1. Electricity and Magnetism
Chapter 23

2. Hans Christian Oersted
1777 – 1851
Best known for observing that a compass needle deflects when placed near a wire carrying a current
First evidence of a connection between electric and magnetic phenomena

3. Magnetic Fields – Long Straight Wire
A current-carrying wire produces a magnetic field
The compass needle deflects in directions tangent to the circle
The compass needle points in the direction of the magnetic field produced by the current

4. Direction of the Field of a Long Straight Wire
Right Hand Rule #1
Grasp the wire in your right hand
Point your thumb in the direction of the current
Your fingers will curl in the direction of the field

5. Magnitude of the Field of a Long Straight Wire
A straight current-carrying wire generates a cylindrical magnetic field in the space surrounding it.
The magnitude of the field at a distance r from a wire carrying a current of I is
µo = 4  x 10-7 T.m / A
µo is called the permeability of free space
The magnetic field of a long straight wire in a vacuum is

6. Example 1
The overhead power cable for a street trolley is strong horizontally 10 m above the ground. A long straight section of it carries 100 amps dc due west. Describe the magnetic field produced by the current, and determine its value at ground level just under the wire. Compare that to the strength of the Earth's field.
Given: r = 10 m and I = 100 A
Find: B

7. Example 1
Solution: We've got a straight current-carrying wire and that produces a known B-field.
See diagram on last slide with the current assumed heading west. At ground level (at a point beneath the westerly current), the Right-Hand-Current Rule tells us that B points due south. We already have an expression for the field of a long straight current-carrying wire in terms of I and r.
m0 = 4p x 10-7 T-m/A.
Which is only 4% of the Earth's field.

8. Force Between Two Conductors
The magnetic fields created by moving charges can interact and create forces much like the electric forces between those charges.
Parallel conductors carrying currents in the same direction attract each other
Parallel conductors carrying currents in the opposite directions repel each other

9. Magnetic Force Between Two Parallel Conductors
The force on wire 1 is due to the current in wire 1 and the magnetic field produced by wire 2
The force per unit length is:

10. Example 2
What is the force per unit length experienced by each of two extremely long parallel wires carrying equal 1.0-A currents in opposite directions while separated by a distance of 1 m in vacuum?
Given: I1 = I2 = 1.0 A and d = 1 m
Find: FM

11. Example 2 Make a drawing. The defining equation is
FM/l = 2 x 10-7N/m, repulsion

12. André-Marie Ampère
1775 – 1836
Credited with the discovery of electromagnetism
Relationship between electric currents and magnetic fields
Mathematical genius evident by age 12

13. Ampère’s Law ***THE MORE CURRENT….THE STRONGER THE MAGNETIC FIELD***
André-Marie Ampère found a procedure for deriving the relationship between the current in an arbitrarily shaped wire and the magnetic field produced by the wire
Ampère’s Circuital Law
Sum over the closed path
***THE MORE CURRENT….THE STRONGER THE MAGNETIC FIELD***

14. Magnetic Field of a Current Loop
The strength of a magnetic field produced by a wire can be enhanced by forming the wire into a loop
All the segments, ∆x, contribute to the field, increasing its strength

15. Magnetic Field of a Current Loop – Total Field

16. Magnetic Field of a Current Loop – Equation
The magnitude of the magnetic field at the center of a circular loop with a radius R and carrying current I is
With N loops in the coil, this becomes

17. Magnetic Field of a Solenoid
If a long straight wire is bent into a coil of several closely spaced loops, the resulting device is called a solenoid
It is also known as an electromagnet since it acts like a magnet only when it carries a current

18. Magnetic Field of a Solenoid, 2
The field lines inside the solenoid are nearly parallel, uniformly spaced, and close together
This indicates that the field inside the solenoid is nearly uniform and strong
The exterior field is nonuniform, much weaker, and in the opposite direction to the field inside the solenoid

19. Magnetic Field in a Solenoid, 3
The field lines of the solenoid resemble those of a bar magnet

20. Example 3
A 20-cm-long solenoid with a 2.0-cm inside diameter is tightly wound on a hollow quartz cylinder. There are several layers with a total of 20 x 103 turns per meter of a niobium-tin wire. The device is cooled below its critical temperature and becomes superconducting. Since the wire is then without resistance, it can easily carry 30 A and not develop any I2R losses. Compute the approximate field inside the solenoid near the middle. What is its value at either end?
Given: n = 20 x 103 m-1, I = 30 A, and D = 2.0 cm
Find: Bz

21. Bz ≈ m0nI =(1.257X 10-6 T-m/A)(20 x 103 m-1)(30A)
Example 3
Solution: We've got a current-carrying solenoid and that produces a known B-field.
The solenoid is long and narrow and will obey the approximations that led to the last equation
Using m0 = X 10-6 T-m/A,
Bz ≈ m0nI =(1.257X 10-6 T-m/A)(20 x 103 m-1)(30A)
Bz ≈ 0.75 T
Which is a formidable field, over 104 times that of the Earth. The field at either end is about half this, 0.38 T

22. Moving Charges and Magnetism
Moving charge creates B
The orientation (direction) of B depends upon the orientation (direction) of I
WITHIN atoms, electrons move in different (often opposing) directions, thus individual B usually cancel out
BETWEEN atoms that DO have B fields, those FIELDS may oppose each other and cancel out
**Only in materials that have BOTH alignments do we see magnetic properties***
FERROMAGNETIC

23. Magnetic Force

24. Right Hand Rule #2 Place your fingers in the direction of
Curl the fingers in the direction of the magnetic field,
Your thumb points in the direction of the force, , on a positive charge
If the charge is negative, the force is opposite that determined by the right hand rule
Maximum force is

25. Finding the Direction of Magnetic Force
Experiments show that the direction of the magnetic force is always perpendicular to both and which form a plane.
Fmax occurs when is perpendicular to
F = 0 when is parallel to

26. Example 4
A conventional water-cooled electromagnet produces a 3.0-T uniform magnetic field in the 4-in. gap between its flat pole pieces. The field is aligned horizontally pointing due north. A proton is fired into the field region at a speed of 5.0 x 106 m/s. It enters traveling in a vertical north-south plane, heading north and downward at 30° below the horizontal. Compute the force vector acting on the proton at the moment it enters the field.
Given: A proton with v = 5.0 X 106 m/s, at 30° below the horizontal in the northerly direction, and B = 3.0 T, north
Find: FM

27. FM = (+1.6 x 10-19 C)(5.0 x 106 m/s)(3.0 T)(sin 30°)
Example 4
Solution: Here a charged particle is moving in a B-field and that should call to mind v x B.
First, make a drawing
The proton has a positive charge of X C and so v x B is due east, FM is due east
The basic force-on-a-moving-charge relationship is
FM = qvB sin q
The angle between v and B is q = 30° and so with q = qe
FM = qe vB sin q
FM = (+1.6 x C)(5.0 x 106 m/s)(3.0 T)(sin 30°)
Fm= 1.2 x N

28. Force on a Charged Particle in a Magnetic Field
Consider a particle moving in an external magnetic field so that its velocity is perpendicular to the field
The force is always directed toward the center of the circular path
The magnetic force causes a centripetal acceleration, changing the direction of the velocity of the particle

29. Force on a Charged Particle
Charged particle entering perpendicular to uniform magnetic field and experiences centripetal acceleration
Equating the magnetic and centripetal forces:
Solving for R:
R is proportional to the momentum of the particle and inversely proportional to the magnetic field
Sometimes called the cyclotron equation

30. Particle Moving in an External Magnetic Field
If the particle’s velocity is not perpendicular to the field, the path followed by the particle is a spiral
The spiral path is called a helix

31. Electromagnets and the Electric Motor
23.2

32. Electromagnets
Electromagnets are created by an electric current travelling through a solenoid. As such, their strength and direction can be controlled.
Strength can be increased by:
Increasing current (I)
Increasing the number of coils (N) – although this means a longer total wire length and thus more resistance.
Increasing the core material’s permeability.

33. Electromagnets

34. Magnetic Force on Current Carrying Conductor
A force is exerted on a current-carrying wire placed in a magnetic field
The current is a collection of many charged particles in motion
The direction of the force is given by right hand rule #2

35. Force on a Wire
The blue x’s indicate the magnetic field is directed into the page
The x represents the tail of the arrow
Blue dots would be used to represent the field directed out of the page
The • represents the head of the arrow
In this case, there is no current, so there is no force

36. Force on a Wire B is into the page The current is up the page
The force is to the left

37. Force on a Wire B is into the page The current is down the page
The force is to the right

38. Force on a Wire, equation
The magnetic force is exerted on each moving charge in the wire
The total force is the sum of all the magnetic forces on all the individual charges producing the current
FM = B I l sin q
q is the angle between and the direction of I
The direction is found by the right hand rule, placing your fingers in the direction of I instead of

39. Example 5
A flat, horizontal rectangular loop of wire is positioned, as shown in Fig a, in a 0.10-T uniform vertical magnetic field. The sides of the rectangle are F-C equal to 30 cm and C-­ equal to 20 cm. Determine the total force acting on the loop when it carries a current of 1.0 A.
Given: B = 0.10 T, F-C = 30 cm, C-D = 20 cm, and I = 1.0 A
Find: FM.

40. Example 5
Solution: We've got a current-carrying loop in a S-field – FM = IlB sin q
First, make a drawing
Current travels from the positive terminal of the battery clockwise around the circuit. The directions of the forces on each segment are arrived at via v x B and are indicated in the diagram. Because the forces on segments F-C and D-E are equal and opposite, they cancel. The total force acting on the loop is the force acting on segment C-D

41. FM = IlB sin q = (1.0 A) (0.20m) (0.10 T) (sin 90°) FM = 0.020 N
Example 5
FM = IlB sin q = (1.0 A) (0.20m) (0.10 T) (sin 90°) FM = N

42. Torque on a Current Loopf
Applies to any shape loop
N is the number of turns in the coil
Torque has a maximum value of NIAB
When f = 90°
Torque is zero when the field is parallel to the plane of the loop

43. Electric Motor
An electric motor converts electrical energy to mechanical energy
The mechanical energy is in the form of rotational kinetic energy
An electric motor consists of a rigid current-carrying loop that rotates when placed in a magnetic field

44. Electric Motor, 2
The torque acting on the loop will tend to rotate the loop to smaller values of q until the torque becomes 0 at q = 0°
If the loop turns past this point and the current remains in the same direction, the torque reverses and turns the loop in the opposite direction

45. Electric Motor, 3
To provide continuous rotation in one direction, the current in the loop must periodically reverse
In ac motors, this reversal naturally occurs
In dc motors, a split-ring commutator and brushes are used
Actual motors would contain many current loops and commutators

46. Electric Motor

47. Electric Motor, final
Just as the loop becomes perpendicular to the magnetic field and the torque becomes zero, inertia carries the loop forward and the brushes cross the gaps in the ring, causing the current loop to reverse its direction
This provides more torque to continue the rotation
The process repeats itself
***This is obviously much easier if the CURRENT itself periodically reverses – AC electricity instead of DC***

48. Induction and the Electric Generator
23.3

49. Electromagnetic Induction
Moving charge(s) creates (induces) a magnetic field (B)
A moving (changing) magnetic field (B) creates (induces) a current (I)
Whether B is increasing or decreasing near a wire determines the current’s direction.

50. Michael Faraday 1791 – 1867 Great experimental scientist
Invented electric motor, generator and transformers
Discovered electromagnetic induction
Discovered laws of electrolysis

51. Faraday’s Experiment – Set Up
A current can be produced by a changing magnetic field
First shown in an experiment by Michael Faraday
A primary coil is connected to a battery
A secondary coil is connected to an ammeter

52. Faraday’s Experiment
The purpose of the secondary circuit is to detect current that might be produced by the magnetic field
When the switch is closed, the ammeter reads a current and then returns to zero
When the switch is opened, the ammeter reads a current in the opposite direction and then returns to zero
When there is a steady current in the primary circuit, the ammeter reads zero

53. Faraday’s Conclusions
An electrical current is produced by a changing magnetic field
The secondary circuit acts as if a source of emf were connected to it for a short time
It is customary to say that an induced emf is produced in the secondary circuit by the changing magnetic field

54. Magnetic Flux (Ф)
The emf is actually induced by a change in the quantity called the magnetic flux rather than simply by a change in the magnetic field
Magnetic flux (Ф) is defined in a manner similar to that of electrical flux – the density of the magnetic field lines in a given area ofspace
Magnetic flux is proportional to both the strength of the magnetic field passing through the plane of a loop of wire and the area of the loop

55. Magnetic Flux, 2 You are given a loop of wire
The wire is in a uniform magnetic field
The loop has an area A
The flux is defined as
ΦM = BA = B A cos θ
θ is the angle between B and the normal to the plane

56. Magnetic Flux, 3
When the field is perpendicular to the plane of the loop, as in a, θ = 0 and ΦB = ΦB, max = BA
When the field is parallel to the plane of the loop, as in b, θ = 90° and ΦB = 0
The flux can be negative, for example if θ = 180°
SI units of flux are T. m² = Wb (Weber)

57. Magnetic Flux, final
The flux can be visualized with respect to magnetic field lines
The value of the magnetic flux is proportional to the total number of lines passing through the loop
When the area is perpendicular to the lines, the maximum number of lines pass through the area and the flux is a maximum
When the area is parallel to the lines, no lines pass through the area and the flux is 0

58. Electromagnetic Induction – An Experiment
When a magnet moves toward a loop of wire, the ammeter shows the presence of a current (a)
When the magnet is held stationary, there is no current (b)
When the magnet moves away from the loop, the ammeter shows a current in the opposite direction (c)
If the loop is moved instead of the magnet, a current is also detected

59. Electromagnetic Induction – Results of Experiment
A current is set up in the circuit as long as there is relative motion between the magnet and the loop (changing magnetic field)
The same experimental results are found whether the loop moves or the magnet moves
The current is called an induced current because is it produced by an induced emf

60. Faraday’s Law and Electromagnetic Induction
The average induced emf for a one turn loop
A single loop of wire will experience an induced voltage that equal the time rate of change of magnetic flux through it at any given instant.
For N turns of wire, the average induced emf is
Which is known as Faraday’s Induction Law
Eav
Eav

61. Example 6
A circular flat coil of 200 turns of wire encloses an area of 100 cm2. The coil is immersed in a uniform perpendicular magnetic field of 0.50 T that penetrates the entire area. If the field is shut off so that it drops to zero in 200 ms, what is the average induced emf? Given that the coil has a resistance of 25 W, what current will be induced in it?
Given: N = 200, A = 100 cm2, Bi = 0.50 T, Bf = 0, ∆t = 200 ms, and R = 25 W
Find: The induced emf and current I

62. Example 6 Solution: Faraday's Law problem
The time rate-of-change of the flux equals the emf.
The B-field links the entire area perpendicularly, the initial flux is
FM = BA = (0.50T)( m2) = T-m2
and so ∆FM = T- m2 since the final flux is zero

63. Example 6
Thus,
From Ohm's Law: E

64. Applications of Faraday’s Law
Electric Guitar
A vibrating string induces an emf in a coil
A permanent magnet inside the coil magnetizes a portion of the string nearest the coil
As the string vibrates at some frequency, its magnetized segment produces a changing flux through the pickup coil
The changing flux produces an induced emf that is fed to an amplifier

65. Applications of Faraday’s Law
Apnea Monitor
The coil of wire attached to the chest carries an alternating current
An induced emf produced by the varying field passes through a pick up coil
When breathing stops, the pattern of induced voltages stabilizes and external monitors sound an alert

66. Application of Faraday’s Law
Motional emf
A straight conductor of length ℓ moves perpendicularly with constant velocity through a uniform field
The electrons in the conductor experience a magnetic force
FM = qvB sin q = qvB
The electrons tend to move to the lower end of the conductor

67. Motional emf
As the negative charges accumulate at the base, a net positive charge exists at the upper end of the conductor
As a result of this charge separation, an electric field is produced in the conductor
Charges build up at the ends of the conductor until the downward magnetic force is balanced by the upward electric force
There is a potential difference between the upper and lower ends of the conductor

68. Motional emf, cont
The potential difference between the ends of the conductor can be found by
E = v B l
The upper end is at a higher potential than the lower end
A potential difference is maintained across the conductor as long as there is motion through the field
If the motion is reversed, the polarity of the potential difference is also reversed
Since emf = E l, the electric field in the wire, which exactly counters the motional emf is
E = v B

69. Example 7
A 1.0-meter-long wire held in a horizontal east-west orientation is dropped at a place where the Earth's magnetic field is 2.0 x T, due north.
B = 2.0 x 10-5 T
1.0 m N S E W
Determine the induced emf 4.0 s after release.
Given: l = 1.0 m, t = 4.0 s. and B = 2.0 X T
Find: The resulting emf

70. Example 7
Solution: Because a wire is moving through a magnetic field and we want the emf, this problem involves the expression E = vBl. We have B and land need v
The speed at t = 4.0 s
v = v0 + gt = 0 + (9.81 m/s2)(4.0 s) = 39.2m/s
E = vBl = (39.2 m/s)(2.0 x 10 -5T)(l.0m) = mV

71. Motional emf in a Circuit
Assume the moving bar has zero resistance
As the bar is pulled to the right with a given velocity under the influence of an applied force, the free charges experience a magnetic force along the length of the bar
This force sets up an induced current because the charges are free to move in the closed path

72. Motional emf in a Circuit
The changing magnetic flux through the loop and the corresponding induced emf in the bar result from the change in area of the loop
The induced, motional emf, acts like a battery in the circuit
E = v B l
and

73. Generators Alternating Current (AC) generator
Converts mechanical energy to electrical energy
Consists of a wire loop rotated by some external means – a Turbine
There are a variety of sources that can supply the energy to rotate the loop
These may include falling water, heat by burning coal to produce steam

74. AC Generators Basic operation of the generator
As the loop rotates, the magnetic flux through it changes with time
This induces an emf and a current in the external circuit
The ends of the loop are connected to slip rings that rotate with the loop
Connections to the external circuit are made by stationary brushes in contact with the slip rings

75. AC Generators, final E Emax
The emf generated by the rotating loop can be found by
E = 2 N B l v sin θ
If the loop rotates with a constant angular speed, ω, and N turns
E = N A B ω sin ω t
E = E max when loop is parallel to the field
E = 0 when when the loop is perpendicular to the field
Emax

76. AC Generators – Detail of Rotating Loop
The magnetic force on the charges in the wires AB and CD is perpendicular to the length of the wires
An emf is generated in wires BC and AD
The emf produced in each of these wires is
E= B l v= B l sin θ

77. AC Generators
Can also consist of a rotating turbine which passes constantly-reversing (and changing) magnetic fields past an electromagnet.
This changes the electromagnet’s magnetic field and induces AC current in its coil

78. DC Generators
Components are essentially the same as that of an ac generator
The major difference is the contacts to the rotating loop are made by a split ring, or commutator

79. DC Generators E The output voltage always has the same polarity
The current is a pulsing current
To produce a steady current, many loops and commutators around the axis of rotation are used
The multiple outputs are superimposed and the output is almost free of fluctuations E

80. Example 8
A simple single-coil dc generator rotates at a constant frequency of 60 Hz in a 0.40-T magnetic field. Given that the coil has 10 turns and encompasses an area of 1200 cm2, what will be its maximum emf?
Given: f = 60 Hz, A = 1200 cm2, B = 0.40 T, and N = 10
Find: Em.

81. Em = (10)(1200X 10-4m2)(0.40T)(376.99 rad/s) = 0.18 kV
Example 8
A coil is rotating in a B-field with a specified frequency – sinusoidal emf
The basic formula for the emf of an ac generator is Eq. (20.7)
The maximum emf (Em) is the amplitude of the oscillating voltage given by
E = NAB w sinwt
namely E m= NAB w
w = 2pf= 2p(60 Hz) = rad/s
Em = (10)(1200X 10-4m2)(0.40T)( rad/s) = 0.18 kV

82. Motors
Motors are devices that convert electrical energy into mechanical energy
A motor is a generator run in reverse
A motor can perform useful mechanical work when a shaft connected to its rotating coil is attached to some external device

83. Transformers
Use EM Induction to control voltages by “Stepping-Up” or “Stepping-Down” the voltage.
Using different numbers of coils of wire wrapped around a common core, the induced emf in one part of the coil is transferred to the other coil.
Larger-N coil has more voltage than the…
Smaller-N coil (for the same B-Field generated in the core
V2/V1 = N2/N1

84. Transformers