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BASIC ELECTRICAL TECHNOLOGY DET 211/3 Chapter 4 – Magnetic Circuits 1 Magnetic Materials and Circuits Introduction Magnet contains a north pole and south pole. Magnet flux leaves the magnet as the north pole and the place where the flux returns to the magnet as the south pole. Two types of magnet: • Permanent magnet • Electromagnet 2 Right Hand Rule and Ampere’s Law When a conductor carries current a magnetic field is produced around it. Fingers– indicate current direction Thumb – indicate the direction of magnetic flux is wrapping around the wire 3 Right Hand Rule and Ampere’s Law The relationship between current and magnetic field intensity can be obtained by using Ampere’s Law. Ampere’s Law states that the line integral of the magnetic field intensity, H around a closed path is equal to the total current linked by the contour. H .dl i H: the magnetic field intensity at a point on the contour dl: the incremental length at that point If θ: the angle between vectors H and dl then Hdl cos i 4 Right Hand Rule and Ampere’s Law Consider a rectangular core with N winding i Ni dl lc Hl c Ni Therefore Ni H lc 5 Relationship between B-H The magnetic field intensity, H produces a magnetic flux density, B everywhere it exists. B H ( weber / m2 ) or Tesla B r 0 H ( wb / m ) or T 2 - Permeability of the medium 0 - Permeability of free space, 4 x 10-7 wb r 0 A.t .m - Relative permeability of the medium For free space or electrical conductor (Al or Cu) or insulators, unity r is 6 MAGNETIC EQUIVALENT CIRCUIT A simple magnetic circuit having a ring shaped magnetic core (toroid) and a coil that extends around the entire circumference When current i flows through the coil of N turns, a magnetic flux is produced. 7 MAGNETIC EQUIVALENT CIRCUIT Assumption: •All fluxes are confined to the core •The fluxes are uniformly distributed in the core The flux outside the toroid (called leakage flux), is so small (can be neglected) Use Ampere’s Law, H .dl Ni Hl Ni H .2r Ni Hl Ni F F = Magnetomotive force (mmf) 8 MAGNETIC EQUIVALENT CIRCUIT B H Ni H ( At / m ) l Ni B (T ) l Where; N – no of turns of coil i – current in the coil H – magnetic field intensity l – mean length of the core 9 Magnetic Flux Density (B) and Magnetizing Curve (B-H Curve) Case 1: Non magnetic material core (Cu, Al, air, plastic, wood,..) B 0 H 0 Ni B (T ) l 10 Magnetic Flux Density (B) and Magnetizing Curve (B-H Curve) Case 2: Ferromagnetic material core (iron, steel, ferrite,..) B H 0 r H 0 r Ni B (T ) l The magnetic flux density, B increases almost linear in the region of low values of magnetic intensity, H. At higher value of H, the change of B is nonlinear. 11 MAGNETIC EQUIVALENT CIRCUIT The flux in the coil, BA( weber ) Where Ni Ni Ni F A l l R R A Ф – flux in the coil (wb) F – magnetomotive force (mmf) R – 1/μA = 1/P ,Reluctance P = permeance A – cross sectional area 12 ANALOGY BETWEEN MAGNETIC CIRCUIT AND ELECTRIC CIRCUIT a) Magnetic equivalent circuit b) Electric equivalent circuit To solve magnetic equivalent circuit – Kirchhoff Voltage and Current Laws (KVL & KCL) Electric circuit Magnetic circuit Driving force EMF (E) MMF (F) Produces Current (i) Flux (Ф) Limited by Resistance (R) Reluctance (R) 13 MAGNETIC CIRCUIT WITH AIR GAP In electric machines, the rotor is physically isolated from the stator by the air gap. Practically the same flux is present in the poles (made by magnetic core) and the air gap. To maintain the same flux density, the air gap will require much more mmf than the core. 14 MAGNETIC CIRCUIT WITH AIR GAP lc Rc c Ac Rg lg g Ag Where Ni Rc Rg Ni H clc H g I g c Bc Ac g Bg Ag lc – mean length of the core lg – the length of the air gap 15 FRINGING EFFECT Fringing Effect: Bulging of the flux lines in the air gap. Effect: The effective cross section area of air gap increase so the reluctance of the air gap decrease. The flux density Bg < Bc, Bc is the flux density in the core. If the air gaps is small, the fringing effect can be neglected. So Ag Ac Bg Bc Ac In practical, large air gap will be divided into several small air gaps to reduce the fringing effect. 16 INDUCTANCE A coil wound on a magnetic core as shown in figure above, is frequently used in electric circuits. This coil may be represented by an ideal circuit element, called inductance, which is defined as the flux linkage of the coil per ampere of its current. Flux linkage N Inductance L i N NBA NHA NHA N 2 N 2 L Hl l i i i R N A 17 Example 15cm 30cm 10cm 15cm i N=200turns l1 l2 15cm 30cm 30cm 15cm 10cm A ferromagnetic core is shown in Figure above. Three sides of this core are of uniform width, while the fourth side is somewhat thinner. The depth of the core (into the page) is 10cm and the other dimensions are shown in figure. There is 200 turn coil wrapped around the left side of the core. Assuming relative permeability r of 2500, how much flux will be produce by a 1A input current? 18 Solution Example The mean path length of region 1 is 45cm and the cross-sectional area is 10 x 10 cm = 100cm2. Therefore, the reluctance in the first region is: l1 l1 R1 A1 r o A1 0.45m (2500)( 4x10 7 )(0.01m 2 ) 14,300 A.turns / Wb The mean path length of region 2 is 130cm and the cross-sectional area is 15 x 10 cm = 150cm2. Therefore, the reluctance in the second region is: R2 l2 l2 A2 r o A2 1.3m (2500)( 4x107 )(0.015m 2 ) 27,600 A.turns / Wb 19 Solution Example Therefore, the total reluctance in the core is: Req R1 R2 14,300 A.turns / Wb 27,600 A.turns / Wb 41,900 A.turns / Wb The total magnetomotive force (MMF) is: F NI (200turns)(1.0 A) 200 A.turns The total flux in the core is given by: F 200 A.turns R 41,900 A.turns / Wb 0.0048Wb 20 Assignment 3 15cm 30cm 15cm 15cm i 30cm N=200turns lc 15cm 30cm 15cm 15cm A ferromagnetic core is shown in Figure above. All side of this core are uniform width. The depth of the core is 10cm. Assuming relative permeability r of 2500, how much flux will be produce by a 1A input current? 21