Sources%20of%20the%20Magnetic%20Field

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Sources of the Magnetic Field
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Magnetic Field of a Point Charge
Given a point charge, q, we know that it
generates an electric field regardless of whether
it is moving or not
If the charge is in fact moving then the charge
also generates a magnetic field
The magnetic field generated by the charge
however does not behave in the same way the
electric field of the charge does
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Magnetic Field of a Point Charge
It is found that the magnetic field is
perpendicular to both the velocity of the charge
and the unit vector from the charge to the point
in question
The magnitude of the field is given by
where f is the angle between the velocity and
unit vector
The magnetic field lines form concentric circles
about the velocity vector
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Magnetic Field of a Point Charge
The direction of the magnetic field is given by
one of the right-hand rules
Point your thumb, of your right hand, in the
direction of the velocity and curl your fingers.
The direction of the field is then the direction
your fingers curl. The magnetic field for a
negative charge will be in the opposite direction.
The vector equation for this is
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Magnetic Field of a Current Element
If instead of a single charge, we have a current
what is the magnetic field then given by
We start with the principle of superposition we
add the magnetic fields of the individual
charges that make up the current
Given n moving charges per unit volume each
carrying a charge q, then the total charge in
this element is
where A is the cross sectional area
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Magnetic Field of a Current Element
We then have for dB
But from our work on currents we also have that
Therefore
Or in vector form
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Biot-Savart Law
For a complete circuit we then use the integral
form of the previous equation
This can be a very difficult integral to evaluate
The level of difficulty depends upon the variable
chosen to integrate over and the presence of any
symmetries that can be exploited
We will develop another, much simpler method,
shortly
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Field of a Straight Wire
We set up the problem as shown
The integral becomes
The result of this integral is
In the limit that a gtgt x
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Magnetic Force Between Two Current Carrying Wires
We have two wires carrying currents I and I
separated by a distance r
The first wire carrying current I sets up a
magnetic field at the second wire given by
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Magnetic Force Between Two Current Carrying Wires
From before we know that a wire carrying a
current that is placed in a magnetic field will
experience a force per unit length that is given
by
This is in fact what the second wire will
experience
And from
This force is directed towards the other wire
the force is attractive
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Magnetic Force Between Two Current Carrying Wires
If we had started with the wire carrying the
current I, we would have ended up with the same
result
If the wires were carrying currents in opposite
directions, then the force between the wires
would be repulsive
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Circular Current Loop
Another circuit for which it is worthwhile to
calculate the magnetic field is the circular
current loop
We start by utilizing the Biot-Savart law
with
a being the radius of the loop
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Circular Current Loop
A current element at the top will yield the
magnetic field as shown
A current element at the bottom will yield a
magnetic field of the same magnitude, but
directed downward
The vertical components will cancel and all that
will be left is a horizontal magnetic field of
incremental value
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Circular Current Loop
The integral is over just the circumference of
the loop as all other variables are constant
If there are N loops, turns, then the result is
just N times above results
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Circular Current Loop
An important point for the field of a current
loop is at the center of the loop where x 0
A current loop consisting of N turns is often
referred to as a Solenoid
A final caution These results for a current loop
only apply on the axis of the loop
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Amperes Law
Just as we developed an easy way to calculate the
electric field in symmetric situations, Gauss
Law, there is a similar technique we can use for
calculating magnetic fields
This is known as Amperes Law
where the integral is around a closed path and I
is the total current passing through the area
bounded by this path
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Amperes Law
There is one big difference to remember between
the two methods
Gauss Law is an integral over a closed surface
Whereas Amperes Law is an integral over a closed
path
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Amperes Law
Given a path around which the integration is to
be done, there is a sign convention for the
current
Determine the direction of the normal, as defined
by the sense of movement around the integration
path
If the current is going in the same direction as
the normal, the current is considered to be
positive otherwise it is negative
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Long Straight Wire
Current in same direction as normal of
integration path
This is the same as we derived from the
Biot-Savart Law
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Long Straight Wire
Current in opposite direction to normal of
integration path
Again this is the same as we derived from the
Biot-Savart Law
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Caution
Amperes Law gives you only the magnetic field
due to any currents that cut through the area
bounded by the integration path
For the indicated integration path, the result of
Amperes Law is zero
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Example
Given three currents and their associated loops
A B C Same
The currents that are used in Amperes Law are
the currents that are within the loop that is
used for integration
Loop A has no current going through it whereas
the other two loops do have currents going
through them, the integral for Loop A is zero
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Example
Given three currents and their associated loops
?
?
2) Now compare loops B and C. For which loop is ?
B dl the greatest?
B C Same
The right hand side of Amperes law is the same
for both loops, therefore the integral for both
loops is the same
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Example
A current I flows in an infinite straight wire in
the z direction as shown. A concentric infinite
cylinder of radius R carries current 2I in the -z
direction.
1) What is the magnetic field Bx(a) at point a,
just outside the cylinder as shown?
The loop used for application of Amperes Law
will be outside the cylinder
The total current going through this loop is the
sum of the two current yielding a current of I.
This gives a magnetic that points in the positive
x direction at a
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Example
A current I flows in an infinite straight wire in
the z direction as shown. A concentric infinite
cylinder of radius R carries current 2I in the -z
direction.
2) What is the magnetic field Bx(b) at point b,
just inside the cylinder as shown?
This time, the Ampere loop only encloses current
I which is in the z direction the loop is
inside the cylinder!
The current in the cylindrical shell does not
contribute to at point b.
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Long Cylindrical Conductor
The problem can be divided into two regions
Outside the conductor
Inside the conductor
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Long Cylindrical Conductor
Outside the conductor
We start with an integration path around the
conductor at an arbitrary distance r from the
central axis of the conductor
Then using Amperes Law
we then have
or
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Long Cylindrical Conductor
Inside the conductor
Since we have not been told otherwise, we assume
that the current is uniformly distributed across
the cross section of the conductor
We define a current density
We now apply Amperes Law around an integration
path centered on the cylindrical axis and radius
r lt R
with I being the current going through the area
defined by the integration path
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Long Cylindrical Conductor
Inside the conductor
and
So after equating, we then have
The magnetic field increases linearly with the
distance from the axis of the cylinder
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Long Cylindrical Conductor
Inside the wire (r lt a)
Outside the wire ( r gt a )
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Solenoid
We choose a path of integration as shown with one
part of the path inside the solenoid and one part
of the path outside
We assume that the path on the outside, cd, is
sufficiently far from the solenoid, compared to
the diameter of the solenoid, that the field is
approximately equal to zero
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Solenoid
We use Amperes Law
First the Left-Hand-Side
a?b Using the right hand rule for currents, the
B field is in the same direction as the path and
it is constant, we then have
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Solenoid
b?c Along the part of the path that is inside
the solenoid, B is perpendicular to dl, so that
contribution is zero, while for the part of the
path that is outside of the solenoid, B 0, so we
then have
The same is true for d?a
c?d Here B is zero so we again have
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Solenoid
Now for the Right-Hand-Side
The total current passing through the area
defined by the integration path is NI where N is
the number of turns and I is the current in the
loop, so the right-hand-side is
Now equating the left and right-hand-sides we have
Or solving for B
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Solenoid
The magnetic field of a solenoid looks like the
following with the exact shape of the field lines
being dependent upon the the number of turns, the
length, and the diameter of the solenoid