Chapter 23. Hans Christian Oersted 1777 – 1851 Best known for observing that a compass needle...
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Electricity and Magnetism Chapter 23
Chapter 23. Hans Christian Oersted 1777 – 1851 Best known for observing that a compass needle deflects when placed near a wire carrying a current First
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
Slide 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
Slide 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
Slide 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 o = 4 x 10 -7 T. m / A o is
called the permeability of free space The magnetic field of a long
straight wire in a vacuum is
Slide 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
Slide 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.
0 = 4 x 10 -7 T-m/A. Which is only 4% of the Earth's field.
Slide 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
Slide 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:
Slide 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: I 1 = I 2 = 1.0 A and d = 1 m Find: F M
Slide 11
Example 2 Make a drawing. The defining equation is F M /l = 2 x
10 -7 N/m, repulsion
Slide 12
Andr-Marie Ampre 1775 1836 Credited with the discovery of
electromagnetism Relationship between electric currents and
magnetic fields Mathematical genius evident by age 12
Slide 13
Ampres Law Andr-Marie Ampre found a procedure for deriving the
relationship between the current in an arbitrarily shaped wire and
the magnetic field produced by the wire Ampres Circuital Law Sum
over the closed path ***THE MORE CURRENT.THE STRONGER THE MAGNETIC
FIELD***
Slide 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
Slide 15
Magnetic Field of a Current Loop Total Field
Slide 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
Slide 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
Slide 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
Slide 19
Magnetic Field in a Solenoid, 3 The field lines of the solenoid
resemble those of a bar magnet
Slide 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 10 3 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 I 2 R losses. Compute
the approximate field inside the solenoid near the middle. What is
its value at either end? Given: n = 20 x 10 3 m -1, I = 30 A, and D
= 2.0 cm Find: B z
Slide 21
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 0
= 1-257 X 10 -6 T-m/A, B z 0 nI =(1.257X 10 -6 T-m/A)(20 x 10 3 m
-1 )(30A) B z 0.75 T Which is a formidable field, over 10 4 times
that of the Earth. The field at either end is about half this, 0.38
T
Slide 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
Slide 23
Slide 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
Slide 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. F max occurs when is perpendicular to F = 0
when is parallel to
Slide 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 10 6 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 10 6 m/s, at 30 below the horizontal in the
northerly direction, and B = 3.0 T, north Find: F M
Slide 27
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 +1.60 X 10 -19 C and so v x B
is due east, F M is due east The basic force-on-a-moving-charge
relationship is F M = qvB sin The angle between v and B is = 30 and
so with q = q e F M = q e vB sin F M = (+1.6 x 10 -19 C)(5.0 x 10 6
m/s)(3.0 T)(sin 30) F m = 1.2 x 10 -12 N
Slide 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
Slide 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
Slide 30
Particle Moving in an External Magnetic Field If the particles
velocity is not perpendicular to the field, the path followed by
the particle is a spiral The spiral path is called a helix
Slide 31
23.2
Slide 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 materials permeability.
Slide 33
Electromagnets
Slide 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
Slide 35
Force on a Wire The blue xs 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
Slide 36
Force on a Wire B is into the page The current is up the page
The force is to the left
Slide 37
Force on a Wire B is into the page The current is down the page
The force is to the right
Slide 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
F M = B I l sin 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
Slide 39
Example 5 A flat, horizontal rectangular loop of wire is
positioned, as shown in Fig. 19.32a, 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: F M.
Slide 40
Example 5 Solution: We've got a current- carrying loop in a
S-field F M = IlB sin 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
Slide 41
Example 5 F M = IlB sin = (1.0 A) (0.20m) (0.10 T) (sin 90) F M
= 0.020 N
Slide 42
Torque on a Current Loop Applies to any shape loop N is the
number of turns in the coil Torque has a maximum value of NIAB When
= 90 Torque is zero when the field is parallel to the plane of the
loop
Slide 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
Slide 44
Electric Motor, 2 The torque acting on the loop will tend to
rotate the loop to smaller values of until the torque becomes 0 at
= 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
Slide 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
Slide 46
Electric Motor
Slide 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***
Slide 48
23.3
Slide 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 currents direction.
Slide 50
Michael Faraday 1791 1867 Great experimental scientist Invented
electric motor, generator and transformers Discovered
electromagnetic induction Discovered laws of electrolysis
Slide 51
Faradays 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
Slide 52
Faradays 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
Slide 53
Faradays 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
Slide 54
Magnetic Flux () 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 () 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
Slide 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 M = B A = B A cos is the angle between B and the normal to the
plane
Slide 56
Magnetic Flux, 3 When the field is perpendicular to the plane
of the loop, as in a, = 0 and B = B, max = BA When the field is
parallel to the plane of the loop, as in b, = 90 and B = 0 The flux
can be negative, for example if = 180 SI units of flux are T. m =
Wb (Weber)
Slide 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
Slide 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
Slide 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
Slide 60
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
Faradays Induction Law E av Faradays Law and Electromagnetic
Induction
Slide 61
Example 6 A circular flat coil of 200 turns of wire encloses an
area of 100 cm 2. 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 ,
what current will be induced in it? Given: N = 200, A = 100 cm 2, B
i = 0.50 T, B f = 0, t = 200 ms, and R = 25 Find: The induced emf
and current I
Slide 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 M = BA =
(0.50T)(0.0100 m 2 ) = 0.0050 T-m 2 and so M = - 0.005 0 T- m 2
since the final flux is zero
Slide 63
Example 6 Thus, From Ohm's Law: E
Slide 64
Applications of Faradays Law 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 Electric Guitar
Slide 65
Applications of Faradays Law 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 Apnea Monitor
Slide 66
Application of Faradays Law A straight conductor of length
moves perpendicularly with constant velocity through a uniform
field The electrons in the conductor experience a magnetic force F
M = qvB sin = qvB The electrons tend to move to the lower end of
the conductor Motional emf
Slide 67
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
Slide 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
Slide 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 10 -5 T, due north. N S EW 1.0 m B = 2.0 x 10 -5 T
Determine the induced emf 4.0 s after release. Given: l = 1.0 m, t
= 4.0 s. and B = 2.0 X 10 -5 T Find: The resulting emf
Slide 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 = v 0 + gt
= 0 + (9.81 m/s 2 )(4.0 s) = 39.2m/s E = vBl = (39.2 m/s)(2.0 x 10
-5 T)(l.0m) = 0.78 mV
Slide 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
Slide 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
Slide 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
Slide 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
Slide 75
AC Generators, final The emf generated by the rotating loop can
be found by E = 2 N B l v sin If the loop rotates with a constant
angular speed, , and N turns E = N A B sin t E = E max when loop is
parallel to the field E = 0 when when the loop is perpendicular to
the field E E max
Slide 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
Slide 77
AC Generators Can also consist of a rotating turbine which
passes constantly-reversing (and changing) magnetic fields past an
electromagnet. This changes the electromagnets magnetic field and
induces AC current in its coil
Slide 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
Slide 79
DC Generators 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
Slide 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 cm 2, what
will be its maximum emf? Given: f = 60 Hz, A = 1200 cm 2, B = 0.40
T, and N = 10 Find: E m.
Slide 81
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 ( E m ) is the amplitude of
the oscillating voltage given by E = NAB sin t namely E m = NAB = 2
f= 2 (60 Hz) = 376.99 rad/s E m = (10)(1200X 10 -4 m 2
)(0.40T)(376.99 rad/s) = 0.18 kV
Slide 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
Slide 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 V 2 /V 1 = N 2 /N 1