find Find the the cur curre rent nt required to establish a magnetizing magnetizing force of 2500 A/m inside the solenoid. solenoid.
(d) (d)
Find Find the the magn magnet etic ic fie field ld str stren engt gth h and the the magn magnet etom omot otiv ivee force force need needed ed to to produce a flux density of 0.33 T in an air-gap of length 15mm.
Question 1-2
(a)
An air-gap between two pole pieces is 20mm in length and the area of the flux path across the gap is 5cm2. 5 cm2. If the flux required in the air-gap a ir-gap is 0.75 mWb find the m.m.f. necessary.
(b)
Find the relative permeability of a material if the absolute permeability is 4.084x10-4 H/m.
(c)
Part of magnetic circuit is made from steel of length 120mm, cross sectional area 15cm2 and relative permeability 800. Calculate (i) the reluctance (ii) the relative permeability of the steel.
(d) A conduct conductor or of length length 15cm 15cm is moved at 750mm/ 750mm/ss at right-an right-angle gless to a unifor uniform m flux density of 1.2 T. Determine the e.m.f induced in the conductor.
ELECTROMAGNETISM
E2063/Unit 1/ 18
(e) Find the speed that a conductor of length 120mm must be moved at right right angles to a magnetic field of flux density density 0.6 T to induce in it it an e.m.f of 1.8 V. (f) A 25 cm long conduc conductor tor moves moves at a unifor uniform m speed speed of 8 m/s throu through gh a uniform uniform magnetic field of flux density 1.2T. Determine the current flowing in the conductor conductor when (i) its ends are open-circui open-circuited ted (ii) its ends are connected to a load of 15 ohms resistance. (g) A conductor of length 0.5 m situated in and at right angles to a uniform magnetic
field of flux density 1 wb/m2 moves with a velocity of 40 m/s. Calculate the e.m.f e.m.f indu induce ced d in the the condu conduct ctor or.. What What will will be the the e.m. e.m.ff induc induced ed if the the conductor moves at an angle 60º to the field. (h) A circular iron ring has a mean circumference of 1.5 m and cross-sectional area of
0.01 m2. A saw saw cut cut of 4 mm wide wide is made made in the the ring ring.. Calc Calcul ulat atee the the magnetizing current required to produce a flux of 0.8 mwb in the gap if the ring is wound with a coil of 175 turns. Assume relative permeability of iron as 400 and leakage factor 1.25. 2
(i) An iron ring has a mean diameter of 15 cm, a cross-sectional area of 20 cm and a
radial gap of 0.5 mm cut in it. It is uniformly wound with 1500 turns of insulated wire and a magnetising current of 1 A produces a flux of 1 mwb. Neglecting the effect of magnetic leakage and fringing, calculate: (i)
relucta ctance of ma magnet netic ci circuit.
(ii (ii)
rel relati ative perm permea eabi bili litty of of iron iron..
ELECTROMAGNETISM
E2063/Unit 1/ 19
FEEDBACK TO SELF-ASSESSMENT 1
Have you tried the questions????? If “YES”, check your answers now. Answer of Question 1-1
(a )
1.57 T
(b)
2.7mWb
(c )
1A
(d)
(i) 262 600 A/m
(ii)3939 A
Answer of Question 1-2
) )
2370A 325 (c) (i) 79580 580/H
(ii) 1 mH/m
(d) (d) 0.13 0.135V 5V (e) (e) 25m/ 25m/ss (f) (i) 0 (ii) 0.16 A (g) (i) 20 V (ii) (ii) 17.32 17.32 V (h) (h) 3.16 3.16 A 5 (i) (i) 15 × 10 AT/wb (ii) 144
CONGRATULATIONS!!!! …..May success be with you always….
