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**Unformatted text preview: **Assignments2/15/11 10:29 AMStudent ViewSummary ViewDiagnostics ViewPrint View withEdit AssignmentAnswersSettings per StudentMP_Assignment#2[ Print ]Due: 8:00am on Monday, February 14, 2011Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading PolicyFaraday's Law and Induced EmfDescription: Discusses Faraday's law; presents a sequence of questions related to finding the induced emfunder different circumstances.Learning Goal: To understand the terms in Faraday's law and to be able to identify the magnitude anddirection 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,whereis the emf induced in a closed loop, andis the rate of change of the magnetic flux through a surface bounded by the loop. For uniform magneticfields the magnetic flux is given by, where is the angle between the magneticfieldand 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 directed opposite tothe "parent" magnetic field, thus countering the increase in flux. If the flux decreases, the induced current'smagnetic field has the same direction as the parent magnetic field, thus countering the 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 sidesuniform magnetic fieldandplaced in a region where aexists (see the diagram). The resistance of the loop is.Initially, the field is perpendicular to the plane of theloop and is directed out of the page. The loop canrotate about either the vertical or horizontal axis,passing through the midpoints of the opposite sides,as shown.http://session.masteringphysics.com/myct/assignmentsPage 1 of 23Assignments2/15/11 10:29 AMPart AWhich of the following changes would induce an electromotive force (emf) in the loop ? When you considereach option, assume that no other changes occur.Check all that apply.ANSWER:The magnitude ofincreases.The magnitude ofdecreases.The loop rotates about the vertical axis (vertical dotted line) shown in thediagram.The loop rotates about the horizontal axis (horizontal dotted line) shown in thediagram.The loop moves to the right while remaining in the plane of the page.The loop moves toward you, out of the page, while remaining parallel to itself.Part BFind the fluxthrough the loop.Express your answer in terms ofANSWER:,, and.=Part CIf the magnetic field steadily decreases fromto zero during a time interval , what is the magnitudeofthe induced emf?Hint C.1Find the change in magnetic fluxWhat is the total change in magnetic fluxExpress your answer in terms ofANSWER:,, andduring this time interval?.=http://session.masteringphysics.com/myct/assignmentsPage 2 of 23Assignments2/15/11 10:29 AMExpress your answer in terms of,,, and .ANSWER:=Part DIf the magnetic field steadily decreases fromto zero during a time interval , what is the magnitudeofthe induced current ?Express your answer in terms of,,, , and the resistanceof the wire.ANSWER:=Part EIf the magnetic field steadily decreases fromto zero during a time interval , what is the direction of theinduced current ?ANSWER:clockwisecounterclockwiseThe flux decreases, so the induced magnetic field must be in the same direction as the original(parent) magnetic field. Therefore, the induced magnetic field is out of the page. Using the right - handrule, we deduce that the direction of the current is counterclockwise.Part FWhich of the following changes would result in a clockwise emf in the loop ? When you consider eachoption, assume that no other changes occur.Check all that apply.ANSWER:The magnitude ofincreases.The magnitude ofdecreases.The loop rotates through 45 degrees about the vertical axis (vertical dottedline) shown in the diagram.http://session.masteringphysics.com/myct/assignmentsPage 3 of 23Assignments2/15/11 10:29 AMThe loop rotates through 45 degrees about the horizontal axis (horizontaldotted line) shown in the diagram.The loop moves to the right while remaining in the plane of the page.The loop moves toward you, out of the page, while remaining parallel to itself.Clockwise emf implies that the induced magnetic field is directed into the page. Therefore, themagnetic flux of the original field must be increasing. Only the first option corresponds to increasingflux.Oscillations in an LC circuit.Description: Derive a differential equation for the current in an LC circuit, guess the solution, and answerconceptual questions about the solution.Learning Goal: To understand the processes in a series circuit containing only an inductor and acapacitor.Consider the circuit shown in the figure.This circuit contains a capacitor of capacitancean inductor of inductanceand. The resistance of allwires is considered negligible.Initially, the switch is open, and the capacitor has a charge . The switch is then closed, and the changesin the system are observed. It turns out that the equation describing the subsequent changes in charge,current, and voltage is very similar to that of simple harmonic motion, studied in mechanics. To obtain thisequation, we will use the law of conservation of energy.Initially, the entire energy of the system is stored in the capacitor. When the circuit is closed, the capacitorbegins to discharge through the inductor. As the charge of the capacitor decreases, so does its energy. Onthe other hand, as the current through the inductor increases, so does the energy stored in the inductor.Assuming no heat loss and no emission of electromagnetic waves, energy is conserved, and at any point intime, the sum of the energy stored in the capacitorand the energy stored in the inductoris aconstant:,whereis the charge on the capacitor andis the current through the inductor (andare functions oftime, of course). For this problem, take clockwise current to be positive.http://session.masteringphysics.com/myct/assignmentsPage 4 of 23Assignments2/15/11 10:29 AMPart AUsing the expression for the total energy of this system, it is possible to show that after the switch isclosed,,whereis a constant. Find the value of the constantHint A.1Since.The derivative of the total energyis a constant,an expression for. Differentiate the expression forin the problem introduction to obtain.Hint A.1.1 A helpful derivative formulaRecall that for any differentiable function, we have. This result is derivedusing the chain rule.ANSWER:=Hint A.2The current in a series circuitIn this one- loop circuit, the current through the capacitor is always equal to that through the inductor.Hint A.3Charge and currentCurrent is defined as the time derivative of charge, but watch the sign. With the conventions adoptedhere (clockwise current and a positive chargeon the upper plate of the capacitor).Therefore,.http://session.masteringphysics.com/myct/assignmentsPage 5 of 23Assignments2/15/11 10:29 AMExpress your answer in terms ofand.ANSWER:=This expression can also be obtained by substitutinginto the equation obtained fromKirchhoff's loop rule. It is always good to solve a problem in more than one way!Part BFrom mechanics, you may recall that when the acceleration of an object is proportional to its coordinate,,such motion is called simple harmonic motion , and the coordinate depends on time as, where , the argument of the harmonic function at, is called the phaseconstant .Find a similar expression for the chargecorrect value ofHint B.1on the capacitor in this circuit. Do not forget to determine thebased on the initial conditions described in the problem.The phase constantInitially, the charge of the capacitor is at a maximum; hence, the argument of the cosine function mustbe zero at the moment the switch is closed,.Hint B.2Find the period of the oscillationsWhat is the periodof these oscillations ?Hint B.2.1 A relation between the period and angular frequencyThe periodand the angular frequencyare related by.Express your answer in terms of,, and other variables given in the introduction.ANSWER:=Express your answer in terms of,, and. Use the cosine function in your answer.ANSWER:http://session.masteringphysics.com/myct/assignmentsPage 6 of 23Assignments2/15/11 10:29 AMANSWER:=Part CWhat is the currentHint C.1in the circuit at timeafter the switch is closed ?Current as a derivative of the chargeNote that. Use the equation you have found forExpress your answer in terms of,,to find., and other variables given in the introduction.ANSWER:=Part DRecall that the top plate of the capacitor is positively charged at. In what direction does the currentin the circuit begin to flow immediately after the switch is closed ?ANSWER:There is no current because there cannot be any current through thecapacitor.clockwisecounterclockwisePart EImmediately after the switch is closed, what is the direction of the EMF in the inductor ? (Recall that thedirection of the EMF refers to the direction of the back - current or the induced electric field in the inductor.)ANSWER:There is no EMF until the current reaches its maximum.clockwisecounterclockwiseIn the remaining parts, assume that the period of oscillations is 8.0 milliseconds. Also, keep in mind thatthe top plate of the capacitor is positively charged at.http://session.masteringphysics.com/myct/assignmentsPage 7 of 23Assignments2/15/11 10:29 AMPart FAt what time does the current reach its maximum value for the first time ?Hint F.1How to approach the problemTry drawing a graph of the current as a function of time, in units of, whereis the period of thecurrent oscillation.Graph ofHint F.2The figure shows a graph of currentANSWER:as a function of time .0.0 ms2.0 ms4.0 ms6.0 ms8.0 msPart GAt what moment does the EMF become zero for the first time ?Hint G.1Relationship between the induced EMF and the current.The EMF is proportional to the rate of change of the current; the EMF becomes zero whenHint G.2.Graph ofhttp://session.masteringphysics.com/myct/assignmentsPage 8 of 23Assignments2/15/11 10:29 AMANSWER:0.02.04.06.08.0Part HAt what moment does the current reverse direction for the first time ?Hint H.1How to approach the problemTry drawing a graph of the current as a function of time, in units of, whereis the period of thecurrent oscillation.Hint H.2Graph ofThe figure shows a graph of current as a function of time.http://session.masteringphysics.com/myct/assignmentsPage 9 of 23AssignmentsANSWER:2/15/11 10:29 AM0.0 ms2.0 ms4.0 ms6.0 ms8.0 msPart IAt what moment does the EMF reverse direction for the first time ?Hint I.1Relationship between the induced EMF and the current.The EMF is proportional to the rate of change of the current. Therefore, the EMF will change directionwhenHint I.2ANSWER:changes sign.Graph of0.0 ms2.0 ms4.0 ms6.0 ms8.0 mshttp://session.masteringphysics.com/myct/assignmentsPage 10 of 23Assignments2/15/11 10:29 AMPart JAt what moment does the energy stored in the inductor reach its maximum for the first time ?Hint J.1Formula for the energy in an inductorThe energyin an inductoris given by,whereHint J.2is the current flowing through the inductor.Graph ofThe figure shows a graph of current as a function of time.Hint J.3Conservation of energyBy conservation of energy, the energy in the inductor will be at a maximum when the energy in thecapacitor is at a minimum. An examination of the function for the charge on the capacitor in Part Breveals that this occurs when the charge on the capicitor is zero.ANSWER:0.0 ms2.0 ms4.0 ms6.0 ms8.0 msPart Khttp://session.masteringphysics.com/myct/assignmentsPage 11 of 23Assignments2/15/11 10:29 AMAt what timeHint K.1do the capacitor and the inductor possess the same amount of energy for the first time ?Use conservation of energySince the total energy is conserved, the energy of the capacitor (like that of the inductor) equals onehalf of the total energy at the time in question.Hint K.2Energy and chargeSince the energy of the capacitor is proportional to the square of its charge, you need to find a timewhen.Hint K.3A useful trigonometric resultRecall that.Express your answer in milliseconds.ANSWER:=Part LWhat is the direction of the current in the circuit 22.0 milliseconds after the switch is closed ?Hint L.1Use the periodic nature of the processThe current would have the same direction as it has 6.0 milliseconds after the switch is closed. (Canyou see why ? )Hint L.2Graph ofThe figure shows a graph of the current as a function of time.http://session.masteringphysics.com/myct/assignmentsPage 12 of 23AssignmentsANSWER:2/15/11 10:29 AMThe current is zero.clockwisecounterclockwisePart MWhat is the direction of the EMF in the circuit 42.0 milliseconds after the switch is closed ?Hint M.1ANSWER:Graph ofThe EMF is zeroclockwisecounterclockwiseSelf - Inductance of a SolenoidDescription: Walks through calculation of self - inductance for a single solenoid with some discussion at theend.Learning Goal: To learn about self - inductance from the example of a long solenoid.To explain self - inductance, it is helpful to consider the specific example of a long solenoid, as shown in thefigure. This solenoid has only one winding, and so the EMF induced by its changing current appears acrossthe solenoid itself. This contrasts with mutual inductance, where this voltage appears across a second coilwound on the same cylinder as the first.Assume that the solenoid has radius , lengthhttp://session.masteringphysics.com/myct/assignmentsPage 13 of 23Assignments2/15/11 10:29 AMalong the z axis, and is wound withturns per unitlength so that the total number of turns is equal to. Assume that the solenoid is much longer than itsradius.As the current through the solenoid changes, theresulting magnetic flux through the solenoid will alsochange, and an electromotive force will be generatedacross the solenoid according to Faraday's law ofinduction:.Faraday's law implies the following relation betweenthe self - induced EMF across the solenoid and thecurrent passing through it:.The "direction of the EMF" is determined with respect to the direction of positive current flow, and representsthe direction of the induced electric field in the inductor. This is also the direction in which the "back - current"that the inductor tries to generate will flow.Part ASuppose that the current in the solenoid ismagnetic fielddue to this current ?Express your answer in terms of(such asANSWER:. Within the solenoid, but far from its ends, what is the, quantities given in the introduction, and relevant constants).=Note that this field is independent of the radial position (the distance from the axis of symmetry) aslong as it is measured at a point well inside the solenoid.Part BWhat is the magnetic fluxthrough a single turn of the solenoid ?Express your answer in terms of the magnetic field, quantities given in the introduction, andany needed constants.ANSWER:=Part Chttp://session.masteringphysics.com/myct/assignmentsPage 14 of 23Assignments2/15/11 10:29 AMSuppose that the current varies with time, so that. Find the electromotive forceinducedacross the entire solenoid due to the change in current through the entire solenoid.Hint C.1Find the flux in terms of the currentIn Part B you found the fluxthrough a single turn of the solenoid. Now find the fluxthrough the entire solenoid.Express your answer in terms ofconstants such as.ANSWER:Hint C.2, other quantities given in the introduction, and various=Find the EMF for the entire solenoidYou now have an expression for the magnetic flux that passes through the solenoid. From this, youshould be able to derive an expression for the EMF in the solenoid. Suppose the total magnetic fluxthrough the solenoid is. What is the electromotive force generated in the solenoid by thechanging flux?Express your answer in terms ofand its derivative, and other variables given in theintroduction.ANSWER:=Hint C.3Putting it togetherYou have now worked out three things:the magnetic field from;the flux from this field;the EMF for the entire solenoid.Put them together and you have the answer!Express your answer in terms of,,, and.ANSWER:=Part Dhttp://session.masteringphysics.com/myct/assignmentsPage 15 of 23Assignments2/15/11 10:29 AMThe self - inductanceis related to the self - induced EMFfor a long solenoid. (Hint: The self - inductanceby the equationwill always be a positive quantity.)Express the self - inductance in terms of the number of turns per lengthand. Find, the physical dimensions, and relevant constants.ANSWER:=This definition of the inductance is identical to another definition you may have encountered:, whereis the magnetic flux due to a current in the inductor. To see thecorrespondence you should differentiate both sides of this equation with respect to time and useFaraday's law, i.e.,.Now consider an inductor as a circuit element. Since we are now treating the inductor as a circuit element,we must discuss the voltage across it, not the EMF inside it. The important point is that the inductor isassumed to have no resistance. This means that the net electric field inside it must be zero when it isconnected in a circuit. Otherwise, the current in it will become infinite. This means that the induced electricfielddeposits charges on and around the inductor in such a way as to produce a nearly equal andopposite electric fieldsuch thatfields produced by charges (like! Kirchhoff's loop law defines voltages only in terms of), not those produced by changing magnetic fields (like). So if wewish to continue to use Kirchhoff's loop law, we must continue to use this definition consistently. That is,we must define the voltagealone (note that the integral is from A to Brather than from B to A, hence the positive sign). So finally,, where we have usedofand the definition.Part EWhich of the following statements is true about the inductor in the figure in the problem introduction,whereis the current through the wire?Hint E.1A fundamental inductance formulaThe self - inductanceis related to the voltageacross the inductor through the equation. Note that unlike the EMF, it does not have a minus sign.ANSWER:Ifis positive, the voltage at end A will necessarily be greater than that atend B.http://session.masteringphysics.com/myct/assignmentsPage 16 of 23Assignments2/15/11 10:29 AMIfis positive, the voltage at end A will necessarily be greater than thatat end B.Ifis positive, the voltage at end A will necessarily be less than that at endB.Ifis positive, the voltage at end A will necessarily be less than that atend B.Part FNow consider the effect that applying an additional voltage to the inductor will have on the currentalready flowing through it (imagine that the voltage is applied to end A, while end B is grounded). Whichone of the following statements is true ?Hint F.1A fundamental inductance formulaThe self - inductanceis related to the voltageacross the inductor through the equation. Note that unlike the EMF, it does not have a minus sign. However, when applyingKirchhoff's rules and traversing the inductor in the direction of current flow, there will be a term, just as traversing a resistor gives a term.ANSWER:Ifis positive, thenwill necessarily be positive andwill bewill necessarily be negative andwill benegative.Ifis positive, thennegative.Ifis positive, thencould be positive or negative, whilewillnecessarily be negative.Ifis positive, thenwill necessarily be positive andwill bepositive.Ifis positive, thencould be positive or negative whilewillnecessarily be positive.Ifis positive, thenwill necessarily be negative andwill bepositive.Note that when you apply Kirchhoff's rules and traverse the inductor in the direction of current flow,you are interested in, just as traversing a resistor gives a term.In sum: when an inductor is in a circuit and the current is changing, the changing magnetic field inthe inductor produces an electric field. This field opposes the change in current, but at the same timedeposits charge, producing yet another electric field. The net effect of these electric fields is that thecurrent changes, but not abruptly. The "direction of the EMF" refers to the direction of the first,induced, electric field.http://session.masteringphysics.com/myct/assignmentsPage 17 of 23Assignments2/15/11 10:29 AMTactics Box 34.1 Using Lenz's LawDescription: Knight Tactics Box 34.1 Using Lenz's Law is illustrated.Learning Goal: To practice Tactics Box 34.1 Using Lenz's Law.Lenzs law is a useful rule for determining the direction of the induced current in a loop. Specifically, it saysthat there is an induced current in a closed conducting loop if and only if the magnetic flux through the loopis changing. The direction of the induced current is such that the induced magnetic field opposes the changein the flux. The following tactics box summarizes the essential steps in using Lenz's law.TACTICS BOX 34.1 Using Lenz's lawDetermine the direction of the applied magnetic field. The field must pass through the loop.Determine how the flux is changing. Is it increasing, decreasing, or staying the same?Determine the direction of an induced magnetic field that will oppose the change in the flux:Increasing flux: The induced magnetic field points opposite the applied magnetic field.Decreasing flux: The induced magnetic field points in the same direction as the applied magnetic field.Steady flux: There is no induced magnetic field.Determine the direction of the induced current. Use the right - hand rule to determine the current direction inthe loop that generates the induced magnetic field you found in Step 3.Follow the steps above to solve the following problem: A wire carries a current in the direction indicated inthe figure. A loop is located next to the wire. If the current in the wire increases, is there a clockwisecurrent around the loop, a counterclockwise current, or no current ?Part ADetermine the direction of the magnetic field of the current - carrying wire.Hint A.1The right- hand rule for fieldsPlace your right thumb in the direction of the current.Curl your fingers around the wire to indicate a circle.Your fingers point in the direction of the magnetic field lines around the wire.http://session.masteringphysics.com/myct/assignmentsPage 18 of 23Assignments2/15/11 10:29 AMANSWER:upwarddownwardinto the plane of the screenout of the plane of the screenPart BAs the current in the wire increases, the magnetic field around the wire increases in magnitude withoutchanging its direction. Does this affect the flux through the loop, and if so, how?ANSWER:As the magnetic field of the wire increases, the flux through the loopincreases as well.As the magnetic field of the wire increases, the flux through the loopdecreases.As the magnetic field of the wire increases, the flux through the loop remainssteady.Now, follow Step 3 in the tactics box above and find the direction of the induced magnetic field.Part CBased on the information found in the previous parts, complete the statement below.Hint C.1The right- hand rule for fieldsPlace your right thumb in the direction of the current.Curl your fingers around the wire to indicate a circle.Your fingers point in the direction of the magnetic field lines around the wire.ANSWER:There isa counterclockwise currentinduced in the loop.Self - Inductance of a Coaxial CableDescription: Find the self - inductance per unit length of a coaxial cable.A coaxial cable consists of alternating coaxial cylinders of conducting and insulating material. Coaxial cablingis the primary type of cabling used by the cable television industry and is also widely used for computernetworks such as Ethernet, on account of its superior ability to transmit large volumes of electrical signal withminimum distortion. Like all other kinds of cables, however, coaxial cables also have some self - inductancethat has undesirable effects, such as producing some distortion and heating.Part Ahttp://session.masteringphysics.com/myct/assignmentsPage 19 of 23Assignments2/15/11 10:29 AMConsider a long coaxial cable made of two coaxial cylindrical conductors that carry equal currentsopposite directions (see figure). The inner cylinder is a small solid conductor of radiusin. The outercylinder is a thin walled conductor of outer radius , electrically insulated from the inner conductor.Calculate the self - inductance per unit lengthcable andof this coaxial cable. (is the inductance of part of theis the length of that part.) Due to whatis known as the "skin effect", the currentflowsdown the (outer) surface of the inner conductingcylinder and back along the outer surface of theouter conducting cylinder. However, you may ignorethe thickness of the outer cylinder.Hint A.1How to approach the problemThe self - inductance of the cable can be found if both the current and the magnetic flux through thecable are known. Using Ampre's law, you can calculate the magnetic field due to the cable; you willfind that there is a magnetic field in the region between the two conductors. Thus, calculate themagnetic flux per unit length through the region between the two cylinders. Since the current in thecable is given, now you should have enough information to calculate the self - inductance of the cable.Hint A.2Find the magnetic field inside the inner cylinderConsider the inner conducting cylinder. It carries a currentmagnitude of the magnetic fieldat a distancealong its outer surface. What is thefrom the axis of the cylinder, in the region inside thecylinder?Hint A.2.1 Ampre's lawAmpre's law states that the line integral of the magnetic fieldcalculated around any closed path isgiven by,whereis the permeability of free space, andis the net current through the area enclosed bythe path. The best loop to use is a circle centered on the cylinders' axis, because the magnetic fieldwill be tangent to the loop at all points, making the dot product easier to compute.Express your answer in terms of some or all of the variables,, and, the permeability offree space.ANSWER:http://session.masteringphysics.com/myct/assignmentsPage 20 of 23Assignments2/15/11 10:29 AMANSWER:=Hint A.3 Find the magnetic field between the cylindersNow consider both conducting cylinders, each carring equal currentsmagnitude of the magnetic fieldat a distancein opposite directions. Find thefrom the axis of the cable in the region between thetwo cylinders.Hint A.3.1 Ampre's lawAmpre's law states that the line integral of the magnetic fieldcalculated around any closed path isgiven by,whereis the permeability of free space, andis the net current through the area enclosed bythe path. The best loop to use is a circle centered on the cylinders' axis, because the magnetic fieldwill be tangent to the loop at all points, making the dot product easier to compute.Express your answer in terms of some or all of the variables,, and, the permeability offree space.ANSWER:=Between the conducting cylinders, the magnetic field is identical to one that would be produced bya wire carrying current located along the cylinders' axis.Hint A.4 Find the magnetic field outside the cableFinally, find the magnitude of the magnetic fieldat a distanceregion outside the cable. Again, assume that a currentfrom the axis of the cable in theflows along the surface of the inner cylinderand flows back on the surface of the outer cylinder.Hint A.4.1 Ampre's lawAmpre's law states that the line integral of the magnetic fieldcalculated around any closed path isgiven by,whereis the permeability of free space, andis the net current through the area enclosed bythe path. The best loop to use is a circle centered on the cylinders' axis, because the magnetic fieldwill be tangent to the loop at all points, making the dot product easier to compute.http://session.masteringphysics.com/myct/assignmentsPage 21 of 23Assignments2/15/11 10:29 AMExpress your answer in terms of some or all of the variables,, and, the permeability offree space.ANSWER:=The fields due to the opposite currents cancel each other out.Hint A.5Find the magnetic flux in the cableFind the magnetic flux per unit lengthinner cylinder andin the cable due to the currents. Letbe the radius of thethat of the outer cylinder.Hint A.5.1 How to approach the problemThe magnetic field of the coaxial cable exists only in the region between the two cylinders. Therefore,you need to calculate the magnetic flux per unit length through this portion of the cable. To do that,think about the direction of the magnetic field (using the right - hand rule) and what area element touse.Hint A.5.2How to calculate the magnetic fluxGiven a finite areain a magnetic field, the magnetic fluxthrough the area is defined by thefollowing integral calculated over the area,,whereis an infinitesimal area element.The magnetic field due to the cable exists only in the region between the cylinders and it isperpendicular to any plane containing the cable axis. Therefore, to calculate the magnetic flux it ismost convenient to choose an area element of length parallel to the cable axis, and of widthlying in the plane containing the axis. The figure below shows such an area element with the crosssection of the cable parallel to the cable axis depicted.Such an area element is perpendicular to themagnetic field, making the dot product easy.http://session.masteringphysics.com/myct/assignmentsPage 22 of 23Assignments2/15/11 10:29 AMExpress your answer in terms of some or all the variables,, , and, the permeability offree space.ANSWER:=Hint A.6Formula for self - inductanceThe self inductanceof a circuit that carries a currentis given by,whereis the magnetic flux through the closed loop of the circuit.Express your answer in terms of some or all the variables,, , and, the permeability of freespace.ANSWER:=The capacitance per unit length, whereof such a coaxial cable is. So the productis the speed of light in vacuum! As you might have guessed, this isnot a coincidence, but a result that is quite generally true for such systems.Score Summary:Your score on this assignment is 0%.You received 0 out of a possible total of 5 points.http://session.masteringphysics.com/myct/assignmentsPage 23 of 23...

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