Mastering Physics - C

Homew ork Chapter 28

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Homework Chapter 28 Due: 11:30pm on Sunday, October 7, 2012 Note: You will receive no credit for late submissions.

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Current Sheet Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current through each wire. wir e. There are

runs in the -y direction

wires wire s per unit unit length in the x direction.

Part A Write Writ e an expression for Use

, the magnetic field a distance

above the xy plane of the sheet.

for the permeability permeability of free space.

Express the magnetic magnetic field as a vector in terms of any or all of the following: following: unit vectors

,

, and/or

,

,

,

,

, and the

.

Hint 1. How to approach the problem You will need to use Ampère's law: . The first step in applying Ampère's law is to choose an appropriate Ampèrean loop. Because you are trying try ing to find the magnetic magnetic field a distance above the sheet, sheet, a good choice for the Ampèrea Ampèrean n loop is is a rectangle rect angle of width widt h

and height height

as shown.



Hint 2. Find How much current

is enclosed by the Ampèrean loop given in the first fir st hint? hint?

Answer in terms of variables given in the problem introduction. ANSWER: ANSWER: =

Hint 3. Determine the direction of the magnetic magnetic field fie ld Above Above the sheet, in which direction does the magnetic magnetic field point? (Be caref ul that that your answer has the correct sign.) Give your answer in terms of the unit vectors

,

, and

.

Hint 1. Direction of o f a field from a single wire The magnetic field generated by a current running through a single wire in the - y direction cannot have any component in what direction? ANSWER: ANSWER:

Hint 2. Direction of the total field From the answer for the field from a single wire we know that each wire generates a magnetic field with components in the x and z directions. In this problem, the magnetic field in one of these directions generated by any wire is canceled out exactly by the magnetic field generated by the other wires. Which component cancels?



Hint 1. A 1. A figure The figure shows the fields due to two wires on opposite sides of a point above the wire.

ANSWER: ANSWER:

ANSWER: ANSWER:



Hint 4. Magnitude of the magnetic field Because the magnetic field points in the the sheet), the line integral in Ampère's law,

direct ion (above the sheet) and the

direction direct ion (below

, does not depend on contributions from fr om the

sides of the loop (which run in the z direction). In addition, the current enclosed by the loop does not depend on the length length of the sides of the loop. This means that the quantity appears nowhere in Ampère's law for this problem, and therefore therefore the magnitude magnitude of the magnetic magnetic field does not vary as a function of height above or below the sheet. By symmetry, the magnitude of the magnetic field also does not vary as a function of xy position. (Because the sheet is infinite, any xy point above the sheet is equivalent to every other.) Following this line of reasoning we conclude conclude that the magnitude of the magnetic field

is constant

everywhere outside the sheet .

Hint 5. Evaluate

What is the value value of Use

evaluated around the Ampèrean loop shown in the figure?

for the (constant) magnitude magnitude of the magnetic field.

ANSWER: ANSWER: =

ANSWER: ANSWER: =



Correct This equation is analogous to

on either side of a infinitely charged sheet. The correspondence corr espondence

seems more obvious if you set the current per unit unit length

. Then the magnetic field you just

calculated is . The electric field, though, points along the perpendicular to the surface.

Do you see why you had to pick the rean loop you used? That is, why would any other loop not have worked. Did you notice that by using Ampère's law you could find the field by using a much simpler integral than Biot-Savart's law? The drawback is that you may not always be able to find a convenient loop in situations where the current distribution is more complicated.

Exercise 28.7 Figure shows two tw o point charges,

and

, moving moving

relative to an observer at point P . Suppose that the lower charge is actually actually negative, with .

Part A Find the magnitude of the magnetic field produced by the two charges at point P if Express your answer in terms of the variables , ANSWER: ANSWER:

,

, and appropriate constants.

.



Correct

Part B What is its direction? ANSWER: ANSWER: out of the page into the page no field

Correct

Part C Find the magnitude of the magnetic field produced by the two tw o charges at point Express your answer in terms of the variables , ANSWER: ANSWER: 0

Correct

Part D What is its direction? ANSWER: ANSWER: out of the page into the page no field

,

if


.



Correct

Part E Find the magnitude of the magnetic field produced by the two tw o charges at point Express your answer in terms of the variables ,

,


ANSWER: ANSWER:

Correct

Part F What is its direction? ANSWER: ANSWER: out of the page into the page no field

Correct

Part G Find the direction direct ion of the magnetic force for ce that ANSWER: ANSWER: upward downward

exerts on

if

.

.



Correct

Part H Find Find the direction of the magnetic magnetic force that

exerts on .

ANSWER: ANSWER: upward downward

Correct

Part I If

, what is the ratio of the magnitude magnitude of the magnetic magnetic force acting on each charge charge

to that of the Coulomb force acting on each charge? ANSWER: ANSWER: =

1.00×10

−6

Correct

Exercise 28.10 A long, long, str aight aight wire, w ire, carrying car rying a current current of 200

, run r uns s through a cubical wooden box, entering entering and leaving leaving through through

holes in the centers of opposite faces (see the figure ). The length of each side of the box is 20.0 . Consider an element element

of the wire 0.100

the box. (Note: Assume that

long long at the center center of

is small in comparison to

the distances from the current element to the points where the magnetic field is to be calculated.)



Part A Compute the magnitude

of the magnetic field produced by this element at the point a. Point a is at the

centers of the face of the cube. ANSWER: ANSWER: = 2.00×10 −6

Correct

Part B Compute the magnitude

of the magnetic field produced by this element at the point b. Point b is at the

midpoint of one edge. ANSWER: ANSWER: = 7.05×10 −7

Correct

Part C Compute the magnitude

of the magnetic field produced by this element at the point c . Point c is at the

centers of the face of the cube. ANSWER: ANSWER: = 2.00×10 −6

Correct

Part D



Compute the magnitude

of the magnetic field produced by this element at the point d . Point d is at the

centers of the face of the cube. ANSWER: ANSWER: = 0

Correct

Part E Compute the magnitude

of the magnetic field produced by this element at the point e. Point e is at a

corner. ANSWER: ANSWER: = 5.45×10 −7

Correct

Exercise 28.13 A wire carrying a 29.0 3.00

current bends bends through a right angle. Consider Consider tw o 2.00

segments segments of wire, each

from the bend (the figure ).

Part A Find the magnitude magnitude of the magnetic field these two tw o segments produce at point them.

, which is midway between betwe en



ANSWER: ANSWER: = 1.82×10 −5

Correct

Part B Find the direction direct ion of the the magnetic field at point

.

ANSWER: ANSWER: into the page out of the page

Correct

Exercise Exercise 28.15: The T he Magnetic Magnetic Field from a Lightning Li ghtning Bolt Lightning Lightning bolts can carry carr y curr currents ents up up to approximately appro ximately 20

. We can model such such a current current as the equivalent equivalent of a

very long, straight wire.

Part A If you were unfortunate unfortunate enough enough to be 4.5

away from such a lightnin lightning g bolt, how large a magnetic magnetic field would would

you experience? Express your answer using two significant significant figures. f igures. ANSWER: ANSWER: = 8.9×10−4

Correct

Part B How does does this this field field (

) compare compare to one one (

) you would would expe experien rience ce by bein being g 4.5

from a lon long, straig straight ht



household household current of 10

?

Express your answer using two significant significant figures. f igures. ANSWER: ANSWER: = 20

Correct

Exercise 28.28 Three parallel wires wir es each carry carr y curre current nt wires is

in the direct ions shown in the figure. The separation separa tion betw een adjacent

.

Part A Calculate the magnitude of the net magnetic force per unit length on the top wire. Express your answer in terms of the variables , ANSWER: ANSWER:

Correct

Part B

, and and appropriate constants.



What is its direction? ANSWER: ANSWER: Upward Downward The force iz zero

Correct

Part C Calculate the magnitude of the net magnetic force per unit length on the middle wire. Express your answer in terms of the variables ,


ANSWER: ANSWER: 0

Correct

Part D What is its direction? ANSWER: ANSWER: Upward Downward The force iz zero

Correct

Part E Calculate the magnitude of the net magnetic force per unit length on the bottom wire. Express your answer in terms of the variables ,




ANSWER: ANSWER:

Correct

Part F What is its direction? ANSWER: ANSWER: Upward Downward The force iz zero

Correct

Exercise 28.44 A toroidal solenoid solenoid (see the figure ) has inner inner radiu r adius s 14.1 and and outer outer radiu radius s 19.5 19.5 . The The sole solen noid oid has has 230 turn turns and and carrie carries s a cu current rrent of 7.00 7.00

.

Part A What What is the the magn magnitud itude e of the the magn magnetic etic field field at 10.0 ANSWER: ANSWER: = 0

from the the center center of the the torus? torus?



Correct

Part B What What is the the magn magnitud itude e of the the magn magnetic etic field field at 16.5


ANSWER: ANSWER: = 1.95×10 −3

Correct

Part C What What is the the magn magnitud itude e of the the magn magnetic etic field field at 20.0


ANSWER: ANSWER: = 0

Correct

Force between an Infinitely Long Wire and a Square Loop A square square loop of wire w ith side length length an infinite wire wir e carrying a current

carries a current

. The center of the loop is located a distance

from

. The infinite wire and and loop are in the same plane; two sides of the square loop

are parallel to the wire and two are perpendicular as shown.



Part A What is is the magnitude,

, of the net force for ce on the loop?

Express the force in terms of

,

,

,

, and

.

Hint 1. How to approach the problem You need to find the total force as the sum of the forces on each straight segment of the wire loop. You'll save some work if you think ahead of time about which forces might cancel.

Hint 2. Determine the direction of force Which of the following diagrams correctly indicates the direction of the force on each individual line segment?



Hint 1. Direction of o f the magnetic magnetic field In the region of the loop, the magnetic field points into the plane of the paper (by the right-hand rule).

Hint 2. Formu Formula la for the force on a current-carrying conductor The magnetic magnetic force on a straight straight wire segment segment of length length , carrying a current current magnetic field

with a uniform uniform

along its length, is ,

where

is a vector along the wire in the direct ion of the current.

ANSWER: ANSWER: a b c d

Hint 3. Determine the magnitude of force Which of the following diagrams correctly indicates the relative magnitudes of the forces on the parallel wire segments? segments?

Hint 1. Find the magnetic magnetic field due to the wire What is the magnitude, magnitude,

, of the wire's wire 's magnetic magnetic field as a function of perpendicular distance



from the wire,

.

Express the magnetic magnetic field magnitude magnitude in terms of

,

, and

.

Hint 1. Am 1. Ampère's père's law Use Ampère's law to obtain the magnetic field. Ampère's law states that , where the line integral can be done around any closed loop.

ANSWER: ANSWER: =

ANSWER: ANSWER: a b c d

Hint 4. Find the force on the section of the loop closest to the wire What is the the magni magnitude tude of the force

on the section of the loop closest to the wire, that is, a distance distance

from it? Express your answer in terms of

,

,

,

, and

.

Hint 1. Formu Formula la for the force on a current-carrying conductor The magnetic magnetic force on a straight straight wire segment segment of length length , carrying a current current magnetic field

along its length, is ,

where

is a vector along the wire in the direct ion of the current.

Hint 2. Find the magnetic magnetic field due to the wire

with a uniform uniform



What is the magnitude, magnitude, from the wire,

, of the wire's wire 's magnetic magnetic field as a function of perpendicular distance

.


,

, and

.


ANSWER: ANSWER: =

ANSWER: ANSWER: =

Hint 5. Find the magnetic field due to the wire What is the magnitude, magnitude, wire,

, of the wire's wir e's magnetic field as a functio function n of perpendicular distance from fro m the

.


,

, and

.


ANSWER: ANSWER: =



ANSWER: ANSWER: =

Correct

Part B The magnetic moment moment

of a current loop is defined as the vecto vectorr whose magnitude equals the area of the

loop loop time times s the the mag magn nitude itude of of the the curre curren nt flowin flowing g in it ( plane in which the current flows. Find the magnitude,

), and and whos whose e direc directio tion n is is perpe perpen ndicu dicula larr to the the , of the force on the loop from fr om Part Part A in terms ter ms of the

magnitude of its magnetic moment. Express

in terms of

,

,

,

, and

.

ANSWER: ANSWER: =

Correct The direction of the net force would be reversed if the direction of the current in either the wire or the loop were reversed. The general result is that "like currents" (i.e., currents in the same direction) attract each other (or, more correctly, cause the wires to attract each other), whereas oppositely directed currents repel. Here, since the like currents were closer to each other than the unlike ones, the net force was attractive. The corresponding situation for an electric dipole is shown in the figure below.

Magnetic Field inside a Very Long Solenoid



Learning Goal: To apply Ampère's law to find the magnetic field inside an infinite solenoid. In this problem we will apply Ampère's law, written , to calculate the magnetic field inside a very long solenoid (only a relatively short segment of the solenoid is shown in the pictures). pictures ). The solenoid has length , diameter , and turns per unit unit length with wit h each carrying carr ying current . It is usual to assume that the component of the current along the z axis is negligible. (This may be assured by winding two layers of closely spaced wires that spiral in opposite directions.) From symmetry considerations it is possible to show that far from the ends of the solenoid, the magnetic field is axial.

Part A Which figure shows the loop that the must be used as the Ampèrean loop for finding

for

inside the

solenoid?

Hint 1. Choice of path for loop integral Which of the following choices are a requirement of the Ampèrean loop that would allow you to use Ampère's law to find ?



a. The path must pass through through the point

.

b. The path must have have enough symmetry symmetr y so that

is constant along large parts part s of

it. c. The path must be a circle. ANSWER: ANSWER: a only a and b a and c b and c

ANSWER: ANSWER: A B C D

Correct

Part B Assume Assume that loop B ( in the Part A figure) figure) has length length

along along

(the

direction). What is t he loop loop integral in

Ampère's law? Assume Assume that the t he top end of the loop is very far from the t he solenoid solenoid (even though though it may not look like it in the figure), so that the field there is assumed to be small and can be ignored. Express your answer in terms of ANSWER: ANSWER: =

Correct

Part C

,

, and other quantities given in the introduction.



What physical physical property does the symbol

represent?

ANSWER: ANSWER: The current along the path in the same direction as the magnetic field The current in the path in the opposite direction from the magnetic field The total current passing through the Ampèrean loop in either direction The net current through the Ampèrean loop

Correct The positive direction of the line integral and the positive direction for the current are related by the right-hand rule: Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction direc tion for and the direction direct ion of your thumb is the positive direct ion for the net current. Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)

Part D What is

, the current passing through the chosen loop?

Express your your answer answer in terms of

(the length of the Ampèrean Ampèrean loop loop along along the axis axis of the solenoid) solenoid)

and other variables given in the introduction. ANSWER: ANSWER: =

Correct

Part E Find

, the z component of the magnetic field inside the solenoid where Ampère's law applies.

Express your answer in terms of ANSWER: ANSWER: =

,

,

,

, and physical constants such as

.



Correct

Part F What is

, the z component of the magnetic field outside the solenoid?

Hint 1. Find the Ampèrean loop to use Which figure shows the loop that the must be used as the Ampèrean loop for finding finding

outside the

solenoid? Note: From symmetry considerations, the field outside (if non-zero) must also be axial and opposite to the field inside. inside.

Hint 1. Choice of path for loop integral Which of the following choices are a requirement of the Ampèrean loop that would allow you to use Ampère's law to find ? a. The path must pass through through the point

.

b. The path must have have enough symmetry symmetr y so that parts of it. c. The path must be a circle. ANSWER: ANSWER:

is constant along large



a only a and b a and c b and c

ANSWER: ANSWER: A B C D E

ANSWER: ANSWER:

0

Correct

Part G The magnetic field inside a solenoid can be found exactly using Ampère's law only if the solenoid is infinitely long. Otherwise, the Biot-Savart law must be used to find an exact answer. In practice, the field can be determined with very little error by using Ampère's law, as long as certain conditions hold that make the field similar to that in an infinitely long solenoid. Which of the following conditions must hold to allow you to use Ampère's law to find a good approximation? a. Consider only only locations where the distance from the the ends is many times b. Consider any location locati on inside the solenoid, as long as c. Consider only locations along the axis of the solenoid.

is much larger than

. for the the solenoid.



Hint 1. Implications of symmetry Imagine that the the solenoid is made of two tw o equal pieces, one extending from fr om the other from

to

to

and

. If both were present the field would have have its normal normal value, value, but if

either is removed the field at

drops to one-half of its previous value. This shows that the field

drops off significantly near the ends of the solenoid (relative to its value in the middle). However, in doing this calculation, you assumed that the field is constant along the length of the Ampèrean loop. So where would this assumption break down?

Hint 2. Off-axis field fi eld dependence dependence You also used symmetry considerations to say that the magnetic field is purely axial. Where would this symmetry argument not hold? Note that far from the ends there cannot be a radial field, because it would imply a nonzero magnetic charge along the axis of the cylinder and no magnetic charges are known to exist (Gauss's Law for magnetic fields and charges). In conjunction with Ampère's law, this allows us to conclude that the z component of the field cannot cannot depend on inside the solenoid.

ANSWER: ANSWER: a only b only c only a and b a and c b and c

Correct

Problem 28.67: Helmholtz Coils The figure is a sectional view of two circular coils with radius , each wound with wit h turns of wire carr ying a current

, circulating in the same direction in both coils.

The coils are separated separated by a distance

equal equal to their

radii. In this configuration the coils are called Helmholtz coils; they produce a very uniform magnetic field in the region between them.



Part A Derive the expression for the magnitude of point point

of the magnetic field at a point on the axis a distance

, which which is midway between the coils.

ANSWER: ANSWER:

Correct

Part B From part (a), obtain an expression for the magnitu magnitude de of the magnetic magnetic field at point point Express Express your answer answer in terms of the variabl variables es ANSWER: ANSWER:

=

Correct

,

,

.

, and appropria appropriate te constant constant (

).

to the right



Part C Calc Calcul ulate ate the magn magnitu itude de of the the magn magnetic etic field field at

if 290 turn turns, s,

= 7.00

, and and

= 7.00

.

ANSWER: ANSWER: = 2.61×10−2

Correct

Part D Calculate

at

.

ANSWER: ANSWER: = 0

Correct

Part E Calculate

at

.

ANSWER: ANSWER: = 0

Correct

Part F Discuss how how your results in parts part s d) and and e) show show that the field is very uniform uniform in the vicinity vicinity of ANSWER: ANSWER:

.



3580 Character(s) remainin remaining g The results in parts D and E s how that that the field is very uniform because the c hange hange in the magnetic f ield with

Submitted, grade pending

Problem 28.88 A wire in the shape shape of a semicircle with radius radius

is oriented in the the yz -plane -plane with its center of curvature at the origin

(see the figure ).

Part A If the current in the wire is

, calculate calculate the the x -component -component of the magnetic field produced at point P , a distance

out along the x -axis. -axis. (Note: Do not forget the contribution from the straight wire at the bottom of the semicircle that runs runs from currents at ANSWER: ANSWER:

=

Correct

Part B

cancel.) cancel.)

to

. You may use the fact that the fields of the two antiparallel antiparallel



Calculate the y -component -component of the magnetic field. ANSWER: ANSWER:

=

Correct

Part C Calculate the z -component -component of the magnetic field. ANSWER: ANSWER: = 0

Correct

Score Summary: Your score on this assignment is 92.4%. You received 12.93 out of a possible total of 14 points.

Mastering Physics - C

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