1. Magnetism & Electromagnetism

2.  Magnets form a magnetic field around them, caused by magnetic “poles.” These are similar to electric “poles” or “charge.”  Magnetic field lines leave the magnet from the north pole and reenter into the south pole

3.  Magnetic field lines are continuous field lines in loops with no beginning or end (not like electric field lines  The symbol for a magnetic field is B

4.  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 and align themselves

5. A compass points to the Earth’s North magnetic pole (which is near the north geographic pole) Is the North Magnetic Pole the north pole of the Earth’s Magnetic field?

6.  Magnetic Monopoles DO NOT EXIST!!! Magnetic poles cannot be separated from each other in the same way that electric poles (charges) can be Electric monopoles exist as either a negatively charged object or a positively charged object

7.  Units: Tesla (SI unit)  N/(C m/s)  N/(A m) Gauss  1 Tesla = 10 4 gauss

8.  Magnetic fields cause the existence of magnetic forces like electric fields cause electric 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

9.  Magnitude: F = qvBsinΘ q = charge in Coulombs (C) v = velocity in m/s B = magnetic field in Tesla Θ = angle between v and B  Direction: Right hand rule if q is positive, left hand rule if q is negative  F B = q v x B (This is a “vector cross product” for those of you who know your math)

10. Direction of the magnetic force? Right Hand Rule To determine the DIRECTION of the force on a POSITIVE charge we use a special technique that helps us understand the 3D/perpendicular nature of magnetic fields. Basically you hold your right hand flat with your thumb perpendicular to the rest of your fingers The Fingers = Direction B-Field The Thumb = Direction of velocity The Palm = Direction of the Force For NEGATIVE charges use left hand!

11.  Calculate the magnitude of the force on a 3.0  C charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed North East South West

12. Sample Problem: – Calculate the magnitude of force exerted on a 3.0 μC charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed a)N, b)E, c)S, d)W a) = qvBsinθ F = qvBsinθ F = 3x10 -6 C ·3x10 5 m/s ∙ 0.2T ∙ sin0 o = 0 N b) = qvBsinθ F = qvBsinθ F = 3x10 -6 C ·3x10 5 m/s ∙ 0.2T ∙ sin90 o = 0.18 N c) = qvBsinθ F = qvBsinθ F = 3x10 -6 C ·3x10 5 m/s ∙ 0.2T ∙ sin180 o = 0 N d) = qvBsinθ F = qvBsinθ F = 3x10 -6 C ·3x10 5 m/s ∙ 0.2T ∙ sin270 o = -0.18 N

13.  Calculate the magnitude and direction of the magnetic force

14. Calculate the magnitude and direction of the magnetic force. v = 300,000 m/s B = 200 mT q = 3.0μC 34 o = qvBsinθ F = qvBsinθ = 3x10 -6 C ∙ 300000 m/s ∙ 0.2T ∙ sin(34 o ) F = 3x10 -6 C ∙ 300000 m/s ∙ 0.2T ∙ sin(34 o ) F =0.101N upward

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

16.  Magnetic Forces CANNOT change the speed or KE of charged particles  Magnetic Forces CANNOT do work on charged particles (F is perpendicular)

17.  Magnetic forces ARE centripetal Remember centripetal acceleration is v 2 /r Centripetal force is mv 2 /r

18.  F = ma F B = F c qvBsin  = mv 2 /r qB = mv/r q/m = v/(rB) B F V F V F V F V

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

20. Magnetic Forces on Charged Particles … …are centripetal. Remember centripetal force is mv 2 /r. For a charged particle moving perpendicular to a magnetic field – F = qvB = mv 2 /r Radius of curvature of the particle – r = mv 2 /qvB = mv/qB

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.  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 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?

24.  An electric field of 2,000 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.

25. 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.

26.  A magnetic field of 2,000 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.

27. 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.

28.  Calculate the force and describe the path of this electron if the electric field strength is 2000 N/C

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

30.  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?

32. Electric and Magnetic Fields Together E B v = E/B e-

33.  It is found that protons when traveling at 20,000 m/s pass undeflected through the velocity filter below. What is the magnetic field between the plates? 0.02 m 400 V e+e+

34. 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? 400 V e 20,000 m/s 0.02 m

35.  F = I L B sin Θ I = current in Amps L = length in m B = magnetic field in Tesla Θ = angle between current and B field

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

37. 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?

38.  What is the magnetic field strength if the current in the wire is 15 A and the force is downward with a magnitude of 40 N/m? What is the direction of the current?

39. 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?

40.  Magnetic Fields affect moving charge F = qvBsinΘ F = ILBsin Θ  Magnetic fields are caused by moving charge

41.  B =  0 I/(2πr)  0 : 4 π x 10 -7 T m/A  Magnetic permeability of free space I: current (A) R: radial distance from center of wire (m)

42. 1. Curve your fingers 2. Place your thumb in the direction of the current 3. Curved fingers represent the curve of the magnetic field 4. Field vector at any point is tangent to the field line

43.  What is the magnitude and direction of the magnetic field at P, which is 3.0 m away from a wire bearing a 13.0 A current?

44. 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? I = 13.0 A P 3.0 m

45.  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? P I 1 = 13.0 A I 2 = 50.0 A 3.0 m

46. 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? I 1 = 13.0 A P 3.0 m I 2 = 50.0 A

47.  Remember this from electrostatics?  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.

48. 16. 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? P I 1 = 13.0 A I 2 = 10.0 A 3.0 m 4.0 m

49. Sample Problem 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 = 13.0 A P 3.0 m I = 10.0 A 4.0 m

50.  You learned that coils with current in them make magnetic fields (electromagnets)  The iron nail was not necessary to cause the field, it only intensified it

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

52. B

53. 1. Curve your fingers 2. Place them along the wire loop so that your fingers point in the direction of the current 3. Your thumb gives the direction of the magnetic field in the center of the loop, where it is straight 4. Field lines curve around and make complete loops

54.  What is the direction of the magnetic field produced by the current I at A? At B?

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

56.  The product of magnetic field and area  Can be thought of as total magnetic “effect” on a coil of wire of a given area

57.  The area is aligned so that a perpendicular to the area vector (orthogonal to area) points parallel to the field

58.  The area is aligned so that a perpendicular to the area vector points perpendicular to the field

59.  The area is neither perpendicular nor parallel

60.  Φ B = B A cos Θ Φ B : magnetic flux in Webers (Tesla meters 2 ) B: magnetic field in Tesla A: area in meters 2 Θ : the angle between the area vector and the magnetic field  Φ B = B A

61.  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 30 degrees

62. 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. a) Assume that the magnetic field is perpendicular to the area vector. b) Assume that the magnetic field is parallel to the area vector. c) Assume that the angle between the magnetic field and the area vector is 30 o.

63.  Assume the angle is 40 degrees, the magnetic field is 50 mT, and the flux is 250 mWb. What is the radius of the loop? (hint: A = πr 2 )

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

65.  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

66.  ε = -NΔΦ B /Δt ε : induced potential (V) N : # loops Φ B : magnetic flux in Wb t : time (s)

67.  If there is only ONE LOOP:  ε = -ΔΦ B /Δt  ε = -Δ(B A cos Θ)/Δt To generate voltage:  Change B  Change A  Change Θ

68.  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?

69. 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?

70.  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 s, the field increases at a uniform rate so that at 11 s, it has a value of 600 mT. At time t = 11 s, the field stops increasing. What is the induced potential: a. At 0.5 s? b. At 3.0 s? c. At 12 s?

71. 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?

72.  The current will flow in a direction so as to oppose the change in flux  Use in combination with the hand rule to predict current direction

73.  The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction fo the current in the wire loop?

74. 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?

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

76. 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?

77.  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 ohms. What is the magnitude and direction of the current?

78. 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 . What is the magnitude and direction of the current?

79.  ε = BLv B: magnetic field L: length of bar moving through field v: speed of bar moving through field  Bar must be “cutting through” the field lines. It cannot be moving parallel to the field  This formula is derivable from Faraday’s Law of Induction

80.  How much current flows through the resistor? How much power is dissipated by the resistor?

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

82.  In which direction is the induced current through the resistor?

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

84.  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?

85. Sample Problem 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 v = 2 m/s 50 cm 3 