2.  Forcing a wire through a magnetic field produces (generates) a current in the wire  Example: Generator  Converts mechanical energy into electrical energy

3.  Generators spin a bundle of wire attached to a motor through magnets causing current or electricity to be produced

4.  Electromagnetic induction- a current is produced in a wire by passing a magnet near it  The magnetic field lines (flux) crossing the wire cause the electrons in the wire to move (electricity)

5.  Force is greatest when the charge moves perpendicular to the magnetic field  At other angles the force is less  The force is 0 when the charge moves parallel to the magnetic field

6.  Magnetic Flux ( Φ ) - The # of magnetic field lines passing through a given area  Magnetic Flux( Φ ) =BA

7.  Where is the greatest concentration of magnetic flux lines?  Near the poles (ends)

8.  A current carrying wire placed in a magnetic field experiences a force  Example: Motors  Converts electrical energy into mechanical energy

9.  1. Keep hand flat  2. Thumb in the direction of the current  3. Fingers in the direction of the magnetic field  4. Palm in direction of the force  All are perpendicular to each other

10.  3 factors affect the magnitude of the force on a current-carrying wire placed in a magnetic field  1. Strength of the magnetic field  2. Amount of current (velocity)  3. Length of the wire

11.  F=B I l  F=Force (N)  B=magnetic field strength (N/am)  I=Current in wire (a)  l=Length of the wire perpendicular to the field (m)

12.  A segment of wire.040 m long is perpendicular to the magnetic field inside a solenoid. When a current of 3.0 amps flows through the wire, it takes a force of 0.020 Newtons to balance the wire. What is the magnetic field inside the solenoid. F= B I L.020N = B (3.0 amps) (.040m) B =.17 N / amp m

13.  The formula can also be written for any single charge moving through a magnetic field  F=Bqv  F=Force (N)  B=Magnetic field strength (N/am)  q=charge (C)  v=velocity of charge (m/s)

14.  When not just any charge, but an electron  F=Bev  e=charge for an electron (1.6 X 10^-19 C)

15.  An electron moves through a magnetic field of.20 N/Am at a speed of 300,000 m/s. What is the force on the charge due to this field? F = B e v F =.20 N/Am (1.6 x 10 -19 C)(300,000 m/s) F = 9.6 x 10 -15 N

16.  Voltage-Energy needed to move a charge

17.  Helps explain how motors work  Motors convert electrical energy into mechanical energy

18.  Faraday’s Law- Voltage (current) induced in a wire is proportional to the rate of magnetic flux cutting across the wire  What happens when you increase voltage?  Increase magnetic flux (the # of magnetic field lines/area)

19.  Relative Motion-The movement of one object with respect to another object

20.  Lenz’s Law-The current induced in a wire is in such a direction that its magnetic field opposes the changing field that induced it

21.  Emf = -B l v  Emf=electromotive force (volts)  B=Magnetic Field Strength  l=length of wire  v=velocity

22.  A wire of length 50 cm is moving at a speed of 2.0 m/s perpendicular to a magnetic field of 0.75 N/A m. What emf is induced in the wire? Emf = -B l v Emf = - (.75 N/A m) (.50 m) (2.0 m/s) Emf = 0.75 Volts

23.  Transformers- A device to increase or decrease voltage  http://phet.colorado.edu/en/simulation/far aday

24.  N1/N2 = V1/V2  N=number of turns of wire  V=voltage (volts)

25.  If the primary coil having 5 turns of wire contains 20 volts, what is the voltage in the secondary coil having 10 turns? N 1 /N 2 = V 1 /V 2 5 / 10 = 20 / V 2 V 2 = 40 Volts

26.  1. A segment of wire 25 cm long is in a magnetic field of.75 N/A m. The force on the wire is 0.30 N. What is the current flowing through the wire?  2. A charge (2.3 X 10^-15 C) moves through a magnetic field at 225,000 m/s. What is the magnetic field strength if the force is 4X10^- 4 N?