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Electrical MachinesElectrical machines are used to convert energy from one form to another form.The magnetic materials are used for the construction of electrical machines. Therefore, it is important to understand the magnetic materials and their application (i.e. magnetic circuits and air gap) used in electrical machines.Benefits of using magnetic materials in the electrical machines are, to obtain high flux density which gives large output and efficiency, also reduced the size of the machine.

CurrentThe rate of change of charge is known as current, and denoted by Ior i and the expression is I=dq/dt. The current flows in the closed circuit in the presence of electromotive force (emf). It is the potential for the source in the circuit.

Magnet and MagnetismMagnet is the property of material that can attract or repel any other iron material. The magnet contains the north and south poles.Magnetism is the amount of force that is created by electric current or the amount of motion of electrons in the atom. Fundamentalmagnetic momentsof elementary particles give rise to a magnetic field. The magnetic moment may be considered to be avectorhaving a magnitude and direction.Magnetic Field and FluxConsider a current is flowing in a current carrying conductor from A to B as shown in figure. Due to this current, a force around the conductor is induced that is called magnet. The force around the magnet where it acts called magnetic field.

The amount of magnetic lines passing through the area near the magnet is called the magnetic flux. The magnetic flux can also be defined as the product of magnetic flux density and the perpendicular area that it penetrates, it is denoted by,= BA Wb

Magnetic Flux Density and Field Intensity It is the ratio of magnetic flux per unit area and expressed as, B= /A, b/m2

Consider a unit pole of a magnet which contains the flux 1b. Also imagine a sphere of radius r meters, which is shown in figure. The surface area of sphere is, A= 4r2The magnetic flux density becomes,B= (1/4r2) The magneto-motive force is the product of current and thenumber of turns of the coil. Mathematically, = (NI) The magneto-motive force per unit length is known as magnetic field intensity and is expressed as, H= (Ni/l) At/m5Relative PermeabilityPermeability offers conductivity for the flux.Relative Permeability is the ratio of permeability of any material () to the permeability of the free space (o). Expression for Relative Permeability is r = /o , (Permeability of non-magnetic material is 1). Relationship between B HMagnetic flux density (B) developed in a magnetic field is directly proportional to the amount of applied magnetic field intensity (H) is,B HB = *H,( is absolute permeability of the material).B = r*o*(H)Practice Problems

The mean circumference of a wooden ring is 0.6m and the uniform cross sectional area is 0.005 m2. The wooden ring is wound by 200 turns and carries a current of 5A. Determine the (a) magnetic field intensity, (b) magnetic flux density and (c) total flux.The mean circumference of a wooden ring is 800mm and the uniform cross sectional area is 600mm2. The wooden ring wound by 400 turns and carries a current of 8A. Determine the (a) magnetic field intensity, (b) magnetic flux density and (c) total flux.

Reluctance It is property of a magnetic material to oppose the development of magnetic flux in a magnetic circuit. It can also be defined as the ratio of magnetomotive force to the flux through any cross section of a magnetic circuit. it can be expressed, = / At/bPERMEANCEAmeasureoftheabilityofamagneticcircuittoconductmagneticflux;thereciprocalofreluctance.It is defined as the ratio of magnetic flux to the magnetomotive force through any cross section of the magnetic circuit. Mathematically, it can be written as,P = / b/AtPractice Problems:A current of 4A is flowing through a wire which consists of 9 turns. The flux around the wire is 0.06wb. Determine the (a) magnetomotive force and (b) reluctance of the circuit.The magnetomotive force of a coil is 54At when carries a current of 6A. Determine the number of turns of the coil & the reluctance of the circuit if the flux is 0.07 wb.NI

R HrAmperes Circuital Law The line integral of the magnetic field intensity around a closed path is equal to the sum of the currents flowing through the area enclosed by the path see Figure.Where N number of conductors carrying current which is entering in to the paper, product NI is called the magnetomotive force at a distance r, and mathematically can be represent as,

Magnetic Flux Density in a Toroidial Coil

Torodal coil (A toroid is a coil of insulated or enameled wire wound on a donut-shaped form made of powdered iron) is wrapped up with N number of turns having a current of I ampere (see figure). For calculation consider R>>> r. Therefore mean length of the coil is, l = 2RAccording to Amperes law it becomes,H*= NI, insert the value of , we get,H = NI/2rThe magnetic flux density, B = *H=NI/2rMagnetic flux can be expressed as, = BA (cross-sectional area of core is(r2) = NIr2/2R

Practice Problems:A 15A current is flowing through a straight wire conductor. The conductor is surrounded by air and is wound by 10 turns. Find the magnetic flux density at a distance of 1.5cm from the conductor.A long straight wire of 15 turns carries a current of 20A. The conductor is surrounded by air. Find the magnetic field intensity and flux density at a distance of 0.2m from the wire.A wooden torodial coil is wound by 200 turns and having the inner diameter of 7cm and the outer diameter of 10cm. The cross-sectional area is 0.005m2. If the current 5A flows through the coil, find the (a) magnetic field strength, (b) flux density, and (c) flux.A wooden torodial coil is wound by 200 turns and having the mean length of 500mm. The cross-sectional area is 0.0008m2. If the current of 2A flows through the coil, find the (a) magnetic field strength, (b) flux density, and (c) flux.

Practice Problems:The dimension of the magnetic circuits shown in the figure. The relative permeability of a magnetic material is 750. Calculate the field intensity.Determine the magnetic flux of the magnetic circuit as shown in figure. The dimensions of the length, width and height are 0.3m, 0.02m and 0.3m respectively. The relative permeability of a magnetic material is 1100.

Parallel Magnetic CircuitTotal flux is divided into two or more sections (see figure). The magnetomotive force is connected to the left side and induced the flux and this flux is divided into 1 and 2 for the mid and right section of the core. The magnetic field intensity for left, right and middle are H, H1, and H2 and length are l1 and l1.

The magnetic potential drop aroundthe equivalent circuit is,F-Hl = H1l1= H2l2 Total flux is, = 1 + 2 As we know,Hl = RUsing above equations, magnomotive force can be written in form of, - R= R11= R22 = R+ R11= R+R22

Total Reluctance, R= R1 R2 / (R1+ R2)Practice Problems:The cross sectional area of a magnetic structure as shown in the figure is 5m2. The left limb is wound by 350 turns and the flux in the right limb is 4Wb. Determine the current. Assume the relative permeability is 300. The cross sectional area of a magnetic structure is 6cm2. The left limb is wound by N turns and the flux in the right limb is 4mWb. Determine N if the 3A current flows in the coil. Assume the relative permeability is 550.

Comparison b/w Magnetic and Electric Circuits The flux moves in magnetic circuits and current moves in electric circuit.The flux is defined as magnetomotive force per reluctance and current is defined as voltage per resistance.The magnetic flux density is defined as the flux per area. The electric current density is defined as the current per area.The magnetomotive force is equal to the product of flux and reluctance. The voltage is equal to the product of current and circuit resistance.The magnetic circuit reluctance is defined as R= l/A. The electrical circuit resistance is defined as R= (l/A).The magnetic field intensity is defined as H=/l and the electric field intensity is defined as E= V/dAir Gap in Magnetic Circuits The gap between two parts of magnetic bar is called Air Gap. This gap is filled by non magnetic material. In machines, rotor moves freely inside the stator through a small air gap. Generally, fluxes cross directly from one bar to other bar at the middle of the air gap, whereas at the edges of an air gap flux bends outward and thereby increasing the effective area of the gap which reduces the magnetic flux density. This reduction process is known as fringing as shown in figure. The fringing will be greater for longer air gap.

Calculation of Magnetic Circuits with Air Gap Consider the mean length, permeability, cross sectional area of the core are lc, and Ac and for air gap lag, ag and Aag respectively. In the composite circuit, the core and the air gap connected in series. As a result same flux will flow through the core and air gap and equivalent circuit can be considered as series circuit as shown in figure l.

Magnetic flux densities for the core & air gap are, Bc= c/Ac & Bag = ag /Aag,Total magneto-motive force,t = c+ ag = Hclc+ Haglagt = Rtwhere, Rt = Rc+ RagIf the permeability is very high than the reluctance of the core is very small and it can be neglected. Therefore, total magneto-motive force is, ag = Rag

Practice ProblemsThe mean length cross sectional area and the air gap length of a iron ring are 35cm, 15cm2 and 0.6 mm respectively. The ring wound by 400 turns and carries a current of 2A, which produces 2.5mWb flux. Determine the reluctance and the relative permeability of the iron ring. The cross sectional area of a magnetic structure is 2cm2 and is wound by 200 turns. The core is made up of iron and the relative permeability of the core is 4000. Determine the current flowing through the coil if the total flux of the circuit is 2.5mWb. The mean length cross sectional area and the air gap length of a iron ring are 55cm, 25cm2 and 0.8 mm respectively. The ring wound by 1200 turns and carries a current of 2A, which produces 6.5mWb flux. Determine the reluctance and the relative permeability of the iron ring.The cross sectional area of a magnetic structure is 2.5cm2 and is wound by 1000 turns. The core is made up of iron and the relative permeability of the core is 4000. Determine the current flowing through the coil if the total flux of the circuit is 3.5mWb.Electromagnetic Force on a Conductor When a current carrying conductor is placed in a magnetic field it experiences a force, known as electromagnetic force or Lorentz force.F = BlI,Where,B, is the magnetic flux density, l is the length of the conductor in meter and I, is the current flowing through the conductor. The force is greater if the length of the conductor increases.Practice ProblemsThe conductor carries a current of 150A. The length of the conductor is 2m and is placed in a magnetic field whose magnetic flux density is 0.35T. Determine the force. The magnetic force of a 3m conductor is 200N.The magnetic flux density of the conductor is 0.5T. Determine the magnitude of the current.

If two conductors carrying currents in same direction, the fluxes are in opposite direction in common region (see figure 1). The total reduction in flux showed the force of attraction.

However, if the conductors carrying current in opposite direction the net flux is increased, which indicates the force of repulsion (see figure 2).Consider figure 2,

Where d distance b/w two conductors,Conductors carrying currents I1 and I2 respectively, the flux density at the conductor 1 due to conductor 2 isB = oI1/2dIf the length of the conductor is l then force either attraction or repulsion is,F = BlI2 Finally, F = oI1I2/2dMagnetic Material and B-H Curve

Ferromagnetic materials: Materials whose permeability thousands times greater than free space i.e. iron, nickel, steel and cobalt.Diamagnetic materials: Materials whose permeability less than free space i.e. copper, gold and silver.Paramagnetic materials: Materials have permeability slightly greater than free space i.e. magnesium, molybdenum, and lithium.B-H CurveThe value of permeability for air and other non magnetic materials are constant with the variation of magnetic field intensity. But the permeability of ferromagnetic materials varies with magnetic field intensity. Therefore, it is important to plot the curve b/w the magnetic flux density and magnetic field intensity. In this curve, the magnetic flux density varies with the variation of magnetic field intensity. This curve is known B-H curve as shown in figure.

InductanceIt is the property of coil that opposes any change of current flowing through the coil itself. If a coil having suitable number of turns is connected across the source, current will flow through it. The flux will be associated due to this current. Therefore, inductance is defined as the flux linkage per unit current.L= /I , The unit of inductance is Henry and the symbol is,

Consider a magnetic circuit as shown in figure, the length of the core is l which contain N number of turns and the cross sectional area is A. The coil carries a current of i which creates the flux b in the core.

According to electromagnetic induction law, the voltage induced across the coil can be written as, v = N (d/dt) = N [(d/di) (di/d)]. As we know, L = N (d/dt) & flux is, = /R = NAi /l d/dt = NA /lFinal expression of inductance, L = N2/R

Hysteresis Loop It is a loop that obtained by variation of magnetic flux density with magnetic field intensity. Consider a un-magnetized core to obtained this loop. The relationship of magnetic field intensity, H*l = NI/lFor zero value of current, H=0 and the curve current will start from the origin of axes. By increasing the value of current the value field intensity also increased accordingly. H*l = NI/l

Retentivity (OC)Coercive force(OD)Coercive force(OC`)The new value of magnetic field intensity at point a is Ha (see figure). Again, increase the value of current until the magnetic field intensity to reach at the saturation point and at this point the value of magnetic field intensity is Hb.Further, increase in current the magnetic field intensity will be increased and the magnitude of magnetic flux density will be same. It means magnetic material has been reached to its saturation region.

In the reverse way, if the magnetic field intensity decreases slowly, the magnetic flux density will also decrease. This decreasing value of magnetic field intensity and magnetic flux density are greater than the previous values and the curve will moves by the other path. There will be some magnetic flux density remaining if the magnetic flux intensity reduced to zero. It means that the core sustained some flux density (OC). This remaining part of the magnetic flux density is known as retentivity of the material.

Again, further increase in magnetic field intensity in the reverse direction then the magnetic flux density will be finished at point D. This value OD is necessary to demagnetize the magnetic material. This amount of magnetic field intensity is known as coercive force. The path will continue up to the point E if we increase the magnetic field intensity. If we magnetize the core in the same direction, the magnetic field density will developed in the opposite direction and it will continue to increase for getting the point C` with increase in the magnetic field intensity. The magnitude of OC` is known as reverse retentivity. The complete hysteresis curve will be obtained if we continue the other steps.

Core Losses There are two types of losses in the magnetic field known as hysteresis and eddy current losses, and combination of these two losses known as core losses.Hysteresis Losses:Consider a coil as shown in figure having N number of turns and carries I ampere of current. The hysteresis loss can be calculated at any time as,

The hysteresis loss in the magnetic field at any time can be calculated from B-H curve as shown in figure,

W = pdt,,W = vidt = (Ni) dW = (Ni) d ------ (1),But, = BA,d= (A) dB,

The current is, i= (Hl)/N, putting all values in equation (1)W = Vcvol H dBWhere, Vcvol = lA is the volume of the core. The energy density per unit volume is, Wv = H dBThe hysteresis loss in the core over one cycle is,Ph= Vcvol Wh fAs understood, magnetic materials in the electrical machines varies in the following way, Area under B-H curve is, A= KBnmIn general, the hysteresis loss can be expressed as, Ph= KhVcvol Bnm fWhere kh, depend on the properties of the magnetic material and volume of the core. The value of the constant n varies b/w 1.5 to 2.5.

Eddy Current Loss The core of the machine normally offers conductive path for the time varying flux. This time varying flux induces a voltage in the body of the core. As a result of this voltage, a small magnitude of current will flow through the core. This current is known as eddy current. The power loss due to this current is known as eddy current loss. The eddy current loss can be expressed as,Pe= KeVcvol t2B2m f2here t is the thickness of the lamination.