EEE1001/PHY1002. Magnetic Circuits. The circuit is of length l=2πr. B andφ circulate

Transcription

1 1 Magnetic Circuits Just as we view electric circuits as related to the flow of charge, we can also view magnetic flux flowing around a magnetic circuit. The sum of fluxes entering a point must sum to zero The sum of MMF drops around any closed loop equals the current enclosed The circuit is of length l=2πr i N turns r B andφ circulate Crosssection of core is A

2 2 Magnetic Circuits We can now introduce a new concept. In analogy to resistence in electrical circuits, we have a quantity which measures the resistance a magnetic material has to the flow of flux: Reluctance, S. The circuit is of length l=2πr i N turns B andφ circulate MMF = φs r Crosssection of core is A C.f. V=iR

3 3 Magnetic Circuits MMF = φs The circuit is of length l=2πr B andφ circulate H=Ni/l B=µH=µNi/l φ=ba=aµh =(µa/l) Ni i r N turns Crosssection of core is A φ (l/µa) = Ni φ S = AA l µa + F - S

4 4 Magnetic Circuits in Series l S iron = AA i µa i l i S gap = AA l g µ 0 A l g The reluctances add like resistances: Ni=φ(S iron +S gap ) F=Ni + F - S φ S S gap S iron

5 Magnetic Circuits in Parallel l 3, A 3 i 1 i 2 l 1, A 1 l 2, A 2 A typical parallel magnetic circuit involves a pair of coils inductively coupled. The net flux can flow in the three arms in either direction depending on the strength of the source of flux, the MMFs C.f. emf sources in parallel circuits. Fields, Materials and Devices 1

6 Magnetic Circuits in Parallel S 1 S F S 1 F 2 φ 1 3 φ 2 - Reluctances: S 1 =l 1 /µa 1 S 2 =l 2 /µa 2 S 3 =l 3 /µa 3 Circuits: N 1 i 1 = φ 1 S 1 + (φ 1 +φ 2 )S 3 N 2 i 2 = φ 2 S 2 + (φ 1 +φ 2 )S 3 Fields, Materials and Devices 1

7 Link between inductance and 7 reluctance in a solenoid The flux linking a coil can be expressed as a function of flux through the core, and a function of the current in the coil: a) Ψ=Nφ b) Ψ=Li But we just saw that MMF=φS

8 Link between inductance and 8 reluctance in a solenoid Combining these: a) Ψ=Nφ b) Ψ=Li c) MMF=φS = Ni Li = Nφ = N (MMF/S) Li = N ( Ni /S ) So L = AA N2 S

9 9 Self-inductance of a magnetic circuit L = AA N2 S

10 10 Voltage and inductance Faraday s law tells us that Again, if the coil carrys a current i, we may use a) Ψ=Nφ b) Ψ=Li V = dψ dt

11 11 Voltage and inductance V = dψ dt Substitution of these expressions for Ψ, we get: V=N(dφ/dt) V=d(iL)/dt V=i(dL/dt)+L(di/dt)

12 12 Voltage and inductance V=N(dφ/dt) V=d(iL)/dt V=i(dL/dt)+L(di/dt) If L is indepentent of time, then V = L di dt We use this in determining stored energy

13 13 Stored energy We shall assume a fixed inductance, L. Then the voltage across the system may be expressed as V = ir + L(di/dt) This may be converted to a power by multiplication by the current iv = i 2 R + il(di/dt)

14 14 Stored energy Ohmic losses Magnetic power iv = i 2 R + il(di/ di dt

15 15 Stored energy The energy is the time integral of the power This can be evaluated rather simply in the limit of a time independent L, as W = ½ Li 2 W= il dt di dt

16 16 Self and mutual inductance Up to this point we have largely only been concerned with single coils. However, we are often in practice interested in coupled coils.

17 17 Self and mutual inductance Suppose two coils are wrapped around a common magnetic path. Coil 1 is driven by a voltage V 1, and coil 2 is disconnected. The flux from the first coil couples the second coil through the magnetic flux circulating in the core. V 1

18 18 Self and mutual inductance The flux-linkage in coil 2 is Ψ 2 =N 2 φ 1 where V 1 φ 1 =L 1 i 1 / N 1 Hence Ψ 2 = N 2 L 1 i 1 / N 1

19 19 Self and mutual inductance We specify the coefficient of fluxlinkage in coil 2 due to the primary current is called the mutual inductance, or V 1 N Ψ 2 = M i 2 L 1 1 M = N 1

20 20 Self and mutual inductance In fact there are two sources of flux linking the coils Self-inductance The flux from the second coil i 1 i 2 Ψ 1 =L 11 i 1 +Ψ 12 Ψ 2 =L 22 i 2 +Ψ 21 Ψ 12 is the flux linking coil 1 due to the current in coil 2, M 12 i 2 Ψ 21 is the flux linking coil 2 due to the current in coil 1, M 21 i 1

21 21 Mutual induction For a system of an arbritrary number of sources of flux, the simplest way to express the set of equations that we need to solve is matrix based: Here, L ii is the self inductance of source i, and M ij =M ji is the mutual inductance of coil i due to coil j.

22 22 Mutual induction V 2 =dψ 2 /dt V 2 =d(n 2 φ 1 )/dt V 2 =N 2 dφ 1 /dt V 2 =N 2 d(ψ 1 /N 1 )/dt V 2 =(N 2 /N 1 ) dψ 1 /dt V 1 V 2 V 2 =(N 2 /N 1 ) V 1

23 23 Mutual induction N 1 V 2 = N 2 V 1 This assumes that all the flux from coil 1 links coil 2. In practice, some flux is lost Flux leakage V 1 V 2

24 24 Mutual induction and energy We shall look at the example of a pair of sources, but the principles apply generally. Taking time derivitives of the coupled equations.

25 25 Mutual induction and energy We suppose that we start with no current and no flux, so that there is no stored energy. We want to find the total energy when there is both current in coil 1 and coil 2. Start with no current in either coil, and increase the current in coil 1 to i 1 before introducing any current into coil 2. The total input power is then given by:

26 26 Mutual induction and energy Since there is no current in coil 2, all terms in i 2 disappear.

27 27 Mutual induction and energy The energy is the time integral of the power:

28 28 Mutual induction and energy We now introduce a current in coil 2 up to i 2, maintaining the current in coil 1 at i 1 (so di 1 /dt=0) The time integral gives the energy

29 29 Mutual induction and energy

30 30 Mutual induction and energy i 1 i 2 The total energy stored in the two coil system is the sum of these two processes. There are two terms related to the selfinductance that we ve seen before, plus an additional term due to their interaction.

31 31 coil Coil sits in radial magnetic field Circumferential current creates axial movement in reaction to magnetic field south north Diaphragm needs to be light and stiff e.g. cardboard south force = i = R 2 B V force v magnet audio + ( ωl) audio il 2 sin ( ωt + φ) if coil induc tan ce L is small Spring e.g. rubber impregnated

32 32 Applications of Electromagnetic Forces Electric motors: Electrical power is input Energy flows in electromagnetic fields Conversion into mechanical power Should it be an electric or magnetic field?

33 33 Applications of Electromagnetic Forces Both H- and E-fields store energy. Force acts in a sense that minimises energy stored in the field Work done = F.dr We equate w.d. to the sum of the loss in stored energy Input electrical energy Losses

34 34 Magnetic case Iron Iron l A coil wound around a piece of iron d.c. current Coil arrangement separated from another piece of iron by an air-gap Magnetic flux flows across gap and exerts a force of attraction

35 35 Magnetic case Area, A Iron Iron Energy density stored in the gap is B B B H db = db = µ 2 l µ 0 0 Energy stored in the gap is the gap volume B 2 E = B2 Al 2µ

36 36 Magnetic case Area, A Iron Iron l Let us now suppose the gap is reduced by an amount δl. The energy stored is reduced by E = B2 Al 2µ δe = B2 Aδl 2µ

37 37 Magnetic case Area, A Iron Iron l For a small enough change, the force is given by (δe/δl), so F = B2 A 2µ The force density is then F = B2 2µ

38 38 Magnetic case Area, A Iron Iron l F = B2 2µ For iron, the magnetic flux density saturates around B=1T What force density does this correspond to?

39 Please make your selection Nm Nm ,000 Nm 2 33% 33% 33% 0.04 Nm2 40 Nm2 400,000 Nm2 :10

40 40 Magnetic force and induction Alternative approach uses the energy as a function of the inductance of the coil producing the field: Stored energy = ½ Li 2 The force is the rate of change with position F = d dx [ ] 2 Li 1 = i 2 dl dx

41 41 Electric case In direct analogy with the magnetic case, the energy stored per unit volume is D E db = D 0 0 D dd ε 2 D εe = = 2ε 2 The force based upon a small distance change is F = ½εE 2 A 2

42 42 Electric case The force per unit area is then F = ½εE 2 = ½DE The maximum value of electric field strength is dictated by the breakdown field. If this is around 3x10 6 V/m for air, what is the maximum force per unit area?

43 Please make your selection Nm Nm ,000 Nm 2 33% 33% 33% 0.04 Nm2 40 Nm2 400,000 Nm2 :10

44 44 Magnetic field based forces are generally greater in practical machines

45 45 Electric field motors There is an exception at the small length scale. This image is of a micro-motor, the rotational forces of which are based upon electrostatics.

46 46 Force of alignment The basic mechanism behind motors is that the fields are arranged so as to bring parts into alignment. force flux lines magnetically permeable material airgap force y x

47 47 Force of alignment We now have the iron offset in the x- direction. The stored energy is reduced if the components are brought closer to vertical alignment. There is a force of alignment

48 48 Force of alignment The force is given by the rate of change of stored energy with movement F x =½i 2 (dl/dx) F y =½i 2 (dl/dy) This is the force present in reluctance motors

49 49 Continuous motion B - - C A rotor A C + + B back-iron stator tooth winding Continuous motion is achieved by careful design of the coils and rotating components. Here coils are energised in sequence to generate rotation.

50 50 Continuous motion

51 51 Example application Dyson 100,000 rpm Vacuum cleaner

52 52 Force from mutual induction Again, the two coil components experience a force of alignment to minimise the magnetic energy. Here it is the mutual inductance in play F x =i 1 i 2 (dm 12 /dx)

53 53

54 54 Application of F=BiL We can look at the field generated by one coil, and calculate the force on the second using F=BiL This may be performed on either set of current carrying coils.

55 55 Application of F=BiL The BiL force induces motion in the direction of the force, which therefore has an associated amount of work. For a small displacement in the x- direction, the work done is W.d. = BiLδx

56 56 Application of F=BiL The conductor of length L moving at a speed u x orthogonal to a magnetic field of flux density B also induces a voltage: V=BLu x V=BL (δx/δt) This voltage opposes the current.

57 57 Application of F=BiL To maintain the current against the induced, opposing voltage, electrical power must be supplied: P = Vi P = BLi (δx/δt) Rearraning yields: Pδt = BLi δx or the energy supplied electrically balances the work done.