Ch 28 HW.pdf

12/6/2016

Ch 28 HW

Ch 28 HW Due: 11:59pm on Thursday, December 8, 2016 You will receive n o credit for items you complete after the assignment is due . Grading Policy

Force between Moving Charges Two point charges, with charges Two and , are each moving with speed toward the origin. At the instant shown is at position (0, ) and is at at ( , 0). (No (Note te that the signs of of the charges are not given because they are not needed to determine the magnitude of the forces between the charges.)

Part A What is the magnitude of the electric force between the two charges? Express

in terms of

,

, , and

.

Hint 1. Which law to use Apply Coulomb's law:

where wher e

is the distance between between the two charges.

Hint 2. Find the value of What is the value of

for the given situati situation? on?

Express your answer answer in terms terms of . ANSWER:

=

ANSWER: =

https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4706985

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Correct

Typesetting math: 63%

Part B What is the magnitude of the magnetic force on

due to the magnetic field caused by

Express the magnitude of the magnetic force in terms of

,

,

, , and

?

.

Hint 1. How to approach the problem First, find the magnetic field gener generated ated by charge on due to the field of .

at the position of charge

. Then evaluate the magnetic force

Hint 2. Magnitude of the magnetic field The Biot-Savart law, which gives the magnetic field produced by a moving charge, can be written , where where is the permea permeability bility of free space and is the vect vector or from the charge to the point wher where e the magnetic field is produ produced. ced. Note Note we have in the numera numerator, tor, not , necessit ating an extra powe powerr of in the denomina denominator. tor. Using this equation find the expression for the magnitude magnitude of the magnetic field experienced by charge charge charge . Express the magnitude of the magnetic field of

(at the location of

) in terms of

,

, , and

due to

.

Hint 1. Determine the cross product What is the magnitude of

?

For any two vectors, , where wher e

is the angle between the vect vectors. ors. Because, in this case,

is 45 45 degr degrees, ees,

Substit ute the appro appropriate priate value of

for this problem, problem, to arrive arrive at a surprisingly simple answer. answer.

Express your answer in terms of

and .

ANSWER: =

ANSWER:

=

Hint 3. Find the direction of the magnetic field https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4706985

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Which of the following best describes the direction of the magnetic field from Biot-Savart law law,, the field must be be perp perpendicular endicular to both Ignore Igno re the effects of the sign of

at

? Remember Remember,, according to the

and .

.


ANSWER:

(along the x axis) (along the y axis) (along the z axis into or out of the screen)

Hint 4. Computing the force You can evaluate the force exerted on a moving charge by a magnetic field using the Lorentz force law: , where wher e

is the force on the moving charge,

is the magnetic field,

the velocit y of the charge. Note that, as long as

and

is the charge of the moving charge, and

are perpe perpendicular ndicular,,

is

.

ANSWER:

=

Correct

Part C Ass uming that the charges are moving mov ing nonr nonrelativis elativis tic tically ally ( magnitudes of the magnetic and electrostatic forces?

), what can you say about t he relationship between the

Hint 1. How to approach the problem Determine which force has a greater magnitude by finding the ratio of the electric force to the magnetic force and then applying the appr approximation. oximation. Recall that .

ANSWER: The magnitude of the magnetic force is greater than the magnitude of the electric force. The magnitude of the electric force is greater than the magnitude of the magnetic force. Both forces have the same magnitude.

Correct This result holds quite generally: Magnetic forces between moving charges are much smaller than electric forces as long as the speeds of the charges are nonrelativistic.

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Magnetic Field at the Center of a Wire Loop A piece of wire is bent to form a circle with radius . It has a steady current as seen frommath: the top Typesetting 63%(looking in the negative direction).

flowing t hroug hrough h it in a counterclockwis counterclockwise e direction

Part A What is

, the z component of

at the center (i.e.,

) of the loop?

Express your answer in terms of , , and constant constants s like

and

.

Hint 1. Specify the integrand The Biot-Savart law, , where wher e

is the permea permeability bility of free space, tells us the magnetic field

current . What is

at a point from a length of wire

at the center of the loop from a small piece of wire

Give your answer in terms of

, ,

, radius ,

carrying

?

, and .

Hint 1. Find the magnitude of the magnetic field What is the magnitude of

at the center of the loop from a small piece of wire

Give your answer in terms of

,

,

, radius , and

?

.

ANSWER:

=

Hint 2. Direction of the magnetic field The field direction is determined from the cross product in the Biot-Savart law. You can use the right-hand rule to find the direction of the cross-product vector.

ANSWER: https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4706985

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=


Hint 2. Perform the integration Now you have to integrate the Biot-Savart expression for the differential field along the length of the wire to find the total magnetic field at the point. If you keep in mind that , , , and are constants in this case, the integral isn't too difficult . What is the value of

around the loop?

Express your answer in terms of variables given in the problem introduction.

Hint 1. Help with the integral The value of this integral is just the length of the circumference of the loop.

Hint 2. Reexpress

to enable integration

The expression is easiest to integrate with respect to the angle piece of the wire

). Express

Express the magnitude of

(the angle from the x axis to the differential

using this idea. in terms of variables such as

and .

ANSWER: =

ANSWER: around the loop =

ANSWER: =

Correct

Current Sheet Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current There are wires per unit length in the x direction.

https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

runs in the -y direction through each wire.

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Part A Write an expression for Use

, the magnetic field a distance

above the xy plane of the sheet.

for the permeability of free space.

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

, ,

,

,

, and the unit vectors ,

, and/or .

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 to find the magnetic field a distance above the sheet, a good choice for the Ampèrean loop is a rectangle of width and height as shown.

Hint 2. Find How much current

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

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


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=


Hint 3. Determine the direction of the magnetic field

Above the sheet, in which direction does the magnetic field point? (Be careful that your answer has the correct sign.) Give your answer in terms of the unit vectors ,

, and .

Hint 1. Direction of 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:

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

ANSWER:


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ANSWER:

Hint 4. Magnitude of the magnetic field Because the magnetic field points in the

direction (above the sheet) and the

direction (below the sheet), the

line integral in Ampère's law, , does not depend on contributions from 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 of the sides of the loop. This means that the quantity appears nowhere in Ampère's law for this problem, and therefore the magnitude of the 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 that the magnitude of the magnetic field outside the sheet .

is constant everywhere

Hint 5. Evaluate What is the value of Use

evaluated around the Ampèrean loop shown in the figure?

for the (constant) magnitude of the magnetic field.

ANSWER: =

ANSWER:

=


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Correct This equation is analogous to on either side of a infinitely charged sheet. The correspondence seems more obvious . Then the magnetic field you just calculated is Typesetting math: 63%if you set the current per unit length . 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.

Forces between a Charge and a Bar Magnet Learning Goal: To understand the forces between a bar magnet and 1. a stationary charge, 2. a moving charge, and 3. a ferromagnetic object. A bar magnet oriented along the y axis can rotate about an axis parallel to the z axis. Its north pole initially points along .

Interaction of stationary charge and bar magnet A positive charge is placed some distance in the magnet.

direction from the magnet. Ass ume that no charges are induced on the

Part A Ass ume that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will experience which of the following? ANSWER:


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A torque due to the charge attracting the north pole of the manget A torque due to the charge attracting the south pole of the magnet A torque only if one magnetic pole is slightly closer to the charge than the other


No torque at all

Correct The fact that a stationary charge produces no torque on a magnet emphasizes a difference between electric and magnetic forces. You may think of the bar magnet as having a positive magnetic charge at the north end and a negative magnetic charge at the south end (even though there exist no magnetic charges in nature). Importantly, these magnetic charges are not the same as electric charges, and they do not interact with stationary electric charges. (This model of the magnetic dipole correctly predicts that the magnetic field lines go from N to S outside the magnet, however.)

Interaction of moving charge and bar magnet Consider a second case in which the charge is again some distance in the

direction from the magnet, but now it is moving

toward the center of the bar magnet, that is, with its velocity along .

Part B Due to its motion in the magnetic field of the bar magnet, the charge will experience a force in which direction?

Hint 1. Determine the magnetic field direction near a charge The bar magnet produces a magnetic field at the position of the moving charge. Given the relative location of the charge with respect to the bar magnet, and the orientation of the bar magnet, in which direction will this field point? Answer in terms of , ,

, or a linear combination thereof.

Hint 1. Dipole magnetic field The magnetic field lines outside of a magnetic dipole go from the dipole's north pole to its south pole.

ANSWER:


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Hint 2. Determine the direction of force on a charge moving in a magnetic field

A charge

moving in a magnetic feld feels a force given by the Lorentz force equation:

, is positiv e here but could be positive or negative in general. The direction of the force is determined by the

where

vector product

and may be found by applying the right-hand rule:

1. Align your right hand so the fingers point along the direction of ________. 2. Rotate your wrist so that the fingers will point along the direction of _______ when they are closed. 3. Your right thumb now points along the direction of the force.

What two vectors should fill the blanks in these statements to correctly determine the direction of the force? ANSWER: / /

ANSWER:

Correct This is the only type of magnetic force: A moving charge will experience a vector force due to a magnetic field. The force between two permanent magnets arises because of the (perpetually) circulating charge of the electrons in the magnets. In reality, electric forces on the bar magnet might well dominate the magnetic forces discussed here. In practice, substantial magnetic forces arise only for permanently magnetized objects and for current-carrying wires that have significant numbers of moving charges in spite of being essentially neutral electrically.

Interaction of iron and bar magnet Now the charge is replaced by an electrically neutral piece of initially unmagnetized soft iron (for example, a nail) that is not moving.


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Part C As a result of the magnetic interaction between the soft iron and the bar magnet, which of the following will occur?

Hint 1. Magnetic induction A piece of soft iron will contain magnetic domains (small dipoles) that have the following properties: 1. They align themselves along the magnetic field beacuse of the torque exerted on them. This is called induction of a magnetic dipole moment. 2. They experience a force due to the magnetic field of the bar magnet. 3. They produce a magnetic field that causes the magnet to experience a force. Consider the extreme case when the magnet is turned either left or right, so that it is facing the nail. Is the force attractive or repulsive?

ANSWER: The magnet will experience a torque due to the iron attracting its north pole. The magnet will experience a torque due to the iron attracting its south pole. Whichever pole of the magnet is closest to the iron will be attracted to the iron. Whichever pole of the magnet is closest to the iron will be repelled from the iron.

Correct The forces here result from magnetic domains (small dipoles) contained within the soft iron that are aligned by the magnetic field of the bar magnet. (These domains may be thought of as small perpetual current loops, and their tendency to be aligned is called permeability.) Whichever pole of the bar magnet is closer will dominate this alignment and will align the dipoles to attract that pole of the bar magnet.

Exercise 28.32 Two long, parallel wires are separated by a distance of 4.00 3.90×10−5

. The force per unit length that each wire exerts on the other is

, and the wires repel each other. The current in one wire is 0.600

.

Part A https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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What is the current in the second wire? ANSWER: = 13.0 Typesetting math: 63%

Correct

Part B Are the two currents in the same direction or in opposite directions? ANSWER: In the same direction In opposite direction

Correct

Exercise 28.41 Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 24.0 clock wise current of 20.0 , as viewed from above, and the outer wire has a diameter of 30.0 .

and carries a

Part A What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires? ANSWER: The current's direction must be clockwise. The current's direction must be counterclockwise.

Correct

Part B What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires? ANSWER: =

25.0

Correct https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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Exercise 28.52 A toroidal s olenoid (see the figure ) has inner radius 13.4 and Typesetting math: 63%. The solenoid has 200 turns and carries a outer radius 18.9 current of 7.80 .

Part A What is the magnitude of the magnetic field at 10.9

from the center of the torus?

ANSWER: =

0

Correct

Part B What is the magnitude of the magnetic field at 16.2


ANSWER: = 1.93×10−3

Correct

Part C What is the magnitude of the magnetic field at 20.3


ANSWER: =

0

Correct

Problem 28.80 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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A circular loop has radius and c arries current above a long, straight wire.

in a clock wise direction (the figure ). The center of the loop is a distance


Part A What is the magnitude of the current

in the wire if the magnetic field at the center of the loop is zero?

Express your answer in terms of the variables

,

,

, and appropriate constants (

and

).

ANSWER: =

Correct

Part B What is the direction of the current

?

ANSWER: The current

must point to the left.

The current

must point to the right.

Correct

Problem 28.62 A long, straight wire c arries a current of 5.20

. An electron is traveling in the vicinity of the wire.

Part A At the instant when the electron is 4.70 from the wire and traveling with a speed of 6.00×104 wire, what is the magnitude of the force that the magnetic field of the current exerts on the electron? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

directly toward the

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ANSWER: =

2.12×10−19


Correct

Part B What is the direction (relative to the direction of the current) of this force? ANSWER: opposite to the current in the same direction as the current perpendicular to the current

Correct

Magnetic Field near a Moving Charge A particle with positive charge located at the origin, around the moving charge.

is moving with speed along the z axis toward positiv e . At the time of this problem it is . Your task is to find the magnetic field at various locations in the three-dimensional space

Part A Which of the following expressions gives the magnetic field at the point

due to the moving charge?

A. B.


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C. D. Typesetting math: 63%

ANSWER:

A only B only C only D only both A and B both C and D both A and C both B and D

Correct The main point here is that the r -dependence is really

. The

results from using

in the numerator rather

than the unit vect or . A second point is that the order of the cross product must be s uch that the right-hand rule works: If your right thumb is along the direction of the current, , your fingers must curl along the direction of the magnetic field.

Part B Find the magnetic field at the point Express your answer in terms of

. , ,

, and

, and use , , and

for the three unit vectors.

Hint 1. Find the magnetic field direction Consider the directions of

and

. What is the direction of the magnetic field at point

Express your answer in terms of ,

?

, and .

Hint 1. Cross products The direction of the cross-product vector can be found by using the right-hand rule. Alternatively, you could use the following relations: . Here is a useful device for remembering these relations. Look at the figure shown . If you go around the circle clockwise (i.e., starting out alphabetically) as in the figure, then the cross-product vector on the righthand side of the equations above appears with a positive sign. For example, suppose that you needed to find the direction of the vector

. Notice that if you start at

in the figure and move to

clock wise, that is, in the direction the arrows point. Therefore, the answer will be Using

, you are going

.

gives the relations: .


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So if you go counterclockwise around the circle in the figure shown, or start out in reverse alphabetical order, then the cross-product vector on the right-hand side appears with a negative sign. Typesetting math:For 63%example, suppose that you needed to find the direction of the vect or . Notice that if you start at in the figure and move to , you are going counterclockwise, that is, against the direction the arrows point. Therefore, the answer will be

.

ANSWER:

ANSWER:

=

Correct

Part C Find the magnetic field at the point Express your answer in terms of

. , ,

,

, and

, and use ,

, and


ANSWER:

=

Correct

Part D Find the magnetic field at the point Express your answer in terms of

. , ,

,

,

, and

, and use ,


, and


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Hint 1. Evaluate the cross product To find the magnetic field for this part, it is convenient to use expression B from Part A: Typesetting math: 63%

Knowing

. and

, find

.

Express your answer in terms of some or all of the variables , three unit vectors.

,

, and

and use ,

, and

for the

ANSWER: =

Hint 2. Find the distance from the charge What is the magnitude of the vector

?

Express your answer in terms of

and

.

ANSWER: =

ANSWER:

=

Correct

Part E The field found in this problem for a moving charge is the same as the field from a current element of length current provided that the quantity is replaced by which quantity?

carrying

Hint 1. Making a correlation If

describes a moving charge, the related expression you find must also describe a moving charge. Remember

that

and that

.

ANSWER:

Correct https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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Magnetic Field due to a Wire Conceptual Question The same amount of current

is flowing through two wires, labeled 1 and 2 in the figure, in the directions indicated by the


arrows. In this problem you will determine the direction of the net magnetic field

at each of the indicated points (A - C).

Part A What is the direction of the magnetic field magnitudes.

at point A? Recall that the currents in the two wires have equal

Hint 1. The magnitude of the magnetic field due to a long, straight current-carrying wire The magnitude of a magnetic field inversely proportional to the distance

is directly proportional to the amount of current from the wire:

flowing in the wire and

.

Hint 2. The direction of the magnetic field due to a long, straight current-carrying wire The magnetic field surrounding a long, straight wire encircles the wire, as shown in the figure:


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The direction of the field is determined by a right-hand rule: Grasp the wire with the thumb of your right hand in the direction of the current flow. The direction in which your fingers encircle the wire is the direction in which the magnetic field encircles the wire. Typesetting math: 63%

Hint 3. How to approach the problem To determine the direction of the magnetic field at point A, you must determine the contribution to the field from both of the wires. The field at point A is the vector sum of these two contributions. Because point A is in the same plane as the wires, the contribution to the net magnetic field at point A from wire 1 or wire 2 will either point into or out of the screen. Keep in mind that if the magnetic fields are in opposite directions, the larger field will decide the direction of the net magnetic field. If they are the same size, the net magnetic field will be zero.

Hint 4. Find the direction of the magnetic field at point A due to wire 1 Is the magnetic field from wire 1 directed into or out of the screen at point A? Be sure to point your thumb in the direction of the current, in this case to the right. ANSWER: in out

Hint 5. Find the direction of the magnetic field at point A due to wire 2 Is the magnetic field from wire 2 directed into or out of the screen at point A? Be sure to point your thumb in the direction of the current, in this case downward. ANSWER: in out

ANSWER: points out of the screen at A. points into the screen at A. points neither out of nor into the screen and

at A.

at A.

Correct

Part B What is the direction of the magnetic field

at point B?

Hint 1. Find the direction of the magnetic field at point B due to wire 1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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Is the magnetic field from wire 1 directed into or out of the screen at point B? ANSWER: Typesetting math: in 63%

out

Hint 2. Find the direction of the magnetic field at point B due to wire 2 Is the magnetic field from wire 2 directed into or out of the screen at point B? ANSWER: in out

ANSWER: points out of the screen at B. points into the screen at B. points neither out of nor into the screen and

at B

at B.

Correct

Part C What is the direction of the magnetic field

at point C?

Hint 1. Find the direction of the magnetic field at point C due to wire 1 Is the magnetic field from wire 1 directed into or out of the screen at point C? ANSWER: in out

Hint 2. Find the direction of the magnetic field at point C due to wire 2 Is the magnetic field from wire 2 directed into or out of the screen at point C? ANSWER:


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in out Typesetting math: 63%

ANSWER: points out of the screen at C. points into the screen at C. points neither out of nor into the screen and

at C.

at C.

Correct

Magnetic Field Generated by a Finite, Current-Carrying Wire A steady current

is flowing t hrough a straight wire of finite length.

Part A Find , the magnitude of the magnetic field generated by this wire at a point P located a distance from the center of the wire. Assume that at P the angle subtended from the midpoint of the wire to each end is as shown in the diagram . Express your answer in terms of ,

,

, and

.

Hint 1. Formula for the magnetic field of a current-carrying wire (Biot-Savart law) The magnetic field

where

of a current-carrying wire (Biot-Savart law) is

,

is the current through the element,

the current in the conductor,

is a vector with length

, in the same direction as

is a unit vector that points from the source point to the field point P,

between the source point and the field point,

is the angle

makes with

length of the entire wire. The figure shows (a) the magnetic field vect ors https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

is the distance

, and the integral is done over the due to a current element

and (b) 23/44

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magnetic field lines in a plane containing the current element

.

Note that both Typesetting math: the wire ( 63% ).

and

are functions of the position along

Hint 2. Simplify the vector integral Consider a small segment of the wire. In what direction is the magnetic field at P due to this segment (given by the Biot-Savart law)? ANSWER: out of the plane of the screen into the plane of the screen along OP--away from O along PO--toward O

Hint 3. Find In this problem,

and

for some general point on the wire can be taken as shown in the figure . What is

? Express your answer in terms of some or all of the variables and and .

Hint 1. How to approach the problem Write and then in terms of and its trignometric functions. Also write an equation for , the distance from the center of the wire to the infinitess imal element , in terms of and . Differentiate this to find . (Note that can be treated as a constant for this differential.) Substitute these quantities into the a equation . https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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Hint 2. Derivative of Note that . Typesetting math: 63%

ANSWER:

=

Hint 4. Find Find an expression for

as shown in the figure.

Express your answer as a function of the variables and/or .

ANSWER: = \large{\frac{x}{{\cos}\left({\theta}\right)}}

ANSWER: \texttip{B_{\rm 1}}{B_1} = \large{\frac{2kI {\sin}\left({\theta}_{m}\right)}{x}}

Correct The magnetic field for an infinitely long wire can be obtained by setting \theta_m=\pi/2 in the previous expression. This gives a magnetic field \large{B = \frac{2kI}{x} = \frac{ \mu_0 I}{2 \pi x}}, which you probably derived in an earlier problem or in lecture using the Biot-Savart law.

Part B Now find \texttip{B_{\rm 2}}{B_2}, the magnetic field generated by this wire at a point P located a distance \texttip{x}{x} from either end of the wire. Assume that at P the angle subtended from the end of the wire to the other end is \texttip{\theta _{\rm end}}{theta_end} as shown in the diagram . https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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Express your answer in terms of \texttip{I}{I}, \texttip{x}{x}, \texttip{\theta _{\rm end}}{theta_end}, and k= \mu_0/(4 \pi).


Hint 1. How to approach the problem When you did the last part, you obtained an integral that you evaluated to find the magnetic field in terms of the angles made by a line perpendicular to the wire passing through the point where you are calculating the field and lines connecting that point and each wire end. To solve this problem, you can simply change the values of these angles (i.e. change your limits of integration.)

Hint 2. Limiting value of \texttip{\theta }{theta} Note that, at the end of the wire near the point chosen, \theta=0. At the other end, use a similar argument for \texttip{\theta }{theta} as you used in the previous part.

ANSWER: \texttip{B_{\rm 2}}{B_2} = \large{\frac{kI {\sin}\left({\theta}_{\rm{end}}\right)}{x}}

Correct Setting \theta_{\rm end}=\pi/2 in the previous expression yields the magnetic field for a semi-infinite wire: \large{B = \frac{kI}{x} = \frac{ \mu_0 I}{4 \pi x}}, which is in fact just half the value of the magnetic field due to an infinitely long wire. This difference results from the point chosen being close to one of the ends of the wire. Such "end effects" for noninfinite wires always change the magneic field near that point.

± Canceling a Magnetic Field Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 50.0 {\rm cm} , as shown in the figure . The currents running through the wires are 8.0 {\rm A}, 20.0 {\rm A}, 10.0 {\rm A}, and \texttip{I}{I}.


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Part A Find the magnitude of the current \texttip{I}{I} that will make the magnetic field at the center of the square equal to zero. Express your answer in amperes.

Hint 1. 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 \texttip{I}{I} whose contribution to the magnetic field will exactly cancel the contribution of the other three wires.

Hint 2. Calculating the contribution from the three known wires What is the magnitude \left|\vec{B}_{\rm c} \right| of the magnetic field at the center of the square due to the wires carrying the 8.0-, 20-, and 10-\rm{A} currents? Be careful with signs when you add the contributions from the three different wires. Express your answer in teslas to three significant figures.

Hint 1. Ampère's law Recall Ampère's law: \mu_0 I=\oint{B\cdot dl}. 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 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.

ANSWER: \left| \vec{B}_{\rm c} \right| = 1.60×10−6

\rm{T}


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ANSWER: \texttip{I}{I} =

2.00

\rm{A}


Correct

Part B What is the direction of the current \texttip{I}{I}?

Hint 1. How to approach the problem If you were able to master the right-hand rule in Part A, you should be able to get this. Your goal is to cancel the magnetic field contributions from the other three wires.

ANSWER: upward downward

Correct

Force between Two Infinite, Parallel Wires You are given two infinite, parallel wires each carrying current \texttip{I}{I}. The wires are separated by a distance \texttip{d}{d}, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.

Part A Is the force between the wires attractive or repulsive?

Hint 1. How to approach the problem First find the direction of the magnetic field at wire 2 due to wire 1 (using the right-hand rule). Next, find the direction of the force on wire 2 due to this magnetic field. The force on a particle with charge \texttip{q}{q} moving with velocity \texttip{\vec{v}}{v_vec} in a magnetic field \texttip{\vec{B}}{B_vec} is \vec{F} = q\vec{v}\times\vec{B}. Don't forget that the electron is negatively charged, so \texttip{q}{q} is negative, and the velocity of the electrons in the wire will be opposite to the direction of the current.

ANSWER: attractive repulsive


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Correct


Part B What is the force per unit length F/L between the two wires? Express your answer in terms of \texttip{I}{I}, \texttip{d}{d}, and constants such as \texttip{\mu _{\rm 0}}{mu_0} and \texttip{\pi }{pi}.

Hint 1. Magnetic field at wire 2 due to wire 1 Use Ampère's law to find the magnitude of the magnetic field of wire 1 at the position of wire 2. Express your answer in terms of \texttip{I}{I}, \texttip{d}{d}, and constants such as \texttip{\m u _{\rm 0}} {mu_0} and \texttip{\pi }{pi}.

Hint 1. Relevant equation Ampère's law s tates that \displaystyle \oint \vec{B}\cdot d\vec{s} = \mu_0 I, where the integral is over a closed loop, \vec{B} is the magnetic field, d\vec{s} is the infinitesimal length element, and \texttip{\mu _{\rm 0}}{mu_0} is the permeability of free space.

ANSWER: \texttip{B}{B} = \large{\frac{{\mu}_{0} I}{2 {\pi} d}}

Hint 2. Force on wire 2 The force on a particle with charge \texttip{q}{q} moving with velocity \texttip{\vec{v}}{v_vec} in a magnetic field \texttip{\vec{B}}{B_vec} is \vec{F} = q\vec{v}\times\vec{B}. The current in a wire with \texttip{n}{n} particles per unit length, where each particle has charge \texttip{q}{q} and velocity \texttip{\vec{v}}{v_vec}, is \vec{I}=nq\vec{v}. Don't forget that the electron is negatively charged, so \texttip{q}{q} is negative, and the velocity of the electrons in the wire will be opposite to the direction of the current.

ANSWER: \large{\frac {F} {L}} = \large{{\mu}_{0}{\frac{I^{2}}{2{\pi}d}}}

Correct

Part C In the SI system, the unit of current, the ampere, is defined by this relationship using an apparatus called an Ampère balance. What would be the force per unit length of two infinitely long wires, separated by a distance 1\;{\rm m}, if 1\;{\rm A} of current were flowing through each of them? Express your answer numerically in newtons per meter. ANSWER:


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\large{\frac {F}{L}} = 2.00×10−7

\large{\frac{\rm N}{\rm m}}

Typesetting math: 63% Correct

In a more practical Ampère balance experiment, the force on a loop of wire between two current-carrrying loops is measured because the two wires create a field that does not vary so much with position.

± PSS 28.2 Ampere's Law Learning Goal: To practice Problem-Solving Strategy 28.2 Ampere's Law. A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius \texttip{R_a} {R_a} = 6.35 {\rm cm} and inner radius \texttip{R_b}{R_b} = 3.55 {\rm cm} . The central conductor and the conducting tube carry equal currents of \texttip{I}{I} = 2.45 {\rm A} in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance \texttip{r}{r} = 5.60 {\rm cm} from the axis of the conducting tube?

Problem Solving Strategy 28.2: Ampere's Law IDENTIFY the relevant concepts: Like Gauss’s law for electricity, Ampere’s law is always true but is most useful in situations where the magnetic field is symmetrical. SET UP the problem using the following steps: 1. Select the integration path you will use with Ampere’s law. If you want to determine the magnetic field at a certain point, then the path must pass through that point. The integration path has to have enough symmetry to make evaluation of the integral possible. 2. Determine the target variable(s). Usually this will be the magnitude of the magnetic field \texttip{\vec{B}}{B_vec} as a function of position. EXECUTE the solution as follows: 1. Carry out the integral \oint \vec{B}\cdot d\vec{l} along your chosen integration path. If \vec{B} is tangent to all or some portion of the integration path and has the same magnitude B at every point, then its line integral equals B multiplied by the length of that portion of the path. If \vec{B} is perpendicular to some portion of the path, that portion makes no contribution to the integral. 2. In the integral \oint \vec{B}\cdot d\vec{l}, \vec{B} is always the total magnetic field at each point on the path. If no net current is enclosed by the path, the field at points on the path need not be zero, but the integral \oint \vec{B}\cdot d\vec{l} is always zero. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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3. Determine the current I_{\rm encl} enclosed by the integration path. The sign of this current is given by a righthand rule. 4. Use Ampere’s law \oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\rm encl} to solve for the target variable. your answer: EVALUATE Typesetting math: 63% If your result is an expression for the field magnitude as a function of position, you can check it by examining how the expression behaves in different limits.

IDENTIFY the relevant concepts You are asked to find the magnetic field at a specific distance from the axis of a current carrying tube. This can be accomplished using Ampere's law, so the above strategy is appropriate.

SET UP the problem using the following steps

Part A Which picture accurately depicts the general pattern of the magnetic field lines created by the current?

Hint 1. Use the right hand rule Use the following steps of the right hand rule to determine the magnetic field. 1. Point your right thumb in the direction of the current. 2. Curl your fingers around the wire to indicate a circle. 3. Your fingers point in the direction of the magnetic field lines around the wire.

ANSWER:


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Correct The current is flowing parallel to the axis of the conducting tube and the inner conducting cylinder. Using the right hand rule to determine the direction of the magnetic field, the magnetic field lines will be concentric circles centered on the axis of the tube and the inner cylinder.

Part B What will be the integration path used in Ampere's law to calculate the magnetic field \vec{B} at a distance \texttip{r}{r} from the axis of the inner cylinder? ANSWER:


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Correct The magnetic field lines are concentric circles around the axis of the current carrying conductors. For this reason the choice of a circular integration path \texttip{\vec{dl}}{dl_vec} that is coaxial with the conductors will greatly simply your calculations.

EXECUTE the solution as follows

Part C What is the value of the magnetic field at a distance \texttip{r}{r} = 5.60 {\rm cm} from the axis of the conducting tube? Recall that \mu_0=4\pi\times 10^{-7}~{\rm T\cdot m/A}. Express your answer numerically in teslas.

Hint 1. How to approach the problem In order to find the value of the magnetic field at a distance \texttip{r}{r} = 5.60 {\rm cm} from the axis of the conducting tube, you must find an expression for the magnetic field using Ampere's law: \oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\rm encl} where I_{\rm encl} is the total current enclosed by the integration path. Use the geometry of the conductors to determine I_{\rm encl}. Keep in mind that you will have to find the current enclosed from both the inner cylinder and the outer tube and add them together to find the total I_{\rm encl}.

Hint 2. Find an expression for \texttip{I_{\rm encl}}{I_encl} https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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What is the expression for the current I_{\rm encl} enclosed in the path of integration in terms of the current \texttip{I}{I}, the outer radius \texttip{R_a}{R_a}, the inner radius \texttip{R_b}{R_b}, and the distance from the axis \texttip{r}{r} where R_a > r > R_b? Express your answer in terms of \texttip{I}{I}, \texttip{R_a}{R_a}, \texttip{R_b}{R_b}, and \texttip{r}{r}.


Hint 1. How to find I_{\rm encl} enclosed by \texttip{r}{r} To find the total enclosed current I_{\rm encl}, you must add together the enclosed current from both the inner cylinder and the outer tube. Since R_a > r > R_b, only a portion of the total current in the tube will be enclosed by the path of integration. To find the current in this enclosed portion of the tube you must find the current density J in the tube. The current density can then be used to find the current I_1 through the enclosed portion of the conducting tube by first finding the area A_1 of the enclosed portion of the tube and then using the expression I_1=JA_1.

Hint 2. Find an expression for I_{\rm encl} for the cylinder What is the expression for the enclosed current I_{\rm encl} for the entire inner cylinder (i.e., for \texttip{r}{r} between the inner cylinder and the outer tube) in terms of the current \texttip{I}{I}? Express your answer in terms of \texttip{I}{I}. ANSWER: \texttip{I_{\rm encl}}{I_encl} = I

Hint 3. Find an expression for the current density J in the conducting tube What is the expression for the current density J in the conducting tube in terms of the current \texttip{I}{I}, the outer radius \texttip{R_a}{R_a}, and the inner radius \texttip{R_b}{R_b}? Express your answer in terms of \texttip{I}{I}, \texttip{R_a}{R_a}, and \texttip{R_b}{R_b}.

Hint 1. How to find the current density in the tube To find the current through a portion of a conducting tube, first calculate the current density J using the expression J=I_O/A_O for the whole object where A_O is the cross sectional area of the object and I_O is the the total current flowing through the object.

Hint 2. How to find an expression for the area The cross sectional area of the conducting tube is the area of a circle with the inner radius \texttip{R_b}{R_b} subtracted from the area of a circle with the outer radius \texttip{R_a}{R_a} as shown in the figure here.

Recall that the area of a circle of radius R is given by the expression A=\pi R^2.

ANSWER:


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\texttip{J}{J} = \large{\frac{-I}{{\pi} \left({R_{a}}^{2}-{R_{b}}^{2}\right)}}


ANSWER: \texttip{I_{\rm encl}}{I_encl} = \large{I \left(1-\frac{r^{2}-{R_{b}}^{2}}{{R_{a}}^{2}-{R_{b}}^{2}}\right)}

Hint 3. Find the expression for the magnetic field What is the expression for the magnetic field B at a distance \texttip{r}{r} from the axis of the conducting tube (R_a > r > R_b) in terms of the current \texttip{I}{I}, the outer radius \texttip{R_a}{R_a}, the inner radius \texttip{R_b}{R_b}, the distance from the axis of the tube \texttip{r}{r}, and \texttip{\mu _{\rm 0}}{mu_0}? Express your answer in terms of \texttip{I}{I}, \texttip{R_a}{R_a}, \texttip{R_b}{R_b}, \texttip{r}{r}, and \texttip{\mu _{\rm 0}}{mu_0}.

Hint 1. Evaluate the integral in Ampere's law What is the expression for the left hand side of Ampere's law in terms of B and \texttip{r}{r} after the path integral is performed? Recall that Ampere's law is \oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\rm encl} and the path \vec{dl} (defined in Part B) is around a circle of radius \texttip{r}{r}. Express your answer in terms of \texttip{B}{B} and \texttip{r}{r}. ANSWER: \mu_0 I_{\rm encl} = B{\cdot}2 {\pi} r

ANSWER: \texttip{B}{B} = \large{\frac{{\mu}_{0} I}{2 {\pi} r} \left(1-\frac{r^{2}-{R_{b}}^{2}}{{R_{a}}^{2}-{R_{b}}^{2}}\right)}

ANSWER: \texttip{B}{B} = 2.83×10−6

\rm T

Correct

EVALUATE your answer

Part D Rank from largest to smallest the values of the magnetic field at the following distances from the axis of the conducting cylinder: \texttip{R_a}{R_a} = 6.35 {\rm cm} , \texttip{R_b}{R_b} = 3.55 {\rm cm} , \texttip{r}{r} = 5.60 {\rm cm} , and r>R_a. Rank from largest to smallest. To rank items as equivalent, overlap them. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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ANSWER:

Reset

Help


B a r r > R_{\rm a }

B a r R_{\rm b} < r < R_ {\rm a }

B a t R_{\rm b}

B a t R_{\rm a}

The correct ranking cannot be determined.

Correct At the inner radius \texttip{R_b}{R_b} of the conducting tube the current enclosed by the integration path for Ampere's law is only the current \texttip{I}{I} through the inner conductor. As the radius of the integration path increases, the total enclosed current decreases since the current in the conducting tube is in the opposite direction as the current in the inner cylinder. At the outer radius \texttip{R_a}{R_a} of the conducting tube, and for all distances larger than \texttip{R_a}{R_a}, the total enclosed current is zero resulting in zero magnetic field.

Exercise 28.2 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3 \times 10^{ - 11} \;{\rm{ m}} with a speed of 2.2 \times 10^6 \; {\rm{ m/s}}.

Part A


If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise, find the magnitude of the electric field that the electron produces at the location of the nucleus (treated as a point). Express your answer using two significant figures. ANSWER: E = 5.1×1011

\rm N/C

Correct

Part B Find the direction of the electric field that the electron produces at the location of the nucleus (treated as a point). ANSWER: from the electron toward the electron out of the page into the page

Correct

Part C Find the magnitude of the magnetic field that the electron produces at the location of the nucleus (treated as a point). Express your answer using two significant figures. ANSWER: B = 13

\rm T

Correct

Part D Find the direction of the magnetic field that the electron produces at the location of the nucleus (treated as a point). ANSWER:


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from the electron toward the electron out of the page


into the page

Correct

Exercise 28.5 A -4.90 {\rm \mu C} charge is moving at a constant speed of 6.80×105 {\rm m/s} in the +x- {\rm{direction}} relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic-field vector it produces at the following points.

Part A x = 0.500\;{\rm{ m}},\; y = 0, z = 0 Enter your answers numerically separated by commas. ANSWER: B_x,\; B_y,\; B_z =

0,0,0

\rm T

Correct

Part B x = 0, y = 0.500\;{\rm{ m}}, z = 0 Enter your answers numerically separated by commas. ANSWER: B_x,\; B_y,\; B_z = 0,0,−1.33×10−6

\rm T

Correct

Part C x = 0.500\;{\rm{ m}}, y = 0.500\;{\rm{ m}}, z = 0 Enter your answers numerically separated by commas. ANSWER: B_x,\; B_y,\; B_z = 0,0,−4.71×10−7

\rm T


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Correct


Part D x = 0, y = 0, z = 0.500\;{\rm{ m}} Enter your answers numerically separated by commas. ANSWER: B_x,\; B_y,\; B_z = 0,1.33×10−6,0

\rm T

Correct

Exercise 28.10 A short current element d\vec l = (0.500 \rm mm)\hat j carries a current of 4.90 {\rm \;A} in the same direction as d\vec l. Point P is located at \vec r = ( -0.730 \rm m)\hat i+ (0.390\,\rm m)\hat k.

Part A Find the magnetic field at P produced by this current element. Enter the x, y, and z components of the magnetic field separated by commas. ANSWER: dB_x, dB_y, dB_z = 1.69×10−10,0,3.16×10−10

{\rm T}

Correct

Exercise 28.22 Currents in dc transmission lines can be 100 {\rm A} or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.

Part A For a line that has current 130 {\rm \;A} and a height of 8.0 {\rm m} above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas. ANSWER: B = 3.3×10−6

\rm T

Correct


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Part B What magnetic field does the line produce at ground level as a percent of the earth's magnetic field, which is 0.50 \rm G. Express your answer using two significant figures.


ANSWER: \large{\frac{B}{B_E}} =

6.5

\%

Correct

Part C Is this value of magnetic field cause for worry? ANSWER: Yes. Since this field is does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth’s field. No. Since this field is does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth’s field. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth’s field. No. Since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth’s field.

Correct

Exercise 28.30 Two long, parallel wires are separated by a distance of 3.70 {\rm {\rm cm}} . The force per unit length that each wire exerts on the other is 5.00×10−5 {\rm {\rm N/m}} , and the wires repel each other. The current in one wire is 0.680 {\rm {\rm A}} .

Part A What is the current in the second wire? ANSWER: I =

13.6 {\rm A}

Correct

Part B Are the two currents in the same direction or in opposite directions? ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4706985

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In the same direction In opposite direction Typesetting math: 63%

Correct

Exercise 28.36 A closely wound, circular coil with radius 2.40 {\rm \;{\rm cm}} has 800 turns.

Part A What must the current in the coil be if the magnetic field at the center of the coil is 0.0750 {\rm \;{\rm T}} ? Express your answer to three significant figures and include the appropriate units. ANSWER: I =

3.58 {\rm A}

Correct

Part B At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center? Express your answer to three significant figures and include the appropriate units. ANSWER: x = 1.84 {\rm cm}

Correct

Exercise 28.41 A closed curve encircles several conductors. The line integral around this curve is \oint \vec{B} \cdot d\vec{l}= 4.45×10−4 {\rm {\rm T \cdot m}} .

Part A What is the net current in the conductors? ANSWER: I =

354

{\rm A}


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Correct


Part B If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? ANSWER: −4.45×10−4

{\rm T} \cdot {\rm m}

Correct

Exercise 28.42 As a new electrical technician, you are designing a large solenoid to produce a uniform 0.120-\rm T magnetic field near the center of the solenoid. You have enough wire for 3600 circular turns. This solenoid must be 58.0 {\rm \;{\rm cm}} long and 4.00 {\rm \;{\rm cm}} in diameter.

Part A What current will you need to produce the necessary field? Express your answer to three significant figures and include the appropriate units. ANSWER: I =

15.4 {\rm A}

Correct

Problem 28.68 In the wire shown in segment BC is an arc of a circle with radius 30.0 \rm cm, and point P is at the center of curvature of the arc. Segment DA is an arc of a circle with radius 20.0 \rm cm, and point P is at its center of curvature. Segments CD and AB are straight lines of length 10.0 \rm cm each.


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Part A Calculate the magnitude of the magnetic field at a point P due to a current 11.0 {\rm \;A} in the wire. Expressmath: your63% answer Typesetting

with the appropriate units

ANSWER: B = 3.84 {\rm {\mu}T}

Correct

Part B What is the direction of magnetic field? ANSWER: out of the page into the page

Correct

Problem 28.73 Long, straight conductors with square cross sections and each carrying current I are laid side by side to form an infinite current sheet (the figure ). The conductors lie in the xy-plane, are parallel to the y-axis, and carry current in the - y-\, {\rm{direction}} There are n conductors per unit length measured along the x-axis.

Part A What are the magnitude of the magnetic field a distance a below the current sheet? Express your answer in terms of the variables I, n, a and appropriate constants (\mu_0 and \pi). ANSWER:


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