Presentation on theme: "Magnetism Magnetism: Permanent and Temporary"— Presentation transcript:
1 Magnetism Magnetism: Permanent and Temporary See FSU’s site for much more information!!1
2 Magnetic Topics Magnetic Fields Electromagnetic Induction Electromagnetism
3 Assignments Read the handout and the Powerpoint on the website. These are in the handout:456/1,6465/1-2,4488/ 1,2b,5,7
4 General Properties of Magnets Like magnetic poles repel; unlike magnetic poles attractMagnetic field lines are directed from north to southMagnetic field lines always form close loopsA magnetic field exists around any wire that carries current
5 Gen’l Properties cont.A coil of wire that carries a current has a magnetic field about a permanent magnet
6 Forces Caused by Magnetic Fields When a current-carrying wire is placed in a magnetic field, a force acts on the wire that is perpendicular to both the field and the wire. Meters operate on this principle.Magnetic field strength is measured in tesla, T (one newton per ampere per meter).b is the symbol for magnetic field
7 Forces cont.An electric motor consists of a coil of wire (armature) placed in a magnetic field. When current flows in the coil, the coil rotates as a result of the force on the wire in the magnetic field.
8 Forces cont.The force a magnetic field exerts on a charged particle depends on the velocity and charge of the particle and the strength of the magnetic field. The direction of the force is perpendicular to both the field and particle’s velocity.
9 E = - N DF / D tAmpere’s Rule for parallel, straight conductors: F = 2k L I1 I2 / dTransformer EquationsPp = Ps VpIp = VsIsIs = Vp = NpIp Vs NsInductionM = -Es / D Ip/ D tL = -E / D I / D t
14 The small picture – how magnetism occurs Domain theory – when enough atoms of a substance line up in the same directionStrong magnets – iron and steelVery strong – Alnico alloyWeak – aluminum, platinumNatural – magnetite or lodestodes formed when rock was molten
15 Magnetic field linesMagnetic flux, (F) – number of field lines passing through a surfaceUnit: weber = 1 nm/ampMagnetic flux density, B = F /AUnit: wb/m2 = nm/a m2 = n/am1 wb/m2 = 1 TeslaEarth, 10–4 T Humans, 10–11 T
16 Hand Rule #1- B field direction around a current carrying wire Point thumb in direction of current in the wireFingers of your hand circle the wire and show the direction of the magnetic fieldKnuckles, NFinger tips, S
18 Hand Rule #2 – Determine the polarity of an electromagnet Wrap the fingers of your right hand around the loops in the direction of the currentExtended thumb points toward the N pole of the electromagnet
19 Solenoid – conducting linear coil which acts like a bar magnet Increase B, magnetic flux density byIncreasing the currentAdding loops of wireInserting an iron core into solenoid – now it is an electromagnet
20 Hand rule #3 – shows force acting on wire in B field Lay right hand flat, palm upExtend thumb 90 degrees to rest of fingersFingers point in direction of B fieldThumb points in direction of current, IImaginary vector coming up perpendicular out of the palm points in the direction of force acting on current carrying wire.
21 Fingers point in direction of b field Thumb - direction of current flowImaginary vector coming from palm is direction that conductor is forced out of the field
23 Moving Charges in a Magnetic Field Right-Hand Rule for Moving Charges
24 Force between 2 current carrying wires. Ampere’s Rule for parallel, straight conductors: F = 2k L I1I2 d F, force, newtons, n k, constant, 10-7 n/a2 L, length of wires, meters, m I, current in wires, amp, a d, distance between wires, meters, m
25 Ampere’s Rule ProblemCalculate the strength of the two equal currents that must flow through two parallel conductors that are 1.5 m long and are 0.25 m apart if a force of attraction of 4.9x10-6 n is present. F = 2k L I1I2 I = Fd F = 4.9x10-6 n d √ 2kL d = 0.25m L = 1.5 m I = 4.9x10-6 n (0.25m) a2 1.22x10-6 √ 2 x10-7n (1.5m) √ 3x10-7 I = 2.02 a
26 20-6 Force Between 2 Parallel Wires Fig 20-23Fig 20-24
27 20-3 Force on Electric Current in Magnetic Field
28 Sources Other physics information: Great diagrams of magnetic rules:
30 Michael Faraday and Joseph Henry around the same time… Discovered that when there is relative motion between a magnetic field and a complete circuit (and the conductor cuts across the magnetic field), that electricity will flow!!! An induced EMF causes electricity to flow.
31 If current flows, there must be an EMF – this is EM induction Here’s the formula:E = - N DF / D tE, emf, volts-N, # of turns of wire (- means the current opposes the change that induced it)DF, change in flux linkage in weber, wbD t, change in time, sec
32 Sample ProblemIf a coil of 200 turns is moved perpendicularly in a magnetic field at a constant rate, find the induced emf. The flux linkage change ( DF / D t) is 4.00 x 10-6 wb in sec.
33 Problem solution E = - N DF / D t E = (-200)(4.00 x 10 –6 wb) 1.00 x 10 –2 sE = x 10 –2 vImagine what thousands of turns would produce!
34 Electric GeneratorsConvert mechanical energy into electrical energy by rotating a looped conductor (armature) in a magnetic fieldAlternating-Current electricity produced is conducted by slip rings and brushes to be used *Direct current can be produced by using split rings *
35 A coil with a wire is wound around a 2. 0 m2 hollow tube 35 times A coil with a wire is wound around a 2.0 m2 hollow tube 35 times. A uniform magnetic field is applied perpendicular to the plane of the coil. If the field changes uniformly from 0.00 T to 0.55 T in 0.85 s, what is the induced emf in the coil?
36 A = 2.0 m2N = 35B = .55 TT = 0.85 sE = -N D F / D t = - NBA / D tE = 35 (0.55 T) (2 m 2)0.85sE = 45.3 v
38 Lenz’s Law -The direction of an induced current is such that the magnetic field resulting from the induced current opposes the change in he field that caused the induced current.When the N pole of a magnet is moved toward the left end of a coil, that end of the coil must become a N, causing induced current flow in opposition.
39 InductanceThe property of an electric circuit by which a varying current induces a back emf in that circuit or a neighboring circuit.Mutual Inductance, MSelf Inductance, L
40 Mutual InductanceEffect that occurs in a transformer when a varying magnetic field created in the primary coil is carried through the iron core to the secondary coil, where the varying field induces a varying emf.
41 M = -Es / D Ip/ D tShows the ratio of induced emf in one circuit to the rate of change of current in the other circuit.M, inductance, HenryEs, average induced emf across secondaryD Ip/ D t, time rate of change in current in primary coil- sign, induced v opposes D I
42 Self InductanceRatio of induced emf across a coil to the rate of change of current in the coilL = -E / D I / D tL, henryI, current, ampT, time, sec
43 TransformerTwo separate coils of wire placed near one another that are used to increase or decrease AC voltages with little loss of energy.It contains a Primary coil and a Secondary coilWhen the primary is connected to AC voltage, the changing current creates a varying magnetic field that is carried through the core to the secondary coil.
44 Transformer, cont.In the secondary coil, the varying field induces a varying emf. This is called mutual inductanceSecondary voltage = secondary #turnsPrimary voltage primary # turnsPower = Voltage x Current
45 Transformers lose no power Pp = Ps VpIp = VsIsTransformer Equation:Is = Vp = NpIp Vs Ns
46 Transformer ProblemA step-up transformer has a primary coil consisting of 200 turns and a secondary coil that has 3000 turns. The primary coil is supplied with an effective AC voltage of 90.0v. A)What is the Vs? B)If Is = 2.00a, find Ip. C) What is the power in the primary circuit?
47 Solution to Transformer Problem Vs = NsVp/Np = 3000(90.0V)/200 = 1.35 kVPp = Ps, VpIp = VsIs Ip = VsIs/Vp =Ip =1350v(2.00a)/90.0v = 30.0aPp = VpIp = 90.0v(30.0a) = 2.70 kW