EE220 ELECTRICAL AND ELECTRONIC PRINCIPLES I UNIT 6 MAGNETIC CIRCUITS Content of Unit 5.1. 5.2. 5.3. 5.4. 5.5. 5.6. 5.7. 5. 8. 5. 9. 3/22/2021 Magnetic flux and flux density Faraday’s Law Lenz’s Law Magnetic circuits Magneto-motive-force (MMF) Magnetic field Strength (H) Permeability Reluctance Electric and Magnetic circuits Analogy Mr B. Kumwenda 2 Content/Objectives 1. State the properties of a magnetic flux 2. Describe the magnetic field due to current in conductor and solenoid 3. State Faraday’s law of electromagnetic induction and Lenz’s law 4. Describe the force on a current carrying conductor 5. Define magnetic density, magnetomotive force, magnetic field strength, permeability, reluctance 6. Analyze magnetic circuits and its analogy with electric circuits 3/22/2021 Mr B. Kumwenda 3 Magnetic Flux and Flux Density • Magnetic flux Φ, measured in weber (Wb), is the amount of magnetic field produced by a magnetic source • Magnetic flux density is the amount of flux passing through a defined area (A) that is perpendicular to the direction of the flux • Magnetic flux density B = Magnetic flux area = Φ 𝐴 Shield from stray field 3/22/2021 Mr B. Kumwenda 4 Properties of Magnetic Flux • The direction of a line of magnetic flux at any point in a non-magnetic medium, such as air, is that of the north-seeking pole of a compass needle placed at that point. i.e. North to South for a magnet • Each line of magnetic flux forms a closed loop • Lines of magnetic flux never intersect • Lines of magnetic flux are like stretched elastic cords, always trying to shorten themselves • Lines of magnetic flux which are parallel and in the same direction repel one another • If a magnetic material, such as soft iron, is placed in the flux path, the flux lines will pass through the soft iron rather than the surrounding air because flux lines pass with greater ease through magnetic materials than through air. • This principle is put to use in the shielding of sensitive electrical elements and instruments that can be affected by stray magnetic fields 3/22/2021 Mr B. Kumwenda 5 Magnetic field due to current in a conductor Conductor top View: Current entering Single turn coil 3/22/2021 Screw rule or righthand thumb rule Mr B. Kumwenda 6 Magnetic field due to current in a current in a solenoid (Coil) Electromagnet Right hand grip rule 3/22/2021 Mr B. Kumwenda 7 Faraday’s and Lenz’s laws of Electromagnetic induction Faraday’s Law: an induced e.m.f. is set up whenever the magnetic field linking that circuit changes and is proportional to the rate of change of the magnetic flux linking the circuit. • 𝒆= 𝒅𝝓 𝑵 𝒅𝒕 = 𝑩𝒍𝒗𝒔𝒊𝒏θ • where 𝑩, the flux density, 𝒍, the length of conductor in the magnetic field, 𝒗, the conductor velocity, θ is the angle the conductor moves to the magnetic field Lenz’s 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. 3/22/2021 Mr B. Kumwenda 8 Lenz’ Law Contd’ A bar magnet passes through a coil: (i) (ii) (iii) (a) Indicate the direction of the induced I in each case. Explain briefly. 3/22/2021 Mr B. Kumwenda 9 When magnet’s N-pole is moving into coil, S N + I e (i) Lenz’s law 3/22/2021 - induced I flows in such a direction as to produce a N-pole to oppose the approaching of magnet. Mr B. Kumwenda 10 When magnet reaches the middle position of the coil, (ii) The induced I become zero I is about to change direction. 3/22/2021 Mr B. Kumwenda 11 When magnet’s S-pole is leaving the coil, N S - e + I (iii) induced I flows in such a direction as to produce a N-pole to oppose the leaving of magnet. 3/22/2021 Mr B. Kumwenda 12 Force on a current carrying conductor Before current flow conductor When current flows conductor Apply Fleming’s Left-Hand Rule or Motor Rule 𝐹 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 ∝ (𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 × 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑐𝑢𝑟𝑟𝑒𝑛𝑡) 𝐹 = 𝐵𝑙𝐼 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 3/22/2021 Mr B. Kumwenda 13 Applications of Electromagnets 3/22/2021 Mr B. Kumwenda 14 Application of Magnetic Induction Magnetic Levitation (Maglev) Trains • Induced surface (“eddy”) currents that produce field in opposite direction Repels magnet Levitates train • Maglev trains today can travel up to 603 km/h (375 mph) in April 2015 – covered 1.8 km in 11 seconds on a 42.8 km magnetic-levitation track • More than twice the speed of conventional trains (150 mph or 241 km/h) 3/22/2021 Mr B. Kumwenda 15 Magnetic circuits Effect of air gap Fringing effect Ideal airgap 3/22/2021 Mr B. Kumwenda 16 Magneto-motive-force (MMF) • In an electric circuit, the current is due to the existence of an electromotive force. • By analogy, we may say that in a magnetic circuit the magnetic flux is due to the existence of a magnetomotive force 𝐹 (m.m.f.) caused by a current flowing through one or more turns. • The value of the m.m.f. is proportional to the current 𝐼 and to the number of turns 𝑁, and is descriptively expressed in ampereturns. mmf 𝐹 = 𝑁𝐼 amperes 3/22/2021 Mr B. Kumwenda 17 Magnetic field Strength (H) • The magnetic field strength is the m.m.f. gradient at any point in a field. • If the magnetic circuit is homogeneous and of uniform cross-sectional area, the magnetomotive force per metre length of the magnetic circuit is termed the magnetic field strength. • 𝐻= 3/22/2021 𝐹 𝑙 = 𝐼𝑁 𝑙 amperes per metre Mr B. Kumwenda 18 Permeability 𝝁 • Measure of the ability of a material to support the formation of a magnetic field within itself • The ratio B/H in a vacuum is termed the permeability of free space and is represented by the symbol μ0 • The value of this is almost exactly the same whether the conductor is placed in free space, in air or in any other non-magnetic material such as water, oil, wood, copper, etc. 𝐵 𝐻 • 𝜇0 = = 4𝜋 × 10−7 H/m for a vacuum and nonmagnetic materials • Relative permeability 𝜇𝑟 = 3/22/2021 𝜇 𝜇0 Mr B. Kumwenda 19 Magnetization Characteristic (or B/H Curve) 3/22/2021 Mr B. Kumwenda 20 Reluctance 𝑆 F Φ • 𝑆= • Derivation: • Using: Φ = BA and F = Hl and • The F ratio: Φ = 𝐻𝑙 𝐵𝐴 = 𝑙 𝜇𝑟 𝜇0 𝐴 𝐵 𝐻 = 𝜇𝑟 𝜇0 =𝑆 • Thus, 𝐹 = Φ𝑆 • The unit of reluctance is the ampere per weber, abbreviated to A/Wb • The reluctance of a magnetic material is proportional to the mean length and inversely proportional to the product of the cross sectional area and the permeability of the magnetic material 3/22/2021 Mr B. Kumwenda 21 Magnetic Circuit Analysis • In order to analyze any magnetic circuit, two steps are mandatory as illustrated below: • Step #1: Find the electric equivalent circuit that represents the magnetic circuit. • Step #2: Analyze the electric circuit to solve for the magnetic circuit quantities. 3/22/2021 Mr B. Kumwenda 22 Step #1: Find the electric equivalent circuit that represents the magnetic circuit. 3/22/2021 Mr B. Kumwenda 23 Step #2: Apply Magnetic Kirchhoff's Laws to solve for the unknown magnetic quantities • First Law : Summation of fluxes entering in a junction is equal to that of leaving • Second Law or Amperes circuital Law : Summation of magneto motive forces in a closed magnetic circuit is zero 3/22/2021 Mr B. Kumwenda 24 Example: Apply Kirchhoff ’s Laws Kirchhoff's First Law a) or Amperes circuital Law 𝐹=0 b) ∴ +𝑁𝐼 − 𝐻𝑎𝑏 𝑙𝑎𝑏 − 𝐻𝑏𝑐 𝑙𝑏𝑐 − 𝐻𝑐𝑎 𝑙𝑐𝑎 = 0 3/22/2021 Mr B. Kumwenda 25 Electric and Magnetic circuits Analogy 3/22/2021 Mr B. Kumwenda 26 Tutorial 1. 3/22/2021 Mr B. Kumwenda 27 Solution 𝒍𝒄 ≫ 𝒍𝒈 Conclusion: 𝑺𝒈 ≫ 𝑺𝒄 thus, 𝑰𝒘𝒊𝒕𝒉 𝒈𝒂𝒑 ≫ 𝑰𝒏𝒐 𝒈𝒂𝒑 Explain? Note: Draw the electric equivalent circuits To represent the cases 3/22/2021 Mr B. Kumwenda 28 Tutorial 2 A rectangular coil measuring 200mm by 100mm is mounted such that it can be rotated about the midpoints of the 100mm sides. The axis of rotation is at right angles to a magnetic field of uniform flux density 0.05T. Calculate the flux in the coil for the following conditions: (a) the maximum flux through the coil and the position at which it occurs; (b) the flux through the coil when the 100mm sides are inclined at 45° to the direction of the flux. 3/22/2021 Mr B. Kumwenda 29 Flux is given by: Φ = 𝐵𝐴𝑠𝑖𝑛𝜃 Solution a. Maximum flux 𝜃 = 90° b. Flux at 𝜃 = 45° 3/22/2021 Mr B. Kumwenda 30 Tutorial 3. A magnetic circuit comprises three parts in series, each of uniform cross-sectional area (c.s.a.). They are: (a) a length of 80mm and c.s.a. 50mm2, (b) a length of 60mm and c.s.a. 90mm2, (c) an airgap of length 0.5mm and c.s.a. 150mm2. A coil of 4000 turns is wound on part (b), and the flux density in the airgap is 0.30 T. Assuming that all the flux passes through the given circuit, and that the relative permeability 𝜇𝑟 is 1300, estimate the coil current to produce such a flux density. 3/22/2021 Mr B. Kumwenda 31 Flux 𝜱: Solution 𝑁𝑜𝑡𝑒: Φ𝑎 = Φ𝑏 = Φ𝑐 Neglecting fringing effects (airgap: part c) Part a. Magnetomotive force Note: Draw magnetic and Equiv.electric circuit: Part a: 80 mm, 50 𝑚𝑚2 Part b: 60 mm, 90 𝑚𝑚2 Part c: 0.5mm, 150 𝑚𝑚2 N = 4000 Apply Ampere circuital Law Part b. Magnetomotive force Part c. Magnetomotive force 3/22/2021 Mr B. Kumwenda ∴ coil current to produce a flux density of 0.3 T is: 𝐹 𝐼= 𝑁 32 END OF LECTURE “Be ready when opportunity comes... Luck is the time when preparation and opportunity meet.” Roy D. Chapin Jr. 3/22/2021 Mr B. Kumwenda 33