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\begin{center}
{\bf Notes on the Magnetic Interaction}
\end{center}

In the last sections, the main interaction we considered was the
electrostatic: Coulomb's Law for point
charges and the principle of superposition.  Why did we call 
Coulomb's Law the electro{\bf static} force?  This is because
the force does not {\bf depend} on the particle's velocity.  Whether
the charged particles are moving or not the force is the
same: $k q_1 q_2/r^2$.  

In this section, we shall see that if the charged particles are both moving, then
there is an additional force between them.  We could call this
force the "velocity dependent part of the electric force", but
this is too wordy.  The name we give it is the {\bf magnetic force}, and
together with the electrostatic force, the complete force is 
called the {\bf electromagnetic interaction}.

We first present some properties of permanent magnetics, then
develop the properties of the electromagnetic interaction.

\bigskip

\centerline{\bf Permanent Magnets}

\bigskip

The magnetic force was discovered long before it was realized that its
source was moving charges.  This is because permanent magnets exert
forces on each other even though the magnet is not moving nor has 
any charge on it.  We list some properties of permanent magnets below:\\

\noindent 1.  Magnets always come with two "poles".  No one has yet to
find a magnet with only one pole, a magnetic monopole.  Whenever there
is a north pole, there always is a south pole.\\

\noindent 2.  Like poles repel, and opposite poles attract.\\

\noindent 3.  If a magnet is cut in half, then two new magnets are formed.
That is if one starts with one North and South pole, and cut it in half, 
then one gets two smaller magnets each with their own North and South pole
pair.\\

You might wonder what would happen if you kept dividing a magnet in half
over and over again.  Eventually you get to a single atom.  Atoms themselves
are little magnets with a north and south pole.  What about the constituents
of the atom: the protons, neutrons and electrons?  The proton, neutron and
electron are each little magnets as well.  This is somewhat perplexing, since
as far as we know the electron has very little, if any, size.  The subject
of quantum mechanics is needed to understand the magnetic properties of
these particles.  In this class we won't get that far, but will cover the
developments of electromagnetism up to around 1860.  We begin with one of the 
most important discoveries in science.

\bigskip

\centerline{\bf Sources of Magnetism}

\bigskip

Many discoveries in science are by accident, and in 1820 a very important
accidental discovery changed the way the world was to be.  The place was
Hans Oerstead's classroom.  Oerstead produced a large current in a wire
that was in the vicinity of a magnetic compass.  When the current flowed
in the wire, he noticed that the magnetic compass needle moved! 
This was the first evidence that electric currents produced magnetic fields.
In summary, he discovered that {\bf electric currents produce magnetic fields},
or stated in microscopic terms: {\bf moving charge produces magnetic fields}.

One cannot understate the importance of this discovery.  Oerstead's discovery
showed that two "supposedly different" forces were somehow related to each 
other.  Over the next 40 years scientists sorted out the connections and
unification of these two forces: electric and magnetic.  It was the beginning
of a quest to see if all the fundamental forces of nature were somehow related to each other.  From 1960-1980,
the "weak" interaction was unified with the electric and magnetic.  Current experiments
are examining if the strong and/or graviational interactions are also part of the
electromagnetic-weak interaction.  In addition to these philosophical implications,
Oerstead's discovery eventually lead to many practical applications.  Before we
can understand the applications, we need to understand the "physics" of the magnetic
field produced by an electric current.

\noindent {\it Initial Definition of the Magnetic Field}\\

We will initially define the magnetic field $\vec{B}$ at a point "P" in space in 
terms of how it affects a little magnetic probe when the probe is located at "P".  If we place 
our little magnetic probe at "P", it will allign itself 
along the direction of the magnetic field.  The strength of the magnetic field
is related to how strong the probe alligns itself with the field.  In lecture we
will use the "magnaprobe" to determine the magnetic field from different sources.

Later we will define the magnetic field in terms of the force it produces on
a small "point" particle, but for now we will define it in terms of its
effect on a small magnetic probe.\\

\noindent {\it Quantitative Description of the Magnetic Field}\\

In lecture we will connect a power supply to a wire and produce an electrical current
in the wire.  Then, using our little "magnaprobe" we will see that the magnetic field
produced by this "current carrying wire" {\it goes around the wire}.  This might seem
a bit strange, but lets try to understand why it is so.  As physicists, we first
start with the simplest situation and try and determine what quantites the 
magnetic field could depend upon.

Let's first consider a small piece of the wire and pose the following
question:  What is the magnetic field $\vec{\Delta B}$ at a point "P" located a
position $\vec{r}$ from the small segment of the wire $\vec{\Delta l}$ that carries
a current $I$?  The vector $\vec{r}$ is a vector from the small wire segment to
the point "P" (where we want to know the magnetic field $\vec{\Delta B}$).

Let the current in the wire be $I$, and the length of the wire be 
$\Delta l$.  The direction of the current will be important, so we will designate the direction of the
current by the direction of $\Delta l$.  This can be done by making $\Delta l$ a vector:
$\vec{\Delta l}$.

What can the magnitude of the magnetic field $|\vec{\Delta B}|$ depend on?  The only quantities
available are: $I$, $\vec{r}$ and $\vec{\Delta l}$.  How could the magnitude depend on
them:   A good guess is the following: if the current $I$
is doubled, then the magnitude of the magnetic field should double.  If the length
of the small line segment is doubled, then so should the magnitude of $\vec{\Delta B}$:

\begin{equation}
  |\vec{\Delta B}| \propto I |\vec{\Delta l}|
\end{equation}

\noindent How might the magnitude of $\vec{\Delta B}$ depend on the distance from
the small wire segment?  It probably decreases with distance, but does it fall off
as $1/r$ or $1/r^2$ or some other power?  We are tempted to be guided by Newtonian
gravity and Coulomb's law and suppose that the inverse square law applys here also. 
So a good guess would be  

\begin{equation}
  |\vec{\Delta B}| \propto {{I |\vec{\Delta l}|} \over r^2}
\end{equation}

\noindent where $r \equiv |\vec{r}|$.  I should mention that this equation
has been tested by experiment, so our guessing was good.  

Now we need to consider the direction of the "little" magnetic field $\vec{\Delta B}$.
You might guess that it points radially away from (or towards) the wire segment, 
like the electrostatic force.  The direction of $\vec{r}$ should have some effect on the direction
of the magnetic field.  However, the source for the magnetic field is 
different than the source for the electrostatic or gravitational forces.  In addition to
$\vec{r}$, another vector enters the picture.
The {\bf source} of magnetism itself {\bf has a direction}: the direction of the current in the wire.  If there
is no current (i.e. no direction) there is no magnetic field.  Remember that the
magnetic field is created by moving charges (current).  So the direction of
$\vec{\Delta l}$ must play a role in determining the direction of $\vec{\Delta B}$.

There are three vectors involved in determining the magnetic field produced by a
small piece of current carrying wire: $\vec{\Delta l}$ which points in the direction
of the current, $\vec{r}$ the vector from the wire segment to the location, and
$\vec{\Delta B}$ the magnetic field at the location in question.  The only mathematical
operation (with the correct mathematical properties) that combines two vectors to 
produce a third vector is the {\it vector cross product}. Consequently, the only
acceptable relationship is:

\begin{equation}
  \vec{\Delta B} \propto {I \over r^2} \; (\vec{\Delta l} \times \hat{r})
\end{equation}

\noindent where $\hat{r} \equiv \vec{r}/r$.  This is a fairly complicated equation.
We restate what the terms mean: $\vec{\Delta B}$ is the magnetic field at the point
"P" produced by a small piece of wire which has a current $I$.  $\vec{\Delta l}$ is
a vector whose magnitude is the length of the "small piece of wire" and whose direction
points in the direction of the current.  $\vec{r}$ is a vector from the piece of wire
to the point "P".

Note that because of the cross product, the direction of $\vec{\Delta B}$ will be
perpendicular to both $\vec{\Delta l}$ and $\vec{r}$.  Thus, the magnetic field
will circulate a straight wire, a phenomena that was demonstrated in lecture.

We can replace the proportionality sign with an equal sign and a constant:

\begin{equation}
  \vec{\Delta B} = {\mu_0 \over {4 \pi}} {I \over r^2} \; (\vec{\Delta l} \times \hat{r})
\end{equation}

\noindent where $\mu_0$ is a constant equal to $4 \pi \times 10^{-7}$ Tm/A.  We will discuss 
units later on in the quarter.  For now accept the fact that $\mu_0 = 4 \pi \times 10^{-7}$ Tm/A.
The "T" stands for Tesla, which is a unit for magnetic field.  We need to determine what
the units for the magnetic field are, which we will do shortly when we examine the force
on a particle possessing charge.  

If you don't like using the unit vector $\hat{r}$, you can write the above equation
in terms of $\vec{r}$, since $\hat{r} = \vec{r}/r$.  In terms of $\vec{r}$ we have

\begin{equation}
  \vec{\Delta B} = {\mu_0 \over {4 \pi}} {I \over r^3} \; (\vec{\Delta l} \times \vec{r})
\end{equation}

Taking the magnitude of the right side of the equation above, we obtain an
expression for the magnitude of $\vec{\Delta B}$:

\begin{equation}
  |\vec{\Delta B}| = {\mu_0 \over {4 \pi}} {{I |\vec{\Delta l}| sin(\theta)} \over r^2}
\end{equation}

\noindent where $\theta$ is the angle between $\vec{\Delta l}$ and $\vec{r}$.  Since
the sin function is maximized when $\theta = 90^\circ$, the magnetic field is maximized
at positions perpendicular to the wire segment.  Since $sin(0) = sin(180^\circ) = 0$ 
the wire segment does not produce any magnetic field at points located on its axis.
The above formula is called the "Biot-Savart" Law in honor of the scientists who
discovered it.  

The Biot-Savart Law relates the magnetic field at a point in space produced by
a small segment of a wire which carries a current.  What if there are more than
one small segments of wire which carry current (there always is).  Then, to find
the net magnetic field $\vec{B}$ at a point in space we might suspect that one
would just add all the vectors $\vec{\Delta B}_i$ from each $i$'th wire segment.
This is the {\bf principle of superposition} applied to the magnetic field where the
sources are segments of wire.  Although we hope the superposition principle is
valid for the "current sources" of the magnetic field, the principle needs to be
verified by experiment.  Experiments show that it is.  Thus, now we have a method
of calculating the magnetic field produced by any finite wire which has an
electrical current flowing through it:\\

\noindent 1.  Divide the wire up into small segments, $\vec{\Delta l}_i$.  The
segments are labeled by the index $i$.\\

\noindent 2.  Use the Biot-Savart law to determine the magnetic field $\vec{\Delta B}_i$ 
at the point "P" from each of the segments.\\

\noindent 3.  Add up all the $\vec{\Delta B}_i$ to find the net magnetic field $\vec{B}$
at the point "P" due to the whole wire (i.e. apply the principle of superposition).  
In adding up the contributions from all the segments, we take the limit as the segment
length $|\vec{\Delta l}|$ goes to zero.  This will result in an integral which we need
to evaluate.\\

\bigskip

Note that the above method is the same procedure we used for the electric field:
Divide the object into small point sources, determine $\vec{E}$ from each point
source in the object, and add up the contributions (i.e. integrate) over the whole object.
We apply the above procedure to two common current distributions: a straight length of
wire of finite size, and a circular loop of wire.\\

\bigskip

\noindent {\it Magnetic field produced by a straight wire}\\

Consider a wire segment that lies on the y-axis.  The lower end starts at
$y = -b$, and the upper end is at $y=a$.  Suppose there is a current $I$ that
flows in the wire.  Lets find the magnetic field at a point "P" on the x-axis
located a distance $d$ from the origin (d,0).

We start by dividing up the wire in to small segments of length $\Delta y$.  
The $i$'th segment will be located at $y_i=i*\Delta y$ where $i$ goes from
$-b/(\Delta y)$ to $a/(\Delta y)$.  The distance from the $i$'th segment
to the point "P" is $r = \sqrt{y_i^2 + d^2}$.  The Biot-Savart Law gives
for the magnitude of the magnetic field $\vec{\Delta B}_i$:

\begin{equation}
 |\vec{\Delta B}_i| = {{\mu_0 I} \over {4 \pi}} {{\Delta y \; sin(\theta)} \over r^2}
\end{equation}

\noindent where $\theta$ is the angle between the direction of the wire segment and the
point "P".  In terms of the geometry, $sin(\theta) = d/r$.:

\begin{equation}
 |\vec{\Delta B}_i| = {{\mu_0 I} \over {4 \pi}} {{\Delta y \; d} \over r^3}
\end{equation}

\noindent In terms of the $y_i$, we have

\begin{equation}
 |\vec{\Delta B}_i| = {{\mu_0 I} \over {4 \pi}} {{\Delta y \; d} \over {(y_i^2 + d^2)^{3/2}}}
\end{equation}

\noindent Note: the direction of all the $\vec{\Delta B}_i$ are in the negative z direction, i.e.
along ($- \hat{k}$).  Adding up all the $\vec{\Delta B}_i$ and taking the 
limit as $\Delta y \rightarrow 0$ 
we have

\begin{equation}
 |\vec{B}| = {{\mu_0 I d} \over {4 \pi}} \int_{-b}^a{{dy} \over {(y^2 + d^2)^{3/2}}}
\end{equation}

\noindent This integral can be solved using the trig substitution $y = d \; tan(\phi)$.  
With this substitution, $dy = d \; sec^2 \phi \; d \phi$.  This substitution also gives
$y^2 + d^2 = d^2 sec^2 \phi$.  Upon substitution, the integral is just an integral 
over $cos(\theta)$:

\begin{equation}
 |\vec{B}| = {{\mu_0 I} \over {4 \pi d}} \int_{-\phi_b}^{\phi_a} cos\phi \; d \phi 
\end{equation}

\noindent Solving the integral and taking the limits yields:

\begin{equation}
 |\vec{B}| = {{\mu_0 I} \over {4 \pi d}} (sin(\phi_a) + sin(\phi_b)) 
\end{equation}

\noindent In terms of $a$ and $b$ the expression becomes

\begin{equation}
 |\vec{B}| = {{\mu_0 I} \over {4 \pi d}} ( {a \over {\sqrt{a^2 + d^2}}} +   
                                    {b \over {\sqrt{b^2 + d^2}}} ) 
\end{equation}

If the wire is of infinite length, the result takes on a simple form.
In this case, $\phi_a$ and $\phi_b$ are both $90^\circ$.  Since
$sin(90^\circ) = 1$, we have

\begin{equation}
 |\vec{B}| = {{\mu_0 I} \over {2 \pi d}}
\end{equation}

\noindent Here we see that the strength of the magnetic field decreases as
the inverse of the perpendicular distance $d$ to the wire.  This $1/d$
decrease was also the case for the magnitude of the electric field
produced by an infinitely long uniformly charged line (or rod) of charge.
In infinitly long source essentially removes one dimension from the
source distribution geometry.  Since the circumference of a circle increases
with radius to the first power, the force fields decrease with $1/d$.
We will use this property of the magnetic field to reformulate the
Biot-Savart Law via closed path line integrals: Ampere's Law.\\

\noindent {\it Magnetic Field produced by a circular current loop}\\

A common wire shape is a circular loop.  Let's use the Biot-Savart Law
to find the magnetic field on the axis of a circular loop of wire that
has a current $I$ flowing through it.  Let the radius of the loop be
$R$, the center of the loop be at the origin, and the loop lie in the
y-x plane.  The x-axis is thus the axis through the center of the loop
Suppose the point "P" is located on the axis of the loop (i.e. the x-axis) a 
distance $x$ from its center.  

We use the same method as before,
we first "chop" up the wire into small wire segments of equal length
$|\vec{\Delta l}|$.  The magnitude of the magnetic field $|\vec{\Delta B}|$
from any of the small segments of wire is

\begin{equation}
  |\vec{\Delta B}| = {{\mu_0 I} \over {4 \pi}} \; {{|\vec{\Delta l}|} \over
                               {x^2 + R^2}}
\end{equation}

\noindent The direction of $\vec{\Delta B}$ is not along the x-axis, but
rather at an angle $\phi$ with respect to the axis.  After integrating around the
circular loop, all components of $\vec{\Delta B}$ perpendicular to the x-axis
cancel out.  Thus, the only component that "survives" is the one along the x-axis
$\Delta B_x$:

\begin{eqnarray*}
  \Delta B_x & = & {{\mu_0 I} \over {4 \pi}} \; {{|\vec{\Delta l}|} \over
                               {x^2 + R^2}} \; cos(\phi)\\
          & = & {{\mu_0 I} \over {4 \pi}} \; {{|\vec{\Delta l}|} \over
                               {x^2 + R^2}} \; {R \over {\sqrt{x^2 + R^2}}}\\
\end{eqnarray*}

\noindent When the $\Delta B_x$ are summed (or integrated) $x$ and $R$ do not
change.  Thus, one only needs to sum up the $|\vec{\Delta l}|$ which is the
circumference of the loop: $2 \pi R$:

\begin{equation}
  \Delta B_x  = {{\mu_0 I} \over {4 \pi}} \; {{2 \pi R} \over
                               {x^2 + R^2}} \; {R \over {\sqrt{x^2 + R^2}}}
\end{equation}

\noindent Collecting terms we have:

\begin{equation}
  B_x = {{\mu_0 I R^2} \over {2 (x^2 + R^2)^{3/2}}}
\end{equation}

\noindent You will test the $x$ dependence of this equation in your laboratory
class.  If you find it to follow the above formula, you will be demonstrating
the validity of the Biot-Savart Law for the current loop.

To find the magnetic field at the center of the loop, we set 
$x = 0$.  Thus at the center of a circular loop we have

\begin{equation}
  B_{center} = {{\mu_0 I} \over {2 R}}
\end{equation}

\noindent with the direction perpendicular to the plane of the loop.\\

\noindent {\it Magnetic Field on the axis of a solenoid}\\

Another interesting current distribution is that of a solenoid.  A solenoid
is essentially wire coiled around a cylinder.  We can use the results of the
last section to find the magnetic field on the axis of a solenoid.  
Suppose the solenoid has a radius $R$ and $n$ turns of wire per length.
Lets use our previous result to calculate the magnetic field on the axis
inside the solenoid.  Let the axis of the solenoid lie on the x-axis, and
suppose the solenoid starts at $x=-b$ and extends to $x=a$, and we will
calculate the magnetic field at the origin.  If the current flows
counter-clockwise looking in the "-x" direction, then the magnetic
field at the origin will point in the +x direction.

Consider first the contribution to the magnetic field, $\Delta B_x$, at the origin due
to a small slice of the solenoid located at position $x$ and of thickness
$\Delta x$.  The amount of current in the slice of thickness $\Delta x$ is
$\Delta I = n \Delta x$.  From the previous section we have:

\begin{equation}
  \Delta B_x = {{\mu_0 R^2 n \; \Delta x} \over {2 (x^2 + R^2)^{3/2}}}
\end{equation}

\noindent To find the net magnetic field at the origin, we need to add
up the $\Delta B_x$ from each thin slice of the solenoid.  This leads to
the integral

\begin{equation}
  B_x = \mu_0 R^2 I n  \int_{-b}^a {{dx} \over {2 (x^2 + R^2)^{3/2}}}
\end{equation}

\noindent We have done this integral before when we calculated the magnetic
field produced by a finite straight current carrying wire.  The trick was to
substitute $x = R tan \phi$.  Then $dx = R sec^2 \phi$, and $(x^2 + R^2) =
R^2 sec^2 \phi$.  After substitution the integral becomes

\begin{equation}
  B_x = {{\mu_0 I n} \over 2}  \int_{-\phi_b}^{\phi_a} cos \phi \; d \phi
\end{equation}

\noindent Evaluating the integral gives

\begin{equation}
  B_x = {{\mu_0 I n} \over 2} (sin(\phi_a) + sin(\phi_b))
\end{equation}

\noindent  $B_x$ can be written In terms of $a$ and $b$:

\begin{equation}
  B_x = {{\mu_0 I n} \over 2} ({a \over {\sqrt{a^2 + R^2}}} + 
                   {b \over {\sqrt{b^2 + R^2}}})
\end{equation}

An interesting case is an infinite solenoid, that is when $a \rightarrow \infty$
and $b \rightarrow \infty$.  Then, one obtains

\begin{equation}
  B_x = \mu_0 I n
\end{equation}

\noindent for points on the axis of the infinite solenoid.  Note here that
the length of the solenoid and hence the total number of turns is infinite,
but the number of turns per length, $n$, is finite.

\bigskip

\centerline{\bf Ampere's Law}

\bigskip

In the last section we discussed the Biot-Savart Law, and applied it to
some examples.  As discussed, one can use the Biot-Savart law to find
the magnetic field at any point in space that is produced by any current
configuration.  To do this, one would follow the same method we have used:
slice up the wires into small segments, determine $\vec{\Delta B_i}$ from
each of the segments, take the limit as the segments approach zero and
integrate over all the wires.  Although this can be a tedious process,
computers can help and in principle one could solve any problem involving
currents in wires.  

We could continue solving other current set-ups, but it is more interesting
to examine if there are any global properties of magnetic fields.  We will
see that a line (or path) integral of the magnetic field
for a closed path yields simple solutions.  This result will help us
determine magnetic fields for some symmetric wire configurations.

The magnitude of the magnetic field produced by an infinitely long wire is
$B = \mu_0 I/(2 \pi d)$, where $d$ is the perpendicular distance to the
wire.  The direction of the magnetic field is circular, around the wire.
The denominator, $2 \pi d$, reminds us of the circumference of a circle.
Consider a circle such that the wire passes through its center and 
whose plane is perpendicular to the wire.
If one were to calculate the line integral (or path integral) of 
$\vec{B} \cdot d \vec{r}$ for a closed path around this circle, one obtains

\begin{equation}
 \oint \vec{B} \cdot d \vec{r} = {{\mu_0 I} \over {2 \pi d}} (2 \pi d)
\end{equation}

\noindent The result is simple since the magnitude of the magnetic field is constant,
$B = \mu_0 I/(2 \pi d)$, and the direction of the magnetic field is parallel
to the path.  In this special case, the circular path with the wire at the center,
the path integral is simply the magnitude of the magnetic field times the path length.  
Note that the factor $2 \pi d$ cancels, and we are left with

\begin{equation}
 \oint \vec{B} \cdot d \vec{r} = \mu_0 I
\end{equation}

\noindent for a circular path of any size.  Since the magnetic field decreases as $1/d$
and the path length increases as $2 \pi d$, the radius of the circle cancels out.
We will show in lecture that this results hold for any path that goes around the
infinitely long wire.

You might be wondering if the direction of the current and/or the direction that
the path is traversed in the closed path integral.  Yes it does.  The convention
is that if your right thumb points in the direction of the current (for the infinite
wire) the magnetic field circles in the direction of your fingers.  Therefore,
if the {\it fingers of your right hand curl in the direction of the path of the line integral}, your
{\it thumb will point in the positive direction for current}.  If the current flows
in the opposite direction as your thumb, then the line integral will be negative.
That is, the path is going in the opposite direction to $\vec{B}$.

What happens if more than infinite wire (carrying current) passes through the closed
path?  Since the principle of superposition applies to the magnetic field, the
closed path line integral will just equal $\mu_0$ times the sum of the currents
that pass through the path.  Whether the current is + or - is determined by the
"right hand rule" described above: If the fingers of your right hand curl in
the direction of the path, then your right thumb will point in the direction of
positive current.

\begin{equation}
 \oint \vec{B} \cdot d \vec{r} = \mu_0 I_{Net \; through \; path}
\end{equation}

\noindent {\it Comments}:\\

\noindent 1.  The little circle in the middle of the integral sign means
that the path integral is closed.  For the equation to apply, {\bf the line
integral must be over a closed path}.\\

\noindent 2.  In lecture we have only justified this equation for straight wires
of infinite length.  Using vector calculus, it can be shown that this result is
valid for any wire configuration as long as the wires (or currents) do not have
an open end.  That is, if all the wires are closed loops or extent to infinity 
then the above equation is valid.  A counter-example is a discharging parallel
plate capacitor.  In this case the current starts at one plate and ends at
the other.  There is a gap in the middle where there is no electrical
current (i.e. no charge flow).\\

\noindent 3.  If the path does not enclose any current, then the closed line
integral on the left is zero.\\

The equation above is refered to as Ampere's Law.  In lecture we will do some classic
examples using Ampere's law: The infinite solenoid, coaxial cable, and the toroid.
In these examples our goal will be to determine the magnetic fields produced by 
these wire configurations.  However, before we can uniquely calculate these
magnetic fields, we need one more global property of magnetic fields: what
is the surface integral over a closed surface for magnetic fields?

For the electrostatic field $\vec{E}$ we showed that Coulombs Law resulted
in $\oint \oint \vec{E} \cdot \vec{dA} = Q_{enclosed}/\epsilon_0$.  This was
because electric fields are produced by a stationary source of charge that
eminate from the source and whose magnitude decreases as $1/r^2$.  If there
are no stationary sources within the closed surface, then the surface integral
equals zero.  For the magnetic field, {\bf there are no stationary sources}.  That is,
there are no magnetic monopoles.  At least until now, none have been discovered.
So until one is discovered, we must have

\begin{equation}
  \oint \oint \vec{B} \cdot \vec{dA} = 0
\end{equation}

\noindent for {\bf any closed surface}!  Whereas $\oint \vec{B} \cdot \vec{dr}$ is only
true when the current sources have no open ends, the surface integral is valid for
any source of magnetic fields.

We can use the two integral properties of the magnetic field:

\begin{eqnarray*}
 \oint \vec{B} \cdot d \vec{r} & = & \mu_0 I_{Net \; through \; path}\\
 \oint \oint \vec{B} \cdot \vec{dA} & = & 0
\end{eqnarray*}

\noindent The first one is valid for current sources with no open ends, and the 
second is always true.  Next I'll briefly present the results for the magnetic
field inside an infinite solenoid and a toroid, with more details given in
lecture.\\

\noindent {\it Infinite Solenoid}\\

We previously showed, using the Biot-Savart Law, that the magnetic field 
on the axis of an infinitely long solenoid was $B_x = \mu_0 I n$, where
$n$ equals the number of turns per length.  What about the magnetic field
at points inside the solenoid but not on the axis?  We will use the fact
that the solenoid has axial symmetry about its central axis.

Consider a point located a distance $r$ from the axis of an infinite solenoid.
At this point, the magnetic field could have three different components:
1) parallel to the axis of the solenoid ($B_x$), 2) radially away from the axis
($B_r$), or 3) circular around the axis ($B_\theta$).  Because of axial symmetry, these
components can only depend on $r$ and not on the position along the axis ($x$)
or the angle around the axis ($\theta$).  That is the components can only
have an $r$ dependence: $B_x(r)$, $B_r(r)$ and $B_\theta (r)$.

If we choose a closed cylindrical surface of radius $r$, length $l$ whose axis
coincides with the axis of the solenoid, then $\oint \oint \vec{B} \cdot \vec{dA}
= 2 \pi $r$ l B_r$.  However, this surface integral must be zero.  Therefore,
$B_r = 0$.  If we choose as a path, a circle of radius $r$ whose axis is
on the axis of the solenoid, then $\oint \vec{B} \cdot \vec{dr} = 2 \pi r B_\theta$.  
Since there is no current flowing through the path, this line integral must be
zero.  Thus, $B_\theta$ must be zero.  Finally, we choose as a path a 
rectangle of dimensions $l \times r$ whose side $l$ lies on the axis of the solenoid.
Then $\oint \vec{B} \cdot \vec{dr} = \mu_0 I n l - B_r l + 0 + 0$.  Since there is
no current flowing through the path this line integral must be zero.  Thus, we
have $l(\mu_0 I n - B_r) =0$ or $B_r = \mu_0 I n$ in the x-direction.  The magnetic field inside
an infinite solenoid is constant and equal to $\mu I n$ everywhere inside the
solenoid.  In lecture we will discuss the magnetic field outside the solenoid.\\

\noindent {\it Toroid}\\

To make a toroid, one could coil wire around a donut, then take the donut out.
Lets consider the magnetic field inside a toroid that has a rectangular
cross section.  Let the inner radius be $a$ and the outer radius $b$, and
the height of the cross section be $h$.  Suppose the toroid has $N$ total turns
of wire.  The toroid has axial symmetry about an axis through its center and
perpendicular to the plane of the toroid.

Consider a point inside the toriod a distance $r$ from the central axis.
That is $a < r  < b$.  The magnetic field at this point could have
three components: 
1) radially away from the axis of symmetry ($B_r$), 2) circular around the axis of
symmetry ($B_\theta$), or 3) "up or down" in the $\pm h$ direction ($B_h$).
We will show in lecture that because of the symmetry (axial and h-direction),
the integral $\oint \oint \vec{B} \cdot \vec{dA} = 0$ leads to
$B_r = 0$ and $B_h = 0$.

If we choose a circular path of radius $r$ that goes around the axis of symmetry
and is inside the toroid.  Then, $\oint \vec{B} \cdot \vec{dr} = 2 \pi r B_\theta$.
From Ampere's law this closed path integral equals the current that passes through
the path, which is $NI$.  Thus, $B_\theta = \mu_0 NI/(2 \pi r)$ for points inside
the toroid.  In lecture we will consider points outside the toroid.

\bigskip

\centerline{\bf Magnetic Force}

\bigskip

Up to now we have discussed some sources for the magnetic field.  The main type
of source we have dealt with has been a wire that has a current flowing through it.
As far as the magnetic force goes, we have only considered how a magnetic field
affects small bar magnets.  In lecture we used our magnaprobe to observe magnetic
fields.  Consider the following question: {\it will an object that has charge
feel a magnetic force}?.  That is, is there any other object besides a small
bar magnet that experience a magnetic force.  We will do two experiments in
lecture:\\

\noindent Experiment 1.  We will place a wire in the presence of a strong magnetic field.
When there is {\bf no current} in the wire, the wire experiences {\bf no force}.  When 
current flows in the wire, the wire jumps.  We will see that the direction of the force
on the wire is {\it perpendicular to both the direction of the magnetic field and the direction of the
current}.\\

\noindent Experiment 2.  We will bring into lecture an "electron gun".  Electrons will be
sent forward in a tube and will hit the screen in front of the tube.  Electrons
have a charge $e$.  When a magnetic field is introduced in the path of the electrons,
the electrons are deflected.  We will see that the force on the electrons is
{\it perpendicular to both the direction of the magnetic field and the direction
of the electrons velocity}.\\

Since current flow is essentially a flow of charge, the same effect is happening in
both experiments.  If an object that has a net charge moves in the presence of a magnetic field,
it feels a force.  The force is perpendicular to both the magnetic field and the
velocity of the object.

This might seem strange at first, since we are used to forces being in towards or away from
a source, or along the direction of a field vector.  The gravitational force is 
always towards the other object.  The electrostatic force is always away or towards 
its source, or along the direction of $\vec{E}$: $F_{electrostatic} = q \vec{E}$.
However, these two forces do not depend on the velocity of the object.  The force
is the same if the object is moving or not.  The magnetic force, on the other
hand, {\bf depends on the velocity of the object}.  If the object is not moving, there is no
force (the wire felt no force if the charges in it had an average velocity equal to zero).
Thus, the magnetic force vector must depend on two other vectors: the magnetic field $\vec{B}$ and
the velocity vector $\vec{v}$ of the object.  The only mathematical operation (in three dimensions)
that takes two vectors and produces a third vector in a mathematically invarient way is 
the cross product.  That is,

\begin{equation}
  \vec{F}_{magnetic} \sim \vec{v} \times \vec{B}
\end{equation}

The net charge also plays a role in the magnetic force on a particle.  
If a particle does not possess charge, it will not feel a magnetic force
(or electric force).  Electrons feel an opposite force that protons
feel if they move with the same velocity in the same magnetic field.
If an object has $N_p$ protons and $N_e$ electrons, then the superposition
principle for forces yields:

\begin{equation}
  \vec{F}_{magnetic} \sim (N_p - N_e)e \vec{v} \times \vec{B}
\end{equation}

\noindent In terms of the net charge on the object, this equation
becomes:

\begin{equation}
  \vec{F}_{magnetic} \sim q \vec{v} \times \vec{B}
\end{equation}

\noindent where $q$ is the net (or total) charge on the object or particle.
From this force equation, we can see that the units of magnetic field
are Force per charge per velocity, or in more basic quantities:
mass/(charge-sec).   As with the electric field, the force equation can be used to define $\vec{B}$,
and hence the $\sim$ can be replaced with an equal sign:

\begin{equation}
  \vec{F}_{magnetic} = q \vec{v} \times \vec{B}
\end{equation}

\noindent Before we discuss the force a wire feels in the presence
of a magnetic field, the above equation deserves some comments:\\

\noindent 1.  The magnitude of the magnetic force is $|\vec{v}|
|\vec{B}| sin(\theta )$ where $\theta$ is the angle between
the magnetic field and the particle's velocity.  Thus, if the
particle moves in the direction of the magnetic field, it does
not feel a magnetic force.  Only the component of the velocity
perpendicular to the magnetic field contributes to the force.\\

\noindent 2. Since the magnetic force is always perpendicular to
the particles velocity, {\bf the magnetic force cannot do any work} on
the particle.  The magnetic force cannot change the speed of a particle,
it can only change the direction of its motion.\\

\noindent 3.  The magnetic field is similiar to the electric field in that
is it a "force per charge".  However, in the case of the magnetic field
it is a (force per charge)/speed.\\

\noindent 4.  It is remarkable that both the electrostatic
force and the magnetic force are proportional to the same quantity: charge.
There is something to be learned from this property.\\

\bigskip

\noindent {\it Magnetic Force on a Straight wire with Current}\\

From the equation for the magnetic force on a particle, we can calculate
the magnetic force on a piece of {\it straight wire} that has a current flowing through
it.  Suppose we have a wire that has $n$ "charge carriers" per unit volume and each
charge carrier has a charge $q$.  The charge carriers are usually electrons.
Suppose each of the charge carriers is moving down the wire with a drift velocity
of $\vec{v}_d$.  Let the wire have a length $L$, a cross sectional area of
$A$, and a current $I$.  If there is a magnetic field $\vec{B}$ at the location
of the wire, then the net force on the straight wire is

\begin{equation}
  \vec{F} = n A L q ( \vec{v}_d \times \vec{B} )
\end{equation}

\noindent since $n$ is the number of charge carries per volume.  Note that $n A L q$
equals the total charge that is moving.  Remembering that the current density
$\vec{J} = n q \vec{v}_d$ we have

\begin{equation}
  \vec{F} = A L ( \vec{J} \times \vec{B} )
\end{equation}

\noindent  This equation is in a nice form.  Remember that the current $I$ equals
$A |\vec{J} |$, so we can replace $A \vec{J}$ with $I$.  However, if we do this
we need to specify the direction of the current.  This is best done by expressing
$L$ as a vector: $\vec{L}$, where the vector $\vec{L}$ points in the direction
of the current.  With this convention, we have

\begin{equation}
  \vec{F} = I ( \vec{L} \times \vec{B} )
\end{equation}

\noindent {\it Comments}\\

\noindent 1.  This equation is only valid for a {\it straight wire}, which has a current $I$.
However, if we have a curved wire, we can break it up into small pieces which can be considered
as straight wires in the limit that the "piece size" goes to zero.  We need then to integrate over the
wire.\\

\noindent 2.  The force on the wire is always perpendicular to the wire.\\

\noindent 3.  The force is the same whether the current is caused by positive charges
moving one way or negative charges moving the other way.\\

In lecture we will do examples of the magnetic force on wires and individual particles.\\

\bigskip

\centerline {\bf Units of the Electromagnetic Interaction}

\bigskip

Now that we understand the electric (velocity independent) force and the magnetic
(velocity dependent) force we can discuss units.  Consider two point particles, one
with a charge $q_1$ and the other a charge of $q_2$ that are separated by a
displacement $\vec{r}$.  Suppose particle "1" has a
velocity $\vec{v}_1$ and particle "2" a velocity of $\vec{v}_2$.

The electric force $\vec{F}_e$ between the two particles is

\begin{equation}
   \vec{F}_e = {1 \over {4 \pi \epsilon_0}} {{q_1 q_2} \over r^2} \hat{r}
\end{equation}

\noindent from Coulomb's law.

To find the magnetic force between the two particles, it is best to start with
the force between charges moving in a wire and a charge outside the wire.  If
charges are moving in a wire of length $\vec{\Delta l}$, then by the Biot-Savart
Law, the magnetic field produced is

\begin{equation}
  \vec{\Delta B} = {\mu_0 \over {4 \pi}} I {{\vec{\Delta l} \times \hat{r}} \over r^2}
\end{equation}

\noindent The current $I$ can be written as $I=|\vec{J}| A$.  $\vec{J}$ is equal to
$n q \vec{v}$.  Since $n A \Delta l$ is equal to the total number of charge carriers $N$ in
the wire segment $\Delta l$.  So the magnetic field produced by the line segment can be
written as

\begin{equation}
  \vec{\Delta B} = {\mu_0 \over {4 \pi}} Nq {{\vec{v} \times \hat{r}} \over r^2}
\end{equation}

\noindent From this equation, we can interpret the magnetic field produced by a single
charge $q_1$ moving with a velocity $\vec{v_1}$ as

\begin{equation}
  \vec{\Delta B} \approx {\mu_0 \over {4 \pi}} q_1 {{\vec{v}_1 \times \hat{r}} \over r^2}
\end{equation}

\noindent I have put in the approximation sign, since there are small corrections to
this equation due to relativity (Einstein's Special Relativity).  If particle "2" is
moving with a velocity of $\vec{v}_2$ at the location $\vec{r}$, then the magnetic
force it feels is

\begin{equation}
   \vec{F}_m = q_2 (\vec{v}_2 \times \vec{\Delta B})
\end{equation}

\noindent using the expression above for $\vec{\Delta B}$ we have

\begin{equation}
  \vec{F}_m \approx {\mu_0 \over {4 \pi}} q_1 q_2 {{\vec{v}_2 \times \vec{v}_1 \times \hat{r}}
                             \over r^2}
\end{equation}

\noindent Note: to have a magnetic force {\it both} charges have to be moving.  Summarizing:

\begin{eqnarray*}
   \vec{F}_e & = & {1 \over {4 \pi \epsilon_0}} {{q_1 q_2} \over r^2} \hat{r}\\
  \vec{F}_m & \approx & {\mu_0 \over {4 \pi}} q_1 q_2 {{\vec{v}_2 \times \vec{v}_1 \times \hat{r}}
                             \over r^2}
\end{eqnarray*}

These equations give us guidance in choosing the units for the electromagnetic interaction.
From mechanics we have the basic units for length, time and mass.  The unit of force is
derived from these three fundamental units.  We now need to consider units for
charge, and corresponding values for $\epsilon_0$ and $\mu_0$.  We have some options,
and we list just two:\\

\noindent Option 1: We choose a standard amount of charge as our unit charge.  Then
by measuring the electrostatic force between two point charges, the first equation allows us to determine
$\epsilon_0$.  By measuring the magnetic force between point charges that are moving,
the second equation allows us to determine $\mu_0$.  One could choose $e$ as the standard unit
of charge.  Then the proton would have a charge of $+1$ and the electron a charge of $-1$ in
these units.\\

\noindent Option 2: We could choose our unit of charge so that one of the constants,
$\mu_0$ or $\epsilon_0$, equals one or another appropriate number in these units.
Then the other equation would give us a way to measure the other constant.\\

Most texts, and the metric system, use the second option.  That is, the unit of charge
is the Coulomb, and one Coulomb is that amount of charge that makes $\mu_0$ exactly
equal to $4 \pi \times 10^{-7}$ in metric units.  That is why you see in the textbooks
that $\mu_0 = 4 \pi \times 10^{-7}$ Tm/A.  The electrostatic equation can be used to
determine $\epsilon_0$ which comes out to be $\epsilon_0 \approx 8.85 \times 10^{-12}$
C$^2$/(Nm$^2$).

There are some very interesting properties of the two equations above:\\

\noindent 1. {\bf No matter what one chooses as the basic unit for charge, the
product $\epsilon_0 \mu_0$ is always the same}.  This can be seen as follows.
Let $q_0$ be the unit of charge that results in a value of $\epsilon_0$ and
$\mu_0$.  Suppose we change the unit of charge to $q = x q_0$, where $x$ is a scale
factor.  The new $\epsilon_0$ would increase by the factor $x$: $\epsilon_0' = 
x \epsilon_0$, and the new $\mu_0$ would decrease by the factor $x$:
$\mu_0' = \mu_0 /x$.  The new product of $\epsilon_0'$ times $\mu_0'$ would
equal the old one $\epsilon_0 \mu_0$ since the factor $x$ would cancel.
Note that the product $\epsilon_0 \mu_0$ has units of $s^2/m^2$ or one over
speed squared!  There are no units of charge in the product.\\

\noindent 2. {\bf The combination $1/ \sqrt{\epsilon_0 \mu_0}$ has units of speed,
which is independent of our choice of charge unit}.  Thus, there is a fundamental
speed associated with the electromagnetic interaction.  In metric units,
$1/ \sqrt{\epsilon_0 \mu_0} \approx 3 \times 10^8$ m/s, the speed of light.
More on this later.\\

\noindent 3.  We can see the relative importance of the two forces for interacting
particles by taking the ratio of the magnetic force to the electrostatic force.
From the above equations we have:

\begin{equation}
  {F_m \over F_e} \sim \epsilon_0 \mu_0 v_1 v_2
\end{equation}

\noindent where the $\sim$ sign is used since we have left out the affect
of the relative angles between the velocities.  Note that $\epsilon_0 \mu_0 = 1/c^2$,
where $c \equiv 3 \times 10^8$ m/s (the speed of light).  Thus, for speeds much less
than the speed of light the magnetic force between individual particles will be
less than the electrostatic force between them.\\

\noindent 4.  The magnetic force $\vec{F}_m$ depends on the particle's velocities.
However, velocity is relative to one's reference frame.  This means that {\bf a
magnetic force for one observer might be an electric force for another}.  The
electric force is also reference frame dependent.  These are important considerations
in formulating the laws of electromagnetism in a way that have the same form
for all inertial observers, and led Einstein to the theory of special relativity.

\bigskip

\centerline {\bf The effect of Changing Magnetic Fields}

\bigskip

We have talked about the sources for electric and magnetic fields,
and the forces that they cause on a charged particle.  We have expressed
the fields in terms of the charge sources using Coulomb's Law and the
Biot-Savart Law:

\begin{eqnarray*}
  \vec{E} & = & k {Q \over r^2} \hat{r}\\
  \vec{\Delta B} & = & {\mu_0 \over {4 \pi}} I {{\vec{\Delta l} \times \hat{r}} \over r^2}
\end{eqnarray*}

\noindent We also used the line and surface integrals, Gauss' Law and Ampere's Law, to 
express the properties of the fields:

\begin{eqnarray*}
  \oint \oint \vec{E} \cdot \vec{dA} & = & {{Q_{net \; enclosed}} \over \epsilon_0}\\
  \oint \oint \vec{B} \cdot \vec{dA} & = & 0\\
  \oint \vec{E} \cdot \vec{dr} &  = & 0\\
  \oint \vec{B} \cdot \vec{dr} & = & \mu_0 I_{net \; through \; path}\\
\end{eqnarray*}

You might think that there is nothing more to understand about the sources of electric
and magnetic fields.  However, there is something missing in the equations
above.  It is due to the velocity dependence of the magetic force.  If one goes into
the reference frame of a moving particle, there is no magnetic force.  However, there
must be some force to take its place.  In this "moving" reference frame, the magnetic
fields are moving.  Therefore, a changing magnetic field must produce some sort
of force that we haven't accounted for yet.  We will demonstrate this phenomena
with an example.

Suppose we have a bar magnet and a closed rectangular loop of conducting wire.  Let the bar magnet be
stationary at the origin and point in the +z direction.  Now suppose the rectangular
loop of wire lies in the x-y plane and moves with constant velocity $v_0$ in the +x direction,
$\vec{v} = v_0 \hat{i}$.  Let the sides of the rectangular wire loop
be $a$ and $b$.   Suppose at $t=0$ one side of the wire ($a$) passes directly 
over the bar magnet.  That is, the front end of the rectangle is at position
$x = vt$.  Suppose the magnet field is strong only near the origin.  We will
view the situations from two different reference frames: one in which the
magnet is at rest and the wire is moving, and the other in which the
wire is at rest and the magnet is moving.\\

\noindent {\it Reference frame in which magnet is at rest}\\

In this reference frame, the the free charges in the front side of the wire 
are moving in the presence of a magnetic field and will feel a force.  However, 
since only the front side of length $a$ is over the magnet, only these charges
will feel a force.  The net result is that overall the charges are forced
around the wire and a current is produced.  We will demonstrate this in class.
We will produce a current in the closed wire loop without batteries!  As far
as the current is concerned, the key quantity is the "voltage" around the wire loop.
The "voltage" is essentially the force/charge times $\Delta l$ integrated around
the wire loop.  In our case, this integral becomes:

\begin{equation}
  \oint_{wire} {\vec{F} \over q} \cdot \vec{dr} =  B v a 
\end{equation}

\noindent The force acts {\bf only on the charges on the front side} of length
$a$ over the magnet.  The force/charge is $B v$ and the distance it acts
is $a$.  So the force per charge times distance around the loop is $Bva$, which is
essentially the voltage around the wire loop.  If the wire has a resistance $R$, then
the current is $I = Bva/R$.\\

\noindent {\it Reference frame in which the wire is at rest}\\

This reference frame is moving with velocity
$\vec{v} = v_0 \hat{i}$ with respect to the other frame.  Here, the loop is at rest and the
magnet is moving with velocity $- v_0 \hat{i}$.  In this frame the free
charges in the wire loop are not moving, i.e. have zero average velocity.
Thus, as far as this observer is concerned the charges do not feel a magnetic
force.  Also, there is no electrical force since there are
no charge sources.  So, if this observer uses the equations of electrodynamics
that we have developed so far, the observer will not predict that a current
is formed in the loop.  When we do the experiment in class, you will
see that keeping the loop fixed and moving the magnet also produces a current
in the loop.  Thus, there is something missing in the equations above.  In this
frame, the wire is at rest but the magnetic field is changing.  Thus, somehow
{\it a changing magnetic field produces some sort of electric field}.\\

To fix the problem, we need to go into the first reference frame, since
we are correctly predicting the experimental outcome, and try and
express $\oint_{wire} \vec{F} \cdot \vec{dl}$ without the velocity $v$ but
with only a changing magnetic field.\\

\noindent {\it Reference frame in which magnet is at rest}\\

Since the position of the front side of the wire is $x=vt$, we can express
the velocity $v$ as $v = \Delta x / \Delta t$.  Substituting for the
force/charge times distance around the loop:

\begin{equation}
  \oint_{wire} {\vec{F} \over q} \cdot \vec{dr} =  B {{\Delta x} \over {\Delta t}} a 
\end{equation}

\noindent However, in this reference frame, $B$ and $a$ are constant, so we can
write the above equation as:

\begin{equation}
  \oint_{wire} {\vec{F} \over q} \cdot \vec{dr} =  {{\Delta (x B a)} \over {\Delta t}} 
\end{equation}

\noindent This is a better form, since there is no velocity in the
expression.  The changing is a combination of the magnetic field times $x$.  In
one frame $B$ is changing and $x$ is not, and in the other frame $x$ is changing
and $B$ is not.  The quantity 
$xaB$ is the magnetic flux through the wire loop.  We can restate the equation
more succinctly as

\begin{equation}
  \oint_{wire} {\vec{F} \over q} \cdot \vec{dr} = - {{\Delta \Phi_B} \over {\Delta t}}
\end{equation}

\noindent Where $\Phi_B$ is the total magnetic flux through the wire loop.  The minus
sign refers to the direction of the current.
When the limit $\Delta t \rightarrow 0$ is taken, the right side becomes the
time derivative of $\Phi_B$.  

\begin{equation}
  \oint_{wire} {\vec{F} \over q} \cdot \vec{dr} = - {{d \Phi_B} \over {d t}}
\end{equation}

\noindent This is a form that both observers can agree on: the right side has the same value in both reference frames.
In one frame the wire is moving (magnetic field not changing) and in the other the 
magnetic field is changing and the wire is not moving.  It can be shown that the
above equation is true any time there is a changing magnetic flux through a 
closed wire loop.  The magnetic flux can change if\\

\noindent a) the wire moves in a stationary magnetic field\\
\noindent b) the wire is stationary and the magnetic field inside the
wire loop changes.\\
\noindent c) the wire's area changes in a stationary magnetic field\\
\noindent d) the wire is moving in an area in which the magnetic field is changing.\\

The closed integral around the wire loop, $\oint_{wire} \vec{F} /q \cdot \vec{dl}$ is
often called the $E.M.F.$ or electromotive force.  It represents the voltage
around the wire loop that is caused by an electric field produced by a (changing) 
magnetic field.  It is often given the symbol $\xi$:

\begin{equation}
 \xi \equiv  \oint_{wire} {\vec{F} \over q} \cdot \vec{dr} = - {{d \Phi_B} \over {d t}}
\end{equation}

\noindent The above equation (or some form of it) is called Faraday's Law after the
discoverer of this rather complicated phenomena.  It was an important insight to
realize that the easiest way to express the effects of a changing magnetic field is
through line and surface integrals, i.e. magnetic flux.  One reason we covered
electric flux and Gauss's Law was was to introduce you to the concept of flux and prepare
you for Faraday's law.  Faraday's discovery was one of the
most important contributions to science.  It revolutionized the world, and we will 
demonstrate its significance with numberous examples in lecture.\\

\noindent {\it Modifying the Field Equations}\\

We are how in a position to change the field equations.  As mentioned, the easiest way to
include the effects of a changing magnetic field is through a line integral:
$\oint \vec{F} /q \cdot \vec{dl}$.  Thus, it is most convenient to use the integral
formulation for the fields, and not the Coulomb or Biot-Savart laws.  In the integral 
forms, {\bf the line and surface integrals
are done over stationary mathematical topologies}.  Although we have no wires, we
must use the result for stationary wires.  If the wire for the closed loop is
stationary, then $\vec{F}/q$ is only caused by an electric field:
$\vec{F}/q = \vec{E}$.  This gives

\begin{equation}
   \oint \vec{E} \cdot \vec{dr} = - {{d \Phi_B} \over {d t}}
\end{equation}

\noindent where the subscript "wire" on the integral is removed.  This equation is
true for any closed mathematical path.  The corrected equations can now be written
as:

\begin{eqnarray*}
  \oint \oint \vec{E} \cdot \vec{dA} & = & {{Q_{net \; enclosed}} \over \epsilon_0}\\
  \oint \oint \vec{B} \cdot \vec{dA} & = & 0\\
  \oint \vec{E} \cdot \vec{dr} & = & - {{d \Phi_B} \over {dt}}\\
  \oint \vec{B} \cdot \vec{dr} & = & \mu_0 I_{net \; through \; path}\\
\end{eqnarray*}

\noindent It is important to note that the path and surface integrals on the left sides of the
equal signs can only be evaluated over paths and surfaces that are not moving.  
The new addition takes into account the effect of changing magnetic fields.
Indirectly it states that a changing magnetic field {\bf induces} an electric
field.  This phenomena is often called induction.  It is an extremely important addition
to the laws of electrodynamics and deserves some comments.\\

\noindent 1.  The time derivative of the magnetic flux, $d \Phi_B /dt$ in the above equation
is often written explicitly as an integral: $\Phi_B = \oint \oint \vec{B} \cdot \vec{dA}$:

\begin{equation}
  \oint \vec{E} \cdot \vec{dr} = - {d  \over {dt}} \oint \oint \vec{B} \cdot \vec{dA} \\
\end{equation}

\noindent The direction of the path for the path integral on the left side defines the 
direction for the area vector on the right.  If the fingers of your right hand curl in
the direction of the path, then your right thumb points in the direction of positive
area.\\

\noindent 2. The minus sign in the induction equation signifies the direction of the
induced electric field or current.  However, it is often easier to use another method
stated by Lenz.  Lenz's Law states that {\it the magnetic field produced by the induced
current opposes the change in the magnetic field causing the induced current.}\\

\bigskip

We conclude the topic of magnetic induction with some examples.\\

\noindent {\it A coil of wire rotating in a constant magnetic field}\\

Suppose there is a constant magnetic field of magnitude $B$.  Suppose we
rotate a closed loop of wire, of area $A$, with the axis of rotation
perpendicular to the magnetic field.  Let the angle between the
area vector and the magnetic field be $\theta$.  When $\theta = 0$
then the magnetic flux through the wire loop is maximized since
$\vec{B}$ is perpendicular to the area: $\Phi_m = B A$.  When
$\theta = 90^\circ$ then the magnetic flux through the loop is zero.
For arbitary angle, $\Phi_m = B A cos\theta$.  Note, in this case the
magnetic flux is just (magnetic field) times (area) {\bf since the
magnetic field is constant inside the wire loop}.  If the loop is
rotating with a constant angular velocity $\omega$, then $\theta = \omega t$:

\begin{equation}
  \Phi_m = B A cos(\omega t)
\end{equation}

\noindent The magnitude of the EMF generated in the wire loop is found by
taking the time derivative of $\Phi_m$:

\begin{eqnarray*}
 |E.M.F.| & = & | {{d \Phi} \over {dt}} | \\
          & = & \omega B A sin(\omega t) 
\end{eqnarray*}

\noindent A wire rotating at a uniform rate in a constant magnetic field will 
produce a sinusoidal "voltage" or current in the loop!  This is a very convenient
way to produce electrical current from mechanical motion.  We refer to this
type of device as an {\bf electric generator}.  The only peculiarity
is that the current produced is sinusoidal in time, that is, it is alternating
current (AC).  This is one reason why we have alternating current delivered
to our homes: it is easy to produce it by mechanical means.\\

\noindent {\it Wire moving in a non-uniform magnetic field}\\

Suppose there is an infinitely long wire with a steady current $I$ flowing
through it.  Let there be another wire which is a rectangular with sides
$a$ and $b$.  Let side "$b$" be parallel to the long wire and located
a distance $x$ from it.  Let side $a$ point directly to the long wire.
Lets first calculate the magnetic flux {\bf through} the rectangular wire {\bf due
to} the magnetic field produced by the infinitely long wire.

The magnetic field produced from the long wire has a magnitude of $\mu_0 I/(2 \pi r)$,
where $r$ is the perpendicular distance from the wire.  The magnetic field is
perpendicular to the closed rectangular wire.  Since the magnetic field is {\bf not constant}
inside the wire loop, the flux is {\bf not equal} to just (Magnetic Field) times (area).
We need to integrate over the loop area.  We divide the area into strips of length $b$ and
width $\Delta r$, where the strip is parallel to the long wire.  Then the flux
through a strip is:

\begin{equation}
 \Delta \Phi_m = {{\mu_0 I} \over {2 \pi r}} b \Delta r
\end{equation}

\noindent Adding up all strips and taking the limit as $\Delta r 
\rightarrow 0$ yields an integral:

\begin{equation}
 \Phi_m = \int_x^{x+a} {{\mu_0 I b} \over {2 \pi r}} d r
\end{equation}

\noindent This integral is easily evaluated to be

\begin{equation}
 \Phi_m = {{\mu_0 I b} \over {2 \pi}} (ln(a+x) - ln(x))
\end{equation}

\noindent  If $I$ remains constant and the rectangular wire does not move, then no current is
induced in it.  This is because the magnetic flux $\Phi_m$ is not
changing.  To change the flux through the wire we need to either
move the one of the wires, or change the current in the long
wire.  Suppose we move the rectanglar wire away with a constant
speed $v = dx/dt$.  To find the change in the flux with time, we
need to differentiate the above equation with respect to time:

\begin{equation}
  {{d \Phi_m} \over {dt}} = {{d \Phi_m} \over {dx}} {{dx} \over {dt}}
\end{equation}

\noindent Since $v = dx/dt$ we obtain

\begin{equation}
  |{{d \Phi_m} \over {dt}}| = {{\mu_0 I b v} \over {2 \pi}} \; {a \over {(a+x)x}} 
\end{equation}

\noindent which is the $E.M.F.$ generated in the rectangular wire loop.

Another way to generate a current in the rectangular loop is to keep it stationary
and change the current in the long wire.  In this case, we have

\begin{equation}
 E.M.F. = {{d \Phi_m} \over {dt}} = {{\mu_0 b} \over {2 \pi}} (ln(a+x) - ln(x)) \; {{dI} \over {dt}}
\end{equation}

\noindent This is an interesting way to generate current in the rectangular wire.  Neither
wire needs to be moving!  {\bf A changing current in one wire produces a current in the other}.
Next we will do another example of (changing current in one wire) $\rightarrow$ current in
another wire.\\

\noindent {\it Changing Current producing Changing Current in stationary wires}\\

Suppose we have a solenoid of length $l$ that has a total of $N_1$ coils of wire.
Let the area of the solenoid be $A$.  Around the solenoid we place a circular
coil of wire that has $N_2$ loops of wire.  Suppose the current in the solenoid
is $I_1$.  What is the magnetic flux through the coil of wire that is around the
solenoid?

This is a simple calculation if we approximate the
solenoid (wire 1) to be infinite: $B \approx \mu_0 n I_1$ where
$n = N_1 /l$, the number of turns per length.  Since the magnetic field
only occupies an area $A$ (of the outer coils of wire), we have

\begin{equation}
 \Phi_2 \approx {{\mu_0 N_1 I_1} \over l} N_2 A
\end{equation}

\noindent where $\Phi_2$ is the magnetic flux though the outer coil
of wire (wire 2).  If $I_1$ changes in time, then the magnetic flux
through the outer coil will change and an E.M.F. will be produced
in this coil:

\begin{equation}
 E.M.F._2 = {{d \Phi_2} \over {dt}} = {{\mu_0 N_1 N_2 A} \over l} \; {{d I_1} \over {dt}}
\end{equation}

Suppose there is a sinusoidal current in the solenoid (wire 1): $I_1 = I_0 sin(\omega t)$.
Then, a sinusoidal current will be generated in the other wire!!:

\begin{equation}
 E.M.F._2 = {{\mu_0 N_1 N_2 A} \over l} \; \omega I_0 cos(\omega t)
\end{equation}

\noindent This is great.  A sinusoidal current in one wire produces a sinusoidal
current in the other.  The nice thing is that by choosing appropriate values
for $N_1$ and $N_2$ we can control the $E.M.F.$ generated in wire "2"
(the outer coil).  We can boost the voltage up or down depending on the
number of turns we have.  This sort of device is called a 
{\bf tranformer}.  Note that the transformer only works with {\bf changing
current}.  It will not work with direct current D.C.  This is another reason
why the electrical voltage that comes to our homes is alternating (A.C.).
It is easy to generate, and it is easy to change (or transform) to a higher
or lower voltage!!  In the U.S., the frequency of the sinusoidal voltage is
$60$ cycles/sec.\\

\bigskip

\centerline{\bf Magnetic Induction in Circuits}

\bigskip

The results of the last two examples will apply to any two closed circuits:
a changing current in one circuit will produce a current in another
circuit.  Let the two circuits be labeled "1" and "2".  If circuit
"1" has a current $I_1$, then a magnetic field $\vec{B}_1$ will
be produced.  This magnetic field will exist at the location of
circuit "2".  There will, therefore, be a magnetic flux through
circuit "2" due to the magnetic field produced by circuit "1".
We label this magnetic flux as $\Phi_{12}$.  

Since the magnetic field $\vec{B}_1$ is proportional to the current
$I_1$ in circuit "1", the magnetic flux through circuit "2" will
also be proportional to $I_1$:

\begin{equation}
  \Phi_{12} \propto I_1
\end{equation}

\noindent We can replace the proportional sign with an equal sign and a
constant:

\begin{equation}
  \Phi_{12} = M_{12} I_1
\end{equation}

\noindent where $M_{12}$ is called the {\bf mutual inductance} of the two wire
configurations.  

You are probably thinking that a current in wire "2" will produce a magnetic
flux through wire "1".  In a similar manner, the magnetic flux through wire "1" will be 
proportional to the current in wire "2":

\begin{equation}
  \Phi_{21} = M_{21} I_2
\end{equation}

\noindent where $\Phi_{21}$ is the flux through wire "1" due to the current
in wire "2".  In our upper division electrodynamics class we show that
the mutual inductances are equal: $M_{12} = M_{21}$.

In the second example above, the mutual inductance between the infinite straight
wire and the rectangular loop is $M_{12} = (\mu_0 b/(2 \pi)) \; ln((a+x)/x)$.
For the solenoid and the outer coil, the nutual inductance is

\begin{equation}
    M_{12} = {{\mu_0 N_1 N_2 A} \over l}
\end{equation}

\noindent Note that:\\

\noindent 1.  The mutual inductance only depends on the geometry of the two wires.
It does not depend on the current flowing in the wires.  There is no $I$ in the
equation for $M_{12}$.  The mutual inductance is simple the
{\it ratio} of the magnetic flux through the surface of one closed wire due to the magnetic field
caused by a current in the other wire.\\

\noindent 2.  To have "mutual inductance" one needs two wires, and both should really be closed
circuits.  However, one wire could be an infinitely straight wire (which carries the current) and the other
one closed.\\

\noindent 3.  Every two closed wire loops will have a mutual inductance, and electrical engineers
will always need to consider this property of circuits.\\

\noindent {\it Self Inductance}:\\

If a wire carries a current, a magnetic field is produced.  If the wire
is closed, then this magnetic field will produce a magnetic flux through
the closed wire itself.  Thus, if a closed wire carries a current,
there is a magnetic flux $\Phi_m$ through the wire circuit that is
caused by the current in the wire itself.  If the current is doubled,
then the magnetic field everywhere is doubled and so is the flux through
the circuit:

\begin{equation}
  \Phi_m \propto I
\end{equation}

\noindent where $I$ is the current in the circuit.  The proportional sign can
be replaced by an equal sign and a constant:

\begin{equation}
  \Phi_m = L I
\end{equation}

\noindent where the constant $L$ is called the {\bf self inductance} of
the wire configuration (or circuit).  Perhaps this property will become
more clear with an example.  Lets find the self-inductance of a 
solenoid.\\

\noindent {\it Note}: The self-inductance of a closed circuit depends only on
the geometry of the wire.  It does not depend on how much current flows in the wire.  In fact,
it is the ratio of the magnetic flux through the area of the wire loop caused by the
current in the wire  divided by the current that is in the wire.\\

\noindent {\it Self-inductance of a solenoid}\\

Suppose we have a solenoid that has a total of $N$ coils (or turns)
of wire.  Let the solenoid have a cross sectional area $A$ and a
length $l$.  To calculate the self-inductance, we first let the
circuit have a current $I$.  Then we calculate the magnetic field
produced by this current.  Finally we calculate the magnetic flux
through the area of the circuit.

If we make the approximation that the magnetic field inside the soleniod
is approximately the same as an infinitely long solenoid, we have

\begin{equation}
  B \approx \mu_0 ({N \over l}) I
\end{equation}

\noindent For an infinite solenoid, the magnetic field is constant throughout
the whole inside of the coil.  The magnetic flux therefore through one loop
of solenoid wire is $\Phi_m \approx A B \approx A \mu_0 N I/l$.  Since there
are $N$ coils of wire, the total magnetic flux through the solenoid is:

\begin{equation}
  \Phi_m \approx {{\mu_0 N^2 A} \over l} \; I
\end{equation}

\noindent From this equation, we see that the magnetic flux through the
circuit is proportional to the current in the circuit.  The constant of
proportionality is the self-inductance $L$ of the circuit:

\begin{equation}
  L \approx {{\mu_0 N^2 A} \over l} 
\end{equation}

\noindent  It is interesting to note that:\\

\noindent a) As with mutual inductance, self-inductance only depends on
the geometry of the closed wire circuit.  It does not depend on the current
$I$ flowing in the wire.\\

\noindent b) Self-inductance has units of $\mu_0$ times distance.\\

\noindent c) There is an $N^2$ dependence for $L$.  The more turns of wire,
the stronger the magnetic field is inside the solenoid.  The more turns
of wire, the more magnetic flux.  This results in a $N^2$ factor for
$L$.\\

Self-inductance is a very important property of a circuit.  In some circuits
inductors are placed in circuits for particular applications.  
Previously, we discussed Kirchoff's law for the sum of the voltage drops around
a closed wire loop, and found the sum to be zero.  This is correct if the
current is not changing in the circuit.  If the current is changing, then
path integral of $\oint \vec{E} \cdot \vec{dl}$ is not zero, but rather
$- d \Phi_m / dt$.  Thus, the Kirchoff Law involving voltage changes
around a closed loop must be modified:

\begin{equation}
  \sum (Voltage \; drops) = - \oint \vec{E} \cdot \vec{dl} = {{d \Phi_m} \over {dt}}
\end{equation}

\noindent where the sum over voltages drops means around a closed path in the circuit.
The direction of positive flux is determined by the right hand rule and the direction
of the path taken to calculate the voltage drops.  That is, if the fingers of your
right hand curl in the path direction, then your thumb points in the direction of
positive area for the magnetic flux.

Lets consider a simple example.  Suppose we have
a circuit that consists of a battery of voltage $V$ and resistor of resistance 
$R$.  Suppose the circuit has a self-inductance of $L$.  The self-inductance
could be caused by an inductor placed in the circuit, or just the self-inductance 
of the wires themselves.  If we apply the modified Kirchoff Law to the circuit
we have

\begin{equation}
  V - IR = {{d \Phi_m} \over {dt}} = L {{dI} \over {dt}}
\end{equation}

\noindent If we rearrainge the terms the equation can be viewed as

\begin{equation}
  V - L {{dI} \over {dt}} = IR
\end{equation}

\noindent In this form, we see that the induction term $L (dI/dt)$ tends
to decrease the voltage in the current loop.  Electrical engineers sometimes
call this term a "back E.M.F.".  The self-induction (or inductive reactance) 
in the circuit tends to impede any change in the current.

In the original form:

\begin{equation}
  V - IR = L {{dI} \over {dt}}
\end{equation}

\noindent the equation can be readily solved.  By "cross-multiplying" we have

\begin{equation}
  \int {{dt} \over L} = \int {{dI} \over {V-IR}}
\end{equation}

\noindent If the switch is closed at $t=0$, that is $I(0) = 0$, we
have

\begin{equation}
  {t \over L} = \int_0^I {{dI'} \over {V-I'R}}
\end{equation}

\noindent whose solution is

\begin{equation}
  I = {V \over R} (1 - e^{-tR/L})
\end{equation}

\noindent Some comments on this result for the "L-R" circuit:\\

\noindent 1.  As $t \rightarrow \infty$ the current approaches $V/R$,
which is the result without any self-inductance.\\

\noindent 2.  The "time constant" $L/R$ is called the LR time constant.
It is the time it takes for the current to reach $(1 - e^{-1})$ ($63 \%$)
of its final value.\\

\noindent {\it L-C Resonance Circuit}\\

An important application is a circuit with an inductor and a capacitor in
the same circuit.  We will consider here only a simple circuit with a
capacitor and inductor in series.  Let the capacitor have a capacitance
of $C$, and the inductor a self-inductance of $L$.  Let $\pm Q$ be the
charge on the plates of the capacitor.  Applying Kirchoff's modified
voltage equation around the circuit gives:

\begin{equation}
  V_c = {{d \Phi_m} \over {dt}}
\end{equation}

\noindent where $V_c$ is the voltage across the capacitor.  If $\pm Q$
are the charges on each plate of the capacitor, $V_c = Q/C$.  If the
self-inductance of the inductor (i.e. the circuit) is $L$, we have

\begin{equation}
  {Q \over C} = L {{dI} \over {dt}}
\end{equation}

\noindent The current is the rate of change of the charge on the capacitor.
If $I$ flows from the positive plate of the capacitor, then $I = - (dQ)/(dt)$.
Differentiating both sides with respect to $t$:

\begin{equation}
 {1 \over C} {{dQ} \over {dt}} = L {{d^2I} \over {dt^2}}
\end{equation}

\noindent Substituting in for $(dQ)/(dt)$ gives

\begin{equation}
  - {I \over C} = L {{d^2I} \over {dt^2}}
\end{equation}

\noindent Rearrainging terms yields the following differential equation:

\begin{equation}
  {{d^2 I} \over {dt^2}} = - {I \over {LC}}
\end{equation}

The solution $I(t)$ is a function whose second derivative is minus itself.
Functions with this property are sinusoidal functions: $I(t) = A sin(\omega t)$.
Since $(d^2 I)/(d t^2) = - \omega^2 A sin(\omega t)$ we see that

\begin{equation}
  \omega = {1 \over {\sqrt{LC}}}
\end{equation}

This is an interesting result.  The current in the L-C circuit oscillates with a frequency
of $f = \omega/(2 \pi) = 1/(2 \pi \sqrt{LC} )$.  This frequency is the resonance frequency
of the circuit.  If an inductor and capacitor are in series in an antenna, signals 
at the resonant frequency have a large amplitude.  By varying $C$ one can tune into
ones favorite radio or TV station.\\

\bigskip

There are many applications of inductors in circuits: producing sparks in 
car motors, etc, and we will discuss some of these in
lecture.  Lets close this section on the magnetic interaction by summarizing
the main points.

\bigskip

\centerline{\bf Summary}

\bigskip

The nature of the magnetic interaction, is quite complicated.  This is because both 
the source and force both depend on the motion of charged particles.  The ideas we have
discussed the past 4 weeks can be broken down to 3 main features:
sources of magnetic fields, the magnetic force on a particle moving in
a magnetic field, and the effect of changing magnetic fields.\\

\noindent {\bf Sources of Magnetic Fields}\\

{\it Moving charges produce magnetic fields}.  The magnetic field produced by a small piece
of wire segment that carries a current $I$ is described by the Biot-Savart law:
$\Delta \vec{B} = ((\mu_0 I) / (4 \pi))(\vec{\Delta l} \times \hat{r})/r^2$.  Ampere's
Law deals with the closed path integral of $\vec{B} \cdot \vec{dl}$:
$\oint \vec{B} \cdot \vec{dl} = \mu_0 I_{through \; path}$. \\

\noindent {\bf Magnetic Force}\\

{\it Particles with charge that are moving in a magnetic field feel a force}.  
The force the particle feels is perpendicular to both the velocity vector
$\vec{v}$ and the magnetic field $\vec{B}$:  $F_m = q \vec{v} \times \vec{B}$.
Applying this force law to the charges moving in a straight wire gives: 
$F_m = I (\vec{l} \times \vec{B})$, where $\vec{l}$ points along the wire
segment in the direction of the current and whose magnitude equals the length
of the wire.\\

\noindent {\bf Changing Magnetic Fields}\\

{\it Changing magnetic fields produce electric fields}.
The magnetic interaction depends on the velocity of both the source and the particle
experiencing the force.  Since velocity is a {\it change} in position, a changing
magnetic (or electric) field should cause some sort of effect.  Faraday 
formulated this effect for changing magnetic fields most succinctly: 

\begin{equation}
  \oint_{wire} {\vec{F} \over q} \cdot \vec{dl} = - {{d\Phi_m} \over {dt}}
\end{equation}

\noindent Or in words: A changing magnetic flux through a wire loop produces 
a "voltage" around the wire loop.

\bigskip

\centerline{\bf Final Thoughts}

There is still one piece missing in the complete picture of the classical
electromagnetic interaction.  If you continue with Phy234 you will cover
the contribution of Maxwell.  He realized that the equations we have so far
are inconsistent with charge conservation, and that a changing electric
field must produce some sort of magnetic interaction.  We state here the
complete set of source equations (in integral form) for the electric and magnetic
fields:

\begin{eqnarray*}
 \oint \oint \vec{E} \cdot \vec{dA} & = & {{Q_{net}} \over \epsilon_0} \; \; (Coulomb)\\
  \oint \oint \vec{B} \cdot \vec{dA} & = & 0  \\
 \oint \vec{E} \cdot \vec{dr} & = & - {{d \Phi_B} \over {dt}} \; \;  (Faraday)\\
  \oint \vec{B} \cdot \vec{dr} & = & \mu_0 I_{net} + \epsilon_0 \mu_0 {{d \Phi_E} \over {dt}}
        \; \; (Ampere + Maxwell) \\
\end{eqnarray*}

These equations are refered to as "Maxwell's Equations" for the classical electrodynamic
fields.  The force equation $F = q \vec{E} + q (\vec{v} \times \vec{B})$ relates the force
to the fields.  These ideas were developed between 1800 and 1864, and led to a radical
change in the lives of humans.  The electric generator and transformer were developed
using Faraday's law, which allowed engineers to bring energy into our homes.  People
could cook food in their kitchens from snow melting high in the mountains.  Electric
motors turned because of the physics contained in these equations.  Maxwell's contribution
connected the electric-magnetic interaction with light (or more generally electromagnetic
radiation).  Radio and television were made possible from an understanding of these
equations.  The discovery of the laws of classical electrodynamics was one of the
most significant achievements of humankind in the 1800's.

The interaction of atoms and molecules with each other are almost entirely electric and 
magnetic.  Thus, all chemical and biological properties derive at their most 
basic level from the laws of electrodynamics.  When you raise your arm, digest your
food, or use gasoline to move your car, the electromagnetic force is at work.
Basically, thinking is electromagnetic in nature.  So when you take your final
exam on the laws of electricity and magnetism, you will be using these same
laws to solve the questions.  As far as I know, the electomagnetic interaction
is the only one which can be used to understand itself.  Think about that!


\end{document}