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Flux Density B × Cross-Sectional Area A) ELL 100 - Introduction to Electrical Engineering LECTURE 31: MAGNETIC CIRCUITS INTRODUCTION A magnetic circuit is made up of one or more closed loop paths containing a magnetic flux ϕ (= magnetic field/flux density B × cross-sectional area A). The flux is usually generated by permanent magnets or electromagnets and confined to a path by magnetic cores consisting of ferromagnetic materials like iron, although there may be air gaps or other materials in the path. Magnetic circuits are employed to efficiently channel magnetic fields in many devices such as electric motors, generators, transformers, relays, solenoids, loudspeakers, hard disks, MRI machines. 2 APPLICATIONS Motors Generators Transformers Circuit Relay Breakers Switches 3 APPLICATIONS Solenoids Hard Disks MRI Machines 4 Magnetomotive force (mmf) In an electric circuit, the current is induced due to the existence of an electromotive force (emf E, battery voltage). By analogy, we say that in a magnetic circuit the magnetic flux is induced due to the existence of a magnetomotive force (mmf F) caused by a current flowing through one or more turns of coil. The value of the mmf F is proportional to the current flowing through the coil and to the number of turns in the coil, and is expressed in units of “ampere-turns” or just amperes (number of turns is dimensionless). F = N.I 5 Magnetic field strength/intensity (H) The magnetomotive force per unit length of the magnetic circuit is termed the magnetic field strength/intensity (H). F H l where, F NI amperes where l is the length of the magnetic circuit or flux loop Units of H are ampere-turns per metre NI H amperes per metre (At/m) or just ampere per metre (A/m) l 6 Permeability of free space μ0 (magnetic constant) The permeability of free space or non-magnetic materials is B for a vaccum and non-magnetic materials 0 H 7 0 4 *10 Hm/ (The units of μ0 are H/m (Henry per meter)) where B (= ϕ/A) is the magnetic flux density (units of Tesla, T), A is the cross-sectional area through which the flux passes, ϕ is the magnetic flux (units of Weber (Wb)) and H is the magnetic field strength (units A/m). 7 SOLVED EXAMPLE Q. A coil of 200 turns is wound uniformly over a wooden ring having a mean circumference of 600 mm and a uniform cross-sectional area of 500 mm2. If the current through the coil is 4.0 A, calculate (a) the magnetic field strength; (b) the flux density; (c) the total flux. Ans: N = 200, I = 4 A, l = 600 mm = 0.6 m, A = 500 mm2 = 5×10-4 m2 (a) H = N.I / l = 200 × 4 / 0.6 = 1333.3 A/m -7 -3 (b) B = μ0H = 4π×10 × 1333.3 = 1.6755×10 T = 1.6755 mT (c) ϕ = B.A = 1.6755×10-3 × 5×10-4 = 8.376×10-7 Wb = 0.8376 μWb 8 SOLVED EXAMPLE Q. Calculate the magnetomotive force required to produce a flux of 0.015 Wb across an air-gap 2.5 mm long, having an effective area of 200 cm2. Ans: ϕ = 0.015 Wb, l = 2.5 mm = 2.5×10-3 m, A = 200 cm2 = 2×10-2 m2 B = ϕ / A = 0.015 / (2×10-2) = 0.75 T -7 5 H = B / μ0 = 0.75 / (4π×10 ) = 5.97×10 A/m F (mmf) = H.l = 5.97×105 × 2.5×10-3 = 1.49×103 A 9 Relative Permeability μr The ratio of the flux density B produced in a material to the flux density produced in vacuum (or in a non-magnetic core) for a particular applied magnetic field strength H. For air and non-magnetic materials, μr = 1 For ferromagnetic materials, e.g. some forms of nickel–iron alloys, the relative permeability can be as large as ~100000 i.e. ~106. For a material having a relative permeability μr, BHr 0 where, = 0 r is the absolute permeability 10 Reluctance S It is the opposition that a magnetic circuit offers to the passage of magnetic flux through it (ratio of mmf applied to the flux induced). BA (1) mmf, F Hl (2) Dividing (1) by (2), BAHA A 0 r F Hl Hl0 r l Thus, S = F / ϕ = l / (μA), where μ = μ0 μr (S has units of A/Wb) The inverse of reluctance is known as permeance (ease of flux passage)11 SOLVED EXAMPLE Q. The simple magnetic circuit shown has a cross-sectional area of 50 cm2 and mean length of 2 m. The relative permeability of the core is 100. The coil has 250 turns and the flux produced is 100 µWb. Find: iron core (a) Reluctance of the magnetic circuit l=2m (b) Current flowing through the coil N=250 50 sq. cm ɸ=100µWb mean length 12 Ans: l (a) Reluctance, S A 75 0 r 4 *10 *100 4 *10 2 S3.18*106 A / Wb 4 *1054 *50*10 iron core mmf NI l=2m (b) Magnetic field strength, H ll Hl I N=250 50 sq. cm N B 100*106 H 159.15 A / m mean 45 ɸ=100µWb A 50*10 *4 *10 length Hl 159.15*2 IA 1.27 N 250 13 SOLVED EXAMPLE Q. The air gap in a magnetic circuit is 1.5 mm long and 2500 mm2 in cross- sectional area. Calculate (a) The reluctance of the air gap (b) The mmf required to set up a flux of 800 µWb in the air gap. l Ans: (a) Reluctance, S A 77 0 rr 4 *10 *1 4 *10 ( of air =1) 1.5*103 mmf MagneticS field strength, H 4.77*105 A / Wb 4 *1076 *2500*10 l mmf Hl B 800*106 (b) H 2.55*105 A / m A 2500*1067 *4 *10 mmf Hl 2.55*1053 *1.5*10 382.5 A 14 EXERCISES Q. A mild steel (relative permeability of 400) ring having a cross-sectional area of 500 mm2 and a mean circumference of 400 mm has a coil of 200 turns wound uniformly around it. Calculate (a) The reluctance of the ring (b) The current required to produce a flux of 800 µWb in the ring. Q. A rectangular shaped core is made of mild steel (relative permeability of 940) plates 15 mm × 20 mm cross-section. The mean length of the magnetic path is 18 cm. The exciting coil has 300 turns with a current of 0.7 A flowing through it. Calculate: (a) Magneto-motive force (b) Flux density (c) Reluctance (d) Magnetic flux 15 MAGNETIC CIRCUIT ANALOGY WITH ELECTRICAL CIRCUIT Flux, ɸ Reluctance, S Current, I Current, I Potential Resistance, R MMF= NI EMF, E difference, V Magnetic circuit Analogous electrical circuit 16 MAGNETIC CIRCUIT ANALOGY WITH ELECTRICAL CIRCUIT 17 SERIES CONNECTION IN MAGNETIC CIRCUIT The applied mmf is equal to the sum of the mmfs dropped across each series element, while the flux through each series element is the same. lc lc ɸϕ ii lc RRc + c A N lg cc N lg NiNi - Rg lg Rg ggA ɸ R1 R2 R3 Rn F FRRRRReq 1 2 3 ........ n (Ri denote reluctances) 18 PARALLEL CONNECTION IN MAGNETIC CIRCUIT The same mmf appears across all the reluctances in parallel, while the total flux is the sum of the individual fluxes in each parallel element. ɸ ɸ1 ɸ2 + R = 1 R R g /(μ A ) 1 R22 = Ni 1 0 1 g /(μ A ) - 2 0 2 where R = R || R = R R / (R + R ) FR eq eq 1 2 1 2 1 2 19 SOLVED EXAMPLE Q. A ring of 30-cm mean diameter is made using a cylindrical iron rod of diameter 2.5 cm. A saw-cut 1-mm wide is made through the ring to create an air- gap. A coil with 500 turns of wire is wound on the ring. Calculate the current required in the exciting coil to produce a flux of 4 mWb in the ring. Assume the relative permeability of iron at this flux density as 800. Neglect any leakage or fringing of the magnetic field. Ans: A = πr2 = 3.14×(0.025 m)2 = 4.91×10-4 m2 B = ϕ / A = (4×10-3 Wb) / (4.91×10-4 m2) 2 = 8.15 Wb/m 20 First consider air gap: -7 H1 = B / μ0 = 8.15/(4π×10 ) A/m = 6.49×106 A/m -3 l1 = 1 mm = 10 m 3 mmf across air gap F1 = H1 l1 = 6.49×10 A Now consider iron ring: Total mmf F = F1 + F2 -7 H2 = B / (μr μ0) = 8.15/(800×4π×10 ) A/m = 1.41×104 A = 8.1×103 A/m Current I = F / N l2 ~ 2πR = 2×3.14×0.15 m = 0.94 m = (1.41×104)/500 = 28.24 A 3 mmf across iron ring F2 = H2 l2 = 7.63×10 A 21 SOLVED EXAMPLE Q. An iron ring of mean circumference 50-cm has an air gap of 0.1-cm length and a winding of 300 turns. If the relative permeability of iron is 400 when a current of 1 ampere flows through the coil, find the flux density in the air gap. Ans: Let B be the flux density in the iron ring. Its value would also be the same in the air gap. First consider iron ring: -7 H1 = B / (μr μ0) = B / (400×4π×10 ) = 1989.44 B l1 = 50 cm = 0.5 m mmf across iron ring F1 = H1 l1 = 995 B 22 Now consider air gap: -7 H2 = B / μ0 = B / (4π×10 ) = 7.96×105 B -3 l2 = 0.1 cm = 10 m mmf across air gap F2 = H2 l2 = 796 B Total mmf F = F1 + F2 = (995 + 796)B = 1791 B But given total mmf F = N.I = 300×1 = 300 2 => B = 300/1791 = 0.1675 Wb/m 23 EXERCISES Q.
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