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Chapter 33 Homework
Due: 8:00am on Wednesday, April 7, 2010
Note: To understand how points are awarded, read your instructor's Grading Policy.
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Canceling a Magnetic Field
Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 46.0
through the wires are 8.0 , 20.0 , 10.0 , and .
, as shown in the figure . The currents running
Part A
Find the magnitude of the current
Hint A.1
that will make the magnetic field at the center of the square equal to zero.
How to approach the problem
Find the magnetic field at the center of the square due to the wires whose current you know. Then, find the current
whose contribution to the magnetic field will exactly cancel the contribution of the other three
wires.
Hint A.2
Calculating the contribution from the three known wires
What is the magnitude
of the magnetic field at the center of the square due to the wires carrying the 8.0-, 20-, and 10-
currents? Be careful with signs when you add the contributions from the three different
wires.
Hint A.2.1
Ampère's law
Recall Ampère's law:
.
You can use this to determine the formula for the magnetic field generated by a long wire. Use a circle centered on the wire as your path of integration.
Hint A.2.2
Getting your signs correct
Recall the right-hand rule: If your thumb, on your right hand, points in the direction in which the current is flowing, your fingers will curl in the direction of the magnetic field.
Express your answer in teslas to three significant figures.
ANSWER:
−6
= 1.74×10
Answer Requested
Good. You should have derived the following equation for the contribution to the magnetic field from one wire:
,
where
is width of the square. Use this formula and the fact that you want the magnetic field at the center to sum to zero to find the current .
Express your answer in amperes.
ANSWER:
= 2.00
Correct
Part B
What is the direction of the current ?
Hint B.1
How to approach the problem
Hint not displayed
ANSWER:
upward
downward
Correct
Magnetic Field due to Semicircular Wires
A loop of wire is in the shape of two concentric semicircles as shown.
The inner circle has radius ; the outer circle has radius . A current
flows clockwise through the outer wire and counterclockwise through the inner wire.
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Part A
What is the magnitude,
Hint A.1
, of the magnetic field at the center of the semicircles?
What physical principle to use
Hint not displayed
Hint A.2
Compute the field due to the inner semicircle
Hint not displayed
Hint A.3
Direction of the field due to the inner semicircle
Hint not displayed
Hint A.4
Compute the field due to the straight wire segments
Hint not displayed
Express
in terms of any or all of the following: , , , and
.
ANSWER:
=
Correct
To see whether
and
makes sense, think of the scaling of different quantities. The size of the current element scales as the radius, whereas the power of
in the denominator is 2 (and equals
the radius also, in this case). So over all, you would expect the magnetic field to scale as 1/radius. Note that such an argument works only because the field due to each point is in the same direction, so you are
doing a much simpler integral.
Part B
What is the direction of the magnetic field at the center of the semicircles?
ANSWER:
into the screen
out of the screen
Correct
Problem 33.11
The magnetic field at the center of a 0.600-cm-diameter loop is 2.50
.
Part A
What is the current in the loop?
ANSWER:
11.9
A
Correct
Part B
A long straight wire carries the same current you found in part a. At what distance from the wire is the magnetic field 2.50
ANSWER:
?
9.55×10−4 m
Correct
Problem 33.16
A 99.0
current circulates around a 1.60-mm-diameter superconducting ring.
Part A
What is the ring's magnetic dipole moment?
ANSWER:
1.99×10−4
Correct
Part B
What is the on-axis magnetic field strength 5.40
ANSWER:
from the ring?
2.53×10−7 T
Correct
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Problem 33.40
A small bar magnet experiences a 2.10×10−2
torque when the axis of the magnet is at
to a 0.150
magnetic field.
Part A
What is the magnitude of its magnetic dipole moment?
ANSWER:
0.198
Correct
Problem 33.25
Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 1.80-mm-diameter superconducting wire.
Part A
What current is needed?
ANSWER:
2150
A
Correct
Magnetic Force on Charged Particles Conceptual Question
For each of the situations below, a charged particle enters a region of uniform magnetic field. Determine the direction of the force on each charge due to the magnetic field.
Part A
Determine the direction of the force on the charge due to the magnetic field.
Hint A.1
Determining the direction of a magnetic force
Hint not displayed
ANSWER:
points into the page.
points out of the page.
points neither into nor out of the page and
.
.
Correct
Part B
Determine the direction of the force on the charge due to the magnetic field.
Hint B.1
Determining the direction of a magnetic force
Hint not displayed
ANSWER:
points out of the page.
points into the page.
points neither into nor out of the page and
.
.
Correct
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Part C
Determine the direction of the force on the charge due to the magnetic field. Note that the charge is negative.
Hint C.1
Effect of a magnetic field on a negative charge
Hint not displayed
ANSWER:
points out of the page.
points into the page.
points neither into nor out of the page and
.
.
Correct
Problem 33.26
A proton moves in the magnetic field
with a speed of
in the directions shown in the figure. For each, what is magnetic force
on the proton?
Part A
Express vector
ANSWER:
in the form
,
,
, where the x, y, and z components are separated by commas.
−13
= 0,5.66×10
Correct
,0 N
Part B
Express vector
ANSWER:
in the form
,
,
, where the x, y, and z components are separated by commas.
= 0,0,0
N
Correct
Problem 33.30
The aurora is caused when electrons and protons, moving in the earth's magnetic field of
, collide with molecules of the atmosphere and cause them to glow.
Part A
What is the radius of the cyclotron orbit for an electron with speed
ANSWER:
?
0.114
m
Correct
Part B
What is the radius of the cyclotron orbit for a proton with speed
ANSWER:
10.4
m
Correct
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Magnetic Force on a Current-Carrying Wire
Learning Goal: To understand the magnetic force on a straight current-carrying wire in a uniform magnetic field.
Magnetic fields exert forces on moving charged particles, whether those charges are moving independently or are confined to a current-carrying wire. The magnetic force
depends on its velocity
on an individual moving charged particle
and charge . In the case of a current-carrying wire, many charged particles are simultaneously in motion, so the magnetic force depends on the total current
The size of the magnetic force on a straight wire of length
carrying current
in a uniform magnetic field with strength
and the length of the wire
.
is
.
Here
is the angle between the direction of the current (along the wire) and the direction of the magnetic field. Hence
this equation can also be written as
refers to the component of the magnetic field that is perpendicular to the wire,
. Thus
.
The direction of the magnetic force on the wire can be described using a "right-hand rule." This will be discussed after Part B.
Part A
Consider a wire of length
= 0.30
that runs north-south on a horizontal surface. There is a current of
shown.) The Earth's magnetic field at this location has a magnitude of 0.50
= 0.50
(or, in SI units,
flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not
) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of
the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that
.
Express your answer in newtons to two significant figures.
ANSWER:
4.6×10−6
Correct
Because the Earth's magnetic field is quite modest, this force is so small that it might be hard to detect.
Part B
Now assume that a strong, uniform magnetic field of size 0.55
pointing straight down is applied. What is the size of the magnetic force on the wire due to this applied magnetic field? Ignore the effect of the Earth's
magnetic field.
Hint B.1
Determining the angle theta
Hint not displayed
Express your answer in newtons to two significant figures.
ANSWER:
8.3×10−2
Correct
This force would be noticeable if the wire were of light weight.
The direction of the magnetic force is perpendicular to both the direction of the current flow and the direction of the magnetic field. Here is a "right-hand rule" to help you determine the direction of the magnetic force.
1. Straighten the fingers of your right hand and point them in the direction of the current.
2. Rotate your arm until you can bend your fingers to point in the direction of the magnetic field.
3. Your thumb now points in the direction of the magnetic force acting on the wire.
Part C
What is the direction of the magnetic force acting on the wire in Part B due to the applied magnetic field?
ANSWER:
due north
due south
due east
due west
straight up
straight down
Correct
Part D
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Which of the following situations would result in a magnetic force on the wire that points due north?
Check all that apply.
ANSWER:
Current in the wire flows straight down; the magnetic field points due west.
Current in the wire flows straight up; the magnetic field points due east.
Current in the wire flows due east; the magnetic field points straight down.
Current in the wire flows due west and slightly up; the magnetic field points due east.
Current in the wire flows due west and slightly down; the magnetic field points straight down.
Correct
As you can see, many current/magnetic field configurations can result in the same direction of magnetic force.
Part E
Assume that the applied magnetic field of size 0.55
Hint E.1
Recall that
is rotated so that it points horizontally due south. What is the size of the magnetic force on the wire due to the applied magnetic field now?
Determining the angle theta
is the angle between the magnetic field direction and the direction of the current flow in the wire. In this situation, what is the value of ?
ANSWER:
0 degrees
38 degrees
90 degrees
180 degrees
Correct
Express your answer in newtons to two significant figures.
ANSWER:
0
Correct
Notice that whenever the current in the wire and the magnetic field point in the same direction (
) or in opposite directions (
), the sine of
is zero, so there is no magnetic force exerted on the wire.
This is consistent with the earlier statement that it is the component of the magnetic field that is perpendicular to the direction of the current that produces the magnetic force.
Also notice that for these two special values of
(when the current is flowing parallel to or antiparallel to the magnetic field) the steps listed for the right-hand rule suggest a unique direction for the magnetic
force. This is another clue that the magnetic force is zero.
Problem 33.36
Part A
What is the force on the first wire in the figure?
ANSWER:
Correct
Part B
What is the force on the second wire in the figure?
ANSWER:
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Correct
Part C
What is the force on the third wire in the figure?
ANSWER:
Correct
Rail Gun
A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass
and electrical resistance rests on parallel horizontal rails (that have negligible electric resistance), which are a distance apart. The rails are also
connected to a voltage source
, so a current loop is formed.
The rod begins to move if the externally applied vertical magnetic field in which the rod is located reaches the value
flattened bottom so that it slides instead of rolling. Use
. Assume that the rod has a slightly
for the magnitude of the acceleration due to gravity.
Part A
Find
, the coefficient of static friction between the rod and the rails.
Hint A.1
How to approach this problem
Hint not displayed
Hint A.2
Force due to the magnetic field
Hint not displayed
Hint A.3
Frictional force
Hint not displayed
Express the coefficient of static friction in terms of variables given in the introduction.
ANSWER:
=
Correct
Score Summary:
Your score on this assignment is 99.2%.
You received 79.43 out of a possible total of 85 points, plus 4.85 points of extra credit.
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