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MP_Assignment#2
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Due: 8:00am on Monday, February 14, 2011
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy
Faraday's Law and Induced Emf
Description: Discusses Faraday's law; presents a sequence of questions related to finding the induced emf
under different circumstances.
Learning Goal: To understand the terms in Faraday's law and to be able to identify the magnitude and
direction of induced emf.
Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.
Mathematically, it can be written as
,
where
is the emf induced in a closed loop, and
is the rate of change of the magnetic flux through a surface bounded by the loop. For uniform magnetic
fields the magnetic flux is given by
, where is the angle between the magnetic
field
and the normal to the surface of area
.
To find the direction of the induced emf, one can use Lenz's law:
The induced current's magnetic field opposes the change in the magnetic flux that induced the current.
For example, if the magnetic flux through a loop increases, the induced magnetic field is directed opposite to
the "parent" magnetic field, thus countering the increase in flux. If the flux decreases, the induced current's
magnetic field has the same direction as the parent magnetic field, thus countering the decrease in flux.
Recall that to relate the direction of the electric current and its magnetic field, you can use the right - hand
rule: When the fingers on your right hand are curled in the direction of the current in a loop, your thumb
gives the direction of the magnetic field generated by this current.
In this problem, we will consider a rectangular loop of wire with sides
uniform magnetic field
and
placed in a region where a
exists (see the diagram). The resistance of the loop is
.
Initially, the field is perpendicular to the plane of the
loop and is directed out of the page. The loop can
rotate about either the vertical or horizontal axis,
passing through the midpoints of the opposite sides,
as shown.
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Part A
Which of the following changes would induce an electromotive force (emf) in the loop ? When you consider
each option, assume that no other changes occur.
Check all that apply.
ANSWER:
The magnitude of
increases.
The magnitude of
decreases.
The loop rotates about the vertical axis (vertical dotted line) shown in the
diagram.
The loop rotates about the horizontal axis (horizontal dotted line) shown in the
diagram.
The loop moves to the right while remaining in the plane of the page.
The loop moves toward you, out of the page, while remaining parallel to itself.
Part B
Find the flux
through the loop.
Express your answer in terms of
ANSWER:
,
, and
.
=
Part C
If the magnetic field steadily decreases from
to zero during a time interval , what is the magnitude
of
the induced emf?
Hint C.1
Find the change in magnetic flux
What is the total change in magnetic flux
Express your answer in terms of
ANSWER:
,
, and
during this time interval?
.
=
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Express your answer in terms of
,
,
, and .
ANSWER:
=
Part D
If the magnetic field steadily decreases from
to zero during a time interval , what is the magnitude
of
the induced current ?
Express your answer in terms of
,
,
, , and the resistance
of the wire.
ANSWER:
=
Part E
If the magnetic field steadily decreases from
to zero during a time interval , what is the direction of the
induced current ?
ANSWER:
clockwise
counterclockwise
The flux decreases, so the induced magnetic field must be in the same direction as the original
(parent) magnetic field. Therefore, the induced magnetic field is out of the page. Using the right - hand
rule, we deduce that the direction of the current is counterclockwise.
Part F
Which of the following changes would result in a clockwise emf in the loop ? When you consider each
option, assume that no other changes occur.
Check all that apply.
ANSWER:
The magnitude of
increases.
The magnitude of
decreases.
The loop rotates through 45 degrees about the vertical axis (vertical dotted
line) shown in the diagram.
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The loop rotates through 45 degrees about the horizontal axis (horizontal
dotted line) shown in the diagram.
The loop moves to the right while remaining in the plane of the page.
The loop moves toward you, out of the page, while remaining parallel to itself.
Clockwise emf implies that the induced magnetic field is directed into the page. Therefore, the
magnetic flux of the original field must be increasing. Only the first option corresponds to increasing
flux.
Oscillations in an LC circuit.
Description: Derive a differential equation for the current in an LC circuit, guess the solution, and answer
conceptual questions about the solution.
Learning Goal: To understand the processes in a series circuit containing only an inductor and a
capacitor.
Consider the circuit shown in the figure.
This circuit contains a capacitor of capacitance
an inductor of inductance
and
. The resistance of all
wires is considered negligible.
Initially, the switch is open, and the capacitor has a charge . The switch is then closed, and the changes
in the system are observed. It turns out that the equation describing the subsequent changes in charge,
current, and voltage is very similar to that of simple harmonic motion, studied in mechanics. To obtain this
equation, we will use the law of conservation of energy.
Initially, the entire energy of the system is stored in the capacitor. When the circuit is closed, the capacitor
begins to discharge through the inductor. As the charge of the capacitor decreases, so does its energy. On
the other hand, as the current through the inductor increases, so does the energy stored in the inductor.
Assuming no heat loss and no emission of electromagnetic waves, energy is conserved, and at any point in
time, the sum of the energy stored in the capacitor
and the energy stored in the inductor
is a
constant
:
,
where
is the charge on the capacitor and
is the current through the inductor (
and
are functions of
time, of course). For this problem, take clockwise current to be positive.
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Part A
Using the expression for the total energy of this system, it is possible to show that after the switch is
closed,
,
where
is a constant. Find the value of the constant
Hint A.1
Since
.
The derivative of the total energy
is a constant,
an expression for
. Differentiate the expression for
in the problem introduction to obtain
.
Hint A.1.1 A helpful derivative formula
Recall that for any differentiable function
, we have
. This result is derived
using the chain rule.
ANSWER:
=
Hint A.2
The current in a series circuit
In this one- loop circuit, the current through the capacitor is always equal to that through the inductor.
Hint A.3
Charge and current
Current is defined as the time derivative of charge, but watch the sign. With the conventions adopted
here (clockwise current and a positive charge
on the upper plate of the capacitor)
.
Therefore,
.
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Express your answer in terms of
and
.
ANSWER:
=
This expression can also be obtained by substituting
into the equation obtained from
Kirchhoff's loop rule. It is always good to solve a problem in more than one way!
Part B
From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate,
,
such motion is called simple harmonic motion , and the coordinate depends on time as
, where , the argument of the harmonic function at
, is called the phase
constant .
Find a similar expression for the charge
correct value of
Hint B.1
on the capacitor in this circuit. Do not forget to determine the
based on the initial conditions described in the problem.
The phase constant
Initially, the charge of the capacitor is at a maximum; hence, the argument of the cosine function must
be zero at the moment the switch is closed,
.
Hint B.2
Find the period of the oscillations
What is the period
of these oscillations ?
Hint B.2.1 A relation between the period and angular frequency
The period
and the angular frequency
are related by
.
Express your answer in terms of
,
, and other variables given in the introduction.
ANSWER:
=
Express your answer in terms of
,
, and
. Use the cosine function in your answer.
ANSWER:
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ANSWER:
=
Part C
What is the current
Hint C.1
in the circuit at time
after the switch is closed ?
Current as a derivative of the charge
Note that
. Use the equation you have found for
Express your answer in terms of
,
,
to find
.
, and other variables given in the introduction.
ANSWER:
=
Part D
Recall that the top plate of the capacitor is positively charged at
. In what direction does the current
in the circuit begin to flow immediately after the switch is closed ?
ANSWER:
There is no current because there cannot be any current through the
capacitor.
clockwise
counterclockwise
Part E
Immediately after the switch is closed, what is the direction of the EMF in the inductor ? (Recall that the
direction of the EMF refers to the direction of the back - current or the induced electric field in the inductor.)
ANSWER:
There is no EMF until the current reaches its maximum.
clockwise
counterclockwise
In the remaining parts, assume that the period of oscillations is 8.0 milliseconds. Also, keep in mind that
the top plate of the capacitor is positively charged at
.
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Part F
At what time does the current reach its maximum value for the first time ?
Hint F.1
How to approach the problem
Try drawing a graph of the current as a function of time, in units of
, where
is the period of the
current oscillation.
Graph of
Hint F.2
The figure shows a graph of current
ANSWER:
as a function of time .
0.0 ms
2.0 ms
4.0 ms
6.0 ms
8.0 ms
Part G
At what moment does the EMF become zero for the first time ?
Hint G.1
Relationship between the induced EMF and the current.
The EMF is proportional to the rate of change of the current; the EMF becomes zero when
Hint G.2
.
Graph of
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ANSWER:
0.0
2.0
4.0
6.0
8.0
Part H
At what moment does the current reverse direction for the first time ?
Hint H.1
How to approach the problem
Try drawing a graph of the current as a function of time, in units of
, where
is the period of the
current oscillation.
Hint H.2
Graph of
The figure shows a graph of current as a function of time.
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0.0 ms
2.0 ms
4.0 ms
6.0 ms
8.0 ms
Part I
At what moment does the EMF reverse direction for the first time ?
Hint I.1
Relationship between the induced EMF and the current.
The EMF is proportional to the rate of change of the current. Therefore, the EMF will change direction
when
Hint I.2
ANSWER:
changes sign.
Graph of
0.0 ms
2.0 ms
4.0 ms
6.0 ms
8.0 ms
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Part J
At what moment does the energy stored in the inductor reach its maximum for the first time ?
Hint J.1
Formula for the energy in an inductor
The energy
in an inductor
is given by
,
where
Hint J.2
is the current flowing through the inductor.
Graph of
The figure shows a graph of current as a function of time.
Hint J.3
Conservation of energy
By conservation of energy, the energy in the inductor will be at a maximum when the energy in the
capacitor is at a minimum. An examination of the function for the charge on the capacitor in Part B
reveals that this occurs when the charge on the capicitor is zero.
ANSWER:
0.0 ms
2.0 ms
4.0 ms
6.0 ms
8.0 ms
Part K
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At what time
Hint K.1
do the capacitor and the inductor possess the same amount of energy for the first time ?
Use conservation of energy
Since the total energy is conserved, the energy of the capacitor (like that of the inductor) equals onehalf of the total energy at the time in question.
Hint K.2
Energy and charge
Since the energy of the capacitor is proportional to the square of its charge, you need to find a time
when
.
Hint K.3
A useful trigonometric result
Recall that
.
Express your answer in milliseconds.
ANSWER:
=
Part L
What is the direction of the current in the circuit 22.0 milliseconds after the switch is closed ?
Hint L.1
Use the periodic nature of the process
The current would have the same direction as it has 6.0 milliseconds after the switch is closed. (Can
you see why ? )
Hint L.2
Graph of
The figure shows a graph of the current as a function of time.
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The current is zero.
clockwise
counterclockwise
Part M
What is the direction of the EMF in the circuit 42.0 milliseconds after the switch is closed ?
Hint M.1
ANSWER:
Graph of
The EMF is zero
clockwise
counterclockwise
Self - Inductance of a Solenoid
Description: Walks through calculation of self - inductance for a single solenoid with some discussion at the
end.
Learning Goal: To learn about self - inductance from the example of a long solenoid.
To explain self - inductance, it is helpful to consider the specific example of a long solenoid, as shown in the
figure. This solenoid has only one winding, and so the EMF induced by its changing current appears across
the solenoid itself. This contrasts with mutual inductance, where this voltage appears across a second coil
wound on the same cylinder as the first.
Assume that the solenoid has radius , length
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along the z axis, and is wound with
turns per unit
length so that the total number of turns is equal to
. Assume that the solenoid is much longer than its
radius.
As the current through the solenoid changes, the
resulting magnetic flux through the solenoid will also
change, and an electromotive force will be generated
across the solenoid according to Faraday's law of
induction:
.
Faraday's law implies the following relation between
the self - induced EMF across the solenoid and the
current passing through it:
.
The "direction of the EMF" is determined with respect to the direction of positive current flow, and represents
the direction of the induced electric field in the inductor. This is also the direction in which the "back - current"
that the inductor tries to generate will flow.
Part A
Suppose that the current in the solenoid is
magnetic field
due to this current ?
Express your answer in terms of
(such as
ANSWER:
. Within the solenoid, but far from its ends, what is the
, quantities given in the introduction, and relevant constants
).
=
Note that this field is independent of the radial position (the distance from the axis of symmetry) as
long as it is measured at a point well inside the solenoid.
Part B
What is the magnetic flux
through a single turn of the solenoid ?
Express your answer in terms of the magnetic field
, quantities given in the introduction, and
any needed constants.
ANSWER:
=
Part C
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Suppose that the current varies with time, so that
. Find the electromotive force
induced
across the entire solenoid due to the change in current through the entire solenoid.
Hint C.1
Find the flux in terms of the current
In Part B you found the flux
through a single turn of the solenoid. Now find the flux
through the entire solenoid.
Express your answer in terms of
constants such as
.
ANSWER:
Hint C.2
, other quantities given in the introduction, and various
=
Find the EMF for the entire solenoid
You now have an expression for the magnetic flux that passes through the solenoid. From this, you
should be able to derive an expression for the EMF in the solenoid. Suppose the total magnetic flux
through the solenoid is
. What is the electromotive force generated in the solenoid by the
changing flux
?
Express your answer in terms of
and its derivative, and other variables given in the
introduction.
ANSWER:
=
Hint C.3
Putting it together
You have now worked out three things:
the magnetic field from
;
the flux from this field;
the EMF for the entire solenoid.
Put them together and you have the answer!
Express your answer in terms of
,
,
, and
.
ANSWER:
=
Part D
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The self - inductance
is related to the self - induced EMF
for a long solenoid. (Hint: The self - inductance
by the equation
will always be a positive quantity.)
Express the self - inductance in terms of the number of turns per length
and
. Find
, the physical dimensions
, and relevant constants.
ANSWER:
=
This definition of the inductance is identical to another definition you may have encountered:
, where
is the magnetic flux due to a current in the inductor. To see the
correspondence you should differentiate both sides of this equation with respect to time and use
Faraday's law, i.e.,
.
Now consider an inductor as a circuit element. Since we are now treating the inductor as a circuit element,
we must discuss the voltage across it, not the EMF inside it. The important point is that the inductor is
assumed to have no resistance. This means that the net electric field inside it must be zero when it is
connected in a circuit. Otherwise, the current in it will become infinite. This means that the induced electric
field
deposits charges on and around the inductor in such a way as to produce a nearly equal and
opposite electric field
such that
fields produced by charges (like
! Kirchhoff's loop law defines voltages only in terms of
), not those produced by changing magnetic fields (like
). So if we
wish to continue to use Kirchhoff's loop law, we must continue to use this definition consistently. That is,
we must define the voltage
alone (note that the integral is from A to B
rather than from B to A, hence the positive sign). So finally,
, where we have used
of
and the definition
.
Part E
Which of the following statements is true about the inductor in the figure in the problem introduction,
where
is the current through the wire?
Hint E.1
A fundamental inductance formula
The self - inductance
is related to the voltage
across the inductor through the equation
. Note that unlike the EMF, it does not have a minus sign.
ANSWER:
If
is positive, the voltage at end A will necessarily be greater than that at
end B.
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If
is positive, the voltage at end A will necessarily be greater than that
at end B.
If
is positive, the voltage at end A will necessarily be less than that at end
B.
If
is positive, the voltage at end A will necessarily be less than that at
end B.
Part F
Now consider the effect that applying an additional voltage to the inductor will have on the current
already flowing through it (imagine that the voltage is applied to end A, while end B is grounded). Which
one of the following statements is true ?
Hint F.1
A fundamental inductance formula
The self - inductance
is related to the voltage
across the inductor through the equation
. Note that unlike the EMF, it does not have a minus sign. However, when applying
Kirchhoff's rules and traversing the inductor in the direction of current flow, there will be a term
, just as traversing a resistor gives a term
.
ANSWER:
If
is positive, then
will necessarily be positive and
will be
will necessarily be negative and
will be
negative.
If
is positive, then
negative.
If
is positive, then
could be positive or negative, while
will
necessarily be negative.
If
is positive, then
will necessarily be positive and
will be
positive.
If
is positive, then
could be positive or negative while
will
necessarily be positive.
If
is positive, then
will necessarily be negative and
will be
positive.
Note that when you apply Kirchhoff's rules and traverse the inductor in the direction of current flow,
you are interested in
, just as traversing a resistor gives a term
.
In sum: when an inductor is in a circuit and the current is changing, the changing magnetic field in
the inductor produces an electric field. This field opposes the change in current, but at the same time
deposits charge, producing yet another electric field. The net effect of these electric fields is that the
current changes, but not abruptly. The "direction of the EMF" refers to the direction of the first,
induced, electric field.
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Tactics Box 34.1 Using Lenz's Law
Description: Knight Tactics Box 34.1 Using Lenz's Law is illustrated.
Learning Goal: To practice Tactics Box 34.1 Using Lenz's Law.
Lenzs law is a useful rule for determining the direction of the induced current in a loop. Specifically, it says
that there is an induced current in a closed conducting loop if and only if the magnetic flux through the loop
is changing. The direction of the induced current is such that the induced magnetic field opposes the change
in the flux. The following tactics box summarizes the essential steps in using Lenz's law.
TACTICS BOX 34.1 Using Lenz's law
Determine the direction of the applied magnetic field. The field must pass through the loop.
Determine how the flux is changing. Is it increasing, decreasing, or staying the same?
Determine the direction of an induced magnetic field that will oppose the change in the flux:
Increasing flux: The induced magnetic field points opposite the applied magnetic field.
Decreasing flux: The induced magnetic field points in the same direction as the applied magnetic field.
Steady flux: There is no induced magnetic field.
Determine the direction of the induced current. Use the right - hand rule to determine the current direction in
the loop that generates the induced magnetic field you found in Step 3.
Follow the steps above to solve the following problem: A wire carries a current in the direction indicated in
the figure. A loop is located next to the wire. If the current in the wire increases, is there a clockwise
current around the loop, a counterclockwise current, or no current ?
Part A
Determine the direction of the magnetic field of the current - carrying wire.
Hint A.1
The right- hand rule for fields
Place your right thumb in the direction of the current.
Curl your fingers around the wire to indicate a circle.
Your fingers point in the direction of the magnetic field lines around the wire.
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ANSWER:
upward
downward
into the plane of the screen
out of the plane of the screen
Part B
As the current in the wire increases, the magnetic field around the wire increases in magnitude without
changing its direction. Does this affect the flux through the loop, and if so, how?
ANSWER:
As the magnetic field of the wire increases, the flux through the loop
increases as well.
As the magnetic field of the wire increases, the flux through the loop
decreases.
As the magnetic field of the wire increases, the flux through the loop remains
steady.
Now, follow Step 3 in the tactics box above and find the direction of the induced magnetic field.
Part C
Based on the information found in the previous parts, complete the statement below.
Hint C.1
The right- hand rule for fields
Place your right thumb in the direction of the current.
Curl your fingers around the wire to indicate a circle.
Your fingers point in the direction of the magnetic field lines around the wire.
ANSWER:
There is
a counterclockwise current
induced in the loop.
Self - Inductance of a Coaxial Cable
Description: Find the self - inductance per unit length of a coaxial cable.
A coaxial cable consists of alternating coaxial cylinders of conducting and insulating material. Coaxial cabling
is the primary type of cabling used by the cable television industry and is also widely used for computer
networks such as Ethernet, on account of its superior ability to transmit large volumes of electrical signal with
minimum distortion. Like all other kinds of cables, however, coaxial cables also have some self - inductance
that has undesirable effects, such as producing some distortion and heating.
Part A
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Consider a long coaxial cable made of two coaxial cylindrical conductors that carry equal currents
opposite directions (see figure). The inner cylinder is a small solid conductor of radius
in
. The outer
cylinder is a thin walled conductor of outer radius , electrically insulated from the inner conductor.
Calculate the self - inductance per unit length
cable and
of this coaxial cable. (
is the inductance of part of the
is the length of that part.) Due to what
is known as the "skin effect", the current
flows
down the (outer) surface of the inner conducting
cylinder and back along the outer surface of the
outer conducting cylinder. However, you may ignore
the thickness of the outer cylinder.
Hint A.1
How to approach the problem
The self - inductance of the cable can be found if both the current and the magnetic flux through the
cable are known. Using Ampère's law, you can calculate the magnetic field due to the cable; you will
find that there is a magnetic field in the region between the two conductors. Thus, calculate the
magnetic flux per unit length through the region between the two cylinders. Since the current in the
cable is given, now you should have enough information to calculate the self - inductance of the cable.
Hint A.2
Find the magnetic field inside the inner cylinder
Consider the inner conducting cylinder. It carries a current
magnitude of the magnetic field
at a distance
along its outer surface. What is the
from the axis of the cylinder, in the region inside the
cylinder?
Hint A.2.1 Ampère's law
Ampère's law states that the line integral of the magnetic field
calculated around any closed path is
given by
,
where
is the permeability of free space, and
is the net current through the area enclosed by
the path. The best loop to use is a circle centered on the cylinders' axis, because the magnetic field
will be tangent to the loop at all points, making the dot product easier to compute.
Express your answer in terms of some or all of the variables
,
, and
, the permeability of
free space.
ANSWER:
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ANSWER:
=
Hint A.3 Find the magnetic field between the cylinders
Now consider both conducting cylinders, each carring equal currents
magnitude of the magnetic field
at a distance
in opposite directions. Find the
from the axis of the cable in the region between the
two cylinders.
Hint A.3.1 Ampère's law
Ampère's law states that the line integral of the magnetic field
calculated around any closed path is
given by
,
where
is the permeability of free space, and
is the net current through the area enclosed by
the path. The best loop to use is a circle centered on the cylinders' axis, because the magnetic field
will be tangent to the loop at all points, making the dot product easier to compute.
Express your answer in terms of some or all of the variables
,
, and
, the permeability of
free space.
ANSWER:
=
Between the conducting cylinders, the magnetic field is identical to one that would be produced by
a wire carrying current located along the cylinders' axis.
Hint A.4 Find the magnetic field outside the cable
Finally, find the magnitude of the magnetic field
at a distance
region outside the cable. Again, assume that a current
from the axis of the cable in the
flows along the surface of the inner cylinder
and flows back on the surface of the outer cylinder.
Hint A.4.1 Ampère's law
Ampère's law states that the line integral of the magnetic field
calculated around any closed path is
given by
,
where
is the permeability of free space, and
is the net current through the area enclosed by
the path. The best loop to use is a circle centered on the cylinders' axis, because the magnetic field
will be tangent to the loop at all points, making the dot product easier to compute.
http://session.masteringphysics.com/myct/assignments
Page 21 of 23
Assignments
2/15/11 10:29 AM
Express your answer in terms of some or all of the variables
,
, and
, the permeability of
free space.
ANSWER:
=
The fields due to the opposite currents cancel each other out.
Hint A.5
Find the magnetic flux in the cable
Find the magnetic flux per unit length
inner cylinder and
in the cable due to the currents
. Let
be the radius of the
that of the outer cylinder.
Hint A.5.1 How to approach the problem
The magnetic field of the coaxial cable exists only in the region between the two cylinders. Therefore,
you need to calculate the magnetic flux per unit length through this portion of the cable. To do that,
think about the direction of the magnetic field (using the right - hand rule) and what area element to
use.
Hint A.5.2
How to calculate the magnetic flux
Given a finite area
in a magnetic field
, the magnetic flux
through the area is defined by the
following integral calculated over the area,
,
where
is an infinitesimal area element.
The magnetic field due to the cable exists only in the region between the cylinders and it is
perpendicular to any plane containing the cable axis. Therefore, to calculate the magnetic flux it is
most convenient to choose an area element of length parallel to the cable axis, and of width
lying in the plane containing the axis. The figure below shows such an area element with the cross
section of the cable parallel to the cable axis depicted.
Such an area element is perpendicular to the
magnetic field, making the dot product easy.
http://session.masteringphysics.com/myct/assignments
Page 22 of 23
Assignments
2/15/11 10:29 AM
Express your answer in terms of some or all the variables
,
, , and
, the permeability of
free space.
ANSWER:
=
Hint A.6
Formula for self - inductance
The self inductance
of a circuit that carries a current
is given by
,
where
is the magnetic flux through the closed loop of the circuit.
Express your answer in terms of some or all the variables
,
, , and
, the permeability of free
space.
ANSWER:
=
The capacitance per unit length
, where
of such a coaxial cable is
. So the product
is the speed of light in vacuum! As you might have guessed, this is
not a coincidence, but a result that is quite generally true for such systems.
Score Summary:
Your score on this assignment is 0%.
You received 0 out of a possible total of 5 points.
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Page 23 of 23
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