Magnetic Fields What causes material to be magnetic? Does just spinning make a magnet?

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Presentation transcript:

Magnetic Fields What causes material to be magnetic? Does just spinning make a magnet?

Why can some material become magnetic? It is the alignment of the spin that causes material to be magnetized. Polarization: opposite spin or different direction of the domain. 1. If the alignment is fixed or constant then the material is a permanent magnet. a. Permanent magnets are made of Al, Ni, and Co. Ferromagnetic material 2. If the alignment is temporary, then the material will make a temporary magnet. (Iron) 3. If the alignment is not obtainable, then the material is non- magnetic. (Plastic)

Magnetic Field Lines Description of the magnetic flux. Magnetic flux acts in the direction of a positive charge. Magnetic field lines go out of north poles and to south poles. The closer together the field lines are, the stronger the field. North attracts to the South and repel to North.

Magnetic Field Electric Field LinesMagnetic Field Lines 1.Vector force 1. Vector force 2.Denser lines = stronger 2. Denser lines = stronger 3.Lines cannot cross 3. Lines cannot cross 4.Created at (+) pole 4. Begins at North pole 5.Destroyed at (-) pole 5. Moves to the South pole 6.Charge does not move 6. Lines move through the through the batterymagnet, continuous flux 7.Charges begin and end 7. Never ends. Always form closed loops

Magnetic Interaction North to South pole attract. North to North or South to South repel. TLS: Page 665 # 59-62

Electromagnetism Han Christian Oersted: first to relate an electric current to magnetism. Findings: 1. Magnetic force act perpendicular to the direction of the current. 2. The direction of the magnetic field line depends on the direction of the current. 3. The strength of the magnetic field lines is inversely proportional to the distance from the conductor. 4. The strength of the field is proportional to the strength of the current. Increase current = increases strength

5. The number of turns of the conductor (solenoid) is proportional to the strength of the field. Increase turning = increase strength 6.The direction of the field can be changed with a change in the direction to the current. 7.Better conductors = better strength

Strength of an Electromagnet 1.Increase current (increase force ‘volts’) 2.Decrease resistance to the current 3.Increase the number of coils (increases the number of lines of force). 4.Better conductors (allows for better alignment of the domain) 5.Decrease temperature (decreases resistance)

Electromagnet Force First Right Hand Rule: determines the direction of a magnetic field relative to the direction of the current.

Electromagnetic Force Second Right Hand Rule: determines the direction of magnetic field relative to the direction of the current in an electromagnet.

Calculating Electromagnetic Force Third Right hand Rule: Thumb is the direction of current, the figures show the direction of the magnetic field lines and the palm show the direction of the force created. F = ILB I = current measured in ampere, A. (C/s) L ( l )= length of the current carrying conductor (m). B = Strength of the magnetic field, Talsa (T) T = N/A m

Electromagnetic Sampler 1. A 45 cm conductive wire carries a current of 7.5 A at a right angle to a T magnetic field. How strong is the force acting on the conductive wire?

Electromagnetic Sampler 2.What is the strength of a magnetic field when a 120 cm conductive wire carrying 6 A exerts a force of 2.3 N on the wire? TLS: #16-20, p 654 Problems Set One.

Calculate the Electromagnetic Force of a Moving Particle The force on a moving particle in a magnetic field is proportional to the charge, the velocity of the particle at a right angle to the field and the magnetic field strength. F = qvB Unit for the force is Newton, N q+ use the right hand rule q- use the left hand rule v velocity in m/s B electric field intensity, Tasla, T

Electromagnetic Sampler 1.What is the force acting on an electron ( 1.6 x C ) moving from left to right in a magnetic field of 0.23 T at a velocity perpendicular velocity of 5.2 x 10 6 m/s? 2.What direction is the force acting?

Electromagnetic Sampler 2. The force acting on a charge, q+, is 5.2 x N. If the charge is of a magnitude equal to 1.55 x C and is traveling at 2.5 x 10 6 m/s, what is the strength of the magnetic field? What direction is the force acting? TLS: #21-25, p 658 Problem Set Two

Electromagnetic Induction EM Induction: the generation of a magnetic field through an electric current. Changing one current induces a change in another current. Michael Farraday: moving a conductive wire through a magnetic field would generate an electric current in the wire. That generated current will cause change in another current.

Fourth Right Hand Rule: Thumb  in the direction of the motion of the wire Finger  in the direction of the magnetic filed N  S Palm  in the direction of the force acting on the wire. The Force = EMF

Electromotive Force EMF: a ‘Force’ applied by the magnetic flux on a charge carried in a conductive wire that is introduced into the magnetic field. EMF = potential difference = voltage EMF  is the influence that causes a current to flow form low to high potential. EMF = BLv B = magnitude of the magnetic field (T) L = length of the conductive wire (m) v = velocity perpendicular to the field (m/s)

Sampler One EMF A conductive wire that is 150 cm long is moved at a constant speed of 4.5 m/s perpendicular to a magnetic field measured at T. What is the EMF? At a resistance of 0.25Ω, what is the current?

Sampler #2 EMF A conductive wire that is 0.45 m in length moves through a magnetic field at 15 m/s. If the wire carries a current of 0.33A and has an internal resistance of the wire is 1.52 Ω, what is the magnitude of the magnetic flux?

Electric Generators Converts mechanical energy to electric energy. Diagram: Strength (generated HP) 1. Increase magnet 2. Increase current (decrease resistance) 3. Increase the number of turnings. Page 675 # 1-4, and page 692 #60-63, 67, 68.

Effective Current/Voltage Effective Current: The amount of current generated by a moving conductor within a magnetic field. I eff = √2 /2 x I max Note page 677 I eff = x I max Effective Voltage: The amount of voltage generated at the effective current. V eff = √2 /2 x V max Note page 678 V eff = x V max

Reinforcement Ohm’s lawI = V/R Power P = I 2 R or P = V 2 /R or P AC = ½ P AC max Page 678, #5-8

Transformers Device used to increase or decrease AC voltage. Components: Iron core primary coil secondary coil conductive turnings

Type of Transformers 1.Step-Up Transformer: increase the voltage by increasing the number of turnings. a. Primary coil has fewer turnings than the secondary coil. 2. Step-Down Transformer: decreases the voltage by decreasing the number of turnings. a. Primary coil has more turnings than the secondary coil.

Transformers The increase or decrease in the voltage is proportional to the number of turnings. From 2 turns  20 turns, there is a 10X increase in the number of turns and therefore a 10 fold increase in voltage. (step up) From 100 turns  50 turns, there is a 2X decrease in the number of turns and therefore a ½ decrease in the voltage (step down).

Transformer Calculation V s / V p = N s /N p The current is inversely proportional to the voltage and therefore to the number of turnings Increase voltage by increasing turning and the results is a decrease in the current. I s / I p = V p / V s = N p / N s

Transformer Sampler 1. A transformer has 200 turns on the primary coil and 3000 turns of the secondary coil. The primary coil is supplied with 90 V. a.What type of transformer is this? b.What is the voltage in the secondary coil? c.The current in the secondary coil is measured at 2A. What is the current in the primary coil?

Transformer Sample 2.In a transformer the primary coil is supplied with 360 V. All you know is that the turnings has been decrease 3X. a. If the primary coil has 30 turnings, how many turning are in the secondary coil? b. What is the voltage in the secondary coil? c. If the current in the primary coil is 0.5 A, what is the current in the secondary coil?