22 2p15v

MasteringPhysics: Assignment Print View

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Homework 2 Due: 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 [Switch to Standard Assignment View]

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

runs in the -y direction

wires per unit length in the x direction.

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.

Hint A.1

How to approach the problem Hint not displayed

Hint A.2

Find Hint not displayed

Hint A.3

Determine the direction of the magnetic field Hint not displayed

Hint A.4

Magnitude of the magnetic field Hint not displayed

Hint A.5

Evaluate

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Hint not displayed Express the magnetic field as a vector in of any or all of the following: the unit vectors

, ,

, ,

, and

, , and/or .

ANSWER: = Correct

This equation is analogous to

on either side of a infinitely charged sheet. The

correspondence seems more obvious if you set the current per 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.

Force between an Infinitely Long Wire and a Square Loop A square loop of wire with side length

carries a current

. The center of the loop is located a distance

from an infinite wire carrying a current

. The infinite wire 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 the magnitude, Hint A.1

, of the net force on the loop?

How to approach the problem

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Hint not displayed Hint A.2

Determine the direction of force Hint not displayed

Hint A.3

Determine the magnitude of force Hint not displayed

Hint A.4

Find the force on the section of the loop closest to the wire Hint not displayed

Hint A.5

Find the magnetic field due to the wire Hint not displayed

Express the force in of

,

, ,

, and

.

ANSWER: = Correct

Part B The magnetic moment

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

loop times the magnitude of the current flowing in it ( plane in which the current flows. Find the magnitude,

), and whose direction is perpendicular to the , of the force on the loop from Part A in of the

magnitude of its magnetic moment. Express

in of

,

, ,

, and

.

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.

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

is flowing through 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 .

Hint A.1

Formula for the magnetic field of a current-carrying wire (Biot-Savart law) Hint not displayed

Hint A.2

Simplify the vector integral Hint not displayed

Hint A.3


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Hint A.4



Express your answer in of ,

,

, and

.

ANSWER: = Correct

The magnetic field for an infinitely long wire can be obtained by setting

in the previous

expression. This gives a magnetic field , which you probably derived in an earlier problem or in lecture using the Biot-Savart law. Part B Now find

, the magnetic field generated by this wire at a point P located a distance

from either end of

the wire. Assume that at P the angle subtended from the end of the wire to the other end is

as shown

in the diagram .

Hint B.1


Hint B.2

Limiting value of Hint not displayed

Express your answer in of , ANSWER:

,

, and

.

=

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All attempts used; correct answer displayed Setting

in the previous expression yields the magnetic field for a semi-infinite wire: ,

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.

Magnetic Field of a Current-Carrying Wire Find the magnetic field a distance current per unit area

from the center of a long wire that has radius

and carries a uniform

in the positive z direction.

Part A First find the magnetic field,

Hint A.1

, outside the wire (i.e., when the distance

is greater than ).

Ampère's law with current density Hint not displayed

Hint A.2

Find the direction of the field Hint not displayed

Hint A.3

Find the left-hand side of Ampère's law Hint not displayed

Hint A.4

Find the right-hand side of Ampère's law Hint not displayed

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Express


in of the given parameters, the permeability constant

(the magnitude of

, the variables ,

), , , and , and the corresponding unit vectors , , and . You may not

need all these in your answer. ANSWER: = All attempts used; correct answer displayed

Part B Now find the magnetic field

Hint B.1

inside the wire (i.e., when the distance

is less than ).

Establish the relationiship to Part A Hint not displayed

Hint B.2

Integrate over the Ampèrean loop Hint not displayed

Express

in of the given parameters, the permeability constant

, the distance

from

the center of the wire, and the unit vectors , , and . You may not need all these in your answer. ANSWER: = Correct

Magnetic Field from Two Wires Learning Goal: To understand how to use the principle of superposition in conjunction with the Biot-Savart (or Ampere's) law.

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From the Biot-Savart law, it can be calculated that the magnitude of the magnetic field due to a long straight wire is given by , where

(

) is the permeability constant,

is the current in the wire, and

is the

distance from the wire to the location at which the magnitude of the magnetic field is being calculated. The same result can be obtained from Ampere's law as well.

The direction of vector

can be found using the right-hand rule. (Take care in applying the right-hand rule.

Many students mistakenly use their left hand while applying the right-hand rule since those who use their right hand for writing sometimes automatically use their "pencil-free hand" to determine the direction of .)

In this problem, you will be asked to calculate the magnetic field due to a set of two wires with antiparallel currents as shown in the diagram . Each of the wires carries a current of magnitude . The current in wire 1 is directed out of the page and that in wire 2 is directed into the page. The distance between the wires is . The x axis is perpendicular to the line connecting the wires and is equidistant from the wires. As you answer the questions posed here, try to look for a pattern in your answers.

Part A Which of the vectors best represents the direction of the magnetic field created at point K (see the diagram in the problem introduction) by wire 1 alone? Enter the number of the vector with the appropriate direction.

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ANSWER:


3 Correct

Part B Which of the vectors best represents the direction of the magnetic field created at point K by wire 2 alone? Enter the number of the vector with the appropriate direction.

ANSWER:

3 Correct

Part C Which of these vectors best represents the direction of the net magnetic field created at point K by both wires? Enter the number of the vector with the appropriate direction.

ANSWER:

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3

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Correct

Part D Find the magnitude of the magnetic field Express your answer in of ,

created at point K by wire 1. , and appropriate constants.

ANSWER: = Correct

Of course,

because point K is equidistant from the wires.

Part E Find the magnitude of the net magnetic field Express your answer in of ,

created at point K by both wires.

, and appropriate constants.

ANSWER: = Correct

This result is fairly obvious because of the symmetry of the problem: At point K, the two wires each contribute equally to the magnetic field. At points L and M you should also consider the symmetry of the problem. However, be careful! The vectors will add up in a more complex way. Part F Point L is located a distance magnetic field Hint F.1

from the midpoint between the two wires. Find the magnitude of the

created at point L by wire 1.


Use the distances provided and the Pythagorean Theorem to find the distance between wire 1 and point L. Express your answer in of ,


ANSWER: = All attempts used; correct answer displayed

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Part G Point L is located a distance magnetic field Hint G.1

from the midpoint between the two wires. Find the magnitude of the net

created at point L by both wires.


Sketch a detailed diagram with all angles marked; draw vectors

and

; then add them using the

parallelogram rule. Hint G.2

Find the direction of the magnetic field due to wire 1

Which of the vectors best represents the direction of the magnetic field created at point L (see the diagram in the problem introduction) by wire 1 alone? Enter the number of the vector with the appropriate direction.

ANSWER:

2 All attempts used; correct answer displayed

Hint G.3

Find the direction of the magnetic field due to wire 2

Which of the vectors best represents the direction of the magnetic field created at point L by wire 2 alone? Enter the number of the vector with the appropriate direction.

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ANSWER:

Hint G.4


Answer not displayed

Find the direction of the net magnetic field

Which of the vectors best represents the direction of the net magnetic field created at point L by both wires? Enter the number of the vector with the appropriate direction.

ANSWER:

Hint G.5

Answer not displayed

Angle between magnetic field due to wire 1 and the x axis

Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to wire 1 at point L and the x axis. Hint G.6

Find the angle between magnetic field due to wire 1 and the x axis

Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to wire 1 at point L and the x axis. Express your answer numerically, in degrees. ANSWER:

Hint G.7

= Answer not displayed

Net magnetic field

Consider the symmetry of the magnetic field at point L due to wire 1 and the magnetic field due to wire 2. You should note that the y components of these two vectors are of equal magnitude but are opposite in

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direction. Therefore they will cancel when added together, leaving you only to worry about the x components. Find the x component of the magnetic field at point L due to wire 1 by using the magnitude of the vector (found in Part F) and the angle between the x axis and the magnetic field vector (found in the previous hint). Because of symmetry, the x component of the magnetic field at point L due to wire 2 is the same size. To find the net magnetic field at point L you need to add together the x components of the magnetic field at point L due to wire 1 and of the magnetic field due to wire 2. Express your answer in of ,



Part H Point M is located a distance magnetic field

from the midpoint between the two wires. Find the magnitude of the

created at point M by wire 1.



ANSWER: = Correct

Part I Find the magnitude of the net magnetic field Hint I.1

created at point M by both wires.


Hint I.2

Find the direction of the magnetic field due to wire 1 Hint not displayed

Hint I.3

Find the direction of the net magnetic field Hint not displayed

Hint I.4

Angle between magnetic field due to wire 1 and the x axis Hint not displayed

Hint I.5

Find the angle between magnetic field due to wire 1 and the x axis Hint not displayed

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Hint I.6


Net magnetic field Hint not displayed




Part J Finally, consider point X (not shown in the diagram) located on the x axis very far away in the positive x direction. Which of the vectors best represents the direction of the magnetic field created at point X by wire 1 alone? Enter the number of the vector with the appropriate direction.

ANSWER:

1 Correct

Part K Which of the vectors best represents the direction of the magnetic field created at point X by wire 2 alone? Enter the number of the vector with the appropriate direction.

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ANSWER:


5 Correct

As you can see, at a very large distance, the individual magnetic fields (and the corresponding magnetic field lines) created by the wires are directed nearly opposite to each other, thus ensuring that the net magnetic field is very, very small even compared to the magnitudes of the individual magnetic fields, which are also relatively small at a large distance from the wires. Thus, at a large distance, the magnetic fields due to the two wires almost cancel each other out! (That is, if point X is very far from each wire, then the ratio is very close to zero.) Another way to think of this is as follows: If you are really far from the wires, then it's hard to tell them apart. It would seem as if the current were traveling up and down, almost along the same line, thereby appearing much the same as a single wire with almost no net current (because the up and down currents almost cancel each other), and therefore almost no magnetic field. Note that this only works for points very far from the wires; otherwsie it's easy to tell that the wires are separated and the currents don't cancel, since they are going up and down at different locations. It comes as no surprise then that one way to eliminate unnecessary magnetic fields in electric circuits is to twist together the wires carrying equal currents in opposite directions.

± Magnetic Field from Current Segments Learning Goal: To apply the Biot-Savart law to find the magnetic field produced on the z axis from current elements in the xy plane. In this problem you are to find the magnetic field component along the z axis that results from various current elements in the xy plane (i.e., at ). The field at a point due to a current-carrying wire is given by the Biot-Savart law, , where

and

, and the integral is done over the current-carrying wire. Evaluating the vector

integral will typically involve the following steps: Choose a convenient coordinate system--typically rectangular, say with coordinate axes Write

in of the coordinate variables and directions ( and

,

, , and .

, etc.). To do this, you must find

. Again, finding the cross product can be done either

geometrically (by finding the direction of the cross product vector first, then checking for cancellations from any other portion of the wire, and then finding the magnitude or relevant component) or algebraically (by using , etc.). Evaluate the integral for the component(s) of interest. In this problem, you will focus on the second of these steps and find the integrand for several different current elements. You may use either of the two methods suggested for doing this. Part A

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The field at the point shown in the figure due to a single current element is given by , where

and

is the variable

Hint A.1

. In this expression, what

in of

and/or

?

Making sense of subscripts Hint not displayed

ANSWER:

Correct

Part B Find

, the z component of the magnetic field at the point

flowing over a short distance

Hint B.1

located at the point

from the current .

Cross product

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Hint not displayed Express your answer in of ,

,

,

, and

. Recall that a component is a scalar; do not

enter any unit vectors. ANSWER:

= 0

Correct

Part C Find



Hint C.1


from the current .

Determine the displacement from the current element Hint not displayed

Hint C.2

Find the direction from the cross product Hint not displayed


,

,

, and


enter any unit vectors. ANSWER: = Correct

Part D Find



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from the current .

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Determine the displacement from the current element

Hint D.1

Hint not displayed Find the direction of the magnetic field vector

Hint D.2


,

,

, and



Part E Find current

Hint E.1

, the z component of the magnetic field at the point P located at flowing over a short distance


from the

.


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Use the cross product to get the direction

Hint E.2


,

,

,

, and


enter any unit vectors. ANSWER:

= 0

Correct

Part F Find current

, the z component of the magnetic field at the point P located at flowing over a short distance

Hint F.1


from the

.


Hint F.2

Determine which unit vector to use Hint not displayed

Hint F.3

Evaluate the cross product Hint not displayed


,

,

,

, and



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Force between Moving Charges Two point charges, with charges and , are each moving with speed toward the origin. At the instant shown

is at position (0,

) and

is at (

, 0).

(Note that the signs 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? Hint A.1

Which law to use Hint not displayed

Hint A.2

Find the value of Hint not displayed

Express

in of

,

,

, and

.

ANSWER: = Correct

Part B What is the magnitude of the magnetic force on Hint B.1

due to the magnetic field caused by

?


Hint B.2

Magnitude of the magnetic field Hint not displayed

Hint B.3

Find the direction of the magnetic field Hint not displayed

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Hint B.4


Computing the force Hint not displayed

Express the magnitude of the magnetic force in of

,

, ,

, and

.

ANSWER: = Correct

Part C Assuming that the charges are moving nonrelativistically (

), what can you say about the relationship

between the magnitudes of the magnetic and electrostatic forces? Hint C.1


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.

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 counterclockwise direction as seen from the top (looking in the negative

flowing through it in a

direction).

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Part A What is Hint A.1

, the z component of

at the center (i.e.,

) of the loop?

Specify the integrand Hint not displayed

Hint A.2

Perform the integration Hint not displayed

Express your answer in of , , and constants like

and

.

ANSWER: = Correct

Magnetic Field due to Semicircular Wires A loop of wire is in the shape of two concentric semicircles as shown. The inner circle has radius ; the outer circle has radius . A current

flows clockwise through the outer

wire and counterclockwise through the inner wire.

Part A What is the magnitude, Hint A.1

, of the magnetic field at the center of the semicircles?

What physical principle to use Hint not displayed

Hint A.2

Compute the field due to the inner semicircle Hint not displayed

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Hint A.3


Direction of the field due to the inner semicircle Hint not displayed

Hint A.4

Compute the field due to the straight wire segments Hint not displayed

Express

in of any or all of the following: , , , and

.

ANSWER: = Correct

To see whether

and

makes sense, think of the scaling of different quantities. The

size of the current element scales as the radius, whereas the power of

in the denominator is 2 (and

equals the radius also, in this case). So over all, you would expect the magnetic field to scale as 1/radius. Note that such an argument works only because the field due to each point is in the same direction, so you are doing a much simpler integral. Part B What is the direction of the magnetic field at the center of the semicircles? ANSWER:

into the screen out of the screen Correct

Magnetic Field near a Moving Charge A particle with positive charge

is moving with speed

problem it is located at the origin,

along the z axis toward positive . At the time of this

. Your task is to find the magnetic field at various locations in

the three-dimensional space around the moving charge.

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

due to the moving charge?

A. B. C. D.

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 vector . A second point is that the order of the cross product must be such 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 Hint B.1

.

Find the magnetic field direction Hint not displayed

Express your answer in of ANSWER:

, , , and

, and use

, , and

for the three unit vectors.

=

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Correct

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

. , , ,

, and

, and use

, , and

for the three unit

vectors. ANSWER: = Correct

Part D Find the magnetic field at the point Hint D.1

.

Evaluate the cross product Hint not displayed

Hint D.2

Find the distance from the charge Hint not displayed

Express your answer in of

, , ,

,

, and

, and use

, , and

for the three unit

vectors. 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 carrying current provided that the quantity is replaced by which quantity? Hint E.1

Making a correlation Hint not displayed

ANSWER: Correct

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± Canceling a Magnetic Field Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 38.0 , as shown in the figure . The currents running through the wires are 8.0

, 20.0

, 10.0

, and .

Part A Find the magnitude of the current

that will make the magnetic field at the center of the square equal to

zero. Hint A.1


Hint A.2

Calculating the contribution from the three known wires Hint not displayed

Express your answer in amperes. ANSWER:

= 2.00 Correct

Part B What is the direction of the current ? Hint B.1


ANSWER:

upward downward Correct

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Torque on a Current Loop in a Magnetic Field Learning Goal: To understand the origin of the torque on a current loop due to the magnetic forces on the current-carrying wires. This problem will show you how to calculate the torque on a magnetic dipole in a uniform magnetic field. We start with a rectangular current loop, the shape of which allows us to calculate the Lorentz forces explicitly. Then we generalize our result. Even if you already know the general formula to solve this problem, you might find it instructive to discover where it comes from. Part A A current

flows in a plane rectangular current loop with height

placed into a uniform magnetic field and there is an angle and

and horizontal sides . The loop is

in such a way that the sides of length

are perpendicular to

,

between the sides of length

.

Calculate , the magnitude of the torque about the vertical axis of the current loop due to the interaction of the current through the loop with the magnetic field.

Hint A.1


Hint A.2

Forces on the parts of the loop that have length Hint not displayed

Hint A.3

Force on a current-carrying wire in a magnetic field

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Hint not displayed Hint A.4

Torque on a loop Hint not displayed

Hint A.5

Forces on the parts of the loop that have length Hint not displayed

Express the magnitude of the torque in of the given variables. You will need a trigonomeric function [e.g., or ]. Use for the magnitude of the magnetic field. ANSWER:

= Correct

Part B Give a more general expression for the magnitude of the torque . Rewrite the answer found in Part A in of the magnitude of the magnetic dipole moment of the current loop vector perpendicular to the plane of the coil and the magnetic field to be complement of angle Hint B.1

. Define the angle between the , noting that this angle is the

in Part A.

Definition of the magnetic dipole moment Hint not displayed

Give your answer in of the magnetic moment ANSWER:

, magnetic field

, and

.

= Correct

The more general vector form of this expression is .

Part C A current magnitude

flows around a plane circular loop of radius , giving the loop a magnetic dipole moment of . The loop is placed in a uniform magnetic field

, with an angle

between the direction of

the field lines and the magnetic dipole moment as shown in the figure. Find an expression for the magnitude of the torque on the current loop.

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Formula for the area of a circle

Hint C.1

Hint not displayed Express the torque explicitly in of , ,

,

the respective vector quantities). Do not use

. You will need a trigonomeric function [e.g..

or

, and

(where

and

are the magnitudes of

].

ANSWER: = Correct

± Magnetic Force on a Current-Carrying Wire Learning Goal: To understand the magnetic force on a straight current-carrying wire in a uniform magnetic field. Magnetic fields exert forces on moving charged particles, whether those charges are moving independently or are confined to a current-carrying wire. The magnetic force on an individual moving charged particle depends on its velocity

and charge . In the case of a current-carrying wire, many charged particles are

simultaneously in motion, so the magnetic force depends on the total current

and the length of the wire

The size of the magnetic force on a straight wire of length

in a uniform magnetic field

with strength

carrying current

.

is .

Here

is the angle between the direction of the current (along the wire) and the direction of the magnetic

field. Hence

refers to the component of the magnetic field that is perpendicular to the wire,

this equation can also be written as

. Thus

.

The direction of the magnetic force on the wire can be described using a "right-hand rule." This will be discussed after Part B. Part A Consider a wire of length = 0.50

= 0.30

that runs north-south on a horizontal surface. There is a current of

flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not

shown.) The Earth's magnetic field at this location has a magnitude of 0.50

(or, in SI units,

) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that .

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Express your answer in newtons to two significant figures.

ANSWER:

4.6×10−6 Correct

Because the Earth's magnetic field is quite modest, this force is so small that it might be hard to detect. Part B Now assume that a strong, uniform magnetic field of size 0.55

pointing straight down is applied. What is

the size of the magnetic force on the wire due to this applied magnetic field? Ignore the effect of the Earth's magnetic field. Hint B.1

Determining the angle theta Hint not displayed

Express your answer in newtons to two significant figures. ANSWER:

8.3×10−2 Correct

This force would be noticeable if the wire were of light weight. The direction of the magnetic force is perpendicular to both the direction of the current flow and the direction of the magnetic field. Here is a "right-hand rule" to help you determine the direction of the magnetic force. Straighten the fingers of your right hand and point them in the direction of the current. Rotate your arm until you can bend your fingers to point in the direction of the magnetic field. Your thumb now points in the direction of the magnetic force acting on the wire.

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Part C What is the direction of the magnetic force acting on the wire in Part B due to the applied magnetic field? ANSWER:

due north due south due east due west straight up straight down Correct

Part D Which of the following situations would result in a magnetic force on the wire that points due north? Check all that apply. ANSWER:

Current in the wire flows straight down; the magnetic field points due west. Current in the wire flows straight up; the magnetic field points due east. Current in the wire flows due east; the magnetic field points straight down. Current in the wire flows due west and slightly up; the magnetic field points due east. Current in the wire flows due west and slightly down; the magnetic field points straight down. Correct

As you can see, many current/magnetic field configurations can result in the same direction of magnetic force. Part E Assume that the applied magnetic field of size 0.55

is rotated so that it points horizontally due south.

What is the size of the magnetic force on the wire due to the applied magnetic field now? Hint E.1

Determining the angle theta Hint not displayed

Express your answer in newtons to two significant figures. ANSWER:

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0

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Correct Notice that whenever the current in the wire and the magnetic field point in the same direction ( or in opposite directions (

), the sine of

)

is zero, so there is no magnetic force exerted on the

wire. This is consistent with the earlier statement that it is the component of the magnetic field that is perpendicular to the direction of the current that produces the magnetic force. Also notice that for these two special values of (when the current is flowing parallel to or antiparallel to the magnetic field) the steps listed for the right-hand rule suggest a unique direction for the magnetic force. This is another clue that the magnetic force is zero.

A Conductor Moving in a Magnetic Field A metal cube with sides of length

is moving at velocity

across a uniform magnetic field

.

The cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vector of two faces are parallel to the direction of motion).

Part A Find

, the electric field inside the cube.

Hint A.1

Net force on charges in a conductor Hint not displayed

Hint A.2

Find the magnetic force magnitude Hint not displayed

Hint A.3

Find the magnetic force direction Hint not displayed

Hint A.4

Determine the force due to the electric field Hint not displayed

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Express the electric field in of ANSWER:

,

, and unit vectors (

, , and/or ).

= Correct

Now, instead of electrons, suppose that the free charges have positive charge . Examples include "holes" in semiconductors and positive ions in liquids, each of which act as "conductors" for their free charges. Part B If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric field point? ANSWER:

Correct

The direction of the electric field stays the same regardless of the sign of the charges that are free to move in the conductor. Mathematically, you can see that this must be true since the expression you derived for the electric field is independent of . Physically, this is because the force due to the magnetic field changes sign as well and causes positive charges to move in the direction (as opposed to pushing negative charges in the direction). Therefore the result is always the same: positive charges on the

side and negative charges on

the

side. Because the electric field goes from positive to negative charges

the

direction (given the original directions of

and

will always point in

).

A Thomson Experiment In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron. With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, , of these particles, repeating the measurements with different cathode materials and different residual gases in the tube. Part A What is the most significant conclusion that Thomson was able to draw from his measurements?

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ANSWER:


He found a different value of

for different cathode materials.

He found the same value of From measurements of

for different cathode materials. he was able to calculate the charge of an

electron. From measurements of

he was able to calculate the mass of an electron.

Correct

That the quantity

was independent of the material provided clear evidence that the particles

comprising the cathode rays were, in fact, a common constituent of matter. Imagine that you set up an apparatus to reproduce Thomson's experiment, as shown in the figure . In an highly evacuated glass tube, a beam of electrons, each moving with speed , es between two parallel plates and strikes a screen at a distance from the end of the plates. First, you observe the point where the beam strikes the screen when there is no electric field between the plates. Then, you observe the point where the beam strikes the screen when a uniform electric field of magnitude is established between the plates. Call the distance between these two points

.

Part B What is the distance and use Hint B.1

and

between the two points that you observe? Assume that the plates have length ,

for the charge and the mass of the electrons, respectively.


Hint B.2

Find the deflection at the end of the plates Hint not displayed

Hint B.3

Find the deflection in the field-free region Hint not displayed


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,

,

,

, and

.

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ANSWER: = Answer Requested

Part C Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude magnetic field is perpendicular to

. The

and directed

straight into the plane of the figure. You adjust the value of so that no deflection is observed on the screen. What is the speed

Hint C.1

of the electrons in this case?


Hint C.2

Find the magnetic force exerted on the electrons Hint not displayed

Hint C.3

Find the electric force exerted on the electrons Hint not displayed


and

.

ANSWER: = Correct

Part D In your experiment, you measure a total deflection of 4.12

when an electric field of

is

established between the plates (with no magnetic field present). When you add the magnetic field as described in Part C, to what value do you have to adjust its magnitude to observe no deflection? Assume that the plates are 6.00

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long and that the distance between them and the screen is 12.0

.

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Hint D.1



Hint D.2

Find the speed of the electrons Hint not displayed

Express your answer numerically in tesla. ANSWER:

−4 = 1.69×10 Correct

Part E Now suppose you carry out a second Thomson experiment with a different beam that contains two types of particles. In particular, both types have the same mass as an electron, but one has charge and the other has charge

. As in the previous experiment, initially only the electric field is imposed and the

deflection of the beam is observed as a spot on the screen; then, in the second phase of the experiment, one attempts to tune the magnetic field to exactly cancel the effect of the electric field. What would you observe on the screen during this experiment? ANSWER:

Two off-centered spots in both the first and the second phases of the experiment Two off-centered spots in the first phase of the experiment; one centered spot in the second phase of the experiment Two off-centered spots in the first phase of the experiment; one centered and one off-centered spot in the second phase of the experiment One off-centered spot in the first phase of the experiment; one centered and one off-centered spot in the second phase of the experiment One off-centered spot in the first phase of the experiment; two off-centered spots in the second phase of the experiment One off-centered spot in both the first and the second phases of the experiment Correct

In the first phase of the experiment, when only the electric field is present, the particles with charge are deflected exactly as the electrons, whereas the particles with charge

experience a greater

force and suffer a larger deflection. Hence, two distinct spots will be observed on the screen. In the second phase of the experiment, when a magnetic field is added, only one spot will be observed on the screen because the balance of the electric force and the magnetic force does not depend on the charge of the particles. Thus, when the magnetic field is tuned to cancel the effect of the electric force, the beam of particles, regardless of the magnitude of their charge, will through the plates undeflected.

Charged Particles Moving in a Magnetic Field Ranking Task

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Five equal-mass particles (A–E) enter a region of uniform magnetic field directed into the page. They follow the trajectories illustrated in the figure.

Part A Which particle (if any) is neutral? Hint A.1

Neutral particles Hint not displayed

ANSWER:

particle A particle B particle C particle D particle E none Correct

Part B Which particle (if any) is negatively charged? Hint B.1

Find the direction of the magnetic force Hint not displayed

ANSWER:

particle A particle B particle C particle D particle E

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none Correct

Part C Rank the particles on the basis of their speed. Hint C.1

Determining velocity based on particle trajectories Hint not displayed

Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER:

View All attempts used; correct answer displayed

Part D Rank the particles A, B, C, and E on the basis of their speed. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER:

View Correct

Part E Now assume that particles A, B, C, and E all have the same magnitude of electric charge. Rank the particles A, B, C, and E on the basis of their speed. Hint E.1

Charged particle trajectories in magnetic fields Hint not displayed

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Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER:

View Correct

Interaction of a Current Loop with a Magnetic Field The effects due to the interaction of a current-carrying loop with a magnetic field have many applications, some as common as the electric motor. This problem illustrates the basic principles of this interaction. Consider a current sides

= 2.00

magnetic field

that flows in a plane rectangular current loop with height

= 4.00

and horizontal

. The loop is placed into a uniform in such a way that the sides of

length


, and there is an

angle

between the sides of length

and

, as

shown in the figures.

Part A

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Will the interaction of the current through the loop with the magnetic field cause the loop to rotate? Hint A.1

Find the direction of the forces on the parts of the loop that have length Hint not displayed

Hint A.2


ANSWER:

Yes, the net torque acting on the loop is negative and tends to rotate the loop in the direction of decreasing angle (clockwise). Yes, the net torque acting on the loop is positive and tends to rotate the loop in the direction of increasing angle (counterclockwise). No, the net torque acting on the loop is zero and the loop is in equilibrium. No, the net force acting on the loop is zero and the loop is in equilibrium. Correct

For parts B and C, the loop is initially positioned at

.

Part B Assume that the current flowing into the loop is 0.500

. If the magnitude of the magnetic field is 0.300

,

what is , the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field? Hint B.1

Torque on a current-carrying loop Hint not displayed

Hint B.2

Find the area of the loop Hint not displayed

Hint B.3

Find the angle between the normal to the loop and the magnetic field Hint not displayed

Express your answer in newton-meters. ANSWER:

−4 = 1.04×10 All attempts used; correct answer displayed

Part C

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What happens to the loop when it reaches the position for which of length

Hint C.1


, that is, when its horizontal sides

(see the figure)?

Find the net torque acting on the loop Hint not displayed

ANSWER:

The direction of rotation changes because the net torque acting on the loop causes the loop to rotate in a clockwise direction. The net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise direction. The net torque acting on the loop is zero; therefore it stops rotating. The net force acting on the loop is zero, so the loop must be in equilibrium. Correct

Part D Now suppose that you change the initial angular position of the loop relative to is placed in such a way that initially the angle between the sides of length

and

, and assume that the loop is

, as shown in

the figure. Will the interaction of the current through the loop with the magnetic field cause the loop to rotate?

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Hint D.1



Hint D.2


ANSWER:

Yes, the net torque acting on the loop is negative and tends to rotate the loop in the direction of decreasing angle (clockwise). Yes, the net torque acting on the loop is positive and tends to rotate the loop in the direction of increasing angle (counterclockwise). No, the net torque acting on the loop is zero and the loop is in equilibrium. No, the net force acting on the loop is zero and the loop is in equilibrium. Correct

Depending on the initial position of the loop relative to different. If initially

, the direction of rotation of the loop will be

, then the net torque acting on the loop will cause the loop to rotate in

the counterclockwise direction. If instead,

, then the net torque will rotate the loop in the

opposite direction.

Mass Spectrometer J. J. Thomson is best known for his discoveries about the nature of cathode rays. Another important contribution of his was the invention, together with one of his students, of the mass spectrometer. The ratio of mass to (positive) charge of an ion may be accurately determined in a mass spectrometer. In essence, the spectrometer consists of two regions: one that accelerates the ion through a potential

and a second

that measures its radius of curvature in a perpendicular magnetic field. The ion begins at potential and is accelerated toward zero potential. When the particle exits the region with the electric field it will have obtained a speed .

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Part A With what speed

does the ion exit the acceleration region?

Suggested general method

Hint A.1

Hint not displayed Initial energy

Hint A.2

Hint not displayed Final energy

Hint A.3

Hint not displayed Find the speed in of

, ,

, and any constants.

ANSWER: = Correct

Part B After being accelerated, the particle enters a uniform magnetic field of strength radius

and travels in a circle of

(determined by observing where it hits on a screen--as shown in the figure). The results of this

experiment allow one to find

in of the experimentally measured quantities such as the particle

radius, the magnetic field, and the applied voltage. What is ? Hint B.1

Cyclotron frequency Hint not displayed

Hint B.2

Relationship of

and Hint not displayed

Hint B.3

Putting it all together Hint not displayed

Express

in of

,

,

, and any necessary constants.

ANSWER: = Correct

5/12/2011 8:00 PM



By sending atoms of various elements through a mass spectrometer, Thomson's student, Francis Aston, discovered that some elements actually contained atoms with several different masses. Atoms of the same element with different masses can only be explained by the existence of a third subatomic particle in addition to protons and electrons: the neutron.

Motion of Electrons in a Magnetic Field An electron of mass

and charge

is moving through a uniform magnetic field

At the origin, it has velocity and

in vacuum.

, where

. A screen is mounted perpendicular

to the x axis at a distance

from the origin.

Throughout, you can assume that the effect of gravity is negligible.

Part A First, suppose Hint A.1

. Find the y coordinate

of the point at which the electron strikes the screen.

Forces acting on electron Hint not displayed

Hint A.2

Two-dimensional kinematics Hint not displayed


and the velocity components

and

.

ANSWER: = Correct

Part B Now suppose

, and another electron is projected in the same manner. Which of the following is the

most accurate qualitative description of the electron's motion once it enters the region of nonzero magnetic field?

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ANSWER:


The electron decelerates before coming to a halt and turning around while always moving along a straight line. The electron's motion will be unaffected. (It will continue moving in a straight line with the same constant velocity.) The electron moves in a circle in the xy plane. The electron moves along a helical path about an axis parallel to the field lines with constant radius and constant velocity in the x direction. Correct

Part C The motion of the electron can be broken down into two parts: 1. constant motion in the x direction plus 2. a periodic part, the projection of which in the yz plane is circular. Find the angular velocity of the electron associated with the circular component of its motion. Hint C.1

Use Newton's 2nd law to determine the force Hint not displayed

Hint C.2

Find the magnitude of the Lorentz force Hint not displayed

Hint C.3

Express

in of Hint not displayed

Hint C.4

Express the acceleration in of Hint not displayed

Express the magnitude of the angular velocity in of

, the magnitude of the electric charge

, and other known quantities. ANSWER: = Correct

A DC Motor A shunt-wound DC motor with the field coils and rotor connected in parallel (see the figure) operates from a 135 DC power line. The resistance of the field windings, , is 4.20

, is 206

. The resistance of the rotor,

. When the motor is running, the rotor

develops an emf

. The motor draws a current of

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4.12


from the line. Friction losses amount to 44.0

.

Part A Compute the field current Hint A.1

.

Voltages in a parallel circuit Hint not displayed

Hint A.2

Current through a resistor Hint not displayed

Express your answer in amperes. ANSWER:

= 0.655 Correct

Part B Compute the rotor current Hint B.1

.

Currents in a parallel circuit Hint not displayed

ANSWER:

= 3.46 Correct

Part C Compute the emf Hint C.1

.


Hint C.2

Relation between voltage and current Hint not displayed

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ANSWER:


= 120 Correct

Part D Compute the rate of development of thermal energy in the field windings. Hint D.1

Equation for thermal energy Hint not displayed

Express your answer in watts. ANSWER:

= 88.5 Correct

Part E Compute the rate Hint E.1

of development of thermal energy in the rotor.

Equation for thermal energy Hint not displayed


= 50.4 Correct

Part F Compute the power input to the motor Hint F.1

.

Equation for total power input Hint not displayed


= 556 Correct

Part G Compute the efficiency of the motor. Hint G.1

Definition of efficiency Hint not displayed

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ANSWER:


0.671 Correct

Changing Energy of a Magnetic Coil A coil with magnetic moment 1.46 magnetic field of magnitude 0.845

is oriented initially with its magnetic moment antiparallel to a uniform .

Part A What is the change in potential energy of the coil when it is rotated 180 degrees, so that its magnetic moment is parallel to the field? Hint A.1

Potential energy for a magnetic dipole Hint not displayed

Hint A.2

Potential energy difference Hint not displayed

Hint A.3

Definition of antiparallel Hint not displayed

Express your answer in joules. ANSWER:

-2.47 Correct

± Determining the Velocity of a Charged Particle A particle with a charge of

5.30

is moving in a uniform magnetic field of

magnetic force on the particle is measured to be

3.80×10−7

1.27 7.60×10−7

) . The .

Part A Are there components of the velocity that cannot be determined by measuring the force? Hint A.1

Magnetic force on a moving charged particle Hint not displayed

ANSWER:

yes

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no Correct

Part B Calculate the x component of the velocity of the particle. Hint B.1

Relation between


Express your answer in meters per second to three significant figures. ANSWER:

= -113 Correct

Part C Calculate the y component of the velocity of the particle. Hint C.1

Relation between


Express your answer in meters per second to three significant figures. ANSWER:

=

-56.5 Correct

Part D Calculate the scalar product

. Work the problem out symbolically first, then plug in numbers after

you've simplified the symbolic expression. Hint D.1

Formula for dot product Hint not displayed

Express your answer in watts to three significant figures. ANSWER:

0 Correct

Part E

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What is the angle between Hint E.1


and

?

Another dot product formula Hint not displayed

Express your answer in degrees to three significant figures. ANSWER:

90 Correct

Notice that the dot product of the velocity and the force is zero. This will always be the case. Since , must be perpendicular to both and . This result is important because it implies that magnetic fields can only change the direction of a charged particle's velocity, not its speed.

Score Summary: Your score on this assignment is 91.3%. You received 91.33 out of a possible total of 100 points.

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