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Magnetism Magnetism: Permanent and Temporary

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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

3 Assignments Read the handout and the Powerpoint on the website.
These are in the handout: 456/1,6 465/1-2,4 488/ 1,2b,5,7

4 General Properties of Magnets
Like magnetic poles repel; unlike magnetic poles attract Magnetic field lines are directed from north to south Magnetic field lines always form close loops A 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 t Ampere’s Rule for parallel, straight conductors: F = 2k L I1 I2 / d Transformer Equations Pp = Ps  VpIp = VsIs Is = Vp = Np Ip Vs Ns Induction M = -Es / D Ip/ D t L = -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 direction Strong magnets – iron and steel Very strong – Alnico alloy Weak – aluminum, platinum Natural – magnetite or lodestodes formed when rock was molten

15 Magnetic field lines Magnetic flux, (F) – number of field lines passing through a surface Unit: weber = 1 nm/amp Magnetic flux density, B = F /A Unit: wb/m2 = nm/a m2 = n/am 1 wb/m2 = 1 Tesla Earth, 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 wire Fingers of your hand circle the wire and show the direction of the magnetic field Knuckles, N Finger 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 current Extended 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 by Increasing the current Adding loops of wire Inserting 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 up Extend thumb 90 degrees to rest of fingers Fingers point in direction of B field Thumb points in direction of current, I Imaginary 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 flow Imaginary 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 Problem Calculate 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 20-23 Fig 20-24

27 20-3 Force on Electric Current in Magnetic Field

28 Sources Other physics information:
Great diagrams of magnetic rules:

29 Electromagnetic Induction

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 t E, emf, volts -N, # of turns of wire (- means the current opposes the change that induced it) DF, change in flux linkage in weber, wb D t, change in time, sec

32 Sample Problem If 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 s E = x 10 –2 v Imagine what thousands of turns would produce!

34 Electric Generators Convert mechanical energy into electrical energy by rotating a looped conductor (armature) in a magnetic field Alternating-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 m2 N = 35 B = .55 T T = 0.85 s E = -N D F / D t = - NBA / D t E = 35 (0.55 T) (2 m 2) 0.85s E = 45.3 v

37 Generator Output

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 Inductance The property of an electric circuit by which a varying current induces a back emf in that circuit or a neighboring circuit. Mutual Inductance, M Self Inductance, L

40 Mutual Inductance Effect 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 t Shows the ratio of induced emf in one circuit to the rate of change of current in the other circuit. M, inductance, Henry Es, average induced emf across secondary D Ip/ D t, time rate of change in current in primary coil - sign, induced v opposes D I

42 Self Inductance Ratio of induced emf across a coil to the rate of change of current in the coil L = -E / D I / D t L, henry I, current, amp T, time, sec

43 Transformer Two 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 coil When 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 inductance Secondary voltage = secondary #turns Primary voltage primary # turns Power = Voltage x Current

45 Transformers lose no power
Pp = Ps  VpIp = VsIs Transformer Equation: Is = Vp = Np Ip Vs Ns

46 Transformer Problem A 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 kV Pp = Ps, VpIp = VsIs  Ip = VsIs/Vp = Ip =1350v(2.00a)/90.0v = 30.0a Pp = VpIp = 90.0v(30.0a) = 2.70 kW

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