Magnetic Force between Two Parallel Conductors

Troubleshoot a SIP Call Between Two Endpoints 1-800-COURSES www.globalknowledge.com Expert Reference Series of White Papers Written and provided by Table of Contents Troubleshoot a SIP Call Between Two Endpoints .1 Document ID: 69467 1 Introduction 1 Prerequisites .1 Requirements 1 Components Used .1 Conventions 1 Configure .1 Network Diagram 2 Configurations 2 Verify .3 Troubleshoot 3 NetPro Discussion Forums − Featured Conversations 11 Related Information .12 Cisco − Troubleshoot a SIP Call Between Two Endpoints i Troubleshoot a SIP Call Between Two Endpoints Document ID: 69467 Introduction Prerequisites Requirements Components Used Conventions Configure Network Diagram Configurations Verify Troubleshoot NetPro Discussion Forums − Featured Conversations Related Information Introduction This document provides a sample configuration of two fax machines in order to demonstrate how a Session Initiation Protocol (SIP) call takes place between two gateways. This document also provides an explanation on the output of the debug ccsip messages command for troubleshooting SIP call failures. Prerequisites Requirements There are no specific requirements for this document. Components Used The information in this document is based on these software and hardware versions: Two fax machines• VG224 that runs Cisco IOS® Software Release 12.4(4)T1• Cisco 3745 router that runs Cisco IOS Software Release 12.3(11)T8• The information in this document was created from the devices in a specific lab environment. All of the devices used in this document started with a cleared (default) configuration. If your network is live, make sure that you understand the potential impact of any command. Conventions Refer to Cisco Technical Tips Conventions for more information on document conventions. Configure In this section, you are presented with the information to configure the features described in this document. Cisco − Troubleshoot a SIP Call Between Two Endpoints Note: Use the Command Lookup Tool ( registered customers only ) to find more information on the commands used in this document. Network Diagram This document uses this network setup: Configurations This document uses these configurations: VG224• Cisco 3745• VG224 vg224#show run Building configuration . ! voice call send−alert voice rtp send−recv ! voice service pots ! voice service voip fax protocol t38 ls−redundancy 0 hs−redundancy 0 fallback cisco sip bind control source−interface FastEthernet0/0 bind media source−interface FastEthernet0/0 ! voice−port 2/0 idle−voltage low ! dial−peer voice 1 pots <fax machine connected to this port> destination−pattern 9000 port 2/0 ! dial−peer voice 100 voip destination−pattern 8000 no modem Magnetic Force between Two Parallel Conductors Magnetic Force between Two Parallel Conductors Bởi: OpenStaxCollege You might expect that there are significant forces between current-carrying wires, since ordinary currents produce significant magnetic fields and these fields exert significant forces on ordinary currents But you might not expect that the force between wires is used to define the ampere It might also surprise you to learn that this force has something to with why large circuit breakers burn up when they attempt to interrupt large currents The force between two long straight and parallel conductors separated by a distance r can be found by applying what we have developed in preceding sections [link] shows the wires, their currents, the fields they create, and the subsequent forces they exert on one another Let us consider the field produced by wire and the force it exerts on wire (call the force F2) The field due to I1 at a distance r is given to be B1 = μ0I1 2πr (a) The magnetic field produced by a long straight conductor is perpendicular to a parallel conductor, as indicated by RHR-2 (b) A view from above of the two wires shown in (a), with one magnetic field line shown for each wire RHR-1 shows that the force between the parallel conductors is attractive when the currents are in the same direction A similar analysis shows that the force is repulsive between currents in opposite directions 1/6 Magnetic Force between Two Parallel Conductors This field is uniform along wire and perpendicular to it, and so the force F2 it exerts on wire is given by F = IlB sin θ with sin θ = 1: F2 = I2lB1 By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2 (Note that F1 = − F2.) Since the wires are very long, it is convenient to think in terms of F / l, the force per unit length Substituting the expression for B1 into the last equation and rearranging terms gives F l = μ0I1I2 2πr F / l is the force per unit length between two parallel currents I1 and I2 separated by a distance r The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions This force is responsible for the pinch effect in electric arcs and plasmas The force exists whether the currents are in wires or not In an electric arc, where currents are moving parallel to one another, there is an attraction that squeezes currents into a smaller tube In large circuit breakers, like those used in neighborhood power distribution systems, the pinch effect can concentrate an arc between plates of a switch trying to break a large current, burn holes, and even ignite the equipment Another example of the pinch effect is found in the solar plasma, where jets of ionized material, such as solar flares, are shaped by magnetic forces The operational definition of the ampere is based on the force between current-carrying wires Note that for parallel wires separated by meter with each carrying ampere, the force per meter is F l 4π × 10 − T ⋅ m/A)(1 A) ( = = × 10 − N/m (2π)(1 m) Since μ0 is exactly 4π × 10 − T ⋅ m/A by definition, and because T=1 N/ (A ⋅ m) , the force per meter is exactly × 10 − N/m This is the basis of the operational definition of the ampere The Ampere The official definition of the ampere is: 2/6 Magnetic Force between Two Parallel Conductors One ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space free of other magnetic fields, causes a force of exactly × 10 − N/m on each conductor Infinite-length straight wires are impractical and so, in practice, a current balance is constructed with coils of wire separated by a few centimeters Force is measured to determine current This also provides us with a method for measuring the coulomb We measure the charge that flows for a current of one ampere in one second That is, C = A ⋅ s For both the ampere and the coulomb, the method of measuring force between conductors is the most accurate in practice Section Summary • The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by F l μ I I = 02πr1 • The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions Conceptual Questions Is the force attractive or repulsive between the hot and neutral lines from power poles? Why? If you have three parallel wires in the same plane, as in [link], with currents in the outer two running in opposite directions, is it possible for the middle wire to be repelled by both? Attracted by both? Explain Three parallel coplanar wires with currents in the outer two in opposite directions Suppose two long straight wires run perpendicular to one another without touching Does one exert a net force on the other? If so, what is its direction? Does one exert a net 3/6 Magnetic Force between Two Parallel Conductors torque on the other? If so, what is its ...[ Team LiB ] Recipe 3.3 Determining the Differences in Data Between Two DataSet Objects Problem You have two DataSet objects with the same schema but containing different data and need to determine the difference between the data in the two. Solution Compare the two DataSet objects with the GetDataSetDifference( ) method shown in this solution and return the differences between the data as a DiffGram. The sample code contains two event handlers and a single method: Form.Load Sets up the sample by creating two DataSet objects each containing a different subset of records from the Categories table from the Northwind sample database. The default view for each table is bound to a data grid on the form. Get Difference Button.Click Simply calls GetDataSetDifference( ) when the user clicks the button. GetDataSetDifference( ) This method takes two DataSet objects with identical schemas as arguments and returns a DiffGram of the differences between the data in the two. The C# code is shown in Example 3-3 . Example 3-3. File: DataSetDifferenceForm.cs // Namespaces, variables, and constants using System; using System.Configuration; using System.IO; using System.Data; using System.Data.SqlClient; // Field name constants private const String CATEGORYID_FIELD = "CategoryID"; DataSet dsA, dsB; // . . . private void DataSetDifferenceForm_Load(object sender, System.EventArgs e) { SqlDataAdapter da; String sqlText; // Fill table A with Category schema and subset of data. sqlText = "SELECT CategoryID, CategoryName, Description " + "FROM Categories WHERE CategoryID BETWEEN 1 AND 5"; DataTable dtA = new DataTable("TableA"); da = new SqlDataAdapter(sqlText, ConfigurationSettings.AppSettings["Sql_ConnectString"]); da.Fill(dtA); da.FillSchema(dtA, SchemaType.Source); // Set up the identity column CategoryID. dtA.Columns[0].AutoIncrement = true; dtA.Columns[0].AutoIncrementSeed = -1; dtA.Columns[0].AutoIncrementStep = -1; // Create DataSet A and add table A. dsA = new DataSet( ); dsA.Tables.Add(dtA); // Fill table B with Category schema and subset of data. sqlText = "SELECT CategoryID, CategoryName, Description " "FROM Categories WHERE CategoryID BETWEEN 4 AND 8"; DataTable dtB = new DataTable("TableB"); da = new SqlDataAdapter(sqlText, ConfigurationSettings.AppSettings["Sql_ConnectString"]); da.Fill(dtB); da.FillSchema(dtB, SchemaType.Source); // Set up Troubleshoot a SIP Call Between Two Endpoints 1-800-COURSES www.globalknowledge.com Expert Reference Series of White Papers Written and provided by Table of Contents Troubleshoot a SIP Call Between Two Endpoints 1 Document ID: 69467 1 Introduction 1 Prerequisites 1 Requirements 1 Components Used 1 Conventions 1 Configure 1 Network Diagram 2 Configurations 2 Verify 3 Troubleshoot 3 NetPro Discussion Forums − Featured Conversations 11 Related Information 12 Cisco − Troubleshoot a SIP Call Between Two Endpoints i Troubleshoot a SIP Call Between Two Endpoints Document ID: 69467 Introduction Prerequisites Requirements Components Used Conventions Configure Network Diagram Configurations Verify Troubleshoot NetPro Discussion Forums − Featured Conversations Related Information Introduction This document provides a sample configuration of two fax machines in order to demonstrate how a Session Initiation Protocol (SIP) call takes place between two gateways. This document also provides an explanation on the output of the debug ccsip messages command for troubleshooting SIP call failures. Prerequisites Requirements There are no specific requirements for this document. Components Used The information in this document is based on these software and hardware versions: Two fax machines• VG224 that runs Cisco IOS® Software Release 12.4(4)T1• Cisco 3745 router that runs Cisco IOS Software Release 12.3(11)T8• The information in this document was created from the devices in a specific lab environment. All of the devices used in this document started with a cleared (default) configuration. If your network is live, make sure that you understand the potential impact of any command. Conventions Refer to Cisco Technical Tips Conventions for more information on document conventions. Configure In this section, you are presented with the information to configure the features described in this document. Cisco − Troubleshoot a SIP Call Between Two Endpoints Note: Use the Command Lookup Tool ( registered customers only ) to find more information on the commands used in this document. Network Diagram This document uses this network setup: Configurations This document uses these configurations: VG224• Cisco 3745• VG224 vg224#show run Building configuration ! voice call send−alert voice rtp send−recv ! voice service pots ! voice service voip fax protocol t38 ls−redundancy 0 hs−redundancy 0 fallback cisco sip bind control source−interface FastEthernet0/0 bind media source−interface FastEthernet0/0 ! voice−port 2/0 idle−voltage low ! dial−peer voice 1 pots <fax machine connected to this port> destination−pattern 9000 port 2/0 ! dial−peer voice 100 voip destination−pattern 8000 no modem passthrough session protocol sipv2 session target ipv4:172.16.184.83 Cisco − Troubleshoot a SIP Call Between Two Endpoints incoming called−number . codec g711ulaw fax protocol t38 ls−redundancy 0 hs−redundancy 0 fallback cisco ! Cisoc 3745 HTTS−VRK1−3745−1#show run Building configuration ! voice service voip sip bind control source−interface FastEthernet0/0 bind media source−interface FastEthernet0/0 ! ! voice−port 4/1/0 ! ! dial−peer voice 9000 voip destination−pattern 9000 session protocol sipv2 session target ipv4:172.16.13.87 incoming called−number . codec g711ulaw fax protocol t38 ls−redundancy 0 hs−redundancy 0 fallback cisco no vad ! dial−peer voice 9 pots destination−pattern 8000 fax rate voice port 4/1/0 forward−digits all Verify There is currently no verification procedure available for this configuration. Troubleshoot Use this section to troubleshoot your configuration. The Output Interpreter Tool ( The paper I have written and know alot about is a contrast on two really well knownand popular guitars. One which is the Gibson Les Paul, and the other which is the Ibanez 453RVC. Both guitars may look alike to some, and to some they may sound alike as well, but arethey really alike? Starting with the bodys, the Gibson Custom Les Paul is made of solid mahogeny wood,which gives the guitar a more Classical look and feel and also a much better and clearer sound.The Gibson is a more luxiourous guitar for which it is better for classical rock and light heavymetal as opposed to the Ibanez which is excellent for hard rock and heavy metal. Going further up the guitar, you have what are called the pickups. A pickup is a reallysensative box that is attached to the body in between the bottom of the neck and the bottom ofthe bridge. The purpose of a pickup is to "pick up" the sounds of the notes or chords that arebeing strummed. There are many different types of pickups; for instance, the Gibson Les Paulhas pickups that are called Humbuckers, which are much higher and of a better quality than thepickups on an Ibanez. The Ibanez comes with regular music store pickups that are not bad butdo not have the quality of the Humbuckers. So having better quality means that the pickups aremore sensative; being more sensative means that the guitar can put out clearer and higher qualitysound.Moving to the lower part of the guitar, both guitars have knobs. The purpose of these knobs are tocontrol the different types of sounds that you want to produce. The Gibson has four controls, butthe Ibanez only has two; having only two knobs instead of four means that the Ibanenz has less ofa selection or variety on the sound that you want opposed to the Gibson having more control overthe sound that you like and the sound that you need. The Gibson having volume and tone foreach pickup allows you to adjust the sound to the way you like it, while the Ibanez has volume andtone for only the one pickup, which controls the sound. The next piece that is connected to every guitar is the neck. Many guitars have manydifferent types of necks varying from length, width, thickness, and different types of wood.The wood on the neck of the Ibanez is poplar wood, which makes a rougher and more ruggedneck. The Gibson, on the other hand, is made of mahogany wood which produces a smootherneck and has a comfortable feel to it. Having a better feel allows the guitar player to increase hisor her speed when just practicing or maybe even accidentally when performing live. The radius ofthe neck on a Gibson is much smaller than the neck of the Ibanez: having a smaller neck also letsthe musician have more control of what they are playing. The fret board is located on the neck of the guitar. The fret board on an Ibanez is madeof rosewood as opposed to the ebony fret board on the Gibson. Having an ebony fret board, thefrets are a lot stronger, more precise (tighter), and better quality. The Ibanez, having rosewood,is not as strong, and from too much playing of the guitar wears the frets down and are dead in acouple of years. So, when the frets are so worn out, you can't just buy a new fret board, youhave to buy a whole new guitar. Then we have the head of the guitar which is located at the very top of the guitar. On thetop of the guitar, we have what are called tuning pegs, used, obviously to tune our guitars. Thetuning pegs on the Gibson are much stronger and they won't go out of tune as easy as the Ibanezwould. The reason the Ibanez goes out of tune is because the pegs are a different type of metal,and the metal that the Ibanez has is much weaker than the Gibsons.So having the Gibson makes it much easier on the musician from constantly having to tune theguitar. Then we have the bridge which is at the bottom of the guitar to hold the strings in place.You connect the string from the tuning peg down along the fret board till you fit them Proc Natl Conf Theor Phys 35 (2010), pp 189-196 THE MAGNETIC CASIMIR EFFECT BETWEEN TWO PARALLEL FERROMAGNETIC PLATES DO PHUONG LIEN Institute of Engineering Physics (IEP) NGUYEN ANH TUAN∗ , NGUYEN TUAN ANH International Training Institute for Materials Science (ITIMS), Hanoi University of Technology (HUT), Dai Co Viet Road, Hai Ba Trung District, Hanoi, Vietnam; ∗ Corresponding author: tuanna@itims.edu.vn Abstract In this work, we study theoretically the long-range magnetic interaction between two parallel ferromagnetic plates, which arises from the magnetic Casimir effect The behavior of the interaction is discussed for two configurations when the magnetization is perpendicular or parallel to the plates The dependence of the Casimir force and Casimir energy on different interplate distances D has been investigated Results of numerical calculations for a Cobalt system have been also presented I INTRODUCTION The Casimir effect, discovered more than 60 years ago in the seminal paper by Casimir [1], is one of the most direct manifestations of the existence of zeropoint vacuum oscillations Casimir predicted the existence of an attractive force between two electrically neutral, infinitely large, parallel conducting planes placed in a vacuum For a long time Casimir’s paper remained relatively unknown, but starting from the 1970s it has rapidly received increasing attention During the last few years significant progress has been made both in the measurement of the Casimir force and in the development of new calculation methods applicable to nontrivial geometries and taking into account real material properties of the interacting bodies Accurate measurements have been difficult to be realized because of the difficulty in aligning the plates in precisely parallel position However, in 2001 a group from the University of Pauda has been able to accurately measure the Casimir force between two parallel plates using micro resonators [2] This force, termed as the Casimir force, is a consequence of the fluctuations of the electromagnetic energy in vacuum, and in the presence of surfaces When changing the boundary condition of the electromagnetic field (e.g by moving a plate with respect to another), the zero-point energy of the system changes and therefore results in an observable force The boundary condition of the electromagnetic field can, however, also be modified without any mechanic displacement, but rather, by changing the order parameter of a collective ordering phenomenon such as ferromagnetism When the two plates are ferromagnetic, the magneto-optical Kerr effect influences 190 DO PHUONG LIEN, NGUYEN ANH TUAN, NGUYEN TUAN ANH the boundary condition of the electromagnetic field so that the Casimir effect manifests the difference in energy (per unit area); ∆E = EAF − EF M (1) between the configurations in which the two plates have their magnetization antiparallel (AF) or parallel (FM) to each other, i.e as a magnetic interaction The dependence of the Casimir force upon the relative orientation of the plates can therefore be derived as: d∆E ∆F = FAF − FF M = − (2) dD where D is the distance between two plates II GENERAL THEORY So far, it exists essentially two kinds of magnetic interaction between magnetic moments or magnetized bodies: the dipole-dipole magneto-static interaction, and the electron-mediated exchange interaction The former, being a relativistic effect, is weak but long-ranged, and therefore mostly effective at the macroscopic or mesoscopic level The latter, being due the combined effect of the Coulomb interaction and the Pauli exclusion principle, is much stronger, but typically short ranged, and is the most important interaction at the atomic scale For two uniformly magnetized ferromagnetic plates held parallel to each other in vacuum, it is shown in paper [3] that the coaction of ... therefore there is never a risk that the two wires will touch and short circuit 4/6 Magnetic Force between Two Parallel Conductors The force per meter between the two wires of a jumper cable being used... the total force on the loop? Find the direction and magnitude of the force that each wire experiences in [link](a) by, using vector addition 5/6 Magnetic Force between Two Parallel Conductors. . .Magnetic Force between Two Parallel Conductors This field is uniform along wire and perpendicular to it, and so the force F2 it exerts on wire is given by