unit 1-electromagnetism

ELECTROMAGNETISM E2063/Unit 1/ 1

ELECTROMAGNETISM

OBJECTIVES

General Objective : To understand the basic principles of electromagnetism.

Specific Objectives : At the end of the 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.

UNIT 1


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 electronic theory. An electromagnet provides the basic of many items of electrical equipment, examples of which include electric bells, relay and telephone receiver.

The distribution of a magnetic field can be investigated. The pattern of magnetic field of bar a magnet shown in Figure 1.1. The direction of a line of flux is from the 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

Figure 1.1 The pattern of magnetic field of bar a magnet.

INPUTINPUT


Source: Electrical And Electrical Principles and Technology by John Bird

Figure 1.2 The magnetic attraction and repulsion by using two bar magnets.

1.1 CURRENT-CARRYING CONDUCTOR AND ELECTROMAGNETISM

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.

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.


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 attraction between them. If closed current-carrying 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).


1.2.1 MAGNETIC FIELD STRENGTH, H (MAGNETISING FORCE)

Magnetic field strength is defined as magnetomotive force, Fm per metre length of measurement being ampere-turn per metre.

ampere turn / metre

where

Fm - magnetomotive forceN - number of turnsI - currentl - average length of magnetic 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 = 0.5AN = 600d = x 10 x 10-2m

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.



The condition of the problem are represented in Fig. 1.5

Fig. 1.5

I = V/ R = 240 / 474 = 0.506 Al = π D = π (25 10-2) = 0.7854 m

H= = = 322.13 AT/m


Activity 1A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

1.1 Find the magnetic field strength applied to a magnetic circuit of mean length 50 cm when a coil of 400 turns is applied to it carrying a current of 1.2 A.

1.2 A current of 2.5A when flowing through a coil produces an m.m.f. of 675 At. Calculate the number of turns on the coil.

1.3 A magnetizing force 8000 A/m is applied to a circular magnetic circuit of 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.


Feedback to the Activity 1A

1. 960A/m

2. 270 turns

3. 10.05 A 4. 2547 AT


1.3 MAGNETIC QUANTITY CHARACTERISTICS AND THEIR RELEVANT FORMULAE

1.3.1 Magnetic Flux and Flux density

Magnetic flux is the amount of magnetic filed produced by a magnetic source. The symbol for magnetic flux is . 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 =

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 150Wb, calculate the flux density.


Magnetic flux, = 150 Wb = 150 x 10-6 WbCross sectional area, A = 200mm x 100mm = 20 000 x 10-6m2

Flux density,

= 7.5 mT

INPUTINPUT


1.3.2 Permeability

For air, or any other non-magnetic medium, the ratio of magnetic flux density to

magnetic field strength is constant , i.e. = a constant. This constant is called the

permeability of free space and is equal to 4 x 10-7 H/m. For air and any other non-magnetic medium, the ratio

For all media other than free space

where r is the relative permeability and is defined as

r varies with the type of magnetic material. From its definition, r for a vacuum is 1. is called the absolute permeability. The approximate range of values of relative

permeability r for some common magnetic materials are :

Cast iron r = 100 – 250Mild steel r = 200 – 800Cast steel r = 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.


= 764

1.3.4 Reluctance


Reluctance,S is the magnetic resistance of a magnetic circuit to presence of magnetic flux.

Reluctance,

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.


Reluctance,

=

= 16 580 H-1

Absolute permeability, = = 5.027 x 10-3 H/m

1.4 ELECTROMAGNETIC INDUCTION


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 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)This induced electromagnetic field is given by E = Blv volts

where

B = flux density, Tl = length of the conductor in the magnetic field, mv = conductor velocity, m/s

If the conductor moves at the angle to the magnetic field, then


E = Blv sin volts


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-circuited(b) its ends are connected to a load of 20 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 ends of the conductor are open circuited no current will flow even though 1.5 V has been induced.

(b) From Ohm’s law


ii. 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


Figure 1.7 Bar magnet move in and move out from a solenoid


Activity 1B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

1.4 The maximum working flux density of a lifting electromagnet is 1.8 T and 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 the magnetic field strength and the m.m.f. required to produce a flux density of 0.25T in an air gap of length 12mm.

1.6 A coil of 300 turns is wound uniformly on a ring of non-magnetic material. The ring has a mean circumference of 40 cm and a uniform cross-sectional area of 4cm2. If the current in the coil is 5 A, calculate (a) the magnetic field strength, (b) the flux density and (c) the total magnetic flux in the ring.

1.7 A mild steel ring has a radius of 50mm and a cross-sectional area of 400mm2. 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 reluctance of the mild steel(b) the number of turns on the 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 of 15m/s at an angle (a) 90 (b) 60and (c) 30 to a magnetic field produced between two square-faced poles of side length 2cm. If the flux leaving a pole face is 5Wb, find the magnitude of the induced e.m.f. in each case.

1.10 A flux density of 1.2wb/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.


Feedback to the activity 1B

1.4 0.1961 m

1.5 2378 A

1.6 (a) 3750 A/m (b) 4.712mT (c) 1.885 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


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) Determine the total flux emerging from a magnetic pole face having dimensions 5 cm by 6 cm, if the flux density is 0.9 T.

(c) A solenoid 200 cm long is wound with 500 turns of wire. Find the current required to establish a magnetizing force of 2500 A/m inside the solenoid.

(d) Find the magnetic field strength and the magnetomotive force needed 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. If the flux required in the air-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 conductor of length 15cm is moved at 750mm/s at right-angles to a uniform flux density of 1.2 T. Determine the e.m.f induced in the conductor.


(e) Find the speed that a conductor of length 120mm must be moved at right angles to a magnetic field of flux density 0.6 T to induce in it an e.m.f of 1.8 V.

(f) A 25 cm long conductor moves at a uniform speed of 8 m/s through a uniform magnetic field of flux density 1.2T. Determine the current flowing in the conductor when (i) its ends are open-circuited (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 induced in the conductor. What will be the e.m.f induced if 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 cut of 4 mm wide is made in the ring. Calculate 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.

(i) An iron ring has a mean diameter of 15 cm, a cross-sectional area of 20 cm 2 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) reluctance of magnetic circuit.(ii) relative permeability of iron.


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

(a) 2370A(b) 325(c) (i) 79580/H (ii) 1 mH/m(d) 0.135V(e) 25m/s(f) (i) 0 (ii) 0.16 A(g) (i) 20 V (ii) 17.32 V(h) 3.16 A(i) (i) 15 105 AT/wb (ii) 144 CONGRATULATIONS!!!!

…..May success be with you always….

unit 1-electromagnetism

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