Download Electric Fields - AP Physics 2 Homework Page

Electricity and Magnetism
AP Physics 2012-13
San Dieguito Academy
Equations we will be using for the following sections
2
Electric Forces
Charles Coulomb suspected that electric forces were
similar to gravitational forces. He set about to find out if
this was so with the use of a torsional balance. He placed
a small charge on a pith ball. This pith ball was placed on
a horizontal bar. The bar was suspended by a thread.
Next, he inserted another slightly charged pith ball near
by. Measuring the deflection of the horizontal bar he was
able to determine that indeed.
where k= 9 x 109 N m2 /C2
The unit of charge “q” is measured in Coulombs. On an elementary level, this is a
large number of protons or electrons.
One Coulomb = 6.24 x 1018 protons or the same number of electrons. Electric
forces compared to gravitational forces are very strong. Let us take for example one
Mole of Hydrogen atoms and individually separate all of the electrons from the
protons. (Remember that a hydrogen atom consists of only one proton and one
electron. We will place the protons in one bucket and the electrons in another.
Sample Problem: These buckets will be placed one meter apart. Lets find the
electrical force acting between the two buckets.
3
Lets do a few more example force problems. Imagine you have two coulombs of
positive charge separated by 300 meters from another 2 coulombs of positive charge.
What is the force on each group of charge?
Electric forces are summative. Find the net force on the 1 Coulomb charge in each
of the examples below.
4
Gravity and Potential Energy
In the past when we have dealt with gravitational
potential energy we looked at small objects being lifted up
a small distance from the surface of the Earth. The PE
gained was just equal to the work done to lift the object.
The Work done was positive because the force applied
and the change in displacement were in the same
direction. We could use the equation for work
W = F d
What happens when the distances are much greater? The
force of gravity is no longer constant but is governed by
Newtons Universal law of gravitation Fg = G m1 m2 / r2. We now must integrate
to find the total work done using W = ∫ Fapplied dx. Notice that the Force we
must apply is a restraining force against gravity. This force is in the opposite
direction to the change in displacement (dx). The angle between the two is 180
degrees .
Before
= After
Definition of work
Fg and Dot product
Cos 180 = -1
Constants move out
5
r with exponents
Integrating
Evaluating
6
A
Suppose a 10 kg ball of krypton falls to Earth from infinitely far away.
How fast is it going upon impact? The Earth has a mass of
6.0 x 1024 kg and a mean radius of 6.38 x 106 meters. Let us use
Conservation of Energy.
Before
=
After
0
=
( KE ) + (
0
=
(
)
) + (
)
vimpact = ____________
It makes sense that PEg must go negative if the KE of the krypton
is to go positive.
B.
Now let us imagine placing this krypton in a very low (read at the surface) circular
orbit. Find the speed of the ball using Fnet and centripetal acceleration
Fnet
Raw Eq
(
=
m
) = (
Substitutions (
) = (
Answer
ac
) (
) (
)
)
v= (
)
What is the net energy for this low flying satellite?
Total Energy = ( PEg ) + ( K E )
Raw equation
Substitutions
Answer
= (
) + (
= (
)
) (
Energy net = (
)
)
7
Electrical Potential Energy
We can also do them using energy. However we have to look potential energy in a
slightly different manner. In the past it was just equal to the work put in to move
an object. This was Fg Let us look at charges first. Imagine that you have two
protons that are infinitely far apart. They have no potential energy. The charge on
the right is pushed closer and closer to the one on the left. The force that is applied
must be equal but opposite that due to the electrical repulsion Fapplied = - Felect
Notice that this charge moves against the repulsive force due to the other. As we do
this we are doing work (W = force x distance) The work we do goes into potential
energy as is shown below.
Important: the force is working with the change in displacement. The Angle
between the two is zero degrees..
8
Here is the math.
Notice: Unlike gravity, if both charges have the same sign, the PE will be
positive.
9
Lets ask some questions.
1. What does it mean when you have a negative answer for PE?
It just means that you have less energy in the system than in your original
state. This energy was transferred into some other form, for example KE
or heat.
2. When we bring two positive charges together they gain potential
energy. Suppose one charge is much greater than the other one.
Does it make any difference which charge is moved?
No! All that counts is how far apart they were to begin with
(PE original) and how far apart they are in the end.
3. What if we bring four charges A, B, C and D together, how do we
calculate the net PE?
You must find all the possible charge combinations. In this case that
includes
PEab
PEac
PEad
PEbc
PEbd
PEcd
10
Lets do a few sample problems. Find the Potential energy in each case below.
11
4. When we bring charges together, where is the energy stored?
This is a tough one. The easiest answer is that it is stored in the
combined electric field that they create.
So, if energy can be stored in fields, we should probably study what a field
looks like. This is very hard to do, as fields such as gravitational, electrical
and magnetic ones are invisible. Almost two hundred years ago, one man by
the name of Michael Faraday tried to this when he came up with what he
called lines of force. His assumption was that all positive charges have
invisible streamers emerging from the positive charge. They went straight
out in all directions. Where they were close together, the field was stronger.
And, the field had a direction that matched these lines of force. He then went
on the state that negative charges were the same, except that the lines of force
would aim toward the negative charges.
From this he formulated the following rules:
1. Lines of electrical force have a beginning (positive charge) and an end (negative
charge)
2. Lines of force can bend but cannot cross each other.
3. Lines of force have a direction at any location that reflects the sum forces acting
on a test charge it that point by all other charges.
12
Let us see what results if two charges are placed near each other.
A positive and a negative charge
Two positive charges
What Faraday showed was that while electric fields are very common, it is not
common that one encounters an electric field that is constant in direction and
intensity. His theory did argue for a way to produce areas that have absolutely no
electric field. This is in what is called a “Faraday cage.” Though he could not prove
it mathematically (Faraday had only three years of formal education) he argued that
if one had a conducting sphere, charges would spread out in such a manner as to be
as far apart as possible. That meant the charges would be evenly distributed on the
outside surface of the sphere.
13
He then argued that there should be no lines of force inside the sphere as they would
be repelled by the charges on the opposite side of the sphere.
He was correct but it takes more than just intuition to demonstrate that something
must be true. As time went by, scientists came to what Faraday was describing
using lines of force as what we call a “field”. In the case of charges, the electric field
and in the case of masses the gravitational field.
Mathematical equations for gravitational and electric fields
Imagine that you, the victim, are standing on the surface of the Earth, the acting
mass. There is a force between the two of you that is given by Newton’s Universal
Law of Gravitation.
Suddenly you disappear leaving the earth all by itself, the Earth would
still have gravity. There just would not be any force. We call this the
14
Earth’s gravitational field. Its strength is measured by the following
equation:
Lets look at the gravitational field strength on the surface of the Earth.
Remember that m = 6 x 1024 kg, r = 6.38 x 106 meters. When we plug into
the field equation we get
(a familiar number)
It is not an accident that “g” the gravitational field strength has the
same units and magnitude as “g” the acceleration due to gravity. Field
can be viewed as the potential to exert a force. The strength of the
force will be the field strength multiplied by your mass.
If
Fgravitational = “g” m
and
Then
“g” m
=
Reducing to
“g” =
a.
15
Fgravitational = m a
ma
Electric Fields
As with mass and gravitational fields, the electric charge has an
associated field. We seldom us the electrical force equation with the
constant “k” rather we use the form below.
Below are two charges. One is +2 Coul and the other is -2 Coul. Find the
field strength and direction at each location.
A ___________ at _____
B ___________ at _____
C ___________ at _____
D ___________ at _____
E ___________ at _____
Most fields are too complex to easily calculate. This is especially true of
electric fields. There are several reasons why this is so.
1. You are often dealing with many, many individual charges. Just one coulomb of
electrons is 6,000,000,000,000,000,000 electrons for which one has to
account.
2. Electrons are in constant motion. It is like herding cats to keep them stationary.
3. Electrons flow over conductors so that they have the least possible potential
energy in any given situation. This means that they are seldom evenly
distributed except in the most simple situations (spheres, sheets, and very long
straight wires.)
16
As a result, we will restrict ourselves to studying cases where
distributions are simple to work with. In spite of this you will find it
impressive how much can be done.
Doing a simple situation the hard way using calculus (Before Gauss)
Lets look a situation where there is a very, very long wire that has a
positive charge distributed along it. This charge is distributed uniformly
with an average charge per unit length of . We want to find the net
electric field at some point P that is a distance D away from the wire. To
approach this by using calculus we will break the wire into really, really
small pieces of length dx. The contribution of just one of these is shown
below.
where dq =  dx. Notice that dE can be broken up into x and y
components.
17
When you add all of the contributions due to each segment dx you get
A very interesting development occurs. The first integral will reduce to
zero since each dx to the right of point P has a corresponding dx to the
left side of P that cancels it
out. The result is the following:
The integral is still pretty messy because we have variables x, R, and .
Cos  = D / R
So R = D / Cos  therefore
R = D sec 
Tan = X / D
X = D Tan 
Differentiate X and d x = D d (tan  )
d x = D sec2  d 
We can now substitute into the equation and we get
18
Giving
Most physics books at this level do not use K = 9 x 109
Rather they use k = 1/(4 eo) where eo is the permittivity of free space
and is equal to 8.85 x 10-12 they do this for reasons that will soon
become apparent. This therefore gives us E at any distance D away from
a long line of charge as
Now, you should get two things from this exercise. The first is that it
appears that the electric field goes down as 1 over the radius as you
move away from a line of charge. Secondly, if this is one of the easiest
situations for which calculus can be used to find an electric field. We
are going to be in for trouble.
There must be a better way!
And there is
It is called Gauss’s Law.
Fredrick Gauss was one of the most gifted mathematicians of all time.
And, although his mathematical prowess was phenomenal his law is
extremely simple.
19
What Gauss did was to imagine that there was something called Flux.
Flux was the amount of field that is intercepted by a surface. Looking at
fields from the point of view of lines of force, the field intensity would
be represented by the density of the lines. The first thing he realized
was that if you had a single charge in space and were to enclose it by a
spherical surface all the lines that emanated from the charge would be
intercepted by the sphere’s surface. Place a bigger sphere and even
though the intensity of the field would be less at its surface, the
increase in intercepting surface would compensate so that the net
recorded flux would be the same as seen by the smaller sphere.
20
Fredrick went one step further. He imagined that a drunk person was
trying to enclose the charge within a sphere, but this person’s sphere was
squished into a peanut shape as shown below.
What he observed was that even where a line emerged and then was later
re-intercepted by the peanut shape, it would eventually have to reemerge once again. Subtracting ingoing flux from outgoing flux and you
are left with the same net flux that a spherical shape saw. The only
factor that actually changed the net flux through a closed surface was
the total net charge inside it. More net charge inside, more net flux .
This is very powerful relationship.
Lets look back for a second at the nature of Flux. The net flux through
any surface is determined by three things.
A.
B.
C.
the size of the surface that is intercepting the field.
the intensity of the field being intercepted.
The angle between the field and the intercepting surface.
21
This third characteristic is very important. Look at the three surfaces
below.
The surface on the left is parallel to the field and intercepts none of it.
The middle one is at an angle to the field and intercepts some of the
field lines. The last one cuts straight across the field. It gets a
maximum amount of the field. For ease of measurement, a surface’s
orientation is given by drawing a normal to the surface. This means that
for maximum flux we want an angle of zero degrees between surface and
field. Gauss therefore defined electric flux by the following equation
 = E dot A
 = E A cos ()
Where
 is the flux
E the electric field intensity
A the surface area
 is the angle between the two
Armed with this, Gauss postulated that for a closed surface
and flux is equal to the net charge inside divided by a special
constant eo which we have seen before.
22
Gauss’s law is extremely easy to use if you recognize one fact, and that is
that shape matters.
1.
pick a shape where E is constant at the surface.
2.
pick a shape maximizes field interception (cos  = 0) or
3.
pick a shape that minimizes field interception (cos  = 90)
Lets look at four simple situations.
1. Field at a location r near a point charge (show that the integral E dA
really gives Q/eo)
We will pick a spherical surface centered over the charge with radius R.
Why? Because E is always coming out of the surface at zero degrees
(cos  = 1) and because the field intensity will be the same at all points
(E is a constant)
23
2. Find the Field at point r near a charged conducting sphere of
radius R Charge density  in coul/m2
To show that outside of it the field is the same as if all the charge were
at a point in the center. We will pick a spherical surface with radius r>R
that is concentric with the conducting sphere.
Why? Because of symmetry, E is always coming out of the surface at
zero degrees (cos  = 1) and because the field intensity will be the same
at all points (E is a constant)
24
3. Use Gauss’ law to show that there is no electric field inside of a
conducting sphere (Faraday Cage)
We will pick a spherical surface with radius r < R and centered over the
charged sphere.
Why? Because of symmetry, E is always coming out of the surface at
zero degrees (cos  = 1) and because the field intensity will be the same
at all points (E is a constant)
25
Find the field at point P above a line of charge
We will pick a Cylindrical surface parallel to the line of charge with flat
circular ends
Why? There are three distinct surfaces. The ends have no net field
going through them ( = 90 and Cos  = 0) and can be cast aside. That
leaves the cylinder. Because of symmetry E is always coming out of the
cylinder at zero degrees (cos  = 1) and the field intensity will be the
same at all points (E is a constant)
26
Field at a point P above a sheet of charge (density  in Coul/m2)
We will pick a Cylindrical surface cutting directly across the sheet of charge with ends
P above and below the surface.
Why? There are three distinct surfaces. The cylinder has no net field going through
it (  = 90 and Cos  = 0) and can be cast aside. That leaves the circular ends.
Because of symmetry E is always coming out of the ends at zero
degrees (cos  = 1) and the field intensity will be the same at all points on
each end (E is a constant)
Notice that the Electric field is constant above and below a large sheet of charge. This will be very
important soon
27
5. Imagine two enormous sheets of charge. One is positively charged, the other negatively charged.
Using Gauss’s Law show that the field between the plates is /e
o
Show that there is not field either above or below the two plates.
28
5. One last situation which we shall encounter often is the insulating sphere (radius
r) of uniform charge density throughout. The sphere has a net charge Q on it.
What is the electric field strength at some point at a radius R inside the sphere?
We will draw a sphere of radius R inside the larger sphere
Why? Because all charge outside this sphere we can neglect. And, because the field due to
the charge inside the sphere will cause a field that is outward directed ( = 0 and E
constant)
We must determine the fraction of the total charge that is inside the Gaussian surface.
Qtotal x volume of small sphere divided by the volume of the large sphere
29
Gravity and Gauss’s Law
Several students have asked if there is a way to determine the gravitational field
strength “g” inside the Earth. We can, but we must modify Gauss’s law for
gravity. First we must find out what the gravitational equivalent for eo is for mass.
Remember that for electricity, we changed from k = 9E9 into 1/(4  eo )
This gave us an eo which was equal to _____________________
We will do the same thing for G =6.67E-11 by converting it to 1/(4  xo )
This gives us xo which is equal to ______________________
Now it is just a matter of substituting into the general equation.
30
Force – Field – Potential – Potential Energy
Definite
Indefinite
Up to this point we have dealt fairly
extensively with forces, fields, and Potential
Energy. The way they are related to each
other is shown below.
Definite
Indefinite
In cases where the masses are point masses
these relationships become:
Definite
Indefinite
In cases where we are dealing with point
charges these become
31
In cases where we are dealing with line charges we use Gauss’ Law to determine the
field strength “E” and then move back to force and PE.
Definite
Indefinite
In cases where we are dealing with sheets of charge we use Gauss’ Law to determine
the field strength and then move back to force and then to PE
Definite
Indefinite
Notice that there is a fourth unassigned ellipse in the diagram. It is about to become very important as we move into
circuits and then into magnetism. Up until now we have dealt with charges at rest. It has been easy to determine
their total charge or their relative charge density. When charge starts to flow, keeping track of every electron is very
hard to do. It becomes well nigh impossible to determine exactly how much Potential Energy is in a system of
moving particles.
32
Fortunately, for most things we do it is not really necessary to know the system’s
total energy. It will suffice just to know relative energy densities at various
locations. We call this electric potential or Voltage. One Volt is what you have
when charges are placed close enough together to have an average energy density of
1 joule per coulomb of charge.
The Relationships between
Potential – Potential Energy – Field – and Force
are given below. We will be using it on the following pages.
F
F = q E
E
Field
Force
V = - E dr
U = F dr
U
U = q V
V
Potential
P.E.
and
F
F = m g
g
Field
Force
U = F dr
Pot =
U
U = m Pot
Pot
Potential
P.E.
33
g dr
Now lets look at some specific examples where we apply what we have learned
F
F = m g
g
Field
Force
U = F dr
Pot =
U
.
U = m Pot
g dr
Pot
Potential
P.E.
Imagine that you have a 10 kg mass that is held stationary in a constant
gravitational field of strength g = 9.8 N/meter which is directed downward. What
is the force that is needed to hold the mass there? Remember that the relationship
between force and field is
F = _____ ______
Substituting we get
F = _____ ______
Or F = ________ directed _____
Now we decide to move the mass upwards a distance of 5 meters. How much
Potential Energy is invested in this move? The relationship between Force and PE
(Ug) is
Ug = ∫ _____ _____
Substituting we get
Ug = ____ ____ ∫ ______
Or Ug = ________
34
Gravitational Potential is by definition the amount of Work (Joules) per unit of mass
(1 kg) that is required to move an object from point A to point B. If from the
preceding information, we want to find the potential needed to move a 1 kg mass
upwards in the gravitational field given we use one the following general equations:
F
F = m g
g
Field
Force
U = F dr
Pot =
U
U = m Pot
Pot
Potential
P.E.
PE to
Potential
g dr
Or
Field to Potential
Ug = ________ x Potential
Potential = ∫______ _______
______ = ________ x Potential
Potential = _______ ∫ _______
Potential = _____________
Potential = ____________
In most cases we will find that it is easier to find potential by using field strength
Potential = ∫______ _______
Where the field is constant we can reduce the above equation to just
Potential = ______ _______
35
The Case of the Constant Electric Field
(what we find in a standard Capacitor).
Imagine that you have a +5 Coulomb charge that is
held stationary in a constant Electrical field of
strength E = 10 N/Coulomb which is directed
downward.
Question one: If this is the field created by two
parallel plates. One above the picture and one
below, what would be the sign of the charge will we
find on the top plate?
_______________ (draw it in the diagram)
What sign of charge will we find on the bottom
plate?
_______________ (draw it in the diagram)
Question two: What is the force that is
needed to hold the charge stationary?
The general relationship between force
and field is:
Question three: We move the charge
upwards a distance of 5 meters, How much
PE (Ue) does this require? The relationship
between force and Ue is:
F = ______ ______
Ue = ∫ ______ ______
Susbstituting we get
Substituting we get
F = ______ _______
Ue = _____ ______ ∫ ______
F = __________ directed _____
Ue = _____________
36
Question four: Using the relationship between field (E) and potential (V) we can
find the potential (Voltage) difference between the charge’s original location and its
new location 5 meters higher. The general equation is
_______ = - ∫ _______ _______
Because E is constant this reduces to a general equation of
_______ = - _______ _______
Substituting values gives
Or
_______ = - _______ _______
Potential Difference = _______________ Volts.
The Electron Volt
Imagine an electron (mass = _______________ and charge of _______________) is
held stationary in a uniform electric field of 0.5 Newtons / Coulomb.
Question one: How far and in what direction would we
have to move the electron until it had gained a potential
of one volt?
The relationship between Field and Potential is
________ =
_________ _________
Because E is constant, this integral reduces to
________ =
_________ _________
which gives a required displacement of
d = __________ meters _______ (direction)
37
After moving this distance the electron has gained PE (Ue) which is given by the
relationship
________ = _______ ________
Substituting
The Energy gained is
________ = _______ ________
______________ Joules is called an electron Volt (eV)
This doesn’t seem to be a lot of energy. However, after moving the electron this
distance we let the electron go. It gains kinetic energy as it looses Potential
Energy. Using conservation of energy we can find the speed of the electron after it
has lost one electron volt.
Before
=
After
_________ = _________
______ Joules = ___ _____ _____
Substituting values
______ Joules = ___ _____ _____
This gives a final velocity of ____________ directed _________
38
Television and Electron Tubes
The old fashioned television set works by firing electrons from the back end of a
vacuum tube to the screen at the front. The screen at the front end is coated with a
phosphorous that glows when it is hit hard enough by these electrons. It takes a lot
of energy for electrons to do this. Most televisions accelerate electrons across
voltages of about 15,000 Volts. Lets look at what is happening.
Screen
Electron gun
15,000 Volts
1. The gun consists of two sheets. One with a hole in the center of it. Draw the
sign of the charges that collect on each sheet.
2. Draw the field that is created by these sheets of charge.
3. The center of the back sheet is heated to a temperature where electrons begin to
boil off. They are attracted by the other sheet How many electron volts do
they have when they leave the back sheet? (trick question)
39
4. How fast are they moving as they impact on the front plate?
a. General equation
b. Substitutions
c. Answer.
5. The electric field (E) created by the two sheets is uniform (constant) if the
distance between the two sheets is two centimeters, what is the strength of
the field that is created?
a. General equation
b. Substitutions
c. Answer.
40
Aiming Electrons in a TV Tube
The way a television paints a picture
is by Accurately aiming the electron
stream that is emerging from the
electron gun at the back end of the
TV tube. Early on it was thought
that the best way was to use the
same type of mechanism as is found
in the electron gun itself. If one
wanted to move the beam upwards it
would be passed between horizontal
plates as shown.
6. What sign charge should be placed on the top plate to do this? ________
7. The electron is in this vertical electric field for a horizontal distance of 4 cm.
By the time it has traveled this horizontal distance we want the electron to be
deflected vertically by a distance of 1 cm. Let us assume that the electrons
have a speed of ______________ m/sec as found in problem 4 and a mass of
9.11E-31 kg.
41
8. How much force must the electrons experience during the 2 cm under which
it is being pulled upwards?
9. How strong of an Electric field do we need to make this deflection?
10. Assuming the plates are 4 cm apart, what voltage is needed to create this
electric field
42
43
Magnetism
AP Physics 2013
San Dieguito Academy
44
Magnetic phenomena have been known in the west for at least two-thousand
years. In ancient Greece naturally occurring magnets were found to attract small
amounts of iron from out of sand. The area from which many of these stones came
was called Magnesia. The stones eventually were given the name magnetite.
The first recorded magnetic compass comes from China in 271 CE The use of
compasses was one of the great reasons for the age of exploration 1300 – 1600 in
Europe. This is especially true as regards sea travel where the compass in
combination with sextants and the first accurate watches allowed sailors to
determine their locations accurately.
Here is what they knew about magnets.
1. Magnets have two poles north and south.
2. A magnet’s north pole attracts another magnet’s south pole (opposites attract).
3. A magnets north pole repels another magnets north pole (likes repel).
4. The Earth itself is a magnet.
5. Iron rods if heated to red hot temperatures and then allowed to cool become
magnetized.
It was also known that iron filings arranged themselves forming curved patterns
at the poles of magnets and curving outward and backward connecting these poles.
During the 1800’s Michael Faraday developed the concept of lines of force based on
these patterns.. These are the rules Faraday used for magnets.
1.
2.
3.
4.
5.
Magnetic field lines emerge from the North Pole of a magnet.
They circle around and enter the South Pole of a magnet.
Where the lines are closer the field is stronger
The direction of the lines represents the direction of the field
Magnetic field lines continue even inside the magnet
Below is a diagram of the Earth’s magnetic field. Along the field lines on the right
side of the diagram are rectangles representing compass magnets.
45
Faraday developed his lines of force for charges as well as for magnets there were
several major differences between the two.
1.
2.
Lines of force start on positive charges and end on negative charges.
With magnets, lines of force have no beginning and no ending.
On this and the next page are several charge and magnetic situations, Draw in the lines of
force.
46
47
In 1800 C.E. Oersted discovered, accidentally, that a current flowing in a wire
created its own magnetic field. This field went did not go in the direction of the
wire. Rather it seemed to be at all points perpendicular to the wire
Faraday developed his lines of force to try to explain this.
48
Below is the rule for the creation of magnetic fields by flowing currents.
1. Thumb
__________________________________________________________
2. Fingers
__________________________________________________________
I call it the hitch-hiker’s rule for creating a field. Pretend you are hitching a ride in
the direction of the current (use your right hand) or in the direction of the moving
electrons (left hand). The direction of the field is given by the curl of your fingers.
.
Interesting Properties of Magnetism
Force a charge to move and the charge creates a magnetic field around itself.
Using what we have seen on the previous page, we can produce a field that looks
very much like that of a bar magnet. To do this we coil wire into what is called a
“solenoid” In the picture below draw the field lines.
49
Move a charge through an existing field and it experiences a force on itself
(and an equal but opposite force on the field is moving through) The strange thing
about this force is that the force is perpendicular to the direction of the charge’s
motion and perpendicular to the magnetic field. To find the direction of the force
we use the following rule:
1.
Arm __________________________________________________________
1.
Fingers ________________________________________________________
1.
Thumb _________________________________________________________
We call this the young fascist rule for charges moving in a B field. Remember use
the right arm for positive charges and the left arm for negative charges. Describe
the path of the moving charge in each of the following.
50
Given a positive charge moving through a magnetic field in each of the cases
below. Find the direction of the missing component
Remember that there is only a force on a charge when it has a component of its
velocity that is perpendicular to the B field.
In the picture to the right electrons are shot through a B field from the left and are
bent downward. What direction is the magnetic field?
______________________________
Hand rules will give you directions but they do not give you magnitudes. The
strength of the force is given below.
A moving charge
a moving current
Fmag = q v
Fmag = I l x B
x
B
q is charge measured in Coulombs
B is magnetic field measured in Tesla
field measured in meters.
v is velocity measured in m/sec
L is the length of wire that is in the B
51
Motors
In 1820 the Danish physics professor Hans Christian Oersted made the
remarkable discovery that an electric current in a wire will deflect a magnetic
compass. Soon afterwards it was demonstrated that the converse was also true: a
strong magnetic field would also exert a force on a wire if the current flowing
through the wire was moving perpendicular to the magnetic field. If a current were
to flow through a wire placed between the two poles of a powerful magnet as shown
below, the wire would be pushed upward. If either the current or the magnetic poles
were to be reversed the force on the wire would then be downward.
North
South
Coil
current in wire moving toward the viewer
Oersted
This relationship between electric currents and magnetic fields has been used
to great advantage in what we call electric motors. One simple motor consists of
just an electromagnet, a disk, and an iron rod. The electromagnet attracts the iron
rod, pulling the disk in a counter clockwise direction until the iron is at its closest
position to the electromagnet. At this point the current in the electromagnet is
turned off and the disk continues turning, pulling the rod upwards, until a complete
rotation has been made. Once again the switch is closed the electromagnet turns on
and the disk is pulled around by the magnetic force.
There are two basic problems with this motor. The first is that you must
manually switch the current on and off once every revolution and the second is that
52
the motor only pulls during half of a turn and then coasts the second half. A remedy
to the switching problem came in the form of a device called a commutator. A
commutator is a strip of conducting metal which is placed around portions of the
motor axle. A copper brush rubs over the axle and alternately passes current to the
axle when touching copper and stops passing current when not touching the copper.
An improvement on this motor is the one shown below. It consists of a
rotating armature with an electromagnet wound on it. This armature is placed
between the poles of a magnet (called the field magnet) and current is passed
through it so that the arm nearest the north pole of the field magnet is also north
and the arm nearest the south pole of the field magnet is also south. Repulsion
pushes the armature around this time in a clockwise motion.
Notice that in figure B not only is this motor working on repulsion between
like poles but that it is also acting on attraction between unlike poles. This motor
needs two brushes and two metal contacts on its commutator. This allows current
to flow into and out of the motor at all times. When the armature makes a 180
degree turn the brushes will change from one metal contact to another effectively
reversing the flow of current and reversing the poles on the armature. This motor
has the added advantage that it is under power during the full cycle of each
revolution.
53
Another similar motor is shown below. Notice that this motor has no bulky
arms to move. Rather its armature windings are around the shaft of the armature.
It actually is an improved version of the wire launcher shown in the very first
picture of this handout. The starter motor of your car is wound this way. Without
the added inertia of the arms of the previous motor, this version can accelerate to
full speed very rapidly.
In the picture above, which direction is the force on the wire that is closest to the
south pole of the magnet?
_______________________
Which direction is the force on the wire closest to the North pole of the magnet?
_______________________
Which direction is the force on the wire on the back side of the motor (that goes
between the previous two wires?
_______________________
54
Lets Do Some Calculations
The rail gun. It has been the ambition of many to accelerate objects to high speeds
using magnetism. One method to do this is the rail gun. Basically it consists of two
tracks that are electrified to a large potential difference. The tracks rest upon very
large magnets as shown below.
If a conducting bar is placed upon the tracks a current will flow. Because the
current is flowing at an angle perpendicular to the B field it experiences a
force. For simplicity sake let us assume that B = 0.5 Tesla, resistance of the
system = 1, the distance between the rails is 1.5 meters, and the mass of the
bar is 5 kg.
a. In what direction will the bar want to travel?
b. What is the net force on the bar?
__________________
Equation:
Substitutions
Answer
__________________
c. If the track is 200 meters long, how fast will the bar be moving when it
reaches the end?
55
2. It has been considered using a rail gun to launch the space shuttle. If the
shuttle has a mass of 6000 kg, How long would the tracks of this gun have to
be? Assume the required launch velocity is 8000 m/sec.
3. During a solar storm electrons stream from the sun covering the distance
between the Sun and the Earth in about 24 hours. They hit the outer layers
of the Earth’s magnetic field where the magnetic field is about 1 x 10-6 Tesla.
At this point answer the following:
In which direction would they turn (right, left, up, down)
___________________
a. What would be the radius of curvature of the electrons?
Equation:
Substitution
Answer
b. How many eV does the average electron have?
Equation:
Substitutions
Answer
4. Not only are electrons shot out in solar storms but so too are protons.
Explain how things would be different and calculate.
In which direction would they turn?
56
__________________
What is the radius of curvature for the protons? __________________
How much energy in eV’s would each proton have?
__________________
5. You are given the diagram of the typical DC motor shown below.
There are 80 windings
around the
armature.
The armature has a
radius of 2 cm. And
a length of 6 cm.
The magnetic field has
a strength of 0.2
Tesla. Find the
following:
Current in the coils is
1 Amp
a. In each picture draw the direction of the magnetic field.
b. Determine the magnitude and direction of the force on the armature.
Equation:
Substitutions
Answer
c. Determine the magnitude of the torque on the armature.
Equation:
Substitutions
57
Answer
d. If the armature is cylindrical and has a mass of 300 grams find its
angular acceleration.
Equation:
Substitutions
Answer
6. A cyclotron is essentially a very powerful electromagnet over which electrons
(and other charged particles) are passed. They orbit in a circular pattern with
a radius based upon their speed. If we want electrons to move at 99 percent of
the speed of light (c = 3 x 108 m/sec) with an orbital radius of 2 meters, how
strong must the magnetic field be?
58
High Energy Particle Physics
(neglecting relativity for the moment)
1. Lets look at a situation where we have particles of unknown mass, charge and
speed. We want to find out everything we can about them. The first thing
we can do is determine if they are charged by passing them through either a
magnetic or electric field as shown below. Explain how you can determine
the charge in each case.
Magnet
If the charge is neutral then
A
B
If the charge is positive then A
If the charge is negative then
Capacitor
B
A
B
2. Now that we know the net charge of the particles, (imagine that they have a
positive charge) the next question is how fast are they moving if we do not
know their mass or exact quantity of charge. This is where a device called the
velocity selector comes in. It is simply the combining of the magnet and the
capacitor in the picture above.
Let us assume that the magnetic
field is fixed with a strength of B
and the Electric field is variable.
Show how by varying the E Field
you can select only particles of
one specific speed?
59
We can go a step further. The E field in the capacitor is related to the d V
across the capacitor by the following equation:
Suppose that the distance between the plates is “d” the equation for the
velocity of the emerging particles is
v=
Notice that all positively charged particles of the same speed will be selected
regardless of their mass or magnitude of their charges
60
The mass spectrometer
1. Imagine a bunch of particles moving of different masses but having the same
velocity how can we determine their individual masses. To do this we use
what is called a mass spectrometer. It is a very simple device. Suppose we
have the positively charged particles from the problem above with a known
velocity. They are shot into a known magnetic field that is perpendicular to
them. There is no electric field present. Less massive particles will land at
point “a” but more massive particles will land further out. For example it
point “b”
What direction is the magnetic field directed? _____________________
The force acting upon the particles by the magnetic field is given by the following
general equation
Since the path of the particles is
circular we can use F = m a to find
the mass of the particles in terms
of known quantities (Spectrometer
magnetic field strength Bs, particle
velocity v, path diameter “D” and
charge q. Show your work below.
61
Maxwell’s Equations
First Law
Second Law
Third Law
Fourth Law
62
Developing Maxwell’s Equations
Maxwell’s equations represent probably the high point of classical physics. They tie
together electricity and magnetism. They predict the speed of light and the nature
of most electromagnetic phenomena: reflection, refraction, diffraction and
interference. They are what allowed us to develop radio and Television, computer
hard drives and motors. They explain alternating current, why an electric
transformer can raise or lower voltage, and how electric generators work.
Interestingly, it only took four equations to do all of this.
Maxwell’s first law we know pretty well already.
It is called Gauss’ Law.
Remember that all this really says is that you can discover the quantity of the net
charge inside of a closed surface by measuring the electric field intercepted at the
surface.
Maxwell’s second law is the magnetic equivalent for Gauss’ Law
It is based
on the assumption that there are no magnetic monopoles. That for every north pole
there must be a south pole. The result is that the sum of all magnetic fields
intercepted by a closed surface must therefore be zero. This doesn’t sound like an
extremely useful statement but it does come in helpful from time to time.
63
Maxwell’s third law is also called Ampere’s Law.
We will start by learning
the short version of it. Basically Ampere’s Law state that any current will create a
magnetic field that is in a plane that is perpendicular to the flow of the charge.
Don’t let this equation intimidate you. Imagine that in the picture below we are
looking at a cross-section of a wire that is carrying a current (I) towards us (out of
the page) This current creates a magnetic field (B) that is oriented in a counterclockwise direction. You can use the hitch hikers rule to show this. Now, the field
will get weaker as we get farther from the wire. This is because the energy in the
field (the field line) must be spread over a greater distance (circumference = 2  r)
Ampere’s law takes advantage of this fact. The left side of the equation is what is
called a line integral. Draw a circle around the wire with some radius r. Add up all
of the small components of the magnetic field B in small increments dl until you
make a complete circle. It does not matter what radius you pick, the answer will be
the same. This answer is directly proportional to the amount of current flowing in
the wire. The proportionality constant is called . In a vacuum it is called o. or
the magnetic permittivity of free space. o.= 4  E 10-7 Tesla Meters/Ampere.
dl
B
B
B
Ampere proved that it wasn’t necessary to draw a perfect circle around the wire.
Any shape will do. Just add up the components of magnetic field that are in the
direction of shape you have drawn and you will get the total current inside the shape
64
you have drawn.
Sample problem 1.
You have a wire that has a current of 10 Amps flowing
through it, What is the magnitude of the magnetic field it creates at a distance of 1
cm from the wire’s center? (ans. approx. 2 x 10-4 Tesla
65
Sample problem 2.
It used to be that to measure the current in a wire one
needed to cut the wire and then run the current through the current measuring
device. New amp meters use a Amperes law, a collar which can wrap around the
wire and a thing called a magnetometer in this collar with which to measure the
magnetic field created by the current in the wire.
Collar
Magnetometer
Wire
B
Hinge
If the magnetometer on your amp meter reads (5.9 x 10-6 Tesla and it is 0.5 cm from
the center of the wire, what current is running through the wire?
66
Sample Problem 3:
The Solenoid
The solenoid is a compact way of creating a pretty strong magnetic field if you do
not have a massive current. A solenoid is just a coil of wire. It may be one layer
thick or many layers. Let us assume that L is the length of the loop. And that N is
the number of turns in the loop. You can use Ampere’s law to get a very good
approximation of the field strength inside the solenoid. What we will do is to draw
a rectangular loop.
1. Notice that in Ampere’s law, the integral B dl is a dot product. Because of this
since the magnetic fields are horizontal, crossing sides A and C at 90 degrees,
there is no contribution here.
2. We have placed side B at a great distance from the solenoid and can assume that
even though the field is parallel to this side, it is so weak that there is almost
no contribution here.
3. this leaves us with side D as the only side that is significant.
Now let us substitute into Ampere’s Law
I is increased by the number of turns
____ _____ _____ = ______________
B is a constant. Remove it from the integral ____ _____ _____ = _____ ______
The integral of dl is just L
____ _____ _____ = _____ ___________
Or
B = ____________________
Notice that N/L is just the number of turns
67
per unit length (n) and the equation
can be reduced to
B = ____________________
Let us see what happens when you have a solenoid that is 10 cm long, with 1000
turns and you decide to run 1 Ampere of current through it.
It turns out that if you were to place an iron core inside the solenoid the magnetic
field is increased by a factor (k) of 200. In the last problem what would the new
magnetic field be?
How would taking into account the relative permeability (k) change the equation for
the magnetic field strength of a solenoid
B=
68
Sample problem 4: The magnetic field created by a toroid.
If you want to make an even more powerful magnet you can bend a solenoid around
on itself so that the magnetic field never leaves iron core inside the solenoid. This is
what is called a toroidal solenoid.
To calculate the field inside of the toroid we will draw a circular loop that is inside
of the toroid itself.
Explain why the loop should be circular
Use Ampere’s law to determine the field (B)
69
Maxwell’s Fourth Law – Magnetic Induction
Maxwell’s fourth law is often known as Faraday’s Law. It asserts that any
change in a magnetic fields near a conductor will cause an induced voltage in the
conductor and a resultant flow of charge through that conductor. One can look at
this several ways. The simplest form of Faraday’s law is
d Fm
V
dt
Remember that  is the amount of magnetic field intercepted a surface.  = B A cos
. The simplest example of how this works is a wire that is being pulled across a
magnetic field as we did with the rail gun back on page 90.
There a 1.5 meter long metal bar was sliding across a magnetic field B = 0.5 Tesla.
If the bar is moving at a speed of 10 m/sec answer the following:
Velocity
A. Which rail will become positively charged?(think hand rule)
B. What will be the potential difference between the two rails? (Faraday’s Law)
70
Notice that this voltage is directly opposed to the voltage that we used to get the
bar moving in the first place. We call this back Emf. It is what limits the maximum
speed that a rail gun can obtain. Let us see how this works. In the rail gun problem
we placed a 12 Volts battery across the rails. In this problem we assumed a net
system resistance of 0.01 Ohm which limited the initial current to _________ Amps.
Since Faraday’s law predicts a counter voltage working against the battery it lowers
the net Emf of the system. At maximum speed, the induced counter Emf will 12
Volts.
5000 V
Velocity
At what speed mill the bar max out?
NASA makes a mistake.
Several years ago the guys at NASA proposed a unique method for generating
electrical power on board the international space station. (see picture below) The
idea employed a direct application of Faraday’s law. What they were going to do
was place a long wire 320 meters that would drop from the station towards the
Earth. As the station and wire orbited hundreds of miles above the equator. Lets
see what happens.
A. Using hand rules, determine whether positive charge will congregate at the
station or the end of the tether.
B. Use Faraday’s law to determine the potential difference between the ship and
the end of the tether.
71
C. It turns out that there are a couple of major problems with the plan. One of
them is electrical in nature. Hint: think circuits.
D. There is also a major mechanical problem. What is it?
The Electric Generator.
In the last problem we used Faraday’s law with a changing flux created as a bar
swept out an area across a magnetic field. Let us consider a situation that on the
surface of it sounds somewhat improbable. Imagine that you have a large magnet
whose south pole is facing you. B = 1.5 Tesla. In front of the magnet is a circular
conducting rubber band as shown below. Now lets imagine that the rubber band
contracts rapidly. Its radius changes from 10 cm to 5 cm in one half of a second.
A. In what direction will the current created by this shrinking go?
B. What average voltage will this create?
C. Now the rubber band suddenly expands again to its original size in the same
amount of time. What direction will the current be going as it expands?
72
The above problem does sound improbable. However it is at the heart of how the
generators that produce most of the electricity in the world today. Let me explain.
Let us imagine a loop of wire as seen below. This loop is free to spin in a circular
motion, around and around. If the loop is in front of a large magnet, from the
magnet’s point of view the area of intercepted magnetic field first decreases,
disappears and then grows again.
1
2
A B
AB
1
2
South
South
3
4
5
BA
3
4
South
South
BA
6
Side View
7
BA
7
South
A
9
South
South
A
A
A
A
AB
8
South
A
A
9
AB
5 Top View 6
South
8
A
A
1. In pictures 1,2,3 which way is the current flowing From A to B or B to A?
__________
2. In pictures 3,4,5, which way is the current flowing From A to B or B to A?
__________
3. In pictures 5,6,7, which way is the current flowing From A to B or B to A?
__________
4. In pictures 7,8,9, which way is the current flowing From A to B or B to A?
__________
5. If the loop is making 60 complete revolutions per second and has a radius of 10
cm and is in a B field of 1 Tesla, What is the average potential produced
during pictures 1,2,3?
73
Inducing currents by changing magnetic field strength
Faraday’s law states induced voltages are given by the equation V = -d /dt where
the flux  = B dot A. Up to this point, we have induced currents in wires by
changing the A (area) in the flux equation not the magnetic field strength B.
However, you can just as effectively change flux by keeping the area constant and
either increasing or decreasing the magnetic field strength that is intercepted. One
way of doing this is to drop a magnet through a loop.
Side View
S
S
N
N
S
N
Top View (ring moving toward you)
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
In the first picture (left) the north end of the magnet is moving downward toward
the ring. From the magnet’s point of view, the ring is moving upwards. The ring
moves upwards through a magnetic field that is pointing outwards. Draw in the
direction of the current that is generated. While the magnet is inside of the loop
(center) there is no net field moving across the loop and there is no net current in
the loop. As the magnet leaves the loop (right), magnetic field lines start crossing
the loop again. Only, this time they are pointed inwards. Draw in the direction of
the current that is generated.
Lenz’s Law (the electromagnetic equivalent of Newton’s third law (for every
action there is an equal but opposite reaction) states that an induced current will
create its own magnetic field. This field will be opposed to the magnetic fields that
created it. Lets go back to the first picture of the dropping magnet. Draw in the
74
magnetic fields created by the current in the loop. They have the same shape as
created by a solid magnet if it were placed at the center of the loop. Draw in this
magnet. Notice that the magnet has the north pole aiming upwards.
What impact do you think this induced magnet will have on the falling of the real
magnet?
Electric Transformers
A transformer is a device that can raise or lower the Voltage of electricity in a
system. It is a simple instrument, just an iron bagel with wires wrapped around it.
The principles upon which it works are just as subtle as the device is simple.
Principle #1: Magnetic fields are current driven. Wrap a wire around a
metal bagel and the current in the wire creates a magnetic field in the bagel.
(if the current flows as is shown, draw the direction of the magnetic field
inside of the bagel.
Principle #2: You can increase the apparent quantity of the current that
is flowing by increasing the number of loops. If one loop creates 1 B then
two loops creates 2 B.
Principle #3: For every action there is an equal but opposite reaction
(Lenz’s Law) If there is a conducting hoop somewhere around the bagel, as
the current in the original loop increases the increasing induced magnetic
field will induce a current in the hoop that will create a magnetic field in
opposition to this. Draw in the direction of the induced current in this hoop.
Principle # 4 Energy is conserved. Let us look at a situation where
there are two loops in the original circuit (left side of bagel) They will create
twice the magnetic field as created by one loop. If there is only one loop on
the right side of the bagel then it must generate twice the current as was in
the original circuit in order to react cancel out the inducing magnetic field.
The hoop on the right side of the bagel is a circuit in itself. This circuit is
called the Secondary circuit. The circuit on the left side of the bagel is called
the Primary circuit. According to the laws of conservation of energy, energy
out (secondary) must be equal to the energy in (primary). The equation for
this is
Pin = Pout
Iin Vin = Iout Vout
75
Remember that if there are twice as many turns in the primary as in the secondary
circuit then there must be twice as much current in the secondary circuit as in the
primary circuit. The power equation reduces to
Vin = 2 Vout
What we have is a voltage reducing transformer. If we want to increase voltage
then we have more turns in the secondary than in the primary. The relationship
between turns and voltage is given by
Tsec / Tpri = Vsec / Vpri
Sample problem. Lets say you want to make a neon sign transformer that raises
120 Volts to 1200 Volts.
A. If the primary(A) has 20 turns, how many turns must the secondary (B)
have?
B. If the secondary draws 60 milliamps how much will the primary draw?
76
Inductance (L)
Inductance ( L )is a measure of a coil’s ability (N = number of turns) to create
magnetic flux as a given current (I)is flowing through it
N m = L I
Remember Faraday’s law? V = - d (Nm) / d t Well if we combine these two the
result is V = - L dI /dt
Or
V
L
dI
dt
where inductance is measured in Henries and a Henry is a Volt Second per Amp
Calculating Inductance is very difficult. Only for a few simple situations can
inductance be relatively easily determined from a theoretical basis. (This is OK
because inductance is easily determined experimentally) For a tightly packed air
core solenoid inductance is
L = o n2 l A
Where
o is the magnetic permeability of free space
n is the number of turns per unit length (turns/meter)
l is the length of the coil (meters)
A is the end view area of the core
77
What is the importance of inductance?
Energy
Inductance has important effects in several ways. First it gives a direct way to
determine the energy stored in a coil’s magnetic field.
UB = 1/2 L I2
See how this compares to energy stored in an electric field in a capacitor
UE = 1/2 C V2
This is sometimes given as
UE = 1/2 q2 / C
Or the energy of mechanical motion
UKE = 1/2 m v2
Or the energy stored in a spring
Usp = 1/2 k x
2
Simple Harmonic Motion
Springs when stretched and set free to move will oscillate in SHM translating
energy between KE and PEspring. It turns out that a conductor-inductor system will
oscillate in SHM as well. These circuits are often referred to as LC circuits. The
energy in these systems alternates between a capacitor and an inductor.
B
E
E
Remember how for a spring  = (k/m). Well for an LC circuit  = (1/LC).
78
It is important here to actually derive this equation.
We start with the Energy equation
U
The general equation
Remember how
And
= UE
+
UB
UE =1/2 C V2
Ub =1/2 C V2
U
Substituting we get
Energy is time dependent
=
1/2 q2/C
dU/dt =
differentiate and we get dU/dt =
Remember L and C
are constants
+ 1/2 L I2
d (1/2 q2/C
q /C dq/dt
1/2 L I2 )/dt
+
+ L I
dI/dt
however U is constant
with time
0 = q /C dq /dt
+ L
I
dI /dt
But wait! di/dt is the
derivative of dq/dt
0 = q /C dq/dt
+ L
I
d2q/dt2
0 =
And I = dq/dt
q /C dq/dt
+
L dq/dt
d2q/dt2
0 =
This all reduces to
q /C
+
L q’’
q’’ = - (1/ LC) q
x’’ = - 2 bx
Which is familiar
For SHM
In this case






 = ______________
Frequency = ___________________
79
B
E
E
Before we leave the topic of Simple Harmonic Motion and LC circuits lets look at one more aspect of how
Faraday’s law plays a part here
In the diagram at the top of the page we see in the figure at the left energy is stored
in a capacitor and that the pressure (Voltage) across the capacitor is trying to push
current through the system in a clockwise motion through the system. The net
resistance of the wire in the coil is negligible. What one might expect is an
extremely rapid discharge of the capacitor. This does not happen. Faraday’s law
below explains why.
d Fm
V
dt
As the current starts to ramp up a back EMF is being created against it. The result
is that current will slowly increases. Remember that current in the coil places
energy in the magnetic field that is produced. This is what you see in the center
picture at the top of the page.
At this point there is no reason for a maintained current. So, how does the capacitor
get re-charged as shown in the diagram at the right at the top of the page? Well,
Faraday’s law comes to play again.
d Fm
V
dt
As the current starts to fall, the decreasing (-) magnetic field creates a positive
voltage across the coil. This positive voltage starts to force current in a clockwise
direction in the circuit. The current slowly declines for two reasons
1. ____________________________________________
1. ____________________________________________
80
81