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Test 2Due: 11:59pm on Sunday, February 27, 2011

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

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Problem 28.62A pair of long, rigid metal rods, each of length , lie parallel to each other on a perfectly smooth table. Their

ends are connected by identical, very light conducting springs of force constant each(the figure ) and

negligible unstretched length. If a current runs

through this circuit, the springs will stretch.

Part A

At what separation will the rods remain at rest? Assume that is large enough so that the separation of

the rods will be much less than .

Express your answer in terms of the variables , , , and appropriate constants ( , and ).

ANSWER:

=

Correct

Test Your Understanding 28.2: Magnetic Field of a Current ElementAn infinitesimal current element located at the origin ( ) carries current in the negative

-direction.

Part A

Rank the following locations in order of the strength of the magnetic field that the current elementproduces at that location, from largest to smallest value.

MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrintView?assignme...

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ANSWER:

View All attempts used; correct answer displayed

The strength d of the magnetic field due to an infinitesimal current element of length carrying

current is

Here is the distance from the current element to the point at which the field is measured, and is

the angle between the direction of the current element and the direction of the vector from the

element to . In this situation points in the negative -direction, so .

, , : Here and . The vectors and are

perpendicular, so , sin , and we get

, , : Here and . The vectors and are

perpendicular, so , sin , and again we get

, , : Here and . The vectors and are opposite to

each other, so , sin , and d = 0.

, , : Here and . The vectors and

are perpendicular, so , sin , and we get

, , and . The vectors and

are at an angle , so sin = . We get

, , : Here and . The vectors and

are at an angle , so sin = . We again get

Exercise 28.22Two long, parallel transmission lines, apart, carry 23.0- and 77.0- currents.

Part A

Find all locations where the net magnetic field of the two wires is zero if these currents are in the samedirection.

Assume that the positive axis is directed from the 77.0- wire to the 23.0- wire perpendicular

to the wires, with the origin on the 77.0- wire.

ANSWER: = 32.3

Correct

Part B

Find all locations where the net magnetic field of the two wires is zero if these currents are in the oppositedirection.

Assume that the positive axis is directed from the 77.0- wire to the 23.0 wire perpendicular

to the wires, with the origin on the 77.0- wire.

ANSWER: =

Correct

Problem 28.55Two identical circular, wire loops 36.0 in diameter each carry a current of 1.20 in the same direction.

These loops are parallel to each other and are 26.0 apart. Line is normal to the plane of the loops andpasses through their centers. A proton is fired at 2050 perpendicular to line ab from a point midway

between the centers of the loops.

Part A

Find the magnitude of the magnetic force these loops exert on the proton just after it is fired.

ANSWER: 1.47×10Correct

Problem 27.67A straight piece of conducting wire with mass and length is placed on a frictionless incline tilted at an

angle from the horizontal (the figure ). There is a

uniform, vertical magnetic field at all points


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(produced by an arrangement of magnets not shown inthe figure). To keep the wire from sliding down theincline, a voltage source is attached to the ends of thewire. When just the right amount of current flowsthrough the wire, the wire remains at rest.

Part A

Determine the magnitude of the current in the wire that will cause the wire to remain at rest.

Express your answer in terms of the variables , , , , and appropriate constants.

ANSWER: =

Correct

Part B

Determine the direction of the current in the wire that will cause the wire to remain at rest.

ANSWER: The current in the wire must be directed from left to right.

The current in the wire must be directed from right to left.

Correct

Part C

In addition viewing the wire from its left-hand end, show in a free-body diagram all the forces that act on thewire.

Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. Theexact length of your vectors will not be graded but the relative length of one to the other will begraded.

ANSWER:

View All attempts used; correct answer displayed


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Problem 27.70A plastic circular loop of radius and a positive charge is distributed uniformly around the circumference of

the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angularspeed .

Part A

If the loop is in a region where there is a uniform magnetic field directed parallel to the plane of the loop,

calculate the magnitude of the magnetic torque on the loop.

Express your answer in terms of the variables , , , and .

ANSWER: =

Correct

Exercise 27.13An open plastic soda bottle with an opening diameter of 2.4 is placed on a table. A uniform 1.85-

magnetic field directed upward and oriented 22 from vertical encompasses the bottle.

Part A

What is the total magnetic flux through the plastic of the soda bottle?

Express your answer using two significant figures.

ANSWER: = −7.8×10−4

Correct

Exercise 27.19: Fusion ReactorIf two deuterium nuclei (charge , mass ) get close enough together, the attraction of the

strong nuclear force will fuse them to make an isotope of helium, releasing vast amounts of energy. The rangeof this force is about . This is the principle behind the fusion reactor. The deuterium nuclei are moving

much too fast to be contained by physical walls, so they are confined magnetically.

Part A

How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse?(Treat the nuclei as point charges, and assume that a separation of is required for fusion.)


ANSWER: = 8.4×106


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All attempts used; correct answer displayed

Part B

What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle ofdiameter 2.20 ?


ANSWER: = 0.16All attempts used; correct answer displayed

Problem 27.84 Derivation of equation tau=mu*B for a Circular Current LoopA wire ring lies in the xy-plane with its center at the origin. The ring carries a counterclockwise current (the

figure ). A uniform magnetic field is in the

-direction, (The result is easily extended to

in an arbitrary direction.) In the figure you can see

that the element .

Part A

Find .

Express your answer in terms of the variables , , , , , , , .

ANSWER: =

Correct

Part B

Integrate around the loop to find the net force.



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ANSWER: = 0Correct

Part C

From part A, find , where is the vector from the center of the loop to

the element . (Note that is perpendicular to


ANSWER: =

Correct

Part D

Integrate over the loop to find the total torque on the loop.


ANSWER: =

Correct

Part E

Show that the result can be written as , where . (Note:

, and .)

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:

Test Your Understanding 28.4: Magnetic Force Between Parallel ConductorsThree long, parallel wires each carry current in the directions shown in the figure. The separation between

adjacent wires is .


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Part A

What is the direction of the net magnetic force on each wire due to the other two wires?

ANSWER: wire #1 upward, wire #2 zero force, wire #3 downward

wire #1 upward, wire #2 zero force, wire #3 upward

wire #1 upward, wire #2 upward, wire #3 downward

wire #1 downward, wire #2 upward, wire #3 downward

wire #1 downward, wire #2 zero force, wire #3 downward

wire #1 upward, wire #2 downward, wire #3 upward

wire #1 downward, wire #2 zero force, wire #3 upward

wire #1 downward, wire #2 downward, wire #3 upward

Correct

The force per unit length on each of two long wires separated by a distance and carrying currents

and has magnitude

The force is attractive if the two currents are in the same direction and repulsive if the currents are inopposite directions. In this situation the current is the same in all three wires, so

Wire #1 feels an upward (repulsive) force from wire #2, for which , and a downward (attractive)

force from wire #3, for which . The net force is upward, since wire #2 is closer to wire #1 and so

exerts a greater force.Wire #2 feels a downward (repulsive) force from wire #1 and an upward (repulsive) force from wire#3. For each force , so the two forces on wire #2 have the same magnitude and hence cancel.

Thus, the net force on wire #2 is zero.Wire #3 feels an upward (attractive) force from wire #1, for which , and a downward (repulsive)

force from wire #2, for which . The net force is downward, since wire #2 is closer to wire #3 and

so exerts a greater force.

Test Your Understanding 28.5: Magnetic Field of a Circular Current LoopA circular loop of wire lies in a horizontal plane and carries a current as shown in the figure. A positively

charged particle moves upward along the axis of the loop, starting from a point below point (at the center of


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the loop).

Part A

What is the direction of the magnetic force that the loop exerts on the charged particle as it moves?

ANSWER: downward at all times during the particle's motion

zero at all times during the particle's motion

downward when the particle is below , upward when the particle is above

upward when the particle is below , downward when the particle is above

upward at all times during the particle's motion

none of the above

Correct

At points along the loop axis, the magnetic field due to the current in the loop is directed along the

axis. Since the velocity of the particle is also along this axis, the vector product equals zero.

Hence, the magnetic force on the particle, , is likewise zero.

Problem 28.65A circular wire loop of radius has turns and carries a current . A second loop with turns of radius

carries current and is located on the axis of the first loop, a distance from the center of the first loop.

The second loop is tipped so that its axis is at an angle from the axis of the first loop. The distance is

large compared to both and .

Part A

Find the magnitude of the torque exerted on the second loop by the first loop.


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ANSWER:

=

Correct

Part B

Find the magnitude of the potential energy for the second loop due to this interaction.

ANSWER: =

Correct

Part C

What simplifications result from having much larger than ? From having much larger than ?

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer:

Exercise 28.1A +6.00 point charge is moving at a constant 8.50×106 in the , relative to a

reference frame. At the instant when the point charge is at the origin of this reference frame, what is themagnetic-field vector it produces at the following points.

Part A

,

Enter your answers numerically separated by commas.

ANSWER: , , = 0,0,−2.04×10−5

Correct

Part B

, ,


ANSWER: , , = 0,0,0Correct


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Part C

, ,


ANSWER: , , = 2.04×10−5,0,0Correct

Part D

, ,


ANSWER: , , = 7.21×10−6,0,0Correct

Score Summary:Your score on this assignment is 69.9%.You received 69.9 out of a possible total of 100 points.


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