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TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS CHAPTER-1: FUNDAMENTAL OF MAGETIC CIRCUITS PERSONAL REMARK : Electrical machines is one of the core subject of the Electrical Engineering. Knowledge of electrical machines is very important because most of the electrical energy (about 90%) is generated and consumed by electrical machines. Electrical machines are classified as Dynamic and static machines. All the Rotating type machines are dynamic machines and transformer is classified as of static machine. All the dynamic machines operate in two modes-Generating and motoring. Generating Mode Power (Electrical) TG Prime Mover Heat Power Electric Generator TM,W Losses Power (Mechanical) Motoring Mode P electrical TG W Load (mechanical) Electric Motor TM Losses P mechanical Transformer Electric power Electric power MAGNETIC CIRCUITS Coupling magnetic field exists in transformers and practically in all the rotating electrical machinery. In low-power electrical machines, magnetic field may be produced by permanent magnets. But in highpower electrical inachinery and transformers, coupling magnetic field is produced by electric current. Thus, an elementary knowledge of the analysis of magnetic circuits is necessary for understanding the working of these devices. NEW DELHI 8860637779 LUCKNOW 9919526958 AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 1 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS The object of this chapter is to develop some basic techniques for PERSONAL REMARK : the analysis of magnetic-field systems and to introduce the concept of induced emf. Introduction to Magnetic Circuits The complete closed path followed by the lines of flux is called a magnetic circuit. Some improtant terms relating to the study of magnetic circuits are discussed here. (I) Magneto Motive Force (MMF) In an electric circuit, the current is due to the presence of electromotive force. By analogy, in a magnetic circuit, the magnetic flux is due to the presence of a magnetomotive force. The mmf is created by a current flowing through one or more turns. This shows that mmf is equal to the product of current and the number of turns in the coil. Mmf = NI ampere–turns (or ATs). Magnetic flux path a b I Area of cross section A N d c Figure 1: A simple magnetic circuit with N turns and current I (II) Magnetic Field Intensity If the magnetic a circuit of figure 1 is homiogeneous and of uniform cross-sectional orea. The magneto motive force per unit length of magnetic circuit is termed as the magnetic field intensity its usual symbol is H. If I is the mean length of magnetic circuit abcd in figure 1, then H IN ampere-turns/metre [or ATs/m] (III) Permeability of Free Space Suppose flux density at C, caused by magnetic field intensity H at x1 is B tesla and if C is one metre away from x1 then permeability of free space µ0 is given by µ0 = B H Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 2 TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS PERSONAL REMARK : Magnetic flux path x X C 1m Figure 2 : Pertaining to permeability of free space Note that conductor x and point c are in vacume (free space), in air or in any other non-magnetic medium. For vacuum or non-magnetic materials µ0 = B = 4 × 10–7 henries/m H (IV) Reluctance. The opposition offered by the magnetic circuit to magnetic flux is called reluctance. Reluctance is analogous to resistance in electric circuit. Just as resistance R = reluctance R is given by, R = . , a AT/Wb µ.A where, l = length of the magnetic path, m A = area of cross-section normal to flux path, m2 µ = µ0, µr = permeability (or absolute permeability) of the magnetic material µr = relative permeability of the magnetic material µ0 = permeability of free space = 4 × 10–7 H/m Since resistance , R = l , absolute permeability µ is analogous to a p 1 conductivity (V) Permeance. Reciprocal of reluctance is called permeaence. Permeance = Wb/AT (VI) Magnetic flux density. magnetic flux is analogous to current in a electric circuit. As current I = = NEW DELHI 8860637779 emf , flux in a magnetic circuit is given by resistance mmf Wb Reluctance LUCKNOW 9919526958 AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 3 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS R = = and mmf = IN ; then magnetic flux is given by, µ0 µ r A PERSONAL REMARK : IN . µ 0 µ r . A Wb Magnetic flux density B is defined as the magnetic flux per unit cross-sectional area of the core and its unit is Wb/m2 or Tesla (T). B= IN . µ0 µr magnetic flux, core area, A I E V/A or we have R = AT/Wb I (VII) Calculation of ampere-turns. In any magnetic circuit, flux Like R = is given by Ampere turns required for the magnetic circuit, I = × Reluctance B H µ φ B . H µA A µ µ =× = Magnetic field intensity H in AT/m in that part of circuit × length of that part normal to flux path. Comparison Between Magnetic and Electric Circuits There are many similarities in magnetic and electric circuits. The reader is usually more conversant with electric circuits. Thus, a comparison of electric and magnetic circuiits in tabular form goes a long way in the better understanding of magnetic circuits. Table 1 Electric Circuit Magnetic Circuit Conductor length I Flux core length l N E Current I I Core area, A Conductor area A A toroidal iron ring of length , A toroidal copper ring of length core area A is excited by a coil of , cross-sectional area A is N turns carrying I amperes so that connected to emf E so that flux is produced. current I flows Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 4 TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS Similarities PERSONAL REMARK : 1. Closed path for the electric current is called an electric circuit. is called a magnetic circuit 2. Driving force is mmf I = IN ATs 2. Driving force is emf E, volts 3. Resistance, R = p.l V/A or A 3. Reluctance, R = + + p R A E – driving force 5. Current, I = resistnce – 5. Magnetic flux, AT/Wb µA driving force reluctance I Wb R 6. Magnetic flux density, B = T A (or Wb/m2) 7. Magnetic field intensity 7. Electric fleld intensity, E V/m l Also R T or E I= A R I 6. Current density, J = A I AT/Wb µA 4. Equivalent circuit 4. Equivalent circuit I or 1. Closed path for the magnetic flux IN AT/m l H = E IR I p . l I . p. l l l A A Also H = IN I .Rl l 1 l l l l µA µ A or . J V/m 8. Conductivity path, of current l or H= 1 BAT/m µ 8. Permeability of magnetic circuit,µ so that B = µH Wb/m2 or T So that , J = A/m2 Dissimilarities 1. The electric current actually flows in an electric circuit. For the existence of this current, energy is drawn from the source continuously. This energy gets dissipated in resistance in the form of heat. 2. Electrical insulator confine the current to well defined paths. NEW DELHI 8860637779 LUCKNOW 9919526958 1. Strictly speaking, magnetic flux does not flow in the magnetic circuit. Energy in needed for estabilishing the required flux. Once the requisite flux is created, no more energy is needed in maintaining it. 2. There are no magnetic irsulators. Even in the best known magnetic insulator some flux can be established. AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 5 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS Ex.1 A mild steel ring has a mean diameter of 20 cm and a cross- PERSONAL REMARK : 2 sectional area of 50 cm . For a relative permeability of 800, calculate (a) reluctance of the ring and (b) current required in 200 turn coil to produce a flux of 1m Wb in the ring. Sol. (a) Reluctance = R = (b) l . Here = D = × 20 cm, A = 50cm2 µ0 . µr A 20 10 –2 4 10 – 7 800 50 10 – 4 = 1.25 × 106 AT/Wb mmf flux. = 1 × 10–3 = 1.25 106 Mmf, I = 1 × 10–3 × 1.25 × 106 = 1250 ATs. Current in the coil, I = I 1250 = 6.25 A N 200 Series and Parallel Magnetic Circuits Like electric circuits, a magnetic circuit may be made up of a series circuit, a parallel circuit or a combination of series-parallel circuits. Such magnetic circuits may be excited by one or more coils. Solution of such magnetic-circuit configurations can be obtained by applying Kirchoff's flux laws. Area A2 Area A1 I 2 1 I 1 g .Area Ag N 3 3 Area A3 Figure 3 : Composite magnetic circuit Series Magnetic Circuit : Just like a series electric circuit, a series magnetic circuit carries the same flux in each part of the circuit configuration. Figure 3 shows a composite magnetic circuit consisting of three different magnetic materials in series. There is also an air gap as shown. The dimensions are indicated in figure 3. Since the same flux exists through each part of the magnetic circuit in figure 3, the total reluctance is equal to the sum of four individual circuit reluctances as under : Total reluctance, R = g 3 1 2 µ 0 µ r1 A1 µ 0 µ r2 . A 2 µ 0 µ r3 . A 3 µ0 . A g Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 6 TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS Note that µr for air is one PERSONAL REMARK : Total mmf = flux × total reluctance of the series circuit g 3 1 2 = µ 0 µ r1 A1 µ 0 µ r2 . A 2 µ 0 µ r3 . A 3 µ 0 . A g = 3 1 2 g A1 µ 0 µ r1 A 2 µ 0 µ r2 A 3 µ 0 µ r3 A g µ 0 = Bg B1 B2 B3 1 2 3 . g µ0 µr1 µ0 µr2 µ0 µr3 µ0 = H1 . 1 + H2 . 2 + H3 . 3 + Hg . g Total mmf, IN = mmf in for [ 1 + 2 + 3 + g] .... (i) .... (ii) Equation (i) may be re-written as under : IN – H1 1 – H2 2 – H3 3 – Hg g = 0 In general, Mmf a closed magnetic circuit = 0 But this statement is similar to Kirchhoff's voltage law for a series electric circuit. Therefore, Kurchhoff's mmf law for a series magnetic circuit is as under Kirchhoff's mmf law (KML) states that algebraic sum of mmfs rises (or falls) taken in a specified direction in a closed magnetic circuit is zero. Parallel Magnetic Circuit : Consider a parallel magnetic circuit shown in figure 4. Coil carrying a current I produces flux in path EFAB. At point B. flux gets divided into two paths, and such that or – – = 0 In general, = 0 The above statement is similar to Kirchhoff's current law. Therefore, for a magnetic circuit also, total magnetic flux towards a junction is equal to the total magnetic flux away from that junction. In other words, algebraic sum of magnetic fluxes at a junction is zero, this is called Kirchhoff's flux law (KFL). I A 1 N F NEW DELHI 8860637779 LUCKNOW 9919526958 B C 2 3 E (a) D AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 7 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS R 1 1 B PERSONAL REMARK : R1 3 2 T R 2 I3 I1 I2 R 3 E (b) E R2 R3 (c) Figure 4.(a) Parallel magnetic circuit. Its equivalent (b) magnetic circuit and (c) electric circuit. From figure 4 (a) if reluctance of parth EFAB = R 1 reluctance of path BF = R 2 and reluctance of path ECDE = R 3, then equivalent magnetic cirucit of figure 4(b) is obtained. From this circuit, it is seen that total coil mmf, I = IN mmf required for path [EFAB + BCDE] IN = 1 Rl1 + 2 Rl2 Also, I = IN = mmf required for path [EFAB + BCDE] IN = 1 Rl1 + 3 Rl3 Figure 4(c) gives the equivalent electric circuit. Leakage Flux and Fringing All practical magnetic circuitcs are so designed that most of the flux produced by an exciting coil is confined to the desired magnetic path of low reluctance. However, a small amount of flux does follow a path through the surrounding air. Figure 5 shows a magnetic circuit with a ferromagnetic core. The flux which returns by such path as a, b, c is called leakage flux. It does not follow the intended path. Therefore, leakage flux may be defined as that flux which does not follow the intended path in a magnetic circuit. Leakage flux does exist in all practical electromagnetic devices. Its effect on the analysis of electrical machinery is carried out by replacing it by an equivalent leakage reactance. It is seen from Figure 5 that total flux thorugh the exciting winding = useful flux + Ieakage flux. A term leakage factor gives a measure of the leakage flux. It is defined as leakage factor = total flux handled by the exciting winding useful flux The value of leakage factor in electrical machinery is in the range of 1.15 to 1.25. Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 8 TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS As the magnetic flux lines cross the air gap, these bulge out as shown PERSONAL REMARK : in figure 5. This bulging out of the flux is called fringing. Larger the air gap, more is the flux fringing. The effect of fringing flux is to increase the effective cross cectional area of the air gap. As a result, flux density in the air gap is not uniform and average flux density gets reduced. Useful flux a + b c Fringing flux – Leakage flux Ferromagnetic core Figure 5: Fringing of the flux across air gap. Useful and leakage fluxes are also shown The effect of fringing can be taken into account empirically by adding one gap length to each of the linear dimensions making up the gap area. For example, if core dimensions are a and b and small gap length is g , then effective gap area taking fringing into consideration is Ag = [(a + lg) × (h + lg)] > core area Ac = a . b. If core is circular of radius r, then effective gap area Ag = [ (r + lg)2] > core area Ac = r2. B-H CURVE A B-H curve, also called magnetisation curve or saturation curve, is the plot of flux density B as the magnetic field intensity H is varied. Figure 6. shows a typical B-H curve of a ferromagnetic material. It has initial non-linear zone OA, zone from A to C is almost linear and zone beyond C is called saturation zone. The flux density B in the saturation zone increase less repidly with H as compared to its change in the linear zone. As B-H curve is not a stratight line, relative permeability r = B of a ferromagnetic material changes with µ0 H the flux density. Therefore, during magnetic circuit calculations, the value of µr and H should correspond to the flux density under consideration. In free space or non-magnetic materials, µ0 is constant, therefore B-H relationship is linear. NEW DELHI 8860637779 LUCKNOW 9919526958 AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 9 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS PERSONAL REMARK : B(T) Saturation zone C Linear zone A Non-linear zone 0 H(AT/m) Figure 6 : Typical magnetization curve of a ferromagnetic material. Ex. 1 For the manetic circuit shown in figure 7, l c = 50 cm, dia = 2.85 cm, l g = 2 mm and N = 500. Air-gap flux = 0.8 mWb. (a) Find the exciting current in the coil in case µr = 500. Neglect magnetic leakage and fringing. Repeat part (a) if flux fringing at air gap is taken into account. (b) Core length c = 50cm I g=2mm N Area Ac = 2.85cm Figure 7 : Pertaining to Example 2. Sol. (a) Air-gap length, l g = 2 mm, µ r = 500, core area Ac = × 2.852 cm2 Gap reluctance, R g = Ig µ0 A c 50 10 –3 4 4 10 – 7 (2.85) 2 10 – 4 = 24.9483 × 105 AT/Wb Iron-core reluctance Ic 50 10 –2 4 Ric = µ 0 . µ r A c 4 10 – 7 500 (2.85) 2 10 – 4 = 12.474 ×105 AT/Wb Total reluctance, Rl = Rlg + Rlc = 37.4223 × 105 AT/Wb But Coil current I = Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material mmf = flux × reluctance . Rl 0.8 10 – 3 37.4223 105 = 5.9876 A N 500 Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 10 TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS (b) When flux fringing is taken into account, the effect gap PERSONAL REMARK : area Ag = 2.85 0.2 = 1.6252 cm2 (d + 2g)2 = (r + g)2 = 2 Ig Air-gap reluctance, Rlg = µ . A 0 g 2 10 –3 4 10 – 7 1.6252 10 – 4 = 19.185 × 105 AT/Wb I= [Rlc + Rlg] N Coil current, I = 0.8 10 –3 [12.474 + 19.185] × 105 = 5.0654A A magnetic circuit of cast steel is arrange as shown in Ex. 2 figure 8. Various dimensions are also indicated in the figure. The exciting coil, with N = 1000 turns, sets up a flux of 1mWb in the central limb. Find the coil current if, for cast steel, Area=2×2cm 15cm 15cm 2mm 25 cm I 2 2mm I N I Area=4×2cm 25 cm 2 15cm Figure 8 (a) µr = and (b) µr = 6000 Neglect fringing and leakage. Sol. (a) When relative permeability µr = , the reluctance offered by cast steel is zero and therefore no mmf (= flux × reluctance) is required by the steel. Reluctance is offered by the air gaps alone. Reluctance R 1 offered by each of the air gaps in outer limbs is R 1 = 2 10 –3 10 – 7 4 10 – 4 = 3.98 × 106 AT/Wb Reluctance R 2 offered by air gap in central limb is NEW DELHI 8860637779 LUCKNOW 9919526958 AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 11 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS PERSONAL REMARK : 2 10 –3 R 2 = 10 – 7 8 10 – 4 = 1.99 × 106 AT/Wb The electrical analog of the magnetic circuit of figure 8 is shown in figure 9. As two reluctances R 1 are in parallel, their resultant is R1 and the circuit gets reduced to that shown in figure. For this 2 circuit, Kirchhoff's mmf law gives Rl1 =0 NI – flux Rl 2 2 or 1000 × I = 1 × 10–3 [1.99 + 1.99] × 106 I= 1 10 –3 (3.98) 10 6 = 3.98 A 1mWb 1mWb R 1 R 1 R 2 NI R 1 R 1 2 NI (b) (b) Figure 9 (a) and (b) : Electrical analog of magnetic circuit of figure 8 (b) Relative permeability, µr = 6000. Here reluctance of cast steel cannot be neglected. The electrical analog of the magnetic circuit of figure 8 is now drawn in figure 10, where Rlc1 is the reluctance of each of the outer two steel limbs and Rlc2 is the reluctance of the central steel core. Reluctance of outer steel core (including the top 15 cm) R c1 = (. 0.15) 10 – 7 6000 4 10 – 4 = 1.33 × 106 AT/Wb For central steel core, Rlc2 = 0.15 4 10 –7 6000 8 10 – 4 = 0.25 × 105 AT/Wb Resultant of outer reluctances = Rl1 Rl c1 (1.33 39.8) 105 = 20.565 × 105 AT/Wb 2 Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 12 TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS PERSONAL REMARK : R 2 R 1 1mWb Rc1 NI R 1 R 2 R 2 Rc2 Rc1 (b) Rc1+R1 2 NI (c) Figure 10 : is redrawn in figure 10(c) for which Kirchhoff's mmf law gives Rl Rl1 Nl – Rl c2 Rl 2 c1 2 IN = 1 × 10–3 [0.25 + 19.9 + 20.565] × 105 Exciting current, I = 10 –3 [40.715 × 105] = 4.072 A 1000 Where e = emf induced, volts; N = number of turns in the coil N = flux linkages with the coil, Wb-turns t = time, seconds LENZ'S LAW Changing magnetic flux linking a coil induces an emf in the coil. When the coil circuit is closed a current begins to flow in the coil. The direction of this induced emf, or induced current, is governed by Lenz's law. According to this law; the induced current develops a flux which always opposes the change responsible for inducing this current. In order to take into consideration the fact that the induced current or induced emf opposes the change in flux, a negative sign is added e=– d d =–N dt dt Lenz's law is the outcome of natural law : for every action, there is an equal and opposite reaction. Lenz's law, in short, is effect opposes the cause. Change in flux linkages induced emf in a coil requires changes in flux or flux linkages with respect to time. This change in the flux linking a coil can be obtained by any of the five methods outlined below : (i) Magnetic flux is constant and stationary, but coil rotates (or conductor moves). Flux-cutting action occurs and emf so induced is known as speed emf, motional emf or rotational NEW DELHI 8860637779 LUCKNOW 9919526958 AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 13 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS emf*. This principle is used in all dc machines and in PERSONAL REMARK : synchronous machines with field winding on the stator. (ii) Coil is stationary, but dc electromagnet (or permanent magnet) is rotated. Here also flux-cutting action generates speed emf. This principle is used in large-sized synchronous generators. (iii) Coil is stationary, but the flux passing through the coil changes with time. induced emf is called transformer emf. (iv) Reluctance of the magnetic circuit varies with rotor rotation. As a consequence, flux variation occurs and emf is generated. This principle is used in alternators. SPEED EMF OR MOTIONAL EMF Consider a single conductor of length l metre moving at right angles to uniform magnetic field between N and S poles. The magnetic field has uniform flux intensity B tesla. Conductor velocity u m/s is perpendicular to its length l and flux density B. Let the conductor travel a small distance dx in a time dt, from position 1 to position 2. Therefore, area swept by the conductor in time dt = l. dx m2 N N N B 2 v B v v sin S B v S S (b) (c) dx Area swept 1 (a) 2 Figure 11: Pertaining to the generation of speed emf in a conductor Flux cut by conductor in time dt = [Uniform flux density] × [area swept by conductor normal to field in time d(t)] = B0. ldx = d Wb As per Faraday's law, magnitude of induced emf e, is e= Since , d B dx dx B . dt dt dt dx = velocity of conductor v in m/s, the emf induced in one dt Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 14 TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS conductor is given by PERSONAL REMARK : e = Blu Now let the conductor move at an angle to the direction of magnetic field B Now component of velocity perpendicular to magnetic field is u sin . e = B u sin In figure 11(c), conductor moves paralle to the flux lines, no flux is cut and therefore induced emf is zero. In case = 0 in figure 1(c), e = 0, Equation can also be expressed as e = B u sin l or E = u B V/m where E = electric field intensity in volts per metre. Here E is in a direction normal to the plane containing u and B . Cross product of u and B indicates the direction of E . For this purpose, stretch four fingers of right hand along u , now move these fingers from u towards B , then outstretched thumb gives the direction of E . In figure 11, u and B are in x – y plane and E is along z-axis, normal to the plane containing u and B . If this rule is applied to figure 12, direction of generated emf would be into the paper which can be denoted by cross. TRANSFORMER EMF The magnitude of transformer emf is given by e = – N d . The coil dt is stationary (or fixed), but flux is time-varying in nature. Note that in motional or speed emf, motion or rotation is involved. Therefore, speed emf is associated with energy conversion from electrical to mechanical or mechanical to electrical. The transformer emf does not involve any motion of the coil. Therefore, transformer emf is not associated with energy conversion; it is, however, associated with energy transfer only. NEW DELHI 8860637779 LUCKNOW 9919526958 AGRA ALLAHABAD 9793424360 9919751941 PATNA 9534284412 NOIDA 8860637779 BHOPAL 9838004479 GORAKHPUR KANPUR JAIPUR 9793424360 9838004494 9838004497 15 TECHGURU TECHGURU CLASSES CLASSES for for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA CHAPTER- 1 : FUNDAMENTALS OF MAGETIC CIRUITS PERSONAL REMARK : z E 90º y v B x Figure 12 : Pertaining to the direction of electric field intensity E GENERAL CASE OF INDUCED EMF Suppose there is a relative movement between coil and the magnetic flux and at the same time, flux is changing with time, Under these conditions, both speed and transformer emfs would be induced in the coil. Thus, total emf, e in the coil would now be given by e = speed or rotational emf, er + transformer emf, et Classroom Study Course LUCKNOW 0522-6563566 LUCK NOW Correspondance Study Material Classroom & Online Test Series Foundation Batches also for 2nd & 3rd Year sturdents Regular & Weekend Batches Interview Guidance 16