1. Magnetism Wednesday, March 28, 2007
The Final Frontier

2. Announcements Tonight’s Homework: Magnetism #1
Electrostatic Exam Corrections: Friday

3. Magnetic Fields
Magnets cause space to be modified in their vicinity, forming a “magnetic field”.
The magnetic field caused by magnetic “poles” is analogous to the electric field caused by electric “poles” or “charges”.
The north pole is where the magnetic field lines leave the magnet, and the south pole is where they reenter.
Magnetic field lines differ from electric field lines in that they are continuous loops with no beginning or end.

4. Magnetic Field, B N S

5. More Magnetic Fields

6. Compasses
If they are allowed to select their own orientation, magnets align so that the north pole points in the direction of the magnetic field.
Compasses are magnets that can easily rotate so that they can align themselves to a magnetic field.
The north pole of the compass points in the direction of the magnetic field.

7. Sample Problem
A compass points to the Earth’s North Magnetic Pole (which is near the North Pole).
Is the North Magnetic Pole the north pole of the Earth’s Magnetic Field?

8. Magnetic “Monopoles” Do not exist!
This is another way that magnetic fields differ from electric fields.
Magnetic poles cannot be separated from each other in the same way that electric poles (charges) can be.

9. Units of Magnetic Field
Tesla (SI)
N/(C m/s)
N/(A m)
Gauss
1 Tesla = 104 gauss

10. Magnetic Force on Particles
Magnetic fields cause the existence of magnetic forces.
A magnetic force is exerted on a particle within a magnetic field only if
the particle has a charge.
the charged particle is moving with at least a portion of its velocity perpendicular to the magnetic field.

11. Magnetic Force on a Charged Particle
magnitude: F = qvBsin
q: charge in Coulombs
v: speed in meters/second
B: magnetic field in Tesla
: angle between v and B
direction: Right Hand Rule
FB = q v x B (This is a “vector cross product”)

12. The Right Hand rule to Determine a Vector Cross Product
Align your hand along the first vector.
Orient your wrist so that you can “cross” your hand into the second vector.
Your thumb gives you the direction of the third vector (which is the result).

13. Sample Problem
Calculate the magnitude force exerted on a 3.0 mC charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed
North.
South.
East.
West.

14. Sample Problem
Calculate the magnitude and direction of the magnetic force.
v = 300,000 m/s
34o
q = 3.0mC
B = 200 mT

15. Motion of Charged Particles in Magnetic Field
Friday, March 30, 2007
Motion of Charged Particles in Magnetic Field

16. Announcements Tonight’s Homework: Magnetism #2
Please pass forward Magnetism #1
Exam Corrections today at Lunch, as well as Monday and Tuesday next week.
AP Reviews start next week 7:15
Monday Kinematics (1 and 2-Dimensional)
Tuesday Energy
Clicker Quiz

17. Magnetic forces…
are always orthogonal (at right angles) to the plane established by the velocity and magnetic field vectors.
can accelerate charged particles by changing their direction.
can cause charged particles to move in circular or helical paths.

18. Magnetic forces cannot...
change the speed or kinetic energy of charged particles.
do work on charged particles.

19. Magnetic Forces… …are centripetal.
Remember that centripetal acceleration is v2/r.
Remember centripetal force is therefore mv2/r.

20. Magnetic Forces are Centripetal
SF = ma
FB = Fc
qvBsin = mv2/r
qB = mv/r
q/m = v/(rB) V F V F V F V F B

21. Sample Problem
What is the orbital radius of a proton moving at 20,000 m/s perpendicular to a 40 T magnetic field?

22. Sample Problem
What must be the speed of an electron if it is to have the same orbital radius as the proton in the magnetic field described in the previous problem?

23. Sample Problem
An electric field of 2000 N/C is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton? Ignore gravitational effects.

24. Sample Problem
A magnetic field of 2000 mT is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton? Ignore gravitational effects.

25. E = 2000 N/C Sample Problem 300,000 m/s e-
Calculate the force and describe the path of this electron.
E = 2000 N/C e-
300,000 m/s

26. B = 2000 mT Sample Problem e- 300,000 m/s
Calculate the force and describe the path of this electron. e-
300,000 m/s
B = 2000 mT

27. Sample problem
How would you arrange a magnetic field and an electric field so that a charged particle of velocity v would pass straight through without deflection?

28. Electric and Magnetic Fields TogetherB E e-
v = E/B

29. Sample Problem
It is found that protons traveling at 20,000 m/s pass undeflected through the velocity filter below. What is the magnitude and direction of the magnetic field between the plates?
20,000 m/s e
0.02 m
400 V

30. Magnetic Force on Current Carrying Wires
Monday, April 1, 2007
Magnetic Force on Current Carrying Wires

31. Announcements Tonight’s Homework: Magnetism #3
(You also will be able to start #4)
Electrostatic Exam Corrections: Monday and Tuesday
Remember review sessions!

32. Magnetic Force on Current-Carrying Wire
F = I L B sin
I: current in Amps
L: length in meters
B: magnetic field in Tesla
: angle between current and field

33. Sample Problem
What is the force on a 100 m long wire bearing a 30 A current flowing north if the wire is in a downward-directed magnetic field of 400 mT?

34. Sample Problem
What is the magnetic field strength if the current in the wire is 15 A and the force is downward and has a magnitude of 40 N/m? What is the direction of the current?

35. Magnetic Fields… Affect moving charge Caused by moving charge!
F = qvBsinq
F = ILBsinq
Hand rule is used to determine direction of this force.
Caused by moving charge!

36. Magnetic Field for Long Straight Wire
B = oI / (2r)
o: 4  10-7 T m / A
magnetic permeability of free space
I: current (A)
r: radial distance from center of wire (m)

37. Right Hand Rule for straight currents• 
Curve your fingers
Place your thumb (which is presumably pretty straight) in direction of current.
Curved fingers represent curve of magnetic field.
Field vector at any point is tangent to field line.

38. I
For straight currents

39. Sample Problem
What is the magnitude and direction of the magnetic field at point P, which is 3.0 m away from a wire bearing a 13.0 Amp current? P
3.0 m
I = 13.0 A

40. Exam (Current) grading
Tuesday, April 3, 2007
Exam (Current) grading

41. Superposition in Magnetic Fields
Wednesday, April 4, 2007
Superposition in Magnetic Fields

42. Announcements Lunch Bunch today – lab HW #3 due today.
HW #4 due tomorrow.
Lunch Bunch #5 due today as well.
Physics Bowl exam tentatively scheduled for April 12
1st period will hear speaker April 12;

43. Principle of Superposition
When there are two or more currents forming a magnetic field, calculate B due to each current separately and then add them together using vector addition.

44. Sample Problem – not in packet
What is the magnitude and direction of the force exerted on a 100 m long wire that passes through point P which bears a current of 50 amps in the same direction?
I2 = 50.0 A P
3.0 m
I1 = 13.0 A

45. Sample Problem 4.0 m 3.0 m I = 10.0 A P I = 13.0 A
What is the magnitude and direction of the electric field at point P if there are two wires producing a magnetic field at this point?
I = 10.0 A
4.0 m P
3.0 m
I = 13.0 A

46. Sample Problem 7.0 m Where is the magnetic field zero? I = 10.0 A

47. Thursday, April 5, 2007
Solenoids

48. Announcements HW #4 due today HW #5 due Monday
Problem 47
HW #5 due Monday
Physics Bowl Thursday, 2nd in A-180
Review sessions (7:15-7:45 AM all next week)

49. In the 4th Grade N S
You learned that coils with current in them make magnetic fields.
The iron nail was not necessary to cause the field; it merely intensified it. B I

50. Solenoid A solenoid is a coil of wire.
When current runs through the wire, it causes the coil to become an “electromagnet”.
Air-core solenoids have nothing inside of them.
Iron-core solenoids are filled with iron to intensify the magnetic field.

51. Magnetic Field Inside a Solenoid
B = on I
o: 4  10-7 T m / A
n: number of coils per unit length
I: current (A)
You are not required to memorize this formula, but only to use it.

52. Magnetic Field around Curved CurrentB

53. Right Hand Rule for magnetic fields around curved wires
Curve your fingers.
Place them along wire loop so that your fingers point in direction of current.
Your thumb gives the direction of the magnetic field in the center of the loop, where it is straight.
Field lines curve around and make complete loops. B I

54. Sample Problem
An air-core 10 cm long is wrapped with copper wire that is 0.1 mm in diameter. What must the current be through the wire if a magnetic field of 20 mT is to be produced inside the solenoid?

55. Sample Problem
What is the direction of the magnetic field produced by the current I at A? At B? I A B

56. Magnetic Field around Curved CurrentB

57. Sample Problem
What is the magnetic field inside the air-core solenoid shown if the resistance of the copper wire is assumed to be negligible? There are 100 windings per cm. Identify the north pole.
120 V I
100-W

58. Monday, April 9, 2007
Work day

59. Tuesday, April 10, 2007
Magnetic Flux

60. Announcements To be exempt from Lunch Bunch this week;
Give me your classwork packet with free response attempt. I will return to you; you’ll correct it based on on-line solutions.
Let me initial your AP review packet.
Review sessions are ongoing. I will be doing a review session tomorrow AM, not lunch bunch.
Tonight: do assignment #6
Pass forward #5
Physics Bowl Thursday, Portable 1B. All other students go to Moreno.

61. Magnetic Flux The product of magnetic field and area.
Can be thought of as a total magnetic “effect” on a coil of wire of a given area. B A

62. Maximum Flux
The area is aligned so that a perpendicular to the area points parallel to the field B A

63. Minimum Flux
The area is aligned so that a perpendicular to the area points perpendicular to the field B A

64. Intermediate Flux
The area is neither perpendicular nor is it parallel B A

65. Magnetic Flux FB = B A cos FB = BA
FB: magnetic flux in Webers (Tesla meters2)
B: magnetic field in Tesla
A: area in meters2.
: the angle between the area and the magnetic field.
FB = BA

66. Sample Problem
Calculate the magnetic flux through a rectangular wire frame 3.0 m long and 2.0 m wide if the magnetic field through the frame is 4.2 mT.
Assume that the magnetic field is perpendicular to the area vector.
Assume that the magnetic field is parallel to the area vector.
Assume that the angle between the magnetic field and the area vector is 30o.

67. Sample Problem
Assume the angle is 40o, the magnetic field is 50 mT, and the flux is 250 mWb. What is the radius of the loop? B A

68. Induced Electric Potential
A system will respond so as to oppose changes in magnetic flux.
A change in magnetic flux will be partially offset by an induced magnetic field whenever possible.
Changing the magnetic flux through a wire loop causes current to flow in the loop.
This is because changing magnetic flux induces an electric potential.

69. Faraday’s Law of Induction
e = -NDFB/Dt
e: induced potential (V)
N: # loops
FB: magnetic flux (Webers, Wb)
t: time (s)

70. Faraday’s Law of Induction
Wednesday, April 11, 2007
Faraday’s Law of Induction

71. Announcements To be exempt from Lunch Bunch this week;
Give me your class work packet with free response attempt. I will return to you; you’ll correct it based on on-line solutions.
Let me initial your AP review packet.
Last mechanics AP review session tomorrow morning. Friday I will host a general session for general questions on mechanics.
Tonight: do assignment #7. Tomorrow I will collect #6.
Physics Bowl Thursday, Portable 1B. All other 2nd period students go to Moreno.
Mock AP exam Monday night, 6:30 PM, in cafeteria. Only excused absences will be allowed a retake, on Thursday morning at 6:15 AM. Free response exams will be given in class, and missing one of those will also require a makeup, to be arranged with me.

72. Faraday’s Law of Induction
e = -NDFB/Dt
e: induced potential (V)
N: # loops
FB: magnetic flux (Webers, Wb)
t: time (s)

73. A closer look … e = -DFB/Dt e = -D(BAcos)/Dt To generate voltage
Change B
Change A
Change 

74. Sample Problem
A coil of radius 0.5 m consisting of 1000 loops is placed in a 500 mT magnetic field such that the flux is maximum. The field then drops to zero in 10 ms. What is the induced potential in the coil?

75. Sample Problem
A single coil of radius 0.25 m is in a 100 mT magnetic field such that the flux is maximum. At time t = 1.0 seconds, field increases at a uniform rate so that at 11 seconds, it has a value of 600 mT. At time t = 11 seconds, the field stops increasing. What is the induced potential
A) at t = 0.5 seconds?
B) at t = 3.0 seconds?
C) at t = 12 seconds?

76. Lenz’s Law
The current will flow in a direction so as to oppose the change in flux.
Use in combination with hand rule to predict current direction.

77. Sample Problem
The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop?

78. Sample Problem
The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop?

79. Sample Problem
The magnetic field is decreasing at a rate of 4.0 mT/s. The radius of the loop is 3.0 m, and the resistance is 4 W. What is the magnitude and direction of the current?

80. Thursday, April 12, 2007
Workday

81. Friday, April 13, 2007
Workday

82. Monday, April 16, 2007
Motional EMF

83. Announcement Tonight’s HW: #8, due Wednesday.
Tonight: Mock AP in cafeteria. Exam starts at 6:30. Bring #2 pencil and eraser. No calculator is necessary.
Tomorrow: More of the Mock AP in class.
Wednesday: More of the Mock AP in class.
Thursday: Mock AP makeup at 6:15 AM. I will be here at 6:00 to set up. Exam starts promptly at 6:15 AM.

84. Work Time Work on Review Packets. Work on Exam Corrections.
Do NOT work on tonight’s homework!

85. Motional emf
e = BLv
B: magnetic field (T)
L: length of bar moving through field
v: speed of bar moving through field.
Bar must be “cutting through” field lines. It cannot be moving parallel to the field.
This formula is easily derivable from Faraday’s Law of Induction

86. Motional emf - derivation
e = DFB/Dt
e = D(BA) /Dt (assume cosq = 1)
e = D(BLx) /Dt
e = BLDx /Dt
e = BLv

87. Sample Problem B = 0.15 T 50 cm 3 W v = 2 m/s
How much current flows through the resistor? How much power is dissipated by the resistor? 
B = 0.15 T
50 cm
3 W
v = 2 m/s

88. Sample Problem B = 0.15 T 50 cm 3 W v = 2 m/s
In which direction is the induced current through the resistor (up or down)? 
B = 0.15 T
50 cm
3 W
v = 2 m/s

89. Sample Problem B = 0.15 T 50 cm 3 W v = 2 m/s
Assume the rod is being pulled so that it is traveling at a constant 2 m/s. How much force must be applied to keep it moving at this constant speed? 
B = 0.15 T
50 cm
3 W
v = 2 m/s

90. Lab: Magnetic Field MapS
Using a compass, map the magnetic field inside and outside your solenoid. Do the following:
Put together 4 sheets of graph paper. Write all group members’ names on paper.
Trace the solenoid (true size)
Draw the Compass Rose
Connect to DC outlet
Map magnetic field lines with compass
Draw North and South Poles of solenoid
Extend field lines through solenoid.