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

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

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Current SheetConsider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current runs in the -y direction

through each wire. There are wires per unit length in the x direction.

Part A

Write an expression for , the magnetic field a distance above the xy plane of the sheet.

Use 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

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

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Hint not displayed

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

the unit vectors , , 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 haveworked.Did you notice that by using Ampère's law you could find the field by using a much simpler integral thanBiot-Savart's law? The drawback is that you may not always be able to find a convenient loop insituations where the current distribution is more complicated.

Force between an Infinitely Long Wire and a Square LoopA 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, , of the net force on the loop?



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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 terms 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 ( ), and whose direction is perpendicular to the

plane in which the current flows. Find the magnitude, , of the force on the loop from Part A in terms of the

magnitude of its magnetic moment.

Express in terms of , , , , and .

ANSWER: =

Correct

The direction of the net force would be reversed if the direction of the current in either the wire or theloop 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 oppositelydirected 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 figurebelow.


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Magnetic Field Generated by a Finite, Current-Carrying WireA 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 Find

Hint not displayed


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Hint A.4 Find

Hint not displayed

Express your answer in terms 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 How to approach the problem

Hint not displayed

Hint B.2 Limiting value of

Hint not displayed

Express your answer in terms of , , , and .

ANSWER: =


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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 differenceresults from the point chosen being close to one of the ends of the wire. Such "end effects" fornoninfinite wires always change the magneic field near that point.

Magnetic Field of a Current-Carrying WireFind the magnetic field a distance from the center of a long wire that has radius and carries a uniformcurrent per unit area in the positive z direction.

Part A

First find the magnetic field, , outside the wire (i.e., when the distance is greater than ).

Hint A.1 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 terms of the given parameters, the permeability constant , the variables ,

(the magnitude of ), , , and , and the corresponding unit vectors , , and . You may not

need all these in your answer.

ANSWER: =


Part B

Now find the magnetic field inside the wire (i.e., when the distance is less than ).

Hint B.1 Establish the relationiship to Part A

Hint not displayed

Hint B.2 Integrate over the Ampèrean loop

Hint not displayed

Express in terms 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 straightwire 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 righthand 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 antiparallelcurrents as shown in the diagram . Each of the wirescarries a current of magnitude . The current in wire

1 is directed out of the page and that in wire 2 isdirected into the page. The distance between thewires is . The x axis is perpendicular to the line

connecting the wires and is equidistant from thewires.As you answer the questions posed here, try to lookfor 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 diagramin the problem introduction) by wire 1 alone?

Enter the number of the vector with the appropriate direction.


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ANSWER: 3Correct

Part B

Which of the vectors best represents the direction of the magnetic field created at point K by wire 2 alone?


ANSWER: 3Correct

Part C

Which of these vectors best represents the direction of the net magnetic field created at point K by bothwires?


ANSWER: 3


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Correct

Part D

Find the magnitude of the magnetic field created at point K by wire 1.

Express your answer in terms of , , 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 created at point K by both wires.


ANSWER: =

Correct

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

Part F

Point L is located a distance from the midpoint between the two wires. Find the magnitude of the

magnetic field created at point L by wire 1.

Hint F.1 How to approach the problem

Use the distances provided and the Pythagorean Theorem to find the distance between wire 1 and pointL.


ANSWER: =



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Part G

Point L is located a distance from the midpoint between the two wires. Find the magnitude of the net

magnetic field created at point L by both wires.

Hint G.1 How to approach the problem

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 thediagram in the problem introduction) by wire 1 alone?


ANSWER: 2All 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 2alone?



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ANSWER: Answer not displayed

Hint G.4 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 bothwires?


ANSWER: Answer not displayed

Hint G.5 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 anglebetween 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: = Answer not displayed

Hint G.7 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 xcomponents. Find the x component of the magnetic field at point L due to wire 1 by using the magnitudeof the vector (found in Part F) and the angle between the x axis and the magnetic field vector (found inthe previous hint). Because of symmetry, the x component of the magnetic field at point L due to wire 2 isthe same size. To find the net magnetic field at point L you need to add together the x components of themagnetic field at point L due to wire 1 and of the magnetic field due to wire 2.


ANSWER: =


Part H

Point M is located a distance from the midpoint between the two wires. Find the magnitude of the

magnetic field created at point M by wire 1.


ANSWER:

=

Correct

Part I

Find the magnitude of the net magnetic field created at point M by both wires.

Hint I.1 How to approach the problem

Hint not displayed

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


ANSWER: =


Part J

Finally, consider point X (not shown in the diagram) located on the x axis very far away in the positive xdirection. Which of the vectors best represents the direction of the magnetic field created at point X by wire1 alone?


ANSWER: 1Correct

Part K

Which of the vectors best represents the direction of the magnetic field created at point X by wire 2 alone?



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ANSWER: 5Correct

As you can see, at a very large distance, the individual magnetic fields (and the correspondingmagnetic field lines) created by the wires are directed nearly opposite to each other, thus ensuring thatthe net magnetic field is very, very small even compared to the magnitudes of the individual magneticfields, which are also relatively small at a large distance from the wires. Thus, at a large distance, themagnetic fields due to the two wires almost cancel each other out! (That is, if point X is very far fromeach 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 themapart. It would seem as if the current were traveling up and down, almost along the same line, therebyappearing much the same as a single wire with almost no net current (because the up and downcurrents almost cancel each other), and therefore almost no magnetic field. Note that this only worksfor points very far from the wires; otherwsie it's easy to tell that the wires are separated and thecurrents 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 circuitsis 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 currentelements in the xy plane.

In this problem you are to find the magnetic field component along the z axis that results from various currentelements 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 , , and .Write in terms of the coordinate variables and directions ( , , etc.). To do this, you must find

and . Again, finding the cross product can be done either

geometrically (by finding the direction of the cross product vector first, then checking for cancellations fromany other portion of the wire, and then finding the magnitude or relevant component) oralgebraically (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 differentcurrent 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 asingle current element is given by

,

where and . In this expression, what

is the variable in terms of and/or ?

Hint A.1 Making sense of subscripts

Hint not displayed

ANSWER:

Correct

Part B

Find , the z component of the magnetic field at the point from the current

flowing over a short distance located at the point .

Hint B.1 Cross product


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Hint not displayed

Express your answer in terms of , , , , and . Recall that a component is a scalar; do not

enter any unit vectors.

ANSWER: = 0Correct

Part C



Hint C.1 Determine the displacement from the current element

Hint not displayed

Hint C.2 Find the direction from the cross product

Hint not displayed



ANSWER: =

Correct

Part D




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Hint D.1 Determine the displacement from the current element

Hint not displayed

Hint D.2 Find the direction of the magnetic field vector

Hint not displayed



ANSWER: =

Correct

Part E

Find , the z component of the magnetic field at the point P located at from the

current flowing over a short distance located at the point .

Hint E.1 Determine the displacement from the current element

Hint not displayed


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Hint E.2 Use the cross product to get the direction

Hint not displayed

Express your answer in terms of , , , , , and . Recall that a component is a scalar; do not


ANSWER: = 0Correct

Part F

Find , the z component of the magnetic field at the point P located at from the

current flowing over a short distance located at the point .

Hint F.1 Determine the displacement from the current element

Hint not displayed

Hint F.2 Determine which unit vector to use

Hint not displayed

Hint F.3 Evaluate the cross product

Hint not displayed

Express your answer in terms of , , , , , and . Recall that a component is a scalar; do not


ANSWER: =

Correct


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Force between Moving Charges

Two point charges, with charges and , are eachmoving with speed toward the origin. At the instantshown is at position (0, ) and is at ( , 0).

(Note that the signs of the charges are not givenbecause they are not needed to determine themagnitude 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 terms of , , , and .

ANSWER: =

Correct

Part B

What is the magnitude of the magnetic force on due to the magnetic field caused by ?


Hint not displayed

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 terms of , , , , and .

ANSWER: =

Correct

Part C

Assuming that the charges are moving nonrelativistically ( ), what can you say about the relationshipbetween the magnitudes of the magnetic and electrostatic forces?

Hint C.1 How to approach the problem

Hint not displayed

ANSWER: The magnitude of the magnetic force is greater than the magnitude of theelectric force.

The magnitude of the electric force is greater than the magnitude of themagnetic force.

Both forces have the same magnitude.

Correct

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

Magnetic Field at the Center of a Wire LoopA piece of wire is bent to form a circle with radius . It has a steady current flowing through it in a

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


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Part A

What is , the z component of at the center (i.e., ) of the loop?

Hint A.1 Specify the integrand

Hint not displayed

Hint A.2 Perform the integration

Hint not displayed

Express your answer in terms of , , and constants like and .

ANSWER: =

Correct

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

wire and counterclockwise through the inner wire.

Part A

What is the magnitude, , of the magnetic field at the center of the semicircles?

Hint A.1 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 terms 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 (andequals the radius also, in this case). So over all, you would expect the magnetic field to scale as1/radius. Note that such an argument works only because the field due to each point is in the samedirection, 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 ChargeA particle with positive charge is moving with speed along the z axis toward positive . At the time of thisproblem it is located at the origin, . 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: Ifyour 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 terms of , , , and , and use , , and for the three unit vectors.

ANSWER: =


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Correct

Part C


Express your answer in terms of , , , , and , and use , , and for the three unit

vectors.

ANSWER: =

Correct

Part D


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 terms 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 FieldFour 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 runningthrough 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 not displayed

Hint A.2 Calculating the contribution from the three known wires

Hint not displayed

Express your answer in amperes.

ANSWER: = 2.00Correct

Part B

What is the direction of the current ?


Hint not displayed

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 thecurrent-carrying wires.

This problem will show you how to calculate the torque on a magnetic dipole in a uniform magnetic field. Westart 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 mightfind it instructive to discover where it comes from.

Part A

A current flows in a plane rectangular current loop with height and horizontal sides . The loop is

placed into a uniform magnetic field in such a way that the sides of length are perpendicular to ,

and there is an angle between the sides of length

and .

Calculate , the magnitude of the torque about thevertical axis of the current loop due to the interactionof the current through the loop with the magneticfield.


Hint not displayed

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 terms of the given variables. You will need a trigonomericfunction [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 interms of the magnitude of the magnetic dipole moment of the current loop . Define the angle between thevector perpendicular to the plane of the coil and the magnetic field to be , noting that this angle is the

complement of angle in Part A.

Hint B.1 Definition of the magnetic dipole moment

Hint not displayed

Give your answer in terms of the magnetic moment , magnetic field , and .

ANSWER: =

Correct

The more general vector form of this expression is.

Part C

A current flows around a plane circular loop of radius , giving the loop a magnetic dipole moment of

magnitude . 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 magnitudeof the torque on the current loop.


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Hint C.1 Formula for the area of a circle

Hint not displayed

Express the torque explicitly in terms of , , , , and (where and are the magnitudes of

the respective vector quantities). Do not use . You will need a trigonomeric function [e.g..

or ].

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 magneticfield.

Magnetic fields exert forces on moving charged particles, whether those charges are moving independently orare 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 aresimultaneously 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 carrying current in a uniform magnetic field

with strength 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, . Thus

this equation can also be written as .

The direction of the magnetic force on the wire can be described using a "right-hand rule." This will bediscussed after Part B.

Part A

Consider a wire of length = 0.30 that runs north-south on a horizontal surface. There is a current of

= 0.50 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 agreementof units, recall that .


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Express your answer in newtons to twosignificant 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 todetect.

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'smagnetic 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 directionof 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 pointthem in the direction of the current.Rotate your arm until you can bend your fingers topoint in the direction of the magnetic field.Your thumb now points in the direction of themagnetic 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 pointsdue east.

Current in the wire flows due west and slightly down; the magnetic field pointsstraight down.

Correct

As you can see, many current/magnetic field configurations can result in the same direction of magneticforce.

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: 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 isperpendicular 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 magneticforce. This is another clue that the magnetic force is zero.

A Conductor Moving in a Magnetic FieldA 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 oftwo 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 terms of , , and unit vectors ( , , and/or ).

ANSWER: =

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 theinduced electric field point?

ANSWER:

Correct

The direction of the electric field stays the same regardless of the sign of the charges that are free tomove in the conductor.Mathematically, you can see that this must be true since the expression you derived for the electricfield is independent of .Physically, this is because the force due to the magnetic field changes sign as well and causes positivecharges 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 will always point in

the direction (given the original directions of and ).

A Thomson ExperimentIn the late 19th century, great interest was directed toward the study of electrical discharges in gases and thenature 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 theratio 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 for different cathode materials.

From measurements of 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 anhighly evacuated glass tube, a beam of electrons,each moving with speed , passes between twoparallel plates and strikes a screen at a distance

from the end of the plates.First, you observe the point where the beam strikesthe screen when there is no electric field between theplates. Then, you observe the point where the beamstrikes the screen when a uniform electric field ofmagnitude is established between the plates. Call

the distance between these two points .

Part B

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

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


Hint not displayed

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

Express your answer in terms of , , , , , 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 . The

magnetic field is perpendicular to and directed

straight into the plane of the figure. You adjust thevalue of so that no deflection is observed on the

screen.What is the speed of the electrons in this case?


Hint not displayed

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

Express your answer in terms of 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 asdescribed in Part C, to what value do you have to adjust its magnitude to observe no deflection?

Assume that the plates are 6.00 long and that the distance between them and the screen is 12.0 .


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Hint D.1 How to approach the problem

Hint not displayed

Hint D.2 Find the speed of the electrons

Hint not displayed

Express your answer numerically in tesla.

ANSWER: = 1.69×10−4

Correct

Part E

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

Two off-centered spots in the first phase of the experiment; one centered spotin the second phase of the experiment

Two off-centered spots in the first phase of the experiment; one centered andone off-centered spot in the second phase of the experiment

One off-centered spot in the first phase of the experiment; one centered andone off-centered spot in the second phase of the experiment

One off-centered spot in the first phase of the experiment; two off-centeredspots in the second phase of the experiment

One off-centered spot in both the first and the second phases of theexperiment

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 observedon the screen because the balance of the electric force and the magnetic force does not depend on thecharge 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 pass through the platesundeflected.

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 followthe 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.


ANSWER:

View Correct

Part E

Now assume that particles A, B, C, and E all have the same magnitude of electric charge. Rank theparticles 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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ANSWER:

View Correct

Interaction of a Current Loop with a Magnetic FieldThe 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 that flows in a plane rectangular current loop with height = 4.00 and horizontal

sides = 2.00 . The loop is placed into a uniform

magnetic field in such a way that the sides of

length are perpendicular to , 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 Find the direction of the forces on the parts of the loop that have length

Hint not displayed

ANSWER: Yes, the net torque acting on the loop is negative and tends to rotate the loopin the direction of decreasing angle (clockwise).

Yes, the net torque acting on the loop is positive and tends to rotate the loopin 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 withthe 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: = 1.04×10−4


Part C


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What happens to the loop when it reaches the position for which , that is, when its horizontal sides

of length are perpendicular to (see the figure)?

Hint C.1 Find the net torque acting on the loop

Hint not displayed

ANSWER: The direction of rotation changes because the net torque acting on the loopcauses the loop to rotate in a clockwise direction.

The net torque acting on the loop is zero, but the loop continues to rotate in acounterclockwise 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 , and assume that the loop

is placed in such a way that initially the angle between the sides of length and is , as shown in

the figure. Will the interaction of the current throughthe loop with the magnetic field cause the loop torotate?


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Hint D.1 Find the direction of the forces on the parts of the loop that have length

Hint not displayed

Hint D.2 Find the direction of the forces on the parts of the loop that have length

Hint not displayed

ANSWER: Yes, the net torque acting on the loop is negative and tends to rotate the loopin the direction of decreasing angle (clockwise).

Yes, the net torque acting on the loop is positive and tends to rotate the loopin 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 , the direction of rotation of the loop will be

different. If initially , 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 SpectrometerJ. J. Thomson is best known for his discoveries about the nature of cathode rays. Another importantcontribution of his was the invention, together with one of his students, of the mass spectrometer. The ratio ofmass 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 theregion with the electric field it will have obtained aspeed .


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Part A

With what speed does the ion exit the acceleration region?

Hint A.1 Suggested general method

Hint not displayed

Hint A.2 Initial energy

Hint not displayed

Hint A.3 Final energy

Hint not displayed

Find the speed in terms of , , , and any constants.

ANSWER:

=

Correct

Part B

After being accelerated, the particle enters a uniform magnetic field of strength and travels in a circle of

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

experiment allow one to find in terms 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 terms of , , , and any necessary constants.

ANSWER:

=

Correct


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

Motion of Electrons in a Magnetic FieldAn electron of mass and charge is moving through a uniform magnetic field in vacuum.

At the origin, it has velocity , where

and . A screen is mounted perpendicular

to the x axis at a distance from the origin.

Throughout, you can assume that the effect of gravityis negligible.

Part A

First, suppose . Find the y coordinate of the point at which the electron strikes the screen.

Hint A.1 Forces acting on electron

Hint not displayed

Hint A.2 Two-dimensional kinematics

Hint not displayed

Express your answer in terms of 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 magneticfield?


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ANSWER: The electron decelerates before coming to a halt and turning around whilealways moving along a straight line.

The electron's motion will be unaffected. (It will continue moving in a straightline 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 lineswith 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. aperiodic part, the projection of which in the yz plane is circular. Find the angular velocity of the electronassociated 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 terms of

Hint not displayed

Hint C.4 Express the acceleration in terms of

Hint not displayed

Express the magnitude of the angular velocity in terms of , the magnitude of the electric charge

, and other known quantities.

ANSWER:

=

Correct

A DC MotorA shunt-wound DC motor with the field coils and rotor connected in parallel (see the figure) operates from a135 DC power line. The resistance of the field

windings, , is 206 . The resistance of the rotor,

, is 4.20 . 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.


Part B

Compute the rotor current .

Hint B.1 Currents in a parallel circuit

Hint not displayed


Part C

Compute the emf .


Hint not displayed

Hint C.2 Relation between voltage and current

Hint not displayed


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ANSWER: = 120Correct

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.


Part E

Compute the rate of development of thermal energy in the rotor.

Hint E.1 Equation for thermal energy

Hint not displayed



Part F

Compute the power input to the motor .

Hint F.1 Equation for total power input

Hint not displayed


ANSWER: = 556Correct

Part G

Compute the efficiency of the motor.

Hint G.1 Definition of efficiency

Hint not displayed


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ANSWER: 0.671Correct

Changing Energy of a Magnetic CoilA coil with magnetic moment 1.46 is oriented initially with its magnetic moment antiparallel to a uniform

magnetic field of magnitude 0.845 .

Part A

What is the change in potential energy of the coil when it is rotated 180 degrees, so that its magneticmoment 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: -2.47Correct

± Determining the Velocity of a Charged ParticleA particle with a charge of 5.30 is moving in a uniform magnetic field of 1.27 ) . The

magnetic force on the particle is measured to be 3.80×10−7 7.60×10−7 .

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 and

Hint not displayed

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 and

Hint not displayed

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: 0Correct

Part E


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What is the angle between and ?

Hint E.1 Another dot product formula

Hint not displayed

Express your answer in degrees to three significant figures.

ANSWER: 90Correct

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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