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Electric Field Due to Increasing Flux

Learning Goal: To work through a straightforward application of Faraday's law to find the EMF and theelectric field surrounding a region of increasing flux

Faraday's law describes how electric fields and electromotive forces are generated from changing magneticfields. This problem is a prototypical example in which an increasing magnetic flux generates a finite lineintegral of the electric field around a closed loop that surrounds the changing magnetic flux through a surfacebounded by that loop. A cylindrical iron rod with cross-sectional area is oriented with its symmetry axis

coincident with the z axis of a cylindrical coordinate system as shown. It has a uniform magnetic field insidethat varies according to . In other words, the magentic field is always in the positive z

direction, and it has no other components.For your convenience, we restate Faraday's law here:

, where

is the line integral of the electric

field, and the magnetic flux is given by

, where is

the angle between the magnetic field and the localnormal to the surface bounded by the closed loop.Direction: The line integral and surface integral reversetheir signs if the reference direction of or is

reversed. The right-hand rule applies here: If thethumb of your right hand is taken along , then the

fingers point along . You are free to take the loop anywhere you choose, although usually it makes sense

to choose it to lie along the path of the circuit you are considering.

Part A

Find , the electromotive force (EMF) around a loop that is at distance from the z axis, where is

restricted to the region outside the iron rod as shown. Take the direction shown in the figure as positive.

Hint A.1 Selecting the loop

Hint not displayed

Hint A.2 Find the magnetic flux

Hint not displayed

Express in terms of , , , , and any needed constants such as , , and .

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

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ANSWER: =

Correct

Part B

Due to the cylindrical symmetry of this problem, the induced electric field can depend only on the

distance from the z axis, where is restricted to the region outside the iron rod. Find this field.

Hint B.1 Calculate the line integral

Hint not displayed

Hint B.2 The z and r components of the electric field

Hint not displayed

Express in terms of quantities given in the introduction (and constants), using the unit

vectors in the cylindrical coordinate system, , , and .

ANSWER: =

Correct

Introduction to Faraday's Law

Learning Goal: To understand the terms in Faraday's law for magnetic induction of electric fields, andcontrast these fields with those produced by static charges.

Faraday's law describes how electric fields and electromotive forces are generated from changing magneticfields. It relates the line integral of the electric field around a closed loop to the change in the total magneticfield integral across a surface bounded by that loop:

,

where is the line integral of the electric field, and the magnetic flux is given by

,

where is the angle between the magnetic field and the local normal to the surface bounded by the closed

loop.Direction: The line integral and surface integral reversetheir signs if the reference direction of or is

reversed. The right-hand rule applies here: If thethumb of your right hand points along , then the

fingers point along . You are free to take the loop


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anywhere you choose, although usually it makes senseto choose it to lie along the path of the circuit you areconsidering.

Part A

Consider the direction of the electric field in the figure. Assume that the magnetic field points upward, asshown. Under what circumstances is the direction of the electric field shown in the figure correct?

Hint A.1 How to approach the problem

Hint not displayed

ANSWER: always

if increases with time

if decreases with time

depending on whether your right thumb is pointing up or down

Correct

Part B

Now consider the magnetic flux through a surface bounded by the loop. Which of the following statementsabout this surface must be true if you want to use Faraday's law to relate the magnetic flux to the lineintegral of the electric field around the loop?

ANSWER: The surface must be the circular disk in the middle of the loop.

The surface must be perpendicular to the magnetic field at each point.

The surface can be any surface whose edge is the loop.

The surface can be any surface whose edge is the loop as long as nomagnetic field line passes through it more than once.

Correct

You are free to take any surface bounded by the loop as the surface over which to evaluate theintegral. The result will always be the same, owing to the continuity of magnetic field lines (they neverstart or end anywhere, since there are no magnetic charges).

It is important to understand the vast differences between electric fields produced by changing magneticfields via Faraday's law and the more familiar electric fields produced by charges via Coulomb's law. Hereare some short questions that illustrate these differences.

Part C


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When can an electric field be measured at any point from the force on a stationary test charge at thatpoint?

Hint C.1 Force on a stationary charge

Hint not displayed

ANSWER: only if the field is generated by the coulomb field of static charges

only if the field is generated by a changing magnetic field

no matter how the field is generated

Correct

In fact, this operation defines an electric field. Similarly, if the test charge is moving, it will measuremagnetic fields.

Part D

When can an electric field that does not vary in time arise?

ANSWER: only if the field is generated by a coulomb field of static charges


in either of the above two cases

Electric fields never vary in time; otherwise, a charge could gain energy fromthe field.

Correct

Part E

When will the integral around any closed loop of the projection of the electric field along that loop

be zero?

ANSWER: only if the field is generated by the coulomb field of static charges

only if the field is generated by the coulomb field of static charges or aconstant current


however the field is generated

The loop integral is always zero; otherwise, a charge moving around the loopwould gain energy.

Correct


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The electric field generated by a static charge or a constant current always has zero loop integral. Aconstant current is a continuous line of evenly-spaced charges moving with constant velocity. Anelectric field generated by any other configuration of moving charges (moving through the loop) wouldhave a non-zero loop integral.

Here is a simple quantitative problem that uses Faraday's law.

Part F

A cylindrical iron rod of infinite length with cross-sectional area is oriented with its axis of symmetry

coincident with the z axis of a cylindrical coordinate system as shown in the figure. It has a magnetic fieldinside that varies according to . Find the theta component of the electric field at

distance from the z axis, where is larger than the radius of the rod.

Hint F.1 Selecting the loop

Hint not displayed

Hint F.2 Find the magnetic flux

Hint not displayed

Hint F.3 Finding the EMF from Faraday's law

Hint not displayed

Hint F.4 Help from symmetry

Hint not displayed

Hint F.5 Find the EMF in terms of

Hint not displayed

Express your answer in terms of , , , , and any needed constants such as , , and .


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ANSWER: =

Correct

A Simple Way to Measure Magnetic FieldsA loop of wire is at the edge of a region of space containing a uniform magnetic field . The plane of the loop

is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area

of the coil inside the magnetic field region is decreasing at the constant rate . That is, , with .

Part A

The induced emf in the loop is measured to be . What is the magnitude of the magnetic field that the

loop was in?

Hint A.1 The formula for the magnetic flux through a loop

Hint not displayed

Hint A.2 How to take the derivative of the product of two functions

Hint not displayed

Hint A.3 The formula for the emf induced in a loop (Faraday's law)

Hint not displayed

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

ANSWER: =

Correct

So you see that in general, there can be contributions to the induced emf in a wire loop both from achanging magnetic field through the loop (about which you may have studied earlier) and from thechange in the area of the loop (within the magnetic field region), as in this problem.

Part B

For the case of a square loop of side length being pulled out of the magnetic field with constant speed

(see the figure), what is the rate of change of area ?


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Hint B.1 How to approach the problem

Hint not displayed

Express your answer in terms of and .

ANSWER: =

Correct

Later, you will learn, if you have not already, that the "motional emf" associated with a rod of length

moving through a uniform magnetic field of magnitude with speed is given by

,

or, equivalently,

.

This is another way of thinking about the result derived above. If you have already studied this, can yousee which sides of the square loop contribute to the motional emf and which do not, and why?

Motion-Induced Electric Fields and Motional EMF

Learning Goal: To understand that the motion of a conductor through a magnetic field generates aperpendicular electric field.

A conducting rod of length is moved at a constant velocity through a uniform magnetic field . This


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field runs perpendicularly out of the page. The end of the rod at is labeled a, and the end of the rod at

is labeled b.

Part A

As a result of the motion through the magnetic field, a charge in the rod will experience a force

: the usual part of the Lorentz force for charges moving through magnetic fields. This force will push thecharge in the rod, and hence this force will be an electromotive force (EMF). For now, we shall say that theforce that moves the charges is due to an induced electric field , which will enable us to calculate

the EMF. The fact that there is an induced electric field at all is rather subtle, because there is no closedloop that encloses some changing flux. Therefore, a method that does not involve Faraday's law must beused to solve this motional EMF problem. In fact, this problem is a good introduction to some of the ideasbehind Faraday's law. Find the y component of the induced electric field .

Hint A.1 Find the force on a charge due to motion in the magnetic field

Hint not displayed

Hint A.2 Find an equivalent electric field

Hint not displayed

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

ANSWER: =

Correct

Part B

To describe the effect of this electric field on the rod, we need to find the EMF . We take as a reference

direction the path from end a to end b (i.e., moving along the positive y axis). The EMF is then negative ifthe induced electric field points in the direction (i.e., like a battery with the positive voltage end at a,

where the positive charge collects due to the magnetic force on the charges).

Hint B.1 What is EMF?

Hint not displayed

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

ANSWER: =

Correct

Part C

There is a big complication in measuring the EMF generated by the moving rod: The wires that connect themeter to the rod also move through the magnetic field, and therefore, there is an electromotive force forthem also. This is a general problem: A voltmeter can measure the EMF produced only in a closed looparound the circuit. In general, the EMF caused by the motion of a rod through a uniform magnetic field willbe canceled by the opposite EMF induced by the motion of the rest of the circuit through this same uniform


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field. The only way to get a nonzero voltmeter reading is to make the field nonuniform, for example, suchthat the bar is moving through a region of nonzero field, but the rest of the circuit is (temporarily) moving in aregion of zero field. For example, consider the arrangement shown in the figure for measuring the EMF inthe moving rod using a voltmeter.In this arrangement, only for and

.

The hookup wires and voltmeter will have to movewith the rod; they are rigid and of the dimensionsand shape shown. The physical setup is that shownat the end of Part B. Which graph shown bestrepresents the magnitude of that will be

measured by the voltmeter? Take to be the

moment pictured in the diagram.

Hint C.1 How to approach the problem

Hint not displayed

Hint C.2 Describe the EMF when only the rod moves through the field

Hint not displayed

Hint C.3 Describe the EMF when the whole circuit is moving through the field

Hint not displayed

ANSWER: a

b

c

d


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e

Correct

Part D

Is the sign of positive or negative? If current flows through the meter from positive to negative, then it

will read a positive voltage.

Hint D.1 Which way is the magnetic force pushing the charge?

Hint not displayed

ANSWER: positive

negative

Correct

It makes little sense to discuss only the EMF generated in the rod. How the wires connect thevoltmeter to the rod is important, too, because they may move through the field (or the field lines maymove across them). The crucial realization (by Michael Faraday) is that EMF is really a property of anentire closed circuit.

Induced EMF and Current in a Shrinking LoopShrinking Loop. A circular loop of flexible iron wire has an initial circumference of 164 , but itscircumference is decreasing at a constant rate of 14.0 due to a tangential pull on the wire. The loop is

in a constant uniform magnetic field of magnitude 1.00 , which is oriented perpendicular to the plane of the

loop. Assume that you are facing the loop and that the magnetic field points into the loop.

Part A

Find the magnitude of the emf induced in the loop after exactly time 5.00 has passed since the

circumference of the loop started to decrease.


Hint not displayed

Hint A.2 An expression for the circumference of the loop as a function of time

Hint not displayed

Hint A.3 An expression for the flux through the loop as a function of its circumference

Hint not displayed


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Hint A.4 A formula for the induced emf in the loop (Faraday's law)

Hint not displayed

Hint A.5 An expression for

Hint not displayed

Express your answer numerically in volts to three significant figures.

ANSWER: = 2.09×10−2

Correct V

Part B

Find the direction of the induced current in the loop as viewed looking along the direction of the magneticfield.

ANSWER: clockwise

counterclockwise

Correct

The induced current flows in the direction that tends to prevent the flux through the coil fromdecreasing. That is, it adds to the magnetic field through the coil as the coil's area is decreasing. Thismeans that the current has to flow clockwise, so that the magnetic field produced by it (right-hand rule)points away from you (you were asked to look at the loop along the direction of the original magneticfield). Alternatively, you could look at how each part of the wire moves toward the center of the loop asit gets smaller. As a result, we can use the standard equation for force on a particle and

the right-hand rule to determine the direction of the current.

An Introduction to EMF and Circuits

Learning Goal: To understand the concept of electromotive force and internal resistance; to understand theprocesses in one-loop circuits; to become familiar with the use of the ammeter and voltmeter.

In order for the current in a conductor to exist continuously, the conductor must be part of a loop, that is, aclosed path through which the charged particles can move without creating a "build-up." Such build-up, if itoccurs, creates its own electric field that cancels out the external electric field, ultimately causing the currentto stop.However, having a loop, or a closed circuit, is not enough to maintain the current; there must also be a sourceof energy. Its necessity is fairly obvious: As charged particles move along the circuit, they lose potentialenergy. In fact, electrostatic forces always push the particles in the direction that leads to a decrease inpotential energy. At some point, each charged particle would reach the location in the circuit where it has thelowest possible potential energy. How can such a particle move toward a point where it would have a higherpotential energy?Such a move requires that nonelectrostatic forces act upon the charged particle, pushing it toward higherpotential energy despite the presence of electrostatic forces. In circuits, such forces exist inside a device


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commonly known as a battery. In a circuit, the battery serves as the energy source that keeps the chargedparticles in continuous motion by increasing their potential energy through the action of some kind ofnonelectrostatic force.The amount of work that the battery does on each coulomb of charge that it "pushes through" is called(inappropriately) the electromotive force (pronounced "ee-em-ef" and abbreviated emf or denoted by ).

Batteries are often referred to as sources of emf (rather than sources of energy, even though they are,fundamentally, sources of energy). The emf of a battery can be calculated using the definition mentionedabove: . The units of emf are joules per coulomb, that is, volts.

The terminals of a battery are often labeled and for "higher potential" and "lower potential," respectively.

The potential difference between the terminals is called the terminal voltage of the battery. If no current isrunning through a battery, the terminal voltage is equal to the emf of the battery: .

However, if there is a current in the circuit, the terminal voltage is less than the emf because the battery hasits own internal resistance (usually labeled ). When charge passes through the battery, the battery doesthe amount of work on the charge; however, the charge also "loses" the amount of energy equal to (

is the current through the circuit); therefore, the increase in potential energy is , and the terminal

voltage is.

In order to answer the questions that follow, you should first review the meaning of the symbols describingvarious elements of the circuit, including the ammeter and the voltmeter; you should also know the way theammeter and the voltmeter must be connected to the rest of the circuit in order to function properly.Note that the internal resistance is usually indicated as a separate resistor drawn next to the "battery" symbol.It is important to keep in mind that this resistor with resistance is actually inside the battery.In all diagrams, stands for emf, for the internal resistance of the battery, and for the resistance of the

external circuit. As usual, we'll assume that the connecting wires have negligible resistance. We will alsoassume that both the ammeter and the voltmeter are ideal: That is, the ammeter has negligible resistance,and the voltmeter has a very large resistance.

Part A

For the circuit shown in the diagram , which potentialdifference corresponds to the terminal voltage of thebattery?

ANSWER: between points K and L

between points L and M

between points K and M

Correct


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Keep in mind that the "resistor" with resistance is actually inside the battery.

The next several questions refer to the four diagrams shown here labeled A, B, C, and D.

Part B

In which diagram(s) (labeled A - D) does the ammeter correctly measure the current through the battery?

Hint B.1 How an ammeter works

Hint not displayed

Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C arecorrect enter AC.

ANSWER: CDCorrect

Part C

In which diagram is the current through the battery nearly zero?

Hint C.1 How to approach the problem

Hint not displayed

ANSWER: A

B

C

D

Correct

Diagram A is the only one in which the current through the battery is the same as the current through


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the voltmeter. Since the latter has a very large resistance, this current is essentially zero.

Part D

In which diagram or diagrams does the ammeter correctly measure the current through the resistor withresistance ?

Hint D.1 How to approach the problem

Note that current is conserved through a wire, and in order for an ammeter to measure the correctcurrent passing through an element, it must be in series with that element.


ANSWER: CDCorrect

Part E

In which diagram does the voltmeter correctly measure the terminal voltage of the battery? Choose the bestanswer.

Hint E.1 How a voltmeter works

Hint not displayed

ANSWER: A

B

C

D

Correct

In diagrams A and B, the voltmeter readings would actually be quite close to the terminal voltage if theammeter has a very low resistance, and the voltmeter, a very high one. However, diagram C clearlyshows the best way to connect the voltmeter in order to measure the terminal voltage.

Part F

In which diagram does the voltmeter read almost zero?


ANSWER: DCorrect

The voltmeter in diagram D is connected to two points that are also connected by a wire that has,


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presumably, very low resistance. Therefore, the charge flowing through that wire will not lose anappreciable amount of potential energy, and the potential difference (voltage) is nearly zero.

Part G

In which diagram or diagrams does the ammeter read almost zero?


ANSWER: ABCorrect

In diagram A, the voltmeter is connected in series with the battery. Since the voltmeter has a very largeresistance there is no (or nearly zero) current in the whole circuit. Therefore, the ammeter reads nocurrent. In diagram B, the current through the ammeter is the same as the current through thevoltmeter. Since the resistance of the voltmeter is very large, the current is nearly zero.

The last group of questions refers to a battery that has emf 12.0 volts and internal resistance 3.00 ohms.

Part H

A voltmeter is connected to the terminals of the battery; the battery is not connected to any other externalcircuit elements. What is the reading of the voltmeter ?

Express your answer in volts. Use three significant figures.

ANSWER: = 12.0Correct

Part I

The voltmeter is now removed and a 21.0-ohm resistor is connected to the terminals of the battery. What isthe current through the battery?

Express your answer in amperes. Use two significant figures.


Part J

In the situation described in Part I, what is the current through the 21.0-ohm resistor?

Express your answer in amperes. Use two significant figures.


Since the battery and the external resistor form one loop, the charge that passes through one must


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pass through another; therefore, the currents must be the same.

Part K

What is the potential difference across the 21.0-ohm resistor from Part I?

Hint K.1 How to approach the problem

The best way to find the potential difference across a resistor when a current is flowing is to use

Ohm's law:



Part L

What is the terminal voltage of the battery connected to the 21.0-ohm resistor from Part I?

Hint L.1 Kirchhoff's voltage law

Kirchhoff's voltage law states that the voltage difference across all the elements in a circuit (in this casejust one resistor) is equal to the voltage at the terminals from the source (in this case a battery).



Since the ends of the resistor with resistance are attached to the terminals of the battery, the

voltage across the resistor is the same as that between the terminals of the battery.

Part M

How much work does the battery connected to the 21.0-ohm resistor perform in one minute?

Hint M.1 How to approach the problem

Find the charge that passes through the battery, and then use the definition of emf.

Hint M.2 Find the charge

How much charge passes through the battery in one minute?

Hint M.2.1 Definition of current


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Recall that current is defined as the number of units of charge that pass through a wire per second.

Express your answer in coulombs. Use three significant figures.

ANSWER: = 30.0

Answer Requested

Express your answer in joules. Use three significant figures.

ANSWER: = 360Answer Requested

A Few Bumps on the Road

Learning Goal: To learn to apply the microscopic theory of conduction.A gauge-12 wire has diameter centimeters and length meters. When the voltage

volts is applied to the ends of the wire, the current is amperes. The concentration of free

electrons in the wire is per cubic meter.

Part A

Find the resistivity of the wire.


Find the electric field and the current density in the wire and use those to calculate the resistivity.

Hint A.2 Definition of resistivity

The resistivity of a material is given by the ratio of the electric field applied to the material and the

current density that then flows through it: .

Hint A.3 What is the current density?

Find the current density flowing through the wire.

Hint A.3.1 Definition of current density

The current density flowing through a material is given by , where is the total current and

is the cross-sectional area in which the current flows.

Hint A.3.2 Find the cross-sectional area

What is the cross-sectional area of the wire?


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Express your answer in square meters. Use three significant figures.

ANSWER: = 3.30×10−6

All attempts used; correct answer displayed

Express your answer in amperes per square meter.

ANSWER: = 7.27×106

Correct

Hint A.4 What is the electric field?

Calculate the electric field along the wire.

Hint A.4.1 Definition of electric field

In a conducting wire, the electric field is constant along the wire, so the electric field is just the

potential difference between the ends of the wire divided by the length of the wire : .

Express your answer in volts per meter.

ANSWER: = 0.200

Correct

Express your answer in ohm-meters. Use two significant figures.

ANSWER: = 2.75×10−8

Correct

Part B

Find the mean time between electron collisions in the wire.


In order to find the mean time between collisions, use the fact that the average velocity of an electronin the presence of an electric field is given by

.

One can then use the definition of the current density and the resistivity to determine the value of .

Hint B.2 Definition of current density


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Recall that current density can be defined as , where is the number of free electrons in

the material, is the electric charge on each electron, and is the drift velocity of all the electrons(average velocity at which they move as a collection of particles).

Hint B.3 Definition of resistivity

Recall that resistivity is defined as

,

where is the applied electric field and is the resulting current density. Use this definition and the

previous hints to determine the mean time between collisions.

Express your answer in seconds. Use two significant figures.

ANSWER: = 2.20×10−14

All attempts used; correct answer displayed

A Five Wire Junction

Learning Goal: To learn to apply the concept of current density and Kirchhoff's junction rule.Consider a junction of five wires, as shown in the figure. The arrows indicate the direction of current flow.

The information about the magnitudes of the current density and the diameters for wires 1, 2, 3, and 4 isgiven in the table. Some of the values are unknown.WireCurrent density ( ) Diameter ( ) Total Current ( )

1 1.6 2.0 ???2 ??? 3.0 2.03 3.0 1.1 ???4 0.8 ??? 4.0

Part A


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Find the current in wire 5.


Hint not displayed

Hint A.2 Kirchhoff's rule

Hint not displayed

Hint A.3 Current density and current

Hint not displayed

Hint A.4 Area of the wire

Hint not displayed

Express your answer in amperes. Use two significant figures. Assume that the current out of thejunction is positive and that the current into the junction is negative.

ANSWER: = -9.9Correct

Note that you did not have to find all the unknown quantities in the table. Separating useful informationfrom the useless (irrelevant) is an important skill that you are expected to develop in studying physics.

A Stretchable ResistorA wire of length and cross-sectional area has resistance .

Part A

What will be the resistance of the wire if it is stretched to twice its original length? Assume that

the density and resistivity of the material do not change when the wire is stretched.

Hint A.1 Formula for the resistance of a wire

Hint not displayed

Hint A.2 Find the cross-sectional area of the stretched wire

Hint not displayed

Express your answer in terms of the wire's original resistance .

ANSWER: =

Correct


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Current and Current Density at a JunctionConsider the juncion of three wires as shown in the diagram.

The magnitudes of the current density and the diameters for wires 1 and 2 are given in the table. Thecurrent directions are indicated by the arrows.

WireCurrent density

( )Diameter

( )

1 3.0 2.02 5.0 3.0

Part A

Find the current in wire 3.


Hint not displayed

Hint A.2 Kirchhoff's rule

Hint not displayed

Hint A.3 Current density and current

Hint not displayed

Hint A.4 Area of the wire

Hint not displayed

Express your answer in amperes to two significant figures. Call current out of the junction positiveand current into the junction negative.

ANSWER: = -26Correct

Part B


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Find the magnitude of the current density in wire 3. The diameter of wire 3 is 1.5 millimeters.

Hint B.1 Current density and current

Hint not displayed

Hint B.2 Area of the wire

Hint not displayed

Express your answer in amperes per square millimeter to two significant figures.

ANSWER: = 15Correct

.

Down To The WireA current of is flowing in a typical extension cord of length . The cord is made of

copper wire with diameter .

The charge of the electron is . The resisitivity of copper is . The

concentration of free electrons in copper is .

Part A

Find the drift velocity of the electrons in the wire.

Hint A.1 Find the current density first

Hint not displayed

Hint A.2 Current density and the drift speed

Hint not displayed

Express your answer in meters per second, to two significant figures.

ANSWER: = 3.30×10−4

Correct

Note that this wire is carrying more current density than is carried by most household wiring ineveryday use. With the given amount of current flowing, the cord would be hot to the touch if it wereunder a rug or had otherwise restricted air flow around it. It would certainly be considered unsafe bystandard electrical safety codes.Even though this wire is carrying a large amount of current for its size, the drift velocity of the electronsis tiny (less than one millimeter per second). This reflects the fact that there is a huge number of free(mobile) electrons in the wire. Let us illustrate this fact with a calculation.

Part B


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The population of the Earth is roughly six billion people. If all free electrons contained in this extension cordare evenly split among the humans, how many free electrons ( ) would each person get?

Hint B.1 Find the volume first

Hint not displayed

Use two significant figures.

ANSWER: = 7.50×1013

Correct

These free electrons undergo frequent collisions with atoms, slowing down and generating heat. Howmany collisions occur in such a conductor? Let us find out.

Part C

Find the total number of collisions ( ) that all free electrons in this extension cord undergo in one second.

Hint C.1 Consider a single electron

Hint not displayed

Hint C.2 Find the time between collisions

Hint not displayed

ANSWER: = 1.80×1037

Correct

Note that does not depend on the applied electric field. The drift speed, however, does.

± How a Real Voltmeter WorksUnlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large.

Part A

A voltmeter with resistance is connected across the terminals of a battery of emf and internal

resistance . Find the potential difference measured by the voltmeter.


Hint not displayed

Hint A.2 How to find the potential between points a and b

Hint not displayed


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Hint A.3 An expression for

Hint not displayed

Hint A.4 Using Kirchhoff's loop rule

Hint not displayed

ANSWER: =

Correct

With a little algebraic manipulation, the answer can also be written as

.

In this form it is easier to see why the voltmeter reading differs from the actual emf it is supposed tomeasure by only a small amount if . It is a good idea to check that the answer gives the

correct result in the limit that .

Part B

If = and , find the minimum value of the voltmeter resistance for which the

voltmeter reading is within 1.0% of the emf of the battery.

Hint B.1 What is meant by "within 1.0%"

Hint not displayed

Express your answer numerically (in ohms) to at least three significant digits.


Typical voltmeters have a range of possible resistances, some of which are much larger than the valueyou just obtained (on the order of megaohms). This allows reasonably accurate measurements ofmuch larger resistances to be made.

Kirchhoff's Rules and Applying Them

Learning Goal: To understand the origins of both of Kirchhoff's rules and how to use them to solve a circuitproblem.

This problem introduces Kirchhoff's two rules for circuits:Kirchhoff's loop rule: The sum of the voltage changes across the circuit elements forming any closed loop iszero.Kirchhoff's junction rule: The algebraic sum of the currents into (or out of) any junction in the circuit is zero.The figure shows a circuit that illustrates the conceptof loops, which are colored red and labeled loop 1 and


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loop 2. Loop 1 is the loop around the entire circuit,whereas loop 2 is the smaller loop on the right. Toapply the loop rule you would add the voltage changesof all circuit elements around the chosen loop. Thefigure contains two junctions (where three or morewires meet)--they are at the ends of the resistorlabeled . The battery supplies a constant voltage

, and the resistors are labeled with their resistances.The ammeters are ideal meters that read and

respectively.The direction of each loop and the direction of eachcurrent arrow that you draw on your own circuits arearbitrary. Just assign voltage drops consistently andsum both voltage drops and currents algebraically and you will get correct equations. If the actual current is inthe opposite direction from your current arrow, your answer for that current will be negative. The direction ofany loop is even less imporant: The equation obtained from a counterclockwise loop is the same as that froma clockwise loop except for a negative sign in front of every term (i.e., an inconsequential change in overallsign of the equation because it equals zero).

Part A

The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits thatare in a steady state.

Hint A.1 At the junction

Hint not displayed

ANSWER: current

voltage

resistance

Correct

Part B

Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance).

Hint B.1 Elements in series

Hint not displayed

Answer in terms of given quantities, together with the meter readings and and the current .

ANSWER:

Correct

If you apply the juncion rule to the junction above , you should find that the ezpression you get is

equivalent to what you just obtained for the junction labeled 1. Obviously the conservation of charge or


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current flow enforces the same relationship among the currents when they separate as when theyrecombine.

Part C

Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuitelement around this loop going in the direction of the arrow. Remember that the current meter is ideal.

Hint C.1 Elements in series have same current

Hint not displayed

Hint C.2 Sign of voltage across resistors

Hint not displayed

Hint C.3 Voltage drop across ammeter

Hint not displayed

Express the voltage drops in terms of , , , the given resistances, and any other given

quantities.

ANSWER:

Correct

Part D

Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changesacross each circuit element around this loop going in the direction of the arrow.

Express the voltage drops in terms of , , , the given resistances, and any other given

quantities.

ANSWER:

Correct

There is one more loop in this circuit, the inner loop through the battery, both ammeters, and resistors and . If you apply Kirchhoff's loop rule to this additional loop, you will generate an extra equation

that is redundant with the other two. In general, you can get enough equations to solve a circuit byeitherselecting all of the internal loops (loops with no circuit elements inside the loop) orusing a number of loops (not necessarily internal) equal to the number of internal loops, with the extraproviso that at least one loop pass through each circuit element.

Batteries in Series or ParallelYou are given two circuits with two batteries of emf and internal resistance each. Circuit A has the

batteries connected in series with a resistor of resistance , and circuit B has the batteries connected in


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parallel to an equivalent resistor.Note that the symbol should be entered in your answers as EMF.

Part A

In which direction does the current in circuit A flow?

Hint A.1 Conventions

Hint not displayed

ANSWER: clockwise

counterclockwise

Correct

Part B

What is the current through the resistor of resistance in circuit A?

Hint B.1 Which formula to use

Hint not displayed

Hint B.2 Total resistance of the circuit

Hint not displayed

Express the current in terms of , , and .

ANSWER: =

Correct

Part C


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Calculate the current through the resistor of resistance for circuit B.

Hint C.1 Which rule to use

Hint not displayed

Hint C.2 What is the emf for loop 1?

Hint not displayed

Hint C.3 What is the emf for loop 2?

Hint not displayed

Hint C.4 Application of Kirchhoff's junction rule (current rule)

Hint not displayed

Express your answer in terms of , , and .

ANSWER: =

Correct

Part D

What is the power dissipated by the resistor of resistance for circuit A, given that ,

, and ?

Hint D.1 What formula to use

Hint not displayed

Calculate the power to two significant figures.

ANSWER: = 0.064Correct W

Part E

For what ratio of and would power dissipated by the resistor of resistance be the same for

circuit A and circuit B?

Hint E.1 Getting started

Hint not displayed

Hint E.2 Finding


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Hint not displayed

ANSWER: = 1Correct

Part F

Under which of the following conditions would power dissipated by the resistance in circuit A be bigger

than that of circuit B?

Hint F.1 How to think about the problem

Hint not displayed

Some answer choices overlap; choose the most restrictive answer.

ANSWER:

Correct

Battery, Ammeter, and ResistorsAn ammeter is connected in series to a battery of voltage and a resistor of unknown resistance . The

ammeter reads a current . Next, a resistor of

unknown resistance is connected in series to the

ammeter, and the ammeter's reading drops to .

Finally, a second resistor, also of resistance , is

connected in series as well. Now the ammeter reads.

Part A


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If , find .


Hint not displayed

Hint A.2 Current through each resistor

Hint not displayed

Hint A.3 Voltage drop across each resistor

Hint not displayed

Hint A.4 Apply Kirchhoff's loop rule:

Hint not displayed


Hint not displayed


Hint not displayed

Hint A.7 Finding a relation between and

Hint not displayed

Express the ratio numerically.

ANSWER: = 0.667

Correct

Brightness of Light Bulbs Ranking Task

Part A

Consider a circuit containing five identical light bulbs and an ideal battery. Assume that the resistance ofeach light bulb remains constant. Rank the bulbs (A through E) based on their brightness.


Hint not displayed

Hint A.2 Comparing bulb A to bulb B


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Hint not displayed

Hint A.3 Comparing bulb D to bulb E

Hint not displayed

Hint A.4 Comparing bulb C to bulb D or E

Hint not displayed

Hint A.5 Comparing bulb C to bulb A or B

Hint not displayed

Rank from brightest to dimmest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

Now consider what happens when a switch in the circuit is opened.

Part B

What happens to the brightness of bulb A?

Hint B.1 How to approach this part

Hint not displayed


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Hint B.2 Consider changes in resistance

Hint not displayed

ANSWER: It gets dimmer.

It gets brighter.

There is no change.

Correct

Part C

What happens to bulb C?

Hint C.1 How to approach this part

Hint not displayed

Hint C.2 Find the current in bulb C earlier

Hint not displayed

Hint C.3 Find the current in bulb C now

Hint not displayed

ANSWER: It gets dimmer.

It gets brighter.

There is no change.

Correct

This is why appliances in your home are always connected in parallel. Otherwise, turning some of themon or off would cause the current in others to change, which could damage them.

Kirchhoff's Loop Rule Conceptual QuestionThe circuit shown belowconsists of four differentresistors and a battery. You don't know the strength ofthe battery or the value any of the four resistances.


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Part A

Select the expressions that will be equal to the voltage of the battery in the circuit, where , for example,

is the potential drop across resistor A.

Hint A.1 Kirchhoff's voltage rule for closed circuit loops

Hint not displayed

Check all that apply.

ANSWER:

Correct

Resistance and Wire LengthYou have been given a long length of wire. You measure the resistance of the wire, and find it to be .

You then cut the wire into identical pieces .

Part A


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If you connect the pieces in parallel as shown ,what is the total resistance of the wires

connected in parallel?

Hint A.1 Find the resistance of the wire segments

Hint not displayed

Hint A.2 Resistors in parallel

Hint not displayed

Express your answer in terms of and .

ANSWER: =

Correct

Series Resistors with Different AreasFour wires are made of the same highly resistive material, cut to the same length, and connected in series.Wire 1 has resistance and cross-sectional area .

Wire 2 has resistance and cross-sectional area .



A voltage is applied across the series, as shown in the figure.


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Part A

Find the voltage across wire 2.

Hint A.1 Resistance of wires

Hint not displayed

Hint A.2 Find the current through wire 2

Hint not displayed

Hint A.3 The voltage across wire 2

Hint not displayed

Give your answer in terms of , the voltage of the battery.

ANSWER: =

Correct

Throw the SwitchIn this problem denotes the emf provided by the source, and is the resistance of each bulb.

Part A

Bulbs A, B, and C in the figure are identical and theswitch is an ideal conductor. How does closing theswitch in the figure affect the potential difference?


Hint not displayed


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Hint A.2 Find the potential difference across bulb C when the switch is closed

Hint not displayed

Hint A.3 Find the potential difference across bulb B when the switch is closed

Hint not displayed

Hint A.4 Find the potential difference across bulb A when the switch is closed

Hint not displayed

Hint A.5 Find the potential difference across bulb A when the switch is open

Hint not displayed


ANSWER: The potential difference across A is unchanged.

The potential difference across B drops to zero.

The potential difference across A increases by 50%.

The potential difference across B drops by 50%.

Correct

Every time the ends of a resistor are joined together, or connected through an ideal conductor, thevoltage across the resistor drops to zero and the resistor is said to be short-circuited.

Part B

One more bulb is added to the circuit and the location of the switch is changed. The new circuit is shown inthe figure. Bulbs A, B, C, and D are identical and theswitch is an ideal conductor. How does closing theswitch in the figure affect the potential difference?


Hint not displayed


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Hint B.2 Find the equivalent resistance of the circuit when the switch is closed

Hint not displayed

Hint B.3 Find the voltage across bulb A when the switch is closed

Hint not displayed

Hint B.4 How to determine whether choice D is correct

Hint not displayed

Hint B.5 Find the voltage across bulb B when the switch is closed

Hint not displayed

Hint B.6 Find the voltage across bulb B when the switch is open

Hint not displayed


ANSWER: The potential difference across A increases.

The potential difference across B doubles.

The potential difference across B drops to zero.

The potential difference across D is unchanged.

Correct

Two ResistorsTwo resistors of resistances and , with , are connected to a voltage source with voltage .

When the resistors are connected in series, the current is . When the resistors are connected in parallel, the

current from the source is equal to .

Part A

Let be the ratio . Find .

Hint A.1 Calculate the source voltage

Hint not displayed

Hint A.2 Find another expresssion for the source voltage

Hint not displayed


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Hint A.3 Equate the two expressions for voltage

Hint not displayed

Hint A.4 Formula for the roots of a quadratic equation

Hint not displayed

Hint A.5 General answer

Hint not displayed

Round your answer to the nearest thousandth.

ANSWER: 0.127Correct

Finding Current by Changing ResistorsA battery provides a voltage of 8.00 and has unknown internal resistance . When the battery is

connected across a resistor of resistance = 7.00 , the current in the circuit is = 1.00 .

Part A

If the external resistance is then changed to = 5.00 , what is the value of the current in the circuit?


Hint not displayed

Hint A.2 Internal resistance explained

Hint not displayed

Hint A.3 Find the internal resistance


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Hint not displayed

Express your answer numerically in amperes.


Measuring the Potential of a Nonideal BatteryA battery with EMF 90.0 has internal resistance = 9.93 .

Part A

What is the reading of a voltmeter having total resistance = 400 when it is placed across the

terminals of the battery?


Hint not displayed

Hint A.2 Series or parallel?

Hint not displayed

Hint A.3 Calculate the current in the circuit

Hint not displayed

Express your answer with three significant figures.


Part B

What is the maximum value that the ratio may have if the percent error in the reading of the EMF of

a battery is not to exceed 2.50 ?


Write an expression for the fraction of error in the reading, which by definition is given as 2.50 , in

terms of the potential of the battery and the potential measured by the voltmeter. Express the potentialsin terms of the current and resistances and solve for the ratio .

Hint B.2 An expression for the error in measurement


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The fractional error in the measurement is simply the difference between the measured potential and the

actual EMF of the battery divided by the actual EMF of the battery, .

Hint B.3 Resistance in the battery

Using Kirchhoff's rules in the previous part, you found that . For the section of the circuit

containing only the voltmeter, .

Express your answer with three significant figures.

ANSWER: = 2.56×10−2

Correct

± Heating a Water BathIn the circuit in the figure, a 20-ohm resistor sits inside112 of pure water that is surrounded by insulatingStyrofoam.

Part A

If the water is initially at temperature 11.8 , how long will it take for its temperature to rise to 58.9 ?


First reduce the system of resistors to a single equivalent resistor; then use this simplified circuit tocalculate the current flowing through the battery. Determine the current flowing through the resistor in thewater and calculate its power output. Finally, use the calculated power output to calculate the timeneeded to heat the water bath.

Hint A.2 Calculate the resistance of the circuit

Calculate the total resistance of the network of resistors shown in the figure.

Hint A.2.1 Reducing a network of resistors to an equivalent resistor


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For any network of resistors, first look at any section (between two junction points) in which allresistors are in series, and combine them appropriately to obtain the equivalent resistance through thatsection of the network. Next, see whether any combined sections are in parallel with each other andcombine them appropriately to to obtain the equivalent resistance through those sections. Continue thisprocess, alternating between sections in series and sections in parallel, until all the resistors have beencombined to make a single equivalent resistor for the system.

Hint A.2.2 Combining the resistors in the middle section

After the current flows through the resistor in the water bath, it splits into three separate paths, eachwith two resistors. What is the equivalent resistance through this section?

Hint A.2.2.1 Series or parallel?

In the middle section there are six resistors. How are they combined in the circuit?

ANSWER: All six resistors are in series.

All six resistors are in parallel.

There are three paths in series and each path consists of two resistorsin parallel.

There are three paths in parallel and each path consists of two resistorsin series.

Correct

Hint A.2.2.2 Resistance in each path

Calculate the resistance of each path, , , , from top to bottom respectively.

Express your answers, separated by commas, using three significant figures.

ANSWER: , , = 20.0,20.0,10.0Answer Requested

Hint A.2.2.3 Resistors in parallel

Hint not displayed

Express your answer in ohms using three significant figures.


Hint A.2.3 Combining the rest of the resistors

Hint not displayed


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Express your answer in ohms using three significant figures.


Hint A.3 Calculate the current in the equivalent resistor

Calculate the current that flows through the battery and the equivalent resistor.

Express your answer in amperes using three significant figures.

ANSWER: = 1.00Answer Requested

Hint A.4 Calculate the current through the resistor in the water bath

Calculate the current that flows through the resistor in the water bath.

Hint A.4.1 Current in the circuit

Note that the first resistor is connected in series with the battery. This means that the current flowingthrough the battery must flow into (and out of) the first resistor before splitting up in the middle section.

Express your answer using three significant figures.


It is possible to find the current that flows through each separate resistor. However, since we areonly looking for the current through the first resistor, that is not necessary: Simple inspection of thecircuit shows that the current through the battery must be the same as the current through the firstresistor (the one in the water bath).

Hint A.5 Calculate the power output of the resistor

Calculate the power dissipated in the resistor immersed in the water bath.

Express your answer in watts using three significant figures.


Recall that one watt is equal to one joule per second. In other words, the power dissipated in theresistor is the same as the energy per second flowing out of the resistor in the form of heat. It is thisheat energy that increases the temperature of the water bath.

Hint A.6 Heating the water


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Recall that for an object of mass , one has the relation , where is the temperature

change of the object, is the heat capacity of the object, and is the heat (or energy) added to the

object to change the temperature. For our system the total heat delivered to the water bath is given by, where is the energy per unit time (power) dissipated in the resistor and is the time interval

during which current flows through the circuit.

Use as the heat capacity of water, and express your answer in seconds using three

significant figures.

ANSWER: = 1110Correct

Comparing brightness of light bulbsConsider five identical light bulbs (A - E) connected to a battery as shown in the circuit below.

Part A

Rank the brightness of all five bulbs from brightess to dimmest.

Hint A.1 Brightness

Hint not displayed

Hint A.2 Compare the brightness of bulbs A and B

Hint not displayed

Hint A.3 Compare the brightness of bulbs D and E

Hint not displayed

Hint A.4 Compare the brightness of bulbs C and D

Hint not displayed

Hint A.5 Compare the brightness of bulbs C and A

Hint not displayed

Hint A.6 Compare the brightness of bulbs D and A

Hint not displayed

Rank the bulbs from brightest to dimmest. To rank items as equivalent, overlap them.

ANSWER:


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View Correct

How does this change effect the circuit?

Suppose bulb E is unscrewed and removed from its socket. (The empty socket remains in the circuit.)

Part B

Does bulb A get brighter, dimmer, or stay the same brightness?


Hint not displayed

Hint B.2 Determine how the current changes

Hint not displayed

ANSWER: Light bulb A gets brighter.

Light bulb A gets dimmer.

Light bulb A stays the same brightness.

Correct

Score Summary:Your score on this assignment is 97.3%.You received 97.3 out of a possible total of 100 points.


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