physics 23 hw4

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Homework 4

Due: 5:00pm on Wednesday, February 8, 2012

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Magnetic Materials

Part A

You are given a material which produces no initial magnetic field when in free space. When it isplaced in a region of uniform magnetic field, the material produces an additional internalmagnetic field parallel to the original field. However, this induced magnetic field disappears whenthe external field is removed.What type of magnetism does this material exhibit?

ANSWER:diamagnetism

paramagnetism

ferromagnetism

Correct

When a paramagnetic material is placed in a magnetic field, the field helps align themagnetic moments of the atoms. This produces a magnetic field in the material that isparallel to the applied field.

Part B

Once again, you are given an unknown material that initially generates no magnetic field. Whenthis material is placed in a magnetic field, it produces a strong internal magnetic field, parallel tothe external magnetic field. This field is found to remain even after the external magnetic field isremoved.Your material is which of the following?

ANSWER:diamagnetic

paramagnetic

ferromagnetic

Correct

Very good! Materials that exhibit a magnetic field even after an external magnetic field isremoved are called ferromagnetic materials. Iron and nickel are the most commonferromagnetic elements, but the strongest permanent magnets are made from alloys thatcontain rare earth elements as well.

Part C

What type of magnetism is characteristic of most materials?



ANSWER: ferromagnetism

paramagnetism

diamagnetism

no magnetism

Correct

Almost all materials exhibit diamagnetism to some degree, even materials that also exhibitparamagnetism or ferromagnetism. This is because a magnetic moment can be induced inmost common atoms when the atom is placed in a magnetic field. This induced magneticmoment is in a direction opposite to the external magnetic field. The addition of all of theseweak magnetic moments gives the material a very weak magnetic field overall. This fielddisappears when the external magnetic field is removed. The effect of diamagnetism isoften masked in paramagnetic or ferromagnetic materials, whose constituent atoms ormolecules have permanent magnetic moments and a strong tendency to align in the samedirection as the external magnetic field.

A Wire and a Compass

You are looking straight down on a magnetic compass that is lying flat on a table. A wire isstretched horizontally under the table, parallel to and a short distance below the compass needle.The wire is then connected to a battery so that a current flows through the wire. This current

causes the north pole of the compass needle to deflect to the left. The questions that follow askyou to compare the effects of different actions on this initial deflection.

Part A

If the wire is lowered farther from the compass, how does the new angle of deflection of thenorth pole of the compass needle compare to its initial deflection?

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Find an expression for the magnetic field's magnitude produced by aconductor

Hint not displayed

ANSWER:It is smaller.

Correct

As the distance between the wire and the compass increases, the magnetic field producedby the wire weakens. Therefore, the north pole of the compass needle will deflect awayfrom its original direction (determined by the magnetic field of the earth) by a smaller angle.

Part B



With the wire back at its initial location, you connect a second identical battery in series withthe first one. When you close the switch, how does the new angle of deflection of the north poleof the compass needle compare to its initial deflection?

Hint B.1 How to approach the problem

Hint not displayed

Hint B.2 Find how the emf delivered to the circuit changes

Hint not displayed

Hint B.3 Find how the current in the wire changes

Hint not displayed

Hint B.4 Magnetic field produced by a conductor

Hint not displayed

ANSWER:It is larger.

Correct

Since the magnitude of the magnetic field produced by the wire increases due to the largercurrent flowing through the wire, the compass needle will detect a stronger magnetic field,and the north pole of the needle will deflect away from its original direction by a largerangle.

Charged Particles Moving in a Magnetic Field Ranking Task

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



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 Correct

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. Rankthe particles A, B, C, and E on the basis of their speed.

Hint E.1 Charged particle trajectories in magnetic fields

Hint not displayed




ANSWER:

View Correct

Direction of the Magnetic Field due to a Wire Conceptual Question

Find the direction of the magnetic field at each of the indicated points.

For the following two questions consider the wire shown in the figure. .

Part A

What is the direction of the magnetic field at Point A?

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

Hint not displayed

Hint A.2 The direction of the magnetic field due to a long, straight, current-carryingwire

Hint not displayed

ANSWER: is out of the page.

is into the page.

is neither out of nor into the page and .

.

Correct

Part B



What is the direction of the magnetic field at Point B?


is into the page.


.

Correct

Now consider the wires shown in this figure. Note that the bottom wire carries a current ofmagnitude .

Part C

What is the direction of the magnetic field at Point C?

Hint C.1 How to approach the problem

Hint not displayed

Hint C.2 Find the direction of the magnetic field at Point C due to wire 1

Hint not displayed

Hint C.3 Find the direction of the magnetic field at Point C due to wire 2

Hint not displayed

ANSWER:



is out of the page.

is into the page.


.

Correct

Part D

What is the direction of the magnetic field at Point D?

Hint D.1 Find the direction of the magnetic field at Point D due to wire 1

Hint not displayed

Hint D.2 Find the direction of the magnetic field at Point D due to wire 2

Hint not displayed


is into the page.


.

Correct

Part E

What is the direction of the magnetic field at Point E?

Hint E.1 Find the direction of the magnetic field at Point E due to wire 1

Hint not displayed

Hint E.2 Find the direction of the magnetic field at Point E due to wire 2

Hint not displayed




is into the page.


.

Correct

Magnetic Force on a Bent Wire Conceptual Question

The bent wire circuit shown in the figure isin a region of space with a uniformmagnetic field in the +z direction. Currentflows through the circuit in the directionindicated. Note that segments 2 and 5 areoriented parallel to the z axis; the otherpieces are parallel to either the x or y axis.

Part A

Determine the direction of the magnetic force along segment 1, which carries current in the -xdirection.

Hint A.1 Magnetic force on a current-carrying wire

Hint not displayed

Hint A.2 Magnetic force on segment 1

Hint not displayed

Enter the direction of the force as a sign (+ or -) followed by a coordinate direction (x,y, or z) without spaces. For instance, if you think that the force points in the positive ydirection, enter +y. If there is no magnetic force, enter 0.

ANSWER: +yCorrect



Part B

Determine the direction of the magnetic force along segment 2, which carries current in the -]direction.

Hint B.1 Magnetic force on segment 2

Hint not displayed


ANSWER: 0Correct

Part C

Determine the direction of the magnetic force along segment 3, which carries current in the +\direction.

Hint C.1 Magnetic field on segment 3

Hint not displayed


ANSWER: +xCorrect

Part D

Determine the direction of the magnetic force along segment 4, which carries current in the +[direction.


ANSWER: -yCorrect

Part E

Determine the direction of the magnetic force along segment 5, which carries current in the +]direction.

Enter the direction of the force as a sign (+ or -) followed by a coordinate direction (x,



y, or z) without spaces. For instance, if you think that the force points in the positive ydirection, enter +y. If there is no magnetic force, enter 0.

ANSWER: 0Correct

Part F

Determine the direction of the magnetic force along segment 6, which carries current in the +[direction.


ANSWER: -yCorrect

Part G

Determine the direction of the magnetic force along segment 7, which carries current in the -\direction.


ANSWER: -xCorrect

Magnetic Force Vector Drawing

For each of the situations below, a charged particle enters a region of uniform magnetic field. Drawa vector to represent the direction of the magnetic force on the particle.

Part A

Hint A.1 The right-hand rule for magnetic force

Hint not displayed

Hint A.2 Apply the right-hand rule

Hint not displayed

Draw the vector starting at the location of the charge. The location and orientation ofthe vector will be graded. The length of the vector will not be graded.



ANSWER:

View Correct

Part B

Hint B.1 Apply the right-hand rule

Hint not displayed


ANSWER:

View Correct

Part C

Hint C.1 Apply the right-hand rule

Hint not displayed


ANSWER:

View



Correct

Rail Gun

A rail gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basicmechanism of acceleration is relatively simple and can be illustrated in the following example. Ametal rod of mass 30.0 and electrical resistance 0.300 rests on parallel horizontal rails that

have negligible electric resistance. The rails are a distance = 10.0 apart. The rails are also

connected to a voltage source providing avoltage of = 5.00 .

The rod is placed in a vertical magneticfield. The rod begins to slide when the fieldreaches the value = 8.82×10−2 .

Assume that the rod has a slightlyflattened bottom so that it slides instead ofrolling. Use 9.80 for the magnitude of

the acceleration due to gravity.

Part A

Find , the coefficient of static friction between the rod and the rails.


Hint not displayed

Hint A.2 Find the force due to the magnetic field

Hint not displayed

Hint A.3 Frictional force

Hint not displayed

Give your answer numerically.

ANSWER: = 0.5

Correct



Problem 20.51You have 1.0 of copper and want to make a practical solenoid that produces the greatest

possible magnetic field for a given voltage.

Part A

Should you make your copper wire long and thin, short and fat, or something else? Considerother variables, such as solenoid diameter, length, and so on.

ANSWER:You should make two solenoids one inside the other connected inseries.

You should make two solenoids one inside the other connected inparallel.

The copper wire should be short and fat.

The copper wire should be long and thin.

Correct

Problem 20.40

A compass needle points 23 E of outdoors. However, when it is placed 12.0 to the east of

a vertical wire inside a building, it points 55 E of .

Part A

What is the magnitude of the current in the wire? The Earth's field there is 0.50 and is

horizontal.

Express your answer using two significant figures.

ANSWER: = 19

Correct

Part B

What is the direction of the current in the wire? The Earth's field there is 0.50 and is

horizontal.

ANSWER:east

down

west

up

Correct



Problem 20.86

Two long straight aluminum wires, each of diameter 0.70 , carry the same current but in

opposite directions. They are suspended by 0.50- -long strings as shown in the figure.

Part A

If the suspension strings make an angle of 3.0 with the vertical, what is the current in the

wires?


ANSWER: 12

Correct

Faraday's Law and Induced Emf

Learning Goal: To understand the terms in Faraday's law and to be able to identify themagnitude and direction of induced emf.

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.Mathematically, it can be written as

,

where is the emf induced in a closed loop, and

is the rate of change of the magnetic flux through a surface bounded by the loop. For uniform

magnetic fields the magnetic flux is given by , where is the angle

between the magnetic field and the normal to the surface of area .



To find the direction of the induced emf, one can use Lenz's law:

The induced current's magnetic field opposes the change in the magnetic flux that induced the current.

For example, if the magnetic flux through a loop increases, the induced magnetic field is directedopposite to the "parent" magnetic field, thus countering the increase in flux. If the flux decreases, theinduced current's magnetic field has the same direction as the parent magnetic field, thus counteringthe decrease in flux.

Recall that to relate the direction of the electric current and its magnetic field, you can use the right-handrule: When the fingers on your right hand are curled in the direction of the current in a loop, your thumbgives the direction of the magnetic field generated by this current.

In this problem, we will consider a rectangular loop of wire with sides and placed in a region

where a uniform magnetic field exists (see the diagram). The resistance of the loop is .

Initially, the field is perpendicular to theplane of the loop and is directed out of thepage. The loop can rotate about either thevertical or horizontal axis, passing throughthe midpoints of the opposite sides, asshown.

Part A

Which of the following changes would induce an electromotive force (emf) in the loop? Whenyou consider each option, assume that no other changes occur.

Check all that apply.

ANSWER:The magnitude of increases.

The magnitude of decreases.

The loop rotates about the vertical axis (vertical dotted line)shown in the diagram.

The loop rotates about the horizontal axis (horizontal dotted line)shown in the diagram.

The loop moves to the right while remaining in the plane of thepage.

The loop moves toward you, out of the page, while remainingparallel to itself.

Correct



Part B

Find the flux through the loop.

Express your answer in terms of , , and .

ANSWER: =

Correct

Part C

If the magnetic field steadily decreases from to zero during a time interval , what is the

magnitude of the induced emf?

Hint C.1 Find the change in magnetic flux

Hint not displayed

Express your answer in terms of , , , and .

ANSWER:

=

Correct

Part D


magnitude of the induced current?

Express your answer in terms of , , , , and the resistance of the wire.

ANSWER:

=

Correct

Part E


direction of the induced current?

ANSWER:clockwise

counterclockwise

Correct



The flux decreases, so the induced magnetic field must be in the same direction as theoriginal (parent) magnetic field. Therefore, the induced magnetic field is out of the page.Using the right-hand rule, we deduce that the direction of the current is counterclockwise.

Part F

Which of the following changes would result in a clockwise emf in the loop? When you considereach option, assume that no other changes occur.


ANSWER:The magnitude of increases.

The magnitude of decreases.

The loop rotates through 45 degrees about the vertical axis(vertical dotted line) shown in the diagram.

The loop rotates through 45 degrees about the horizontal axis(horizontal dotted line) shown in the diagram.

The loop moves to the right while remaining in the plane of thepage.

The loop moves toward you, out of the page, while remainingparallel to itself.

Correct

Clockwise emf implies that the induced magnetic field is directed into the page. Therefore,the magnetic flux of the original field must be increasing. Only the first option correspondsto increasing flux.

Transformers

Learning Goal: To understand the concepts underlying the operation of transformers.

One of the advantages of alternating current (ac) over direct current (dc) is the ease with whichvoltage levels can be increased or decreased. Such a need is always present due to the practicalrequirements of energy distribution. On the one hand, the voltage supplied to end users must bereasonably low for safety reasons (depending on the country, that voltage may be 110 volts, 220volts, or some other value of that order). On the other hand, the voltage used in transmittingelectric energy must be as high as possible to minimize losses in the transmission lines. A

device that uses the principle of electromagnetic induction to increase or decrease the voltage bya certain factor is called a transformer.

The main components of a transformer are two coils (windings) that are electrically insulated from eachother. The coils are wrapped around the same core, which is typically made of a material with a verylarge relative permeability to ensure maximum mutual inductance. One coil, called the primary coil, isconnected to a voltage source; the other, the secondary coil, delivers the power. The alternating currentin the primary coil induces the changing magnetic flux in the core that creates the emf in the secondarycoil. The magnitude of the emf induced in the secondary coil can be controlled by the design of thetransformer. The key factor is the number of turns in each coil.

Consider an ideal transformer, that is, one in which the coils have no ohmic resistance and the



magnetic flux is the same for each turn of both the primary and secondary coils. If the number

of turns in the primary coil is and that in the secondary coil is , then the emfs induced in

the coils can be written as

,

and therefore,

.

If the coils have zero resistance (as we have assumed), then for each coil the terminal voltage willbe equal to the induced emf. We can then write

.

Note that if , then . In this case we have a step-up transformer. Conversely, if

, then , and we are dealing with a step-down transformer. Without energy losses,

the power in the primary and secondary coils is the same:

.

In answering the questions below, consider the transformer ideal unless otherwise noted.

Part A

The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primarycoil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. Whatvoltage would be measured across the secondary coil?

ANSWER:zero

0.75

1.5

3.0



Correct

In order for an emf to be induced in the secondary coil, the flux through it must bechanging; therefore, the current in the primary coil must also be changing. If a constantvoltage is supplied to the primary coil, no emf will be induced in the secondary coil, and,therefore the secondary voltage will be zero.

Part B

A transformer is intended to decrease the value of the alternating voltage from 500 volts to 25volts. The primary coil contains 200 turns. Find the necessary number of turns in the

secondary coil.

ANSWER: = 10

Correct

This is a step-down transformer: The voltage decreases.

Part C

A transformer is intended to decrease the value of the alternating current from 500 amperes to25 amperes. The primary coil contains 200 turns. Find the necessary number of turns in the

secondary coil.


Hint not displayed

ANSWER: = 4000

Correct

This is a step-up transformer: The voltage increases by the same factor by which thecurrent decreases.

Part D

In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns.If the primary current is 2.5 amperes, what is the secondary current ?

Express your answer numerically in amperes.

ANSWER: = 12.5

Correct

Part E

The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. Thepower supplied to the primary coil is 400 watts. What is the power generated in the secondary



coil if it is terminated by a 20-ohm resistor?

Hint E.1 In an ideal world…

Hint not displayed

ANSWER:

Correct

In the case of an ideal transformer, the power in the primary circuit is the same as that inthe secondary circuit.

Part F

A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil isconnected to a 120-volt ac source. What is the current in the primary coil?

Hint F.1 How to approach the problem

Hint not displayed

Express your answer in amperes.

ANSWER: = 0.5

Correct

Part G

The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes.What is the efficiency of the transformer? The efficiency of a transformer is defined as the

ratio of the output power to the input power, expressed as a percentage: .

Express your answer as a percentage.

ANSWER: = 95.4

Correct %



There is always some energy loss across a transformer. However, modern transformershave very high efficiencies, usually well exceeding 90 .

A Magnet and a Coil

When a magnet is plunged into a coil at speed , as shown in the figure, a voltage is induced in

the coil and a current flows in the circuit.

Part A

If the speed of the magnet is doubled, the induced voltage is ________ .


Hint not displayed

ANSWER:twice as great

Correct

Part B

The same magnet is plunged into a coil that has twice the number of turns as before. Themagent is shown before it enters the coil in the figure. If the speed of the magnet is again , the

induced current in the coil is _______ .




Hint not displayed

Hint B.2 Find the induced emf in the coil

Hint not displayed

Hint B.3 Find the resistance of the coil

Hint not displayed

Hint B.4 Induced current

Hint not displayed

ANSWER:unchanged

Correct

By increasing the number of turns in the coil, the induced emf increases, but so does theresistance of the coil. Since those two quantities increase by the same factor, their ratioremains constant, and the induced current in the circuit is unchanged.

Magnetic Flux through a Coil

You hold a wire coil perpendicular to a magnetic field .

Part A



If the magnitude of increases while its direction remains unchanged, how will the magnetic

flux through the coil change?


Hint not displayed

Hint A.2 Magnetic flux

Hint not displayed


ANSWER:The flux is unchanged because the position of the coil with

respect to is unchanged.

The flux increases because the magnitude of increases.

The flux decreases because the magnitude of increases.

The flux is unchanged because the surface area of the coil isunchanged.

Correct

The magnetic flux through a coil is directly proportional to the magnitude of the magneticfield.

Part B

If is kept constant but the coil is rotated so that it is parallel to , how will the magnetic flux

through the coil vary?


Hint not displayed

Hint B.2 Magnetic flux

Hint not displayed

Hint B.3 Find the cosines of the angles

Hint not displayed




ANSWER:The flux is unchanged because the magnitude of is constant.

The flux increases because the angle between and the coil's

axis changes.

The flux decreases because the angle between and the coil's

axis changes.

The flux is unchanged because the area of the coil is unchanged.

Correct

As the orientation of the coils changes, the magnetic flux through the coil decreases. Itreaches its minimum value (zero) when the coil is parallel to the field.

Understanding Changing Flux

In this problem, you will use Lenz's law to explore what happens when an electromagnet isactivated a short distance from a wire loop.

You will need to use the right-hand rule to find the direction of the induced current.

Consider the arrangement shown in the figure.

Part A

When the switch is open, which of the following statements about the magnetic flux through thewire loop is true? Assume that the direction of the vector area of the wire loop is to the right.

ANSWER:There is no magnetic flux through the wire loop.

There is a positive flux through the wire loop.

There is a negative flux through the wire loop.

Correct



When the switch is open, there is no current flowing in the circuit. Thus, the electromagnetdoes not produce a magnetic field, and the flux is zero.

Part B

What is the direction of the induced current in the wire loop (as seen from the left) when theswitch is open?

ANSWER:There is no induced current.

The induced current is clockwise.

The induced current is counterclockwise.

Correct

Part C

Now the switch on the electromagnet is closed. What is the direction of the induced current inthe wire loop immediately after the switch is closed (as seen from the left)?


Hint not displayed

Hint C.2 The field produced by the electromagnet

Hint not displayed

Hint C.3 Apply Lenz's law

Hint not displayed




Correct

Part D

Finally, the switch on the electromagnet is reopened. The magnitude of the external magneticflux through the wire loop ______ (A. increases, B. decreases, C. remains constant), and thereis _______ (A. zero, B. a clockwise, C. a counterclockwise) current induced in the loop (asseen from the left).

Hint D.1 How to approach the problem



Hint not displayed

Hint D.2 The field produced by the electromagnet

Hint not displayed

Hint D.3 Apply Lenz's law

Hint not displayed

Enter the letters corresponding to the responses that correctly complete the statementabove. For example, if the correct answers are A and C, type A,C

ANSWER: B,BCorrect

Now consider the new arrangement shown in the figure. Note that the orientation of the battery isreversed with respect to the firstarrangement you considered. Answer thefollowing questions related to thearrangment with the new batteryorientation.

Part E

The switch on the electromagnet, initially open, is closed. What is the direction of the inducedcurrent in the wire loop (as seen from the left)?

Hint E.1 How to approach the problem

Hint not displayed

Hint E.2 The field produced by the electromagnet

Hint not displayed

Hint E.3 Apply Lenz's law



Hint not displayed




Correct

Part F

Now the switch on the electromagnet is reopened. The magnitude of the external magnetic fluxthrough the wire loop ______ (A. increases, B. decreases, C. remains constant), and there is_______ (A. zero, B. a clockwise, C. a counterclockwise) current induced in the loop (as seenfrom the left.

Hint F.1 How to approach the problem

Hint not displayed

Hint F.2 The field produced by the electromagnet

Hint not displayed

Hint F.3 Apply Lenz's law

Hint not displayed

Enter the letters corresponding to the responses that correctly complete the statementabove. For example: A,C

ANSWER: B,CCorrect

Video Tutor: Eddy Currents in Different Metals

First, launch the video below. You will be asked to use your knowledge of physics to predict the outcomeof an experiment. Then, close the video window and answer the question at right. You can watch thevideo again at any point.



Part A

Electric rail cars often use magnetic braking. The brake consists of a set of electromagnets thatare held just above the rails. To brake the train, the electromagnets are switched on, creating amagnetic field that induces eddy currents in the metal rails passing beneath them.

In the figure, which of the choices correctly represents the eddy currents induced in the rails? Thediagrams represent a view from above, looking down at the rail through the electromagnet. Theelectromagnet moves to the right, and the magnetic field points into the screen.

Hint A.1 How to approach the problem.

Hint not displayed

ANSWER:A

B

C

D

Correct

The magnetic flux increases under the leading edge of the electromagnet and decreases



under its trailing edge. Therefore, by Lenz’s law, the induced magnetic field will point out ofthe screen under the leading edge and into it under the trailing edge. The eddy currents inchoice B have the right directions to induce such magnetic fields, according to the right-hand rule.

Magnetic Flux and Induced EMF in a Coil

A coil has turns enclosing an area of . In a physics laboratory experiment, the coil is rotated

during the time interval from a position in which the plane of each turn is perpendicular to

Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitudeof Earth's magnetic field at the lab location is .

Part A

What is the total magnitude of the magnetic flux through the coil before it is rotated?

Hint A.1 Formula for the magnetic flux through a wire loop

Hint not displayed

Hint A.2 The initial angle between the magnetic field and the area vector

Hint not displayed

ANSWER: =

Correct

Part B

What is the magnitude of the total magnetic flux through the coil after it is rotated?

Hint B.1 The angle between the magnetic field and the area vector

Hint not displayed

Express your answer numerically, to three significant figures.

ANSWER: = 0

Correct

Part C

What is the magnitude of the average emf induced in the coil?



Hint C.1 Formula for the average emf induced in a coil (Faraday's law)

Hint not displayed

ANSWER:

average induced emf =

Correct

PhET Tutorial: Faraday’s Electromagnetic Lab

Learning Goal: To understand the conditions that cause the largest induced currents due to Faraday'slaw of induction.

For this tutorial, use the PhET simulation Faraday’s Electromagnetic Lab. This simulation allows you toexperiment with a pickup coil or a transformer to study Faraday’s law of induction.

Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file.Choose to run or open it.

You can click on any of the five tabs labeled Bar Magnet, Pickup Coil, Electromagnet, Transformer,and Generator. For the Bar Magnet, you can drag a compass around to get a feel for the magneticfield. You can also select Show Field Meter to measure the strength of the magnetic field. For PickupCoil, you can move the magnet through a coil loop to either light a bulb or affect a voltmeter. For theTransformer, you can move an electromagnet near a coil loop and see what makes the light bulb shinemost brightly.

Feel free to play around with the simulation. When you are done, click Reset All before beginning Part A.

Part A

Faraday’s law of induction deals with how a magnetic flux induces an emf in a circuit. Recall thatmagnetic flux depends on magnetic field strength and the effective area the field is passing through.We’ll start our investigation by looking at the field strength around a bar magnet.

Select the Bar Magnet tab. Deselect Show Compass, and drag the bar magnet to the center of the screen.Then, select Show Field Meter, which allows you to measure the strength of the magnetic field at anylocation.



For which of the regions shown in the figure is the magnetic field the strongest?

ANSWER:Region C

The magnetic field strength is the same for all three regions.

Region A

Region B

Correct

The magnetic field is strongest near either of the two poles (the south pole or the north pole).

Part B

Select the Pickup Coil tab, and place the bar magnet inside the coil containing two loops.

Try to find a location where the stationary magnet induces a current in the coil and causes the light bulb toshine. Which of the following is correct?

ANSWER:The light bulb shines due to an induced current if the magnet is insidethe coil.

There is no induced current in the coil, so the light bulb does not shine,if the magnet is stationary (for any location of the magnet).

The light bulb shines due to an induced current if one pole of themagnet is near the middle of the coil.

Correct



Apparently, a magnetic field that isn’t changing in time doesn’t induce a current in the coil.

Part C

Now, let’s look at a situation with changing flux. Starting from the far left of the screen, move themagnet to the right so it goes through the middle of the current loop at a constant speed and out to theright of the loop.

Roughly where is the magnet when the light bulb is the brightest? (The brightness of the light bulb isdepicted by the length of the rays emanating from it.)

ANSWER:The light bulb is brightest when the middle of the magnet is in themiddle of the coil.

The brightness of the light bulb is the same, regardless of the location ofthe magnet (as long as it is moving).

The light bulb does not shine since the magnet is moving at a constantspeed.

The light bulb is brightest when either end of the magnet is in themiddle of the coil.

Correct

This is consistent with Faraday’s law of induction, which states that the induced emf in the coil isequal to the time rate of change of the magnetic flux through the coil. Since the magnetic field isstrongest near the poles of the magnet (as we saw in Part A) and is weaker near the middle of themagnet, the magnetic flux changes the quickest when the poles are going through the coil loop.This causes the greatest induced emf, which causes the greatest current through the light bulb.



Part D

How does the brightness of the light bulb change if the magnet is moved through the coil more quickly?

ANSWER:The bulb shines less brightly.

The bulb shines more brightly.

The bulb shines with the same brightness, but for a shorter duration.

Correct

As the magnet is approaching the coil, the magnetic flux is increasing since the field strength isstronger closer to the magnet. The faster the magnet approaches, the quicker the magnetic fluxincreases, which causes a greater induced emf and a greater current.

Part E

Select the Transformer tab, which contains an electromagnet at left (containing a DC battery) and apickup coil attached to a light bulb at right. The electromagnet produces a magnetic field very similar tothat of the bar magnet.

Try to find a location where the stationary electromagnet (with a DC current) induces a current in the pickupcoil and causes the light bulb to shine. Which of the following is correct?

ANSWER:The light bulb shines due to an induced current if one pole of theelectromagnet is near the middle of the coil.

There is no induced current in the coil, so the light bulb does not shine,if the electromagnet is stationary (for any location of the electromagnet).

The light bulb shines due to an induced current if the electromagnet isinside the coil.

Correct

Just as we saw earlier (in Part B), a constant magnetic flux does not cause an induced emfthrough the pickup coil. According to Faraday’s law of induction, an emf is induced only when themagnetic flux changes with time.

Part F

Now, select an AC current source at top right (instead of the DC battery). You can adjust the frequencyof oscillation of the current using the horizontal slider bar (in the blue AC Current Supply window), andyou can adjust the peak current with the vertical slider bar.

Place the electromagnet to the left of the pickup coil (as shown below).



How does the average brightness of the light bulb depend on the AC frequency?

ANSWER:The brightness increases when the frequency increases.

The brightness remains the same when the frequency increases.

The brightness decreases when the frequency increases.

Correct

A higher frequency causes the magnetic field produced by the current to change more rapidly,which causes a greater rate of change of the magnetic flux through the pickup coil. This greateremf causes a greater current.

Part G

With the electromagnet still situated just to the left of the pickup coil, turn the frequency all the waydown to its lowest setting (5 ). Watch how the brightness changes in time and how it correlates with

the instantaneous current (which is shown by the position of the red vertical line in the AC currentsupply meter graph). When is the brightness the greatest?

ANSWER:When the current is zero

The brightness doesn’t change with time!

When the current is a minimum

When the current is a maximum

Correct



The induced emf that drives the current through the light bulb is proportional to the rate of changeof the flux. The magnetic flux is proportional to the current in the electromagnet, so the emf islargest when the current is varying the most quickly. This occurs when the current is zero. Noticethat the instantaneous rate of change of the current is zero when the current is at a maximum orminimum value, which explains why the brightness of the bulb is zero at those moments.

Part H

Adjust the frequency to 100 so the bulb is always “on.”

With the electromagnet still just to the left of the pickup coil, how does the average brightness of thelight bulb depend on the number of loops of the pickup coil? (You can change the number of loops inthe Pickup Coil properties box at bottom right.)

ANSWER:The brightness increases when the number of loops increases.

The brightness does not depend on the number of loops.

The brightness decreases when the number of loops increases.

Correct

The total magnetic flux through the pickup coil is equal to the magnetic flux through each looptimes the number of loops. So when you increase the number of loops, you increase the flux,which makes the flux change more quickly in time as the current oscillates. This greater rate ofchange of the magnetic flux causes a greater induced emf in the pickup coil.

Part I

With the electromagnet still just to the left of the pickup coil, how does the average brightness of thelight bulb depend on the loop area of the pickup coil? (You can change the loop area (as a percentage)using the slider bar at bottom right.)

ANSWER:The brightness decreases when the loop area of the pickup coilincreases.

The brightness increases when the loop area of the pickup coilincreases.

The brightness does not depend on the loop area.

Correct

The average rate of change of the magnetic flux, for a given frequency, is proportional to themaximum magnetic flux. For this configuration, the maximum magnetic flux increases withincreasing loop area (since essentially all of the field lines from the electromagnet are directedtoward the right).

Part J

Now, place the electromagnet centered inside the pickup coil, as shown below.



How does the average brightness of the light bulb depend on the loop area of the pickup coil?

ANSWER:The brightness increases when the loop area of the pickup coilincreases.

The brightness decreases when the loop area of the pickup coilincreases.

The brightness does not depend on the loop area.

Correct

This somewhat surprising result is due to the magnetic flux actually decreasing with an increasingloop area (for a particular instantaneous current in the electromagnet) since the magnetic fieldfrom the electromagnet reverses its direction outside its coil. This field (outside the coil) producesa flux that in part cancels out the magnetic flux due to the field inside the coil. The smallermagnetic flux causes a slower rate of change of the magnetic flux as the current oscillates.

You can see this by looking at the direction of the electromagnet’s field in the following two situations.

Pickup coil loop area 20 : The

electromagnet’s field inside the pick-up coil is all directed to the right.

Pickup coil loop area 100 : The

electromagnet’s field inside the pick-up coil is partly directed to the right(at the center) and partly directed tothe left (outside the electromagnet).



PhET Interactive SimulationsUniversity of Coloradohttp://phet.colorado.edu

Problem 21.18

A 10.8 diameter coil consists of 50 turns of circular copper wire 1.0 in diameter. A

uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.21×10−3 .

Part A

Determine the current in the loop.


ANSWER: = Answer not displayed

Part B

Determine the rate at which thermal energy is produced.




ANSWER: =

Answer not displayed

Problem 21.20

A simple generator is used to generate a peak output voltage of 21.0 . The square armature

consists of windings that are 5.9 on a side and rotates in a field of 0.460 at a rate of 56.0

.

Part A

How many loops of wire should be wound on the square armature?

Express your answer as an integer.

ANSWER: = 37

Correct loops

Problem 21.28

The back emf in a motor is 81 when the motor is operating at 1000 .

Part A

How would you change the motor's magnetic field if you wanted to reduce the back emf to 45

when the motor was running at 2900 ?


ANSWER: = 0.19

Correct

Score Summary:

Your score on this assignment is 89.3%.You received 25.89 out of a possible total of 29 points.

physics 23 hw4

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