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Advanced electromagnetism. E.m.f. in a conductor Moving a conductor through a magnetic field can induce an emf. The faster the conductor moves through.

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Presentation on theme: "Advanced electromagnetism. E.m.f. in a conductor Moving a conductor through a magnetic field can induce an emf. The faster the conductor moves through."— Presentation transcript:

1 Advanced electromagnetism

2 E.m.f. in a conductor Moving a conductor through a magnetic field can induce an emf. The faster the conductor moves through the field the greater the emf and hence the greater the current NS

3 E.m.f. in a conductor The bigger the length of the conductor moving through the field the greater the emf and hence the greater the current N S

4 E.m.f. in a conductor The greater the flux density (B) of the field, the greater the emf and hence the greater the current N S

5 Magnitude of induced e.m.f. in a conductor E = Blv volts (B = flux density in Teslas, l = length in metres and v = velocity in m/s) E.m.f. in a conductor

6 It is the rate at which the conductor “cuts” through the magnetic field = dФ/dt volts E.m.f. in a conductor

7 Right hand rule The direction of the current can be found using Fleming’s right hand rule

8 Example A conductor, 800 mm long, is moved at a uniform speed at right-angles to a magnetic field of density 0.8Tesla. Calculate the velocity required to generate an e.m.f. in the conductor of 10 V.

9 Example A conductor, 800 mm long, is moved at a uniform speed at right-angles to a magnetic field of density 0.8Tesla. Calculate the velocity required to generate an e.m.f. in the conductor of 10 V.

10 Example From: E = B ℓ v V = e / B ℓ = 10 / (0.8 x 800 x ) = m/s

11 E.m.f induced in a rotating coil E = - d / dt (NΦ)

12 example A coil of 3000 turns, when energised, produces a magnetic flux of 3.5 mWb. If the energising current is reversed in 0.3 seconds, determine the direction and average value of e.m.f. induced in the coil.

13 example From: E = -N x (dΦ) / dt E = x (-2 x 3.5 x ) / 0.3 = 70V Flux reversal hence +ve

14 Flux linkage Flux relates to the number of field lines Φ = BA Flux linkage takes into consideration the number of turns in a coil Flux linkage = NΦ

15 Induction in terms of Flux linkage (NΦ) L = NΦ/I = N.dΦ/dI Rate of change in flux linkage

16 Example A coil of 250 turns is wound on a non-magnetic ring. If a current of 5 A produces a magnetic flux of 0.4mWb, calculate: a) Inductance of coil b) Average e.m.f. induced in the coil when switching on if the current takes 3 ms to rise to its final value.

17 Example From: L = NΦ / I L = 250 x (0.4 x ) / 5= 20 mH b) From: e.m.f.av = - NΦ./ t e.m.f.av = x (0.4 x ) / (3 x ) = V.

18 N S N S Pushing a magnet into a coil induces a current in the coil wire Pulling the magnet out of the coil induces a current in the opposite direction

19 Current growth in an inductive circuit I = I o (1-e Rt/L ) I t

20 Current decay in an inductive circuit I = I o (e -Rt/L ) I t

21 120V 120 Ω 40 Ω

22 Example A non-reactive resistor of 120 Ω is connected in parallel with a coil of inductance 4 H and resistance 40 Ω. Calculate the current flowing in the coil 0.06 seconds after the circuit is disconnected from a 120 V dc supply.

23 Example Initial current through coil = V/RI= 120 / 40 = 3 A From i = I x e -Rt/L I = 3 x e -(120 x 0.06) / 4 = 3 x e e -1.8 = I= 3 x = mA

24 Energy stored in an inductor W(energy) = 1 / 2 LI 2

25 Example A current of 5 A flows through a coil of 3 H. Calculate the amount of energy stored in the coil.

26 Example From: energy stored = 1/2 x LI 2 Energy stored = 0.5 x 3 x (5 2 ) = 37.5 Joules

27 Mutual Inductance E = - MdI/dt = -N 2 dΦ/dt M = N 2 Φ 2 /I 1

28 Example Two coils have a mutual inductance of 300 μH. Calculate the e.m.f. produced in one coil when the current in the other coil changes at the rate of 20 x10 3 A/s.

29 Example From: E = -M x dI / dt E = -300 x x = -6 V

30 Magnetism The relationship between magnetic field strength and magnetic flux density is: B = H × µ where µ is the magnetic permeability of the substance

31 Magnetic field strength equation in a coil H = (NI) / l where: H = magnetic field strength (ampere per metre) I = current flowing through coil (amperes) N = number of turns in coil l = length of magnetic circuit Magnetism

32 The magnetomotive force in an inductor or electromagnet consisting of a coil of wire is given F = NI where N is the number of turns of wire in the coil and I is the current in the wire.

33 Magnetism Permeability Is a measure of how easily a magnetic field can set up in a material It is the ratio of the flux density of the magnetic field within the material to its field strength µ =B/H Permeabilty of free space µ o is 4π x10 -7 H/m

34 Magnetism Relative Permeablity µ r This is how much more permeable the material is compared to free space (a vacuum). The permeability of the material can be calculated by multiplying its relative permeability by the permeability of free space. µ = µ o x µ r

35 Magnetism Magnetic Flux The rate of flow of magnetic energy across or through a (real or imaginary) surface. The unit of flux is the Weber (Wb)

36 Magnetic Flux Density A measure of the amount of magnetic flux in a unit area perpendicular to the direction of magnetic flow, or the amount of magnetism induced in a substance placed in the magnetic field. The SI unit of magnetic flux density is the Tesla, (T). One Tesla, (1T), is equivalent to one weber per square metre (1 Wb/ m 2 ).

37 Magnetism To summarise The magnetic flux density, B, multiplied by the area swept out by a conductor, A, is called the magnetic flux, Φ. Φ = BA. Unit of flux: weber, Wb Unit of flux density: Tesla, T

38 Example A coil of400 turns is wound uniformly over a wooden ring of mean circumference 200 mm and cross- sectional area 150 mm 2. If the coil carries a current of 2 A, calculate: a) Magneto-motive-force b) Magnetic Field Strength c) Flux Density d) Total Flux

39 Example a) From: m.m.f = NI = 400 x 2 = 800 AT b) From: H = m.m.f. / ℓ = 800 / 200 x = 4 x 10 3 A/m c) From: B / H = μ 0. B = μ 0 x H = 4π x x 4 x 10 3 = x Tesla d) From: Φ = B x area = x x 150 x = μWeber

40 ‘Hard’ magnetic materials Hard magnets, such as steel, are magnetised, but afterwards take a lot of work to de-magnetise. They're good for making permanent magnets..

41 ‘Soft’ magnetic materials Soft magnets are the opposite. With an example being iron, they are magnetised, but easily lost their magnetism, be it through vibration or any other means. These are best for things that only need to be magnetised at certain points, eg magnetic fuse/trip switch

42 Magnetism Solenoid

43 Magnetism The magnetomotive force in an inductor or electromagnet consisting of a coil of wire is given by: F = NI where N is the number of turns of wire in the coil and I is the current in the wire. The unit is amp.turns (AT)

44 Magnetism Magnetic field strength in a coil = mmf/ length of the coil

45 Magnetic field strength equation in a coi H = (NI) / l where: H = magnetic field strength (ampere per metre) I = current flowing through coil (amperes) N = number of turns in coil l = length of magnetic circuit

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47 Retentivity – A measure of the residual flux density corresponding to the saturation induction of a magnetic material. In other words, it is a material's ability to retain a certain amount of residual magnetic field when the magnetizing force is removed after achieving saturation

48 Residual Magnetism or Residual Flux - the magnetic flux density that remains in a material when the magnetizing force is zero. Coercive Force - The amount of reverse magnetic field which must be applied to a magnetic material to make the magnetic flux return to zero. (The value of H at point c on the hysteresis curve

49 Magnetism Example Starting with the concept of molecular magnets in a magnetic material, explain a) Relative permeability of a material b) Loss of magnetisation in a ‘soft’ material c) Magnetic saturation

50 Magnetism a) Relative permeability of a material, how easily molecular magnets align with applied field b) Loss of magnetisation in a ‘soft’ material, how easily molecular magnets take up random alignment c) Magnetic saturation, molecular magnets all aligned in field direction

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53 Example A coil, uniformly wound over a mild steel ring, produces a magnetic field strength of 3000 A/m when energised. Using an appropriate curve : i) Flux density ii) Relative permeability of mild steel under the stated conditions

54 Example From Curve - Flux Density v Magnetic field strength Flux density = 1.5Tesla ii) From B / H = μ 0 x μ r. Then μ r = B / H μ 0 = 1.5 / (3000 x 4π x ) = 398

55 Example  (relative) =  (material) /  air)

56 Example Reluctance of a magnetic circuit (S) S = mmf/ Ф S = ℓ / μ 0 μ r a

57 Example A mild steel ring has a cross- sectional area of 400 mm 2 and mean circumference of 300 mm. If a coil of 250 turns is wound uniformly over the ring, calculate: i) Flux density in the ring assuming a total flux of 600 μWb ii) Reluctance of the ring assuming μ r iii) Current to produce the required flux density Example

58 i) From Φ = B x area. Flux density B = 600 x / 400 x = 1.5 Tesla ii) From curve – μ r = 960 From: S = ℓ / μ 0 μ r a = 300 x / (4π x x 960 x 400 x ) = 6.2 x 10 5 A/Wb

59 Example From: m.m.f. = ΦS = 600 x x 6.2 x 10 5 = 372 Ampere-turns. Current m.m.f/ turns =372/ 250 = 1.48 Amperes


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