University Physics Mastering Physics Ch 29

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Chapter 29 Assignment

Due: 11:00pm on Tuesday, April 10, 2012

Note: To understand how points are aw arded, read your instructor's Grading Policy.

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Exercise 29.7

The current in the long, straight wire AB shown in the figure is upward and is increasing steadily at

a rate .

Part A

At an instant when the current is , what is the magnitude of the field at a distance to the

right of the wire?

Express your answer in terms of the appropriate quantities.

ANSWER:

=

Correct

Part B

At an instant when the current is , what is the direction of the field at a distance to the

right of the wire?

ANSWER:into the page

out of the page

Correct

Part C

[ Print ]



What is the flux through the narrow shaded strip?


ANSWER:

=

Correct

Part D

What is the total flux through the loop?


ANSWER:

=

Correct

Part E

What is the induced emf in the loop?


ANSWER:

=

Correct

Part F

Evaluate the numerical value of the induced emf if , , ,

and .

ANSWER: = 5.06×10−7

Correct

Exercise 29.13

The armature of a small generator consists of a flat, square coil with 150 turns and sides with alength of 1.30 . The coil rotates in a magnetic field of 7.70×10−2 .

Part A

What is the angular speed of the coil if the maximum emf produced is 2.80×10−2 ?

ANSWER: = 14.3

Correct



Exercise 29.20

A 1.90-m-long metal bar is pulled to the right at a steady 4.6 perpendicular to a uniform,

0.800-T magnetic field. The bar rides on parallel metal rails connected through = 24.9- , as

shown in the figure , so the apparatusmakes a complete circuit. You can ignorethe resistance of the bar and the rails.

Part A

Calculate the magnitude of the emf induced in the circuit.

Express your answer using two significant figures.

ANSWER: = 7.0

Correct

Part B

Find the direction of the current induced in the circuit.

ANSWER:clockwise

counterclockwise

Correct

Part C

Calculate the current through the resistor.


ANSWER: = 0.28

Correct



Exercise 29.27

A 1.50- bar moves through a uniform, 1.70- magnetic field with a speed of 3.00 (the figure

). In each case, find the emf inducedbetween the ends of this bar and identifywhich, if any, end (a or b) is at the higherpotential.

Part A

The bar moves in the direction of the -axis.

Find the emf induced between the ends of the bar.

ANSWER: = 4.60

Correct

Part B

Which end of this bar, if any, (a or b) is at the higher potential?

ANSWER:a

b

none

Correct

Part C



ANSWER: = 6.11

Correct

Part D




ANSWER:a

b

none

Correct

Part E



ANSWER: = 0

Correct

Part F


ANSWER:a

b

none

Correct

Part G

How should this bar move so that the emf across its ends has the greatest possible value withb at a higher potential than a ?

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

ANSWER: My Answer:The bar should move the the upper left because the Emf reaches its maximumvalue at sin(90) and cos(0). This gives us just Emf=BVL.

Part H

What is this maximum emf?

ANSWER: = 7.65

Correct

Exercise 29.31



A long, thin solenoid has 360 turns per meter and a radius of 1.09 . The current in the solenoid

is increasing at a uniform rate . The magnitude of the induced electric field at a point which is

near the center of the solenoid and a distance of 3.57 from its axis is 7.60×10−6 .

Part A

Calculate .

ANSWER: = 10.1

Correct

Exercise 29.38

In the figure the capacitor plates have area5.00 and separation 2.00 mm. The

plates are in vacuum. The charging current

has a constant value of 1.80 mA. At t = 0 thecharge on the plates is zero.

Part A

Calculate the charge on the plates when 0.500 .

ANSWER: = 9.00×10−10

Correct

Part B

Calculate the electric field between the plates when 0.500 .

ANSWER: = 2.03×105

Correct

Part C

Calculate the potential difference between the plates when 0.500 .



ANSWER: = 407Correct

Part D

Calculate , the time rate of change of the electric field between the plates.

ANSWER: = 4.07×1011

Correct

Part E

Calculate the displacement current density between the plates.

ANSWER: = 3.60

Correct

Part F

Calculate the total displacement current .

ANSWER: = 1.80×10−3

Correct

Exercise 29.42

At temperatures near absolute zero, approaches 0.142 for vanadium, a type-I

superconductor. The normal phase of vanadium has a magnetic susceptibility close to zero.

Consider a long, thin vanadium cylinder with its axis parallel to an external magnetic field in

the +x-direction. At points far from the ends of the cylinder, by symmetry, all the magnetic vectorsare parallel to the x-axis.

Part A

At temperatures near absolute zero, what is the magnitude of the resultant magnetic field

inside the cylinder for 0.130 ?

ANSWER: = 0

Correct

Part B



What is the direction of the resultant magnetic field inside the cylinder for this case?

ANSWER:in the -direction

in the -direction

perpendicular to the -axis

the field is zero

Correct

Part C

What is the magnitude of the resultant magnetic field outside the cylinder (far from the ends)

for this case?

ANSWER: = 0.130

Correct

Part D

What is the direction of the resultant magnetic field outside the cylinder (far from the ends)

for this case?


in the -direction


the field is zero

Correct

Part E

What is the magnitude of the magnetization inside the cylinder for this case?

ANSWER: = 1.03×105

Correct

Part F

What is the direction of the magnetization inside the cylinder for this case?

ANSWER:



in the -directionin the -direction


the magnetization is zero

Correct

Part G

What is the magnitude of the magnetization outside (far from the ends) the cylinder for this

case?

ANSWER: = 0

Correct

Part H

What is the direction of the magnetization outside the cylinder (far from the ends) for this

case?


in the -direction



Correct

Part I

At temperatures near absolute zero, what is the magnitude of the resultant magnetic field

inside the cylinder for ?

ANSWER: = 0.260

Correct

Part J

What is the direction of the resultant magnetic field inside the cylinder for this case?




in the -direction


the field is zero

Correct

Part K

What is the magnitude of the resultant magnetic field outside the cylinder (far from the ends)

for this case?

ANSWER: = 0.260

Correct

Part L

What is the direction of the resultant magnetic field outside the cylinder (far from the ends)

for this case?


in the -direction


the field is zero

Correct

Part M

What is the magnitude of the magnetization inside the cylinder for this case?

ANSWER: = 0

Correct

Part N

What is the direction of the magnetization inside the cylinder for this case?


in the -direction





Correct

Part O

What is the magnitude of the magnetization outside the cylinder for this case?

ANSWER: = 0

Correct

Part P

What is the direction of the magnetization outside the cylinder for this case?


in the -direction



Correct

Problem 29.45

In the circuit shown in the figure , thecapacitor has capacitance 20 and

is initially charged to 100 with the

polarity shown. The resistor has

resistance 10 . At time 0 the switch

is closed. The small circuit is notconnected in any way to the large one.The wire of the small circuit has aresistance of 1.0 and contains 25

loops. The large circuit is a rectangle 2.0 by 4.0 , while the small one has

dimensions 10.0 and 20.0 .

The distance is 5.0 . (The figure is

not drawn to scale.) Both circuits are held stationary. Assume that only the wire nearest the smallcircuit produces an appreciable magnetic field through it.

Part A



Find the current in the large circuit after S is closed.


ANSWER: = 3.7

Correct

Part B

Find the current in the small circuit 200 after S is closed.


ANSWER: = 5.4×10−5

Correct

Part C

Find the direction of the current in the small circuit.

ANSWER:clockwise

counterclockwise

Correct

Part D

Justify why we can ignore the magnetic field from all the wires of the large circuit except for thewire closest to the small circuit.

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

ANSWER: My Answer:We can ignore the magnetic field from the wires that are far away from the smallcircuit because the current is small enough that it will not matter. So themagnetic field through them can be assumed to be zero. Therefore the problemstates that only the wire nearest to the small circuit produces an appreciablemagnetic field.

Problem 29.50

Consider the loop in the figure . The area is = 610 , and it spins with angular velocity

= 30.0 in a magnetic field of strength = 0.460 .



Part A

What is the maximum induced emf if the loop is rotated about the y-axis?

ANSWER: = 0.842

Correct

Part B

What is the maximum induced emf if the loop is rotated about the x-axis?

ANSWER: = 0

Correct

Part C

What is the maximum induced emf if the loop is rotated about an edge parallel to the z-axis?

ANSWER: = 0.842

Correct

Problem 29.52: Make a Generator?

You are shipwrecked on a deserted tropical island. You have some electrical devices that youcould operate using a generator but you have no magnets. The earth's magnetic field at yourlocation is horizontal and equal to , and you decide to try to use this field for a

generator by rotating a large circular coil of wire at a high rate. You need to produce an emf of and estimate that you can rotate the coil at by turning a crank handle. You also

decide that to have an acceptable coil resistance, the maximum number of turns the coil can haveis .

Part A



What area must the coil have?


ANSWER: = 18

Correct

Part B

If the coil is circular, what is the maximum translational speed of a point on the coil as itrotates?


ANSWER: = 7.5

Correct

Problem 29.66

A metal rod with a length of 26.0 lies in the xy-plane and makes an angle of 36.9 with the

positive x-axis and an angle of 53.1 with the positive y-axis. The rod is moving in the

x-direction with a speed of 4.00 . The rod is in a uniform magnetic field

0.160 0.220 4.00×10−2 .

Part A

What is the magnitude of the emf induced in the rod?

ANSWER: = 2.50×10−2

Correct

Score Summary:

Your score on this assignment is 99.8%.You received 21.96 out of a possible total of 22 points.

University Physics Mastering Physics Ch 29

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