Electricity and magnetism 1


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Electricity and magnetism 1

  1. 1. ELECTRICITY AND MAGNETISMELECTROSTATICSIf a rod of ebonite rubbed with fur, or with a coat-sleeve, it gains the power to attract lightobjects (bodies), such as pieces of papaer or tin-foil or a suspended glass beads. The discoverythat a body could be made atttractive by rubbing is attributed to Thales (640-548 B.C) butdiscovery is made by chinese long before than f.He seems to have been led to it through theGreeks practice of spinning silk with an amber spindle; the rubbing of a spindle in itsbearings caused the silk to adhere to it. The Greek word for amber is elektron, and a bodymade attractive by rubbing is said to be ―electrified‖. This branch of Electricity, the earliestdiscovered, is called Electrostatics.Little progress was made in the study of electrification until the sixteenth century A.D. ThenGilbert, who was physician-in-ordinary to Queen Elizabeth, found that other substancesbesides amber could be electrified: for example, glass when rubbed with silk. He failed toelectrify metals, however, and concluded that to do so was imposibble.More than 100 years later-in 1734- he was shown to be wrong, by du Fay ; du Fay foundthat a metal could be electrified by rubbing with fur or wool or silk, but only if it wereheld in a handle of glass or amber; it could not be electrified if it were held directly in thehand. His experiments followed the discovery, by Gray in 1729, that electric charges couldbe transmitted through the human body,water,and metals. These are examples ofconductors; glass and amber are examples of insulators.Positive and Negative ElectricityIn the course of his experiments du Fay also discovered that there were two kinds ofelectrification in nature: he showed that electrified glass and amber tended to opposeone anothers attractiveness. To illustrate how he did so, we may use ebonite instead ofamber, which has the same electrical properties. We suspend a pith-ball, and attract it with anelectrified ebonite rod E (fig.1(i); we then bring an electrified glass rod G towards the eboniterod, and the pith-ball falls away (fig.1(ii). Benjamin Franklin, a pioneer of electrostatics, gavethe name of ―positive electricity‖ to the charge on a glass rod rubbed with silk, and ―negativeelectricity‖ to that on an ebonite rod rubbed with fur.Fig.1 (i) and (ii) 1
  2. 2. ElectronsElectric charge is a fundamental property of matter. Charge comes in two types, arbitrarilycalled positive and negative. Charge is quantized, with one elementary charge—themagnitude of the electron or proton charge— equal to 1.60X10 19 C, where the coulomb (C)is the SI unit of charge. Charge is also conserved, in that the algebraic sum of the charges in aclosed system never changes Towards the end ofthe 19 th century J.J.Thomson dicoverd the existence of theelectron. It is about 1/1840 th of the mass of the Hydrogen atom-and experiments show that itcarries a tiny quantity of negative electricity. Later experiments showed that electrons arepresent in all atoms. Electrons exist round nucleus which carry positive electricity. Normallyatoms are electricly neutral which means the total negative charge on the electrons is equalto the positive charge on the nucleus. Robert Millikan (1868-1953), in1909 discovered that electric charge is quantized:any positive or negativecharge can be written as q=ne, where n is positive or negative integer,e is a constant of nature called the elementary charge.( 1.60x 10 19 C). The elementary charge e is one of the important constants of nature q=ne n= 1 , 2 , 3, . . . Electric charge is quantized. Electric charge isconserved: the algebraic net charge of any isolated systemcannot changeGold-leaf Electroscope One of the earliest instruments used for testing positive and negative charges consistedof a metal rod A to which gold leaves L were attached (Fig..2). The rod was fitted with acircular disc or cap B, and was insulated with a plug P from metal case C which screened Lfrom outside influences other than those brought near to B.When B is touched by an ebonite rod rubbed with fur, some of the negative charge on the rodpasses to the cap and L; and since like charges repel, the leaves diverge ( Fig.2 (i). If anunknown charge X is now brought near to B, an increased divergence implies that X isnegative(fig.2(ii). A poistive charge is tested in a similar way; the electroscope is first given apositive charge and an increased divergence indicates a positive charge. Figure 2 2
  3. 3. Induction and Charging By InductionWe shall now show that it is possible to obtain charges, called induced charges, without anycontact with another charge. An experiment on electrostatic induction, as the phenomenon iscalled, is shown in fig.3(i). Two insulated metal spheres A,B are arranged so that they touchone another, and a negatively charged ebonite rod C is brought near to A. The spheres arenow separated, and then the rod is taken away. Tests with a charged pit-ball now show that Ahas a positive charge and B a negative charge fig.3(ii). If the spheres are placed together sothat they touch, it is found that they now have no effect on a pith-ball held near. Their chargesmust therefore have neutralized each other completely thus showing that the induced positiveand negative charges are equal. This is explained by the movement of electrons from A to Bwhen the rod is brought near. B has then a negative charge and A an equal positive charge. 3
  4. 4. Fig.3 An experiment on electrostatic induction Fig 4 Electrostatic induction Fig 3 shows how a conductor can be given a permanent charge by induction, without dividingit in two. We first bring a charged ebonite rod, say, near to the conductor, (i);next we connectthe conductor to earth by touching it momentarily (ii); finally we remove the ebonite. We thenfind that the conductor is left with a positive chargeThis phenomenon of induction can again be explained by the movement of electrons. If theinducing charge is negative, then, when we touch the conductor, electrons are repelled from itto earth, as shown in fig 3(ii) and a positive charge is left on the conductor. If the inducingcharge is positive, then the electrons are attracted up from the earth to the conductor, whichthen becomes negatively charged.Conductors and Insulators In some materials, such as metals,tap water and the human body,outer electrons ofatoms are loosly bound. therefore negative charge can move rather freely. We call suchmaterials conductors. İn other materials, such as glass, chemically pure water and plastic nonof the charge can move freely. We call these materials nonconductors,or insulators. 4
  5. 5. İf you rub a copper rod with a wool while holding the rod in your hand you will not beable to charge the rod,because both you and the rod, are conductors Again conductors are material in which a significant number of charged particles(electrons in metals) are free to move. The charged particles in nonconductors,orinsulators are not free to moveLaw of Force between two ChargesThe magnitude of the force between two electrically charged bodies was studied by Coulombin 1875. he showed that, if the bodies were small compared with the distance between them,then the force F was inversely proportional to the square of the distance r, 1F (1) r2This result is known as the inverse square law, or Coulomb’s law.It is not possible to verify the law accurately by direct measurement of the force between twocharged bodies. In 1936 Plimton and Lawton showed, by an indirect method, that the powerin the law cannot differ from 2 by more than 2 10-9. We have no reason to suppose,therefore, that the inverse square law is other than exactly true.Quantity of ChargeThe SI unit of charge is the coulomb (C). The ampere(A), the unit of current, is defined later.The coulomb is defined as that quantity of charge which passes a section of a conductor inone second when the current flowing is one ampere.By measuring the force F between two charges when their respective magnitudes q 1 and q 2are varied, it is found that F is proportional to the product q 1 q 2Thus F q 1 q 2 (2)Law of ForceCombining (1) and (2), we have q1q 2 q1q 2 F F= k (3) r2 r2where k is a constant. For reasons explained later, k is written as 1/4 o, where o is aconstant called the permittivity of free space if we suppose the charges are situated in avacuum. Thus 1 q1q 2F= 4 0 r2In this expression, F is measured in newtons (N), q in coulombs (C) and r in metres (m).Now, from (4), 5
  6. 6. q1 q 2 o = 4 Fr 2Hence the units of o are coulomb2 newton -1 metre-2 (C2N-1m-2).PermittivitySo far we have considered charges in a vacuum. If charges are situated in other media suchas water, then the force between the charges is reduced. Equation (4) is true only in a vacuum.In general, we write 1 q1q 2F= 4 r2Where is the permittivity of the medium. The permittivity of air at 1.005 times that, o,of a vacuum. For most purposes, therefore, we may assume the value of o for thepermittivity of air. The permittivity of water is about eighty times that of a vacuum. Thus theforce between charges situated in water is eighty times less than if they were situated the samedistance apart in a vacuum. The unit of o, more widely used, is farad metre-1 ( Fm-1). We shall see later that ohas the numerical value of 8.854 10-12, and 1/4 o then has the value 9 109 approximately. The elementary charge e is one of the important constants of nature and its charge ( 1.60x 10 19 C). Particles Charge (C) Mass (kg) Electron (e) -1,6021917x10-19 9,1095x10-51 Proton (p) +1,6021917x10-19 1,67261x10-27 Neutron (n) 0 1,67492x10-27Table:1Fig.3 Vectoral force to show vector form of Coulomb‘s LawVector form of Coulomb’s Law 6
  7. 7. q1q 2 F= k û r (4) ûr =r r2If there are more than 2 charges resultant charges acted on one charge isF1 = F21 + F31 + F41 (5)PROBLEMS1-a)Calculate the value of two equal charges if they repel one another with a force of 0.1 N 1when situated 50 cm apart in a vacuum. =9 109 4 0 b) What would be the size of the charges if they were situated in an insulating liquid whosepermitivity was 10 times that of a vacuum?SOLUTION: 1 q1q 2 From F= 4 0 r2 Since q =q here, 9 x109 q 2 0.1= (0.5) 2 2 0.1x(0.5) 2 or q = 9 x10 9 q=2.7x10 6 C (coulomb),approx. =2.7 C ( microcoulomb). b) The permitivity of the liquid =10 0 1 q1q 2 F= 4 0 r2 1 q2 = 2 10 (4 0) r (0.1) (0.5)2 10 q2 = 9 10 (9) q= 5 .3 10 6 C=5.3 C -------------------------------------------------- 7
  8. 8. 2) Three charged particles are arranged in aline as shown below. Distance between q 1and q 2 0.30m. Total distance between q 1 and q 3 is 0.50m. k=9.0x 109 N. m 2 / C 2 ;q 1 = -8 C q 2 =+3 C q 3 = -4 CCalculate the each electrostatic force on particle 3 (4 C) due to other two chargesSOLUTION: (9.x109 )(4 x10 6 )(8 x10 6 ).F 31 = =1.2N repulsive (0.50) 2 (9 x109 )(4 x10 6 )(3x10 6 )F32 = =2.7N attractive (0.20) 2 THE ELECTRIC FIELDElectric Intensity or Field-strengthAn ―electric field‖ can be defined as a region where an electric force is experienced. As ingravitation or magnetism,The force exerted on a charged body in an electric field depends on the charge of thebody and on the intensity or strength of the field. If we wish to explore the variation inintensity of an electric field, then we must place a test charge q 0 at the point concernedwhich is small enough not to upset the field by its introduction. The intensity E of anelectrostatic field at any point is defined as the force per unit charge which it exerts atthat point. Its direction is that of the force exerted on a positive charge. q 0From this definition, FE= (6) q0 lim FE= q0 0 q0F= E q 0Since F is measured in newtons and q 0 in coulombs, it follows that intensity E has unitsof newton per coulomb (NC-1). We shall see later that a more practical unit of E is voltmetre-1 (Vm-1)jThe electric field at a point charge 8
  9. 9. q E=k (7) r2 The electric field at a point is a vector giving the electric force per unit chargethat would be experienced by a charge at that point: F E= q0 The electric field of a point charge q is therefore (8) We can easily find an expression for the strength E of the electric field due to a point charge situated in a vacuum .For the force between two such charges. 1 Qq0 F Q F= E= = 4 0 r2 q0 4 0 r 2 The direction of E is away from the point charge if the charge Q is positive; it is radially inward if the charge Q is negative. Continious Charge Distributions With continuous distributions of charge, the sum over all point charges becomes anintegral, giving qi ∆E= k e ri i ri 2 9
  10. 10. (9) Q Volume charge density ρ= (10) V Q Surface charge density (11) A Q Line charge density (12)  For any finite charge distribution with nonzero net charge, the field approaches that ofa point charge at large distances. LİNES OF FORCE A convanient way to visualize the electric field produced by a charge or charge distribution is to construct a map of the field lines of the force of the field. By other words electric fields can be mapped out by electrostatic lines of force, which may be defined as a line such that the tangent to it is in the direction of the force on a small positive charge at that point. Arrows on the lines of force show the direction of the force on a positive charge; the force on a negative charge is in the opposite direction. 10
  11. 11. Fig.5Fig.63) An electron of charge 1.6 10-19 C is situated in a uniform electric field of intensity1200 volt cm-1. Find the force on it, its acceleration, and the time it takes to travel 2 cmfrom rest (mass of electron, m,=9.10 10-31 kg )Force on electron F= eENow E= 1200 volt cm-1= 120 000 volt m-1 F = 1.6 10-19 1.2 10 5 =1.92 10-14 N (newton) F 1.92 x10 14Acceleration, a= = 31 = 2.12 10 16 m s-2 ( metre second-2) m 9.1x10 11
  12. 12. Time for 2 cm travel is given by 1 2 s= at 2 2s 2 0.02 t= = = 1.3 10-9 seconds. a 2.12 10(16) -------------------------------------------------4) Calculate the magnitude and direction of the electric field at a point P which is 30 cm tothe right of a point charge q=-3x 10 6 C k=9.0x 109 N m 2 /C 2 30 cm------ • P 6 q=-3x 10 C ESOLUTION: the magnitude of the electric field due to sıngle poınt charge ıs gıven q (9.0 X 109 )(3.0 X 10 6 ) E=k 2 = 2 =3.0X 105 N/C r (0.30) Direction of the electric field is toward the charge q--------------------------------------- 5) Determine the magnitude of the electric force on the electron of a hydrogen atom exerted by the proton Asume the distance between electron orbits and proton is r= 0.53x 10 10 m., k=9.0x 109 N. m 2 / C 2 ;SOLUTION: 1 q1q 2 From F= 4 0 r2 19Since q 1 =q 2 =1.6x10 C here (9 x109 )(1.6 x10 19 )(1.6 x10 19 )F= 10 2 =8.2x10 8 N (0.53x10 )-----------------------------------------------------------6) The Earth, which is an electrical conductor,carries a net charge of Q= - 4.3x10 5 Cdistributed approximately uniformly over its surface . Find the surface charge density andcaculate the electric field at Earth‘s surface. Earth‘s radius R=6.37x10 6 mSOLUTION: Q Q 4.3x105Earth‘s surface charge density = = = =-8.43x10 10 C/m 2 A 4 R 2 4 (6.37 x106 ) 2 12
  13. 13. .A 8.43 x10 10E.A= E = =-95 N/C 8.85 x10 12Where the minus(-) sign indicates that the field direction is downwardELECTRIC FLUXBefore discussing Gauss‘s law itself we first discuss the concept of fluxFlux from a Point ChargeWe have already shown how electric fields can be described by lines of force. It can besaid that the density of the lines increases near the charge where the field intensity is high.The intensity E at a point can thus be represented by the number of lines per unit areathrough a surface perpendicular to the lines of force at the point considered. Asurface area A through which auniformelectric field E passesFig.7Fig.8The electric flux , through this surface is defined as the product. =ExA (13)"number of field lines crossing a surface." We call this quantity the electric flux, Ф,through the surfaceIf the area is not perpendicular to E but rather makes an angle θ fewer field lines willpass the area. In this case the electric flux through the surface as =ExA= ExAcos θ (14) 13
  14. 14. Fig.97) +1 C charge in the center of sphere whose radius 1m..Find the flux emerging fromthe sphere. k=8.89x 109 N. m 2 / C 2SOLUTION:Electric field at the surface of the sphere Q 1x106 CE=k 2 =(8.89x 109 N. m 2 / C 2 ) 2 =8.99x 103 N/C r (1m)ФE =E.A= E(4 r 2 ) A=4 r 2 =12.6 m 2ФE =E.A= E(4 r 2 )=(8.99x 103 N/C)12.6 m 2 =1.13x 105 N. m 2 / C 28)Three charges -1 µC, 2 µC and 3 µC are placed respectively at the corners A, B, C of anequilateral triangle of side 2 metres. Calculate (a) the potential, (b) the electric field, at a pointX which is half-way along BC. = = 18 x 103 V. 14
  15. 15. VA = = = - 5 x 103 V (approx.) = (18+27-5) x 103 V = 40 x 103 V.EB = = = 18 x 103 V m-1.EC = 27 x 103 V m-1.EA = = = 3 x 103 V m-1.E2 = EA2 + (EC - EB)2 = (9 + 81) x 106 = 90 x 106E =3 x 103 V m-1 = 9.5 x 103 V m-1.Tan = = = = 18º 25´. GAUSS’S LAW We can in preciple determine theelectric field due to any given distribution of electric charge using Coulmb‘s law. The total electric field at any point will be the vector sum(or integral) of contributions from all charges present E= E1 + E 2 +….. E= dE Except for some simple cases, the sum or integral can be quite complicated to evaluate. In some cases , theelectric field due to given charge distribution can be calculated more easily using Gauss‘s law.The major importance of Gauss‘s law is that it gives us additional insight into the nature of electrostatic fields and more general relationship 15
  16. 16. between charge and field. If a surface is curved, then we divide the surface into manysmall patches, each small enough that its essentially flat and that the field is essentiallyuniform over each. If a patch has area dA, then Equation below gives the flux through it: dФ = E • dA. (15)where the vector dA is normal to the patch (Fig. 10). The total flux through the surface isthen the sum over all the patches. If we make the patches arbitrarily small that sumbecomes an integral, and the flux isFig.10The magnitude of electeric field on the surface of sphere is the same as E = keq/r2For ΔAi It can be writtenNet flux through Gauss surfaceIn many cases (in particular) we deal with the electric flux through closed surface = E.dA Gauss’s Law (16)Consider a sphere of radius r drawn in space concentric with a point charge. The value ofelectric field E. It can be written at this place. dA.=A= 4πr2The total flux through the sphere is, =ExA 1The net flux through the sphere is ke= therefore 4 0 16
  17. 17. q qФE = E(4 r 2 )= (4 r 2 )= 4 r2 0 0 q q ch arg e.inside.sphereФE = ФE = = (1)Gauss’s Law 0 permittivity (17)The electric flux through any closed surface is proportional to the charge enclosed bythat surface.This demonstrates the important fact that the total flux crossing any sphere drawn outsideand concentrically around a point charge is a constant. It does not depend on the distancefrom the charged sphere. It should be noted that this result is only true if the inversesquare law is true.Field due to Charged Sphere qThe flux passing through any closed surface whatever its shape, is always equal towhereq is the total charge enclosed by the surface. This relation, called Gauss‘s Law, can beused to find the value of E in other common cases.(1).Outside a charged sphereThe flux across a spherical surface of radius r, concentric with a small sphere carrying acharge q is given by, q qFlux= E 4 r2 = E=q /4 r 2 (18)Fig.11 17
  18. 18. This is the same answer as that for a point charge. This means that outside a charged sphere, the field behaves as if all the charge on the sphere were concentrated at the centre. Fig.12 Two gaussian surfaces surrounding a spherical charge distribution. Surface 1 lies outside the distribution, and encloses all the charge q. Surface 2 lies inside the distribution, and encloses only part of the charge (2)Inside a charged empty sphere Fig.13 Flux and Electric field inside a charged empty sphere Suppose a spherical surface A is drawn inside a charge sphere, as shown in fig. 12 Inside this sphere the flux is zero.. Hence. E must be zero everywhere inside a charged sphere. E=0 (19) The electric field is zero inside a conductor ( E=0 ) in electrostatic equilibrium. Ifa conductor in electrostatic equilibrium carries a net charge, all excess charge resides onthe conductor surface. 18
  19. 19. (3) Outside a charged Plane Conductor Now consider a charged plane conductor with a surface charge density of coulomb metre-2. Applying equation ch arg e.inside.surface E area Now by symmetry, the intensity in the field must be perpendicular to the surface. Further, the charges which produce this field are those in the projection of the area on the surface . The total charge here is thus .A coulomb .A E.A= E (20)Field Round Points and The Action of Points, we saw that the surface-density of charge (charge per unit area) round a point of a conductoris very great. Consequently, the strength of the electric field near the point is very great. Theintense electric field breaks down the insulation of the air, and sends a stream of chargedmolecules away from the point,by other words charge leaks away from a sharp point throughthe air. This mechanism of the breakdown, is called a ‗corona discharge’. Coronabreakdown starts when the electric field strength is about 3 million volt metre-1. thecorresponding surface-density is about 2.7 10-5 coulomb metre-2In all types of high-voltage equipment sharp corners and edges must be avoided, exceptwhere points are deliberately used as electrodes. Otherwise, corona discharges may break outfrom the sharp places. All such places are therefore enlarged by metal globes, these are calledstress-distributors. POTENTIAL IN FIELDS When an object is held at a height above the earth it is said to have potential enrgy. A heavy body tends to move under the force of attraction of the earth from a point of great 19
  20. 20. height to one of less, and we say that points in the earth‘s gravitational field have potentialvalues depending on their height.Electric potential is analogous to gravitational potential, but this time we think of points inan electric field. Thus in the field round a positive charge, for example, a positive chargemoves from points near the charge to points further away. Points round the charge are saidto have an electric potential.Electric potential in the space is described potantial energy per unit of charge at that point E V= P (21) qPOTENTIAL DIFFERENCELet us consider two points A and B in an electrostatic field, and let us suppose that theforce on a positive charge q has a component in the direction BA. Then if we move apositively charged body from B to A, we do work against this component of the field E.We define the potential difference between A and B as the work done in moving a unitpositive charge from B to A. We denote it by the symbol VAB W AB VAB = V B - V A =(22) qFig.14 The work required to move a charge q from A to B in a uniform electric field isqE  . We define the potential difference ∆V between two points in an elektrik field asWORK AND ENERGYThe work done will be measured in joules(J). The unit of potential differences is called thevolt and may be defined as follows: the potential difference between two points A and Bis one volt if the work done in taking one coulomb of positive charge from B to A is onejoule.From this definition, if a charge of q coulombs is moved through a p.d. of V volt, then thework done W in joules is given by WW= qV (23) V= qLet us consider two points A and B in an electrostatic field, A being at a higher potentialthan B. The potential difference between A and B we denote as usual by VAB.If we take apositive charge q from B to A, we do work on it of amountW=q.VAB : the charge gains this amount of potential energy. 20
  21. 21. If we now let the charge go back from A to B , it loses that potential energy: work is done on it by the electrostatic force, in the same way as work is done on a falling stone by gravity. This work may become kinetic energy, if the charge moves freely, or external work if the charge is attached to some machine, or a mixture of the two. The work which we must do in first taking the charge from B to A does not depend on the path along which we carry it, just as the work done in climbing a hill does not depend on the route we take. The fact that the potential differences between two points is a constant, independent of the path chosen between the points, is the most important property of potential in general. POTENTIAL DIFFERENCE FORMULA To obtain a formula for potential difference, let us calculate the potential difference between two points in the field of a single point positive charge, q in fig.14. for simplicity we will assume that the points, A and B, lie on a line of force respectively (24) Generally, you may see our ΔVA→B written as VAB or VBA or VB — VA. We use the Δ here to show explicitly that were talking about a change or difference from one point to another, and we use the subscript A→B to make it clear that this the potential difference going from A to B. In the next section well show how our notation is equivalent to the commonly used VB — VA, and in subsequent chapters well sometimes use just the symbol V for potential difference. In the special case of a uniform field, Equation above reduces to (25) There  is a vector from A to B. Figure 14 shows the special case when the field Eand path  are in opposite directions. 21
  22. 22. The potential difference can be positive or negative, depending on whether the pathgoes against or with the field. Moving a positive charge through a positive potential differenceis like going uphill: We must do work on the charge, and its potential energy increases.Moving a positive charge through a negative potential difference is like going down hill: Wedo negative work or, equivalently, the field does work on the charge, and its potential energydecreases. In both cases the opposite is true for a negative charge; even though the potentialdifference remains the same, the work and potential energy reverse because of the negativesign on the charge.9) At the back end of TV picture tube the field, between Aand B points which is 5 cm, isuniform with of 600 kN/C. Find the potential difference between Aand B points .SOLUTION:Potential differenceV A B = V=Eℓ=(600x 103 N/C)(0.05m)=30 kVİf( +) B is potantially higher than A, if (-) B is potantially lower than A The Volt and the Electron Volt CE The definition of potential difference shows that its units are joules/coulomb. Potentialdifference is important enough that 1 J/C has a special name—the Voltt (V). To say that a carhas a 12-V battery, for example, means that the battery does 12 J of work on every coulombthat moves between its two terminals. We often use the term voltage to speak of potential difference, especially in describingelectric circuits. Strictly speaking the two terms are not synonymous, since voltage is usedeven in nonconservative situations that arise when fields change with time. But in commonusage this subtle distinction is usually not bothersome. Potential Difference Depends on Two Points Specifically, it is the energy per unit charge involved in moving between those points.Always think of potential difference in terms of two points.. In molecular, atomic, and nuclear systems its often convenient to measure energy inelectronvolts (eV), defined as follows: One electron volt is the energy gained by a particlecarrying one elementary charge when it moves through a potential difference of onevolt. 22
  23. 23. Since one elementary charge is 1.6x10-19 C, 1 eV is 1.6x 10-19 J. Energy in eV is particularly easy to calculate when charge is given in elementary charges.CAPACITORSA capacitor is a device for storing electricity. The earliest capacitor was invented almostaccidentally by van Musschenbroek of Leyden, in about 1746, and became known as aLeyden jar. A present day, all capacitors consist of two metal platesinsulator in differentgeometrical forms separated by an The insulator is called the dielectric; in somecapacitors it is oil or airFig 15CARGING A CAPACITORTo study the action of a capacitor we need two plates which have the same shape .Twoplates are spaced apart by a fixed distance We connect the batteries in series, a two waykey(K as shown in fig 16) and a voltmeter to measure their total voltage. If we close the keyat A (switch on), the capacitor is connected via the resistor to the batterry, and the potentialdifference across the capacitor, V, which is measured 23
  24. 24. Fig 16by the voltmeter, The potential difference becomes steady when it is equal to the batteryvoltage V. If we now open the key(switch of), the voltmeter reading stays unchanged (unlessthe capacitor is leaky) the capcitor is said to be charged, to the battery voltage: its conditiondoes not depend at all on the resistor, whose only purpose was to slow down the chargingprocess, so that we could follow it on the voltmeter.DISCHARGING A CAPACITORWe can show that the charged capacitor is storing electricity by discharging it. , if we put apiece of wire across its terminals, a fat spark passes just as the wire makes contact, and thevoltmeter reading falls to zero.If we now recharge the capacitor and then close the key at B, in fig 16, we allow thecapacitor to discharge through the resistor R. The potential difference across it now falls tozero as slowly as it rose during charging.CHARGING AND DISCHARGING PROCESSESWhen we connect a capacitor to a battery, electrons flow from the negative terminal of thebattery on to the plate A of the capacitor connectedto it (fig17) and , at the same rate,electrons flow from the other plate B of the capacitor towards the positive terminal of thebattery. Positive and negative charges thus appear on the plates, and oppose the flow ofelectrons which causes them. As the charges accumulate, the potential difference between theplates increases, and the charging current falls to zero when the potential difference becomesequal to the battery voltage V.Fig 17 24
  25. 25. When the battery is disconnected and the plates are joined together by a wire, electrons flowback from plate A to plate B until the positive charge on B is completely neutralized. Acurrent thus flows for a time in the wire, and at the end of the time the charges on the platesbecome zero.CAPACITANCE DEFINITION, AND UNITSExperiments with a ballistic galvanometer, which measures quantity of electricity, show that,when a capacitor is charged to a potential difference V, the charges stored on its plates Q ,are proportional to V. The ratio of the charge on either plate to the potential differencebetween the plates is called the capacitance, C, of the capacitor. QThus C= ………………………………………………….(1) V Q=CV…………………………………………….........(2) Qand V= ………………………………………………......(3) CWhere Q is coulombs, V is in volts (V) then capacitance C is in farads (F) .One farad (1F)is the capacitance of an extremely large capacitor. In practical circuits,i such as in radioreceivers, the capacitance of capacitors used are therefore expressed in microfarads. Onemicrofarad is one millionth part of a farad, that is , 1 F 10 -6F. It is also quite usual toExpress small capacitors, such as those used, in picofarads (pF). A picofarad is one millionthpart of a microfarad, that is, 1pF= 10-6 F = 10-12F.FACTORS DETERMINING CAPACITANCE, VARIABLE CAPACITORWe shall now find out by experiment what factors influence capacitance. To interpret ourobservations we shall require the formula for potential difference: Q V= CThis shows that, when a capacitor is given a fixed charge, the potential difference betweenits plates is inversely proportional to its capacitance.. The capacitance of a capacitor decreases when the seperation of its plates is increased, thecapacitance is inversely proportional to the separation.Dielectric. Let us now put a dielectric- a sheet of glass or ebonite between the plates. Whenwe increasethe dielectric by using several sheets together we see that(1) the capacitanceincreasewith the dielectric thickness.. In practical capacitors the dielectric completely fillsthe space between the plates. (2)The capacitance is also directly proportional to the areaof the plates. 25
  26. 26. Fig 18A capacitor in which the effective area of the plates can be adjusted is called a variablecapacitor. In the type shown in fig. 18, the plates are semicircular and one set can be swunginto or out of the other. The capacitance is proportional to the area of overlap of the plates.The plates are made of brass or aluminium, and the dielectric may be air, oil, or mica.We shall see shortly that a capacitor with paralel plates, having a vacuum ( or air, if weassume the permittivity of air is the same as a vacuum) between them, has a capacitance givenby 0A C= dWhere C = capacitance in farads (F), A= area of overlap of plates in m2 , d= distance betweenplates in m and 0 = 8.854×10-12 farad metre -1.If a material of permittivity completely fills the space between the plates, then thecapacitance becomes: A C= dCAPACITANCE VALUES, ISOLATED SPHERESuppose a sphere of radius r metre situated in air is given a charge of Q coulombs. We assumethat the charge on a sphere gives rise to potentials on and outside the sphere as if all thecharge were concentrated at the centre. The surface of the sphere has a potential given by: Q V= 4 0 r Q Q =4 0 r C= =4 0 r V V Capacitance, C= 4 0 rCONCENTRIC SPHERES 26
  27. 27. Faraday used two concentric spheres to investigate the dielectric constant of liquids. Supposea,b are the respective radii of the inner and outer spheres. Let +Q be the charge given to theiner sphere and let the outer sphere be earthed, with air between them( Fig 19) Fig 19The induced charge on the outer sphere is -Q. The potential Va of the iner sphere= potential Q Q Qdue to --Q. =+ since the potential due to the charge -Q is - everywhere 4 0 a 4 0b 4 0binside the larger sphere.But Vb = 0 , as the outer sphere is earthed. 1 Q Q Q b a potential difference, V, = Va – Vb = V= 4 a b 4 0 abQ 4 0 ab 4 0 ab = C=V b a b aPARALLEL PLATE CAPACITOR Fig 20Suppose two paralel plates of a capacitor each have a charge numerically equal to Q fig.20. Qthe surface density is then where A is the area of either plate, and the intensity between Vthe plates, E, is given, by Q Q A E= = C= = A V dNow the work done in taking a unit charge from one plate to the other Work=force×distance=E×d where d is the distance between plates. But the work done per unit Qcharge=V, the p.d. between the plates. V= d = d AProblem:10 27
  28. 28. The plates of a paralel-plate capacitor are 5 mm apart and 2 m2 in area. The plates are invacuum. A potential difference of 10,000 volts is applied across the capacitor. Compute (a)the capacitance (b) the charge on each plate, and (c) the electric intensity in the space betweenthem.SOLUTION:C= = = 3.54 x 10-9 C2 N-1 m-1. 1 C2 N-1 m-1 = 1 C2 J-1 = 1 C (J/C)-1 = 1 C V-1 = 1 F,C = 3.54 x 10-9 F = 0.00354 F.Q = CVab = (3.54 x 10-9 C V-1) (104 V) = 3.54 x 10-5 C.E= = = = 20 x 105 N C-1;E= = = 20 x 105 V m-1. o and its measurement:We can now see how the units of o may be stated in a more convinient manner and how itsmagnitude may be measured. 0A CdUnits. From C= , we have o = d A 28
  29. 29. faradxmete rThus the unit of o= 2 = farad metre-1 (meter )Now from previous , C= OA/d. Thus on charging, the charge stored, Q, is given by Q=CV= 0 VA/dARRANGEMENTS OF CAPACITORSIn radio circuts capacitors often appear in arrangements whose resultant capacitances must beknown. To derive expressions for these, we need the equation defining capacitance in its threepossible forms. Q QC= V= Q=CV V CIn paralel. Fig.21 shows three capacitors, having all their left hand plates connectedtogether, and all their right hand plates likewise. They are said to be connected in paralel. Ifa battery is now connected across them they all have the same potential difference V. Thecharges on the individual capacitors are respectively.Q 1 =C 1 V Fig.21Q 2 =C 2 VQ 3 =C 3 VThe total charge on the system of capacitors isQ=Q 1 +Q 2 +Q 3 =(C 1 +C 2 +C 3 )VAnd the system is therefore equivalent to a single capacitor,of capacitance 29
  30. 30. QC= =C 1 +C 2 +C 3 VThus when capacitors are connected in parallel,their resultant capacitance is the sum oftheir individual capacitances.It is greater than the greatest individual one.In series.. Fig 22Fig 22 shows three capacitors having the right-hand plate of one connected to the left-hand ofthe next,and so on –connected in series.When a battery is connected across the ends of thesystem,a charge Q is transferred from the plate H to the plate A,a charge,-Q being left onH.This charge induces a charge +Q on plate G ;similarly,charges appear on all the othercapacitor plates,as shown in the figure.(The induced and inducing charges are equal becausethe capacitor plates are very large and very close together,in effect,either may be said toenclose the other.)The potential differences across the individual capacitors are,therefore,given by Q Q QV AB = V DF = V GH = C1 C2 C3The sum of these is equal to the applied potential difference V because the work done intaking a unit charge from H to A is the sum of the work done in taking it from H to G,from Fto D,and from B to A.Therefore 1 1 1V=V AB +V DF +V GH = Q( + + ) C1 C 2 C 3The resultant capacitance of the system is the ratio of the charge stored to the appliedpotential difference,V. The resultant capacitance is therefore given by1 1 1 1 = + +C C1 C 2 C3Thus,to find the resultant capacitance of capacitors in series,we must add the reciprocalsof their individual capacitances.The resultant is less than the smallest individual.PROBLEM:11 30
  31. 31. Find the charges on the capacitors in figure shown belove and the potential differences acrossthem.SOLUTION:C´ = C2 + C3 = 3µF.C= = = 1.2µF.= Q1 + Q2 + Q3 = CV = 1.2 x 10-6 x 120 = 144 x 10-6 coulombV1 = = 72 volt,V2 = V - V1 = 120 – 72 = 48 volt,Q2 = C2V2 = 2 x 10-6 x 48 = 96 x10-6 coulomb,Q3 = C3V2 = 10-6 x 48 = 48 x10-6 coulomb.Energy of a Charged CapacitorIn general, the energy stored by a capacitor of capacitance C,carrying a charge Q,at a potential 1 Q Qdifference V,is W= (Q2/C) As we know C= V= Q=CV therefore 2 V C 1 = QV 2 1 = CV2 2If C is measured in farad, Q in coulomb and V in volt,then the expressions derived in (1) willgive the energy W in joules.PROBLEM:12A 12-V battery is connected to a 20 F capacitor.How much electric energy can be stored inthe capacitor? 31
  32. 32. 1 1SOLUTION: Energy W= CV2 = (20x10−6F)(12V)2=1.4x10−3J 2 2---------------------------------------PROBLEM:13) A circular parallel-plate capacitor with a spacing d=3 mm is charged toproduce an electric field strength of 3x 106 V/m. Find the potantial difference between theplates and the capacitance value.SOLUTIONThe potantial differece between the plates is V=E.d=(3x 106 V/m)(3x 10 3 m)=9x 103 VThe capacitance value is Q 1x10 6 CC= = 3 =1.11x 10 10 F V 9 x10 VPROBLEM:14)The plates of a paralel plate capacitor are separated by a distanced=1mm.What must be the plate area if the capacitance is to be 1F? 0 =8.85x 10 12 F/m Cd (1F )(1x10 3 m)A= = 12 =1.1x 108 m 2 0 8.85 x10 F / m-------------------------------------------------PROBLEM:15) A storage capacitor on a random access memory(RAM) chip has acapacitance of 55x 10 15 F. If the capacitor is charged to 5.3V, how many excess electrons areon its negative plate? e=1.6x 10 19 C q CV (55 x10 15 F )(5.3)n= == = 19 =1.8x 106 electrons e e 1.6 x10Concentric capacitorAn example, suppose b= 10cm and a=9cm 4 0 ab C= b a 4 8.85 10 ( 12 ) 0.1 0.09 = (0.1 0.09 ) = 100pF (approx)-------------------------------------------------------PROBLEM:16) Determine the equivalent capacitance of combination between points a and bas shown below. Take C1 =6.0 F , C 2 =4.0 F , C 3 =8.0 F . If the capacitors are charged by a12 V battery Determine the charged on each capacitor. 32
  33. 33. SOLUTION C 2 and C 3 connected parallel C 23 = C 2 + C 3 =4 F +8 F .=12. F C 23 is in series with C1 , So the equivalent capacitance C of the entire cmbination is given by 1 1 1 1 1 3 = + = + = Hence C eq = 4 F C eq C1 C 23 6 12 12 The total charge that flows from the battery is Q=CV=(4x 10 6 )(12)=4.8x 10 5 C Both C1 and C 23 carry the same charge of this Q Q 4.8 x10 5 The voltage across C1 then V1 = = =8 V C1 6 x10 6 The voltage across the combination C 23 is Q 4.8 x10 5 V23 = = =4 V C 23 12 x10 6 The actual capacitors C 2 and C 3 are in parallel,this represent the voltage across each of them V 2 = V 3 =4V The charges on C 2 and C 3Q2 = C 2 V 2 =(4x 10 6 )(4)=1.6x 10 5 CQ3 = C 3 V 3 =3.2x 10 5 C---------------------------PROBLEM:17) The fig. below shows a network of capacitors between the terminals A and B. a) Reduce this network to a single equivalent capacitor b) Find the total charge supplied by the 12V battery to the capacitor networkSOLUTION 33
  34. 34. a) First, we can reduce the two paralel capacitors between B and Y to asingle equivalent capacitor C1 : C1 =4 F +8 F =12 F Next, we can reduce the two paralel capacitors between Y and X to a single equivalent capacitor C 2 : C 2 =6 F +2 F =8 F C1 , C 2 and the last 24 F capacitors are connected in series and finally,the three capacitors can be reduced to a single equivalent capacitor C: 1 1 1 1 1 = + + = C=4 F C 12 8 24 4 b) Q=CV=(4 F )(12V)=48 C -------------------------- PROBLEM18) Find the equivalent capacitance of the combination in fig as shown below First, C 3 and C 4 are in parallel,their equivalent capacitance C34 = C 3 + C 4 =1 F +3 F =4 F C C 12x4 Then C34 in series with C 2 C 234 = 2 34 = =3 F C 2 C 34 12 4 Now C1 and the combination C 234 are in paralel connected to A and B. The equivalent capacitance of entire circuit is C= C1 + C 234 =4+3=7 F ----------------------------------------- . PROBLEM:19) Two capacitors, of capacitance 4 F and 2 F respectively, are joined in series with a battery of e.m.f. 100 volts. The connections are broken and the like terminals of the capacitors are then joined. Find the final charge on each capacitor. The combined capacitance, C, of the capacitors is given by 1 1 1 = + C C1 C 21 1 1 3 4 = + = or C= FC 4 2 4 3 34
  35. 35. charge on each capacitor = charge on equivalent capacitor 4 400=CV = 100 = C 3 3When like terminals are joined together, the p.d across each capacitor, which is different atfirst, becomes equalized. Suppose it reaches a p.d. V. Then, as the total charge remainsconstant, 400 400İnitial total charge = + = final total charge = 4V + 2V. 3 3 800 6V 3 400 V V, 9 400 1600 charge on larger capacitor= 4 = C 9 9 400 800And charge on smaller capacitor = 2 = C 9 9------------------------------------------------------------PROBLEM:20) A 100 F capacitor can tolerate a maximum potantial difference of 20 V,while a 1 F capacitor can tolerate 300 V. Which capacitor can store the most energy? Themost charge?SOLUTION: 1 1W 100 = CV2 = (100 F )(20V ) 2 =20x 103 J=20mJ 2 2 1 1W 1 = CV2 = (1 F )(300V ) 2 =45x 103 J=45mJ 2 2The smaller capacitor can store more energy. On the other hand, the larger capacitor storemore chargeQ 100 =CV== (100 F )(20V)=2000 C=2 CQ 1 =CV=(1 F )(300V)=300 C =0.30mC--------------------------------------------------CEPROBLEM:21) A wafer of titanium dioxide has an area of 1 cm 2 and 0.10mmthickness.Aluminum is evaporated on the paralel faces to form paralel plate capacitor.a) Calculate the capacitanceb)When the capacitor is charged with a 12 V battery,what is the magnitude of chargedelivered to each plate?c)What is the free surface charge density?Dielectric constant of titanium dioxide at room temperature k=173 ; 0 =8.85x 10 12 35
  36. 36. SOLUTION: a) k A (173 )(8.85 x10 12 F / m)(10 4 m 2 )C= 0 = 3 =1.53x 10 9 F d 0.10 x10 m b) The battery delivers the free charge Q:Q=CV=(1.53x 10 9 F)(12V)=18.4x 10 9 C . c) The surface density of free charge Q 18.4x 10 -9 C = = =1.84x 10 4 C/ m 2 A 10 4 m 2----------------------------------------------------- I.ENG.PROBLEM:22) Two insulated spherical conductors of radii 5.00 cm and 10.00cm are chargedto potentials of 600 volts and 300 volts respectively. Calculate the total energy of the system.Also calculate the energy after the spheres have been connected by a fine wire. Comment on 1the difference between the two results. (N) k= =9.0x 109 N. m 2 / C 2 ; 4 0 Capacitance C of a sphere of radius r= 4 o r QFor 5 cm radius, or 5 10-2m, C1 = =4 0 r V -2 5 x10 2 5 11 C1 = 4 o 5 10 = = x 10 F 9 x10 9 9 Using 4 0=1 / (9 10 9). 10 For 10 cm radius, C2= 10-11 F. 9 1 Since energy, W= CV2 2 1 5 1 15 total energy = 10-11 x (600)2 + 10-11 3002 2 9 2 9 = 15 10 -7 J.CURRENT AND RESISTANCE 36
  37. 37. Electric CurrentSo far we have been studying static electricity,( electric charges at rest- assumption ofelectrostatic equilibrium). Now consider situations in which charges are moving.We call a flow of charge an electric currentand an electric currentis defined as chargescrossing the given area in time t.Accordingly,its units are coulombs per second (C/s).This units is given the special nameampere(A) after André Marie AmpèreWhen a light bulb is connected between the terminals of a battery by theconductingwire.When such a circuit is formed, charge can flow through the wires of the circuit, fromone terminal of the battery to the other Fig 23 .Conventionally from the positive terminal tonegative terminal but electron flow from the negative terminal to positive terminal Fig 24 Fig 23 Fig 24The electric current QI= tWhere Q is the amount of charge that passes through the conductor at any location duringthe time interval t. Fig 25Connsider a conductor containing n charges per unit volume,each carrying charge q andmoving with drift speed d . If A is the conductor‘s cross-sectional area,then a length  hasvolume A  and therefore contains nA  charges(Fig 25. Since each carries charge q. Thetotal charge is Q = nA  q and t=  / d Q nAq I= = =nAq d t / dThe instantaneous currentWhere we consider the ratio of charge to small time intervalHere dq is the positive charge that passes in time dt through a surface of conductor 37
  38. 38. PROBLEM: 23) How many electrons are contained in 3.20x10-6 Coulombs of charge?SOLUTION: q=ne n=q/n n= =2.0x1013 elektronRESISTANCE AND OHM‘S LAWGeorg Simon Ohm(1787-1854) who established experimentally that the current in a metalwire is proportional to te potential difference V applied to its two endsI VAnd the ratio of potential difference V to current I is the Resisatance R of a wire VR= Ohm’s Law I, Ohm‘s Law often written in the equivalent form VV=IR I= ROhm found experimentally that in metal conductor (for ohmic materials), R is costantindependent of V. But R is not a constant for many subtances nor for devices which arecalled nonohmic materials.such as diods,vacuum tubes,transistors and so on. Thus ―Ohm‘sLaw‖ is not a fundamental law.24) The voltage in typical houshold wiring is 120 V. How much current does a 100W lightbulb draw? What is the bulb‘s resistance under these conditions? P 100W V 120VI= = =0.833 A From Ohm‘s law R= = =144 V 120V I 0.833A V2We have seen that P= I 2 R and P= R 2 2 V (120 V )R= = =144 P 100 W 38
  39. 39. ResistivityIt is found experimentally that the resıstance Rof a metal wire is directly proportional to itslength and inversly proportional to its cross-sectional area A. That is R= AWhere ,the constant of proportionality is called resistivityand depends on materials used.Typical values of ,whose units are .m are given for various materials in the middlecolumn of table 2Table 2The resistivity of a material depends somewhat on temperature.In general, the resıstance of ametals increases with temperature. T = 0 1 (T T0 )Where is the temperature coefficient of resistivity which is shown in the last column oftable 2. 1Reciprocal of the resistivity, called the conductivity (sigma),is = and has unit of 1( .m)PROBLEM: 25) Calculate the resistance at 0 0 C of a 2m length of copper wire with adiameter of 1mm. At 0 0 C is =1.56x 10 8 .mSOLUTION: 39
  40. 40.  (1.56 x10 8 )(2)R= = = A (0.5 x10 3 )PROBLEM:26) A copper wire 0.50 cm in diameter and 70cm long is used to connect a carbattery to the starter motor.What is the wire‘s resistance? The resistivity of copper at roommtemperature 1.68x 10 8 .mSOLUTION:  (1.68x10 8 )(0.70m) 4R= = =6x 10 A (0.25x10 2 m)Electric PowerElectrical enrgy is useful to us beccause it can be easily transformed into other forms ofenergy. P=IVThis relation gives us the power transformed by any device( such as motors,electricheaters,toaster,hair dryers….and so on),where I is the current passing through it and V is thepotential difference acroross it. The SI unit of electric power is wattThe rate ofenrgy tranformation in a resisatance R can be written,usingV=IR in two other waysP=IV =I(IR)=I 2 R V V2 =( )V= R R ELECTRİC CIRCUITSElectric circuits are basic parts of all electronic device from radio and TV sets to computersand automobiles.To have current in an electric circuits first we need a device such as batteryor an electric generator that transforms one type of energy( chemical,mechanical,light…so on)into electric enrgy. such a device is called source of electromotive force or of emf.The symbol εis usually used for emf .(do notconfuse it withE for electric field) A battery hasitself some resistance,which is called its internal resistance;it is usually desinated r.When no current is drown from the battery the terminal voltage(V ab = V a -V b )equals theemf. When a current I flows from the battery there is an internal drop in voltage equal to Ir.Fig 26. Thus the terminal voltage V ab (the actual voltage delivered) is V ab =ε - Ir. 40
  41. 41. Fgig 26Resistor in Series Fig 27When two or more resistors are connected end to end as shown in fig 27 ,they are said to beconnected in series. We let V represent the voltage across all 3 resistors.and V1 , V 2 and V 3 bethe potential difference across each of the resistors R1 , R 2 and R3 As shown in fig.From V=IR V1 =I R1 V 2 =I R 2 V 3 =I R3 VV= V1 + V 2 + V 3 =I R1 +I R 2 +I R3 V= I( R1 + R 2 + R3 ) R= = R1 + R 2 + R3 IReq = R1 + R 2 + R3Resistor in ParallelAnother simple way to connect resistors isa in parallel so that the total current from thesource splits into separate branches as shown in Fig.28.Let I 1 , I 2 and I 3 be the currentsthrough each of the resistors R1 , R 2 ,and R3 respectively. 41
  42. 42. Fig 28 V V VI = I1 + I 2 + I 3 I1 = I2 = I3 = R1 R2 R3V V V V = + +R eq R1 R2 R3When devide out the V from each term 1 1 1 1 = + +R eq R1 R2 R3Kirchhoff’s RulesSome circuits cannot be simplified using series and paralel combinations.When there is morethan one source of emf or when circuit element are connected in complicated ways. There aretwo Kirchhoff‘s Rules.Kirchhoff’s first or junction rulestates that‖at any junction point,the sum of all currentsentering the junction must equal the sum of all currents leaving the junction‖Fig 29For example,at the junction point ain fig. 29 I 1 and I 2 are entering at a point A whereas I 3 isleaving.Thus Kirchhoff‘s junction rule states that I 1 + I 2 = I 3 or I 1 + I 2 - I 3 =0Kirchhoff’ssecond or closed loop rule is based on on the conservation of energy.It statesthat ―the sum of the changes in potential around any closed path of a circuit must be 42
  43. 43. zero‖ In other words, round a loop‖ The algebraic sum of the voltage drop is equal to thealgebraic sum of the emf IR =∑εor ∑∆V=027) Two batteries in oposite direction and two external resistors are connected with a wire ofnegligible resistance as shown below.a) Determine the current Ib)Calculate the power dissipated or generated in each of the circuit elementsSOLUTION:As you see batteriesare connected to drive current in oposite direction aroun the circuit. Asε ε , the current I flows counterc lockwise which is indicated in the diagram. The battery 1 2of ε provides power to the circuit ,whereas 2 absorbspower. 1a) In the circuit. ε -ε 1 2 =I( r1 + r2 + R3 + R 4 ) 10V 6V I= =0.1681 A 0.6 0.2 15 8The first battery is converting chemical energy into electric energy at a rateP1 = ε 1 x I=10Vx0.168A=1.681 WBattery 2 converting electric energy into chemical energy at a rateP2 =ε 2 x I=6Vx0.1681A=1.008 WIn other words, the first battery is discharging,whereas battery 2 is being chargedThe power dissipated in the various resistive elements isPr1 = I 2 r1 = (0.1681 A) 2 x0.60 =0.017WPr 2 = I 2 r2 = (0.1681 A) 2 x0.20 =0.006WPR 3 = I 2 R3 = (0.1681 A) 2 x15 =0.424W 43
  44. 44. PR 4 = I 2 R 4 = (0.1681 A) 2 x8 =0.226WCEN+IENG+EEE+MTB MAGNETISM The word of magnetism came from the region of Magnesia which is called Manisanow in Anatolia. People in this region observed that certain stones (magnetite) attracted bitsof ordinary iron. They also found that the iron itself became magnetized by touching themagnetic mineral. Fig 30 shows a type of a magnet. One of the most striking features of magnet is that ifa bar magnet is cut in half, you do not obtain isolated north and south poles. Instead, two newmagnets are produced. If the cutting operation is repeated, more magnets are produced, eachwith a north and a south pole. We can continue subdividing each half into two, right down tothe subatomic scale, elementary particles (electron, proton, neutron) themselves act like tinymagnets. In the eleventh century A.D. the Chinese invented the magnetic compass. Thisconsisted of a magnet, floating on a buoyant support in a dish of water. The magnet has twoends which are called poles. The pole of a freely suspended magnet that points toward geo-graphic north is called the north pole of the magnet. The other pole points toward the southpole and is called the south pole Fig 30 It is a familiar fact that when two magnets are brought near one another, each exerts aforce on the other. The force can be either attractive or repulsive and can be felt even whenthe magnets dont touch. If the north pole of one magnet is brought near the north pole of asecond magnet, the force is repulsive. Similarly, if two south poles are brought close, theforce is repulsive. But when a north pole is brought near a south pole, the force is attractive.These results are shown in Fig. 31 44
  45. 45. Fig 31 Fig 32 Magnetic Fields The region around a magnet where a magnetic force experienced, is called amagnetic field.As in gravitation or electric field surrounding an electriccharge. Theappearance of a magnetic field is quickly obtained by iron filings and accurately plotted witha small compass. The grain of iron line up along field lines. Of course, we do not get thedirection of the magnetic field, only the pattern of the field. Figure 32 shows how thin ironfilings reveal the magnetic field lines. The direction of the magnetic field at a given point can be defined as the direction thatthe north pole of a compass needle would point. Figure 33 (a) shows how one magnetic fieldline around a bar magnet is found using compass needles. The magnetic field determined inthis way for the field outside a bar magnet is shown in Fig 33 (b). Notice that the lines alwayspoint out from the north pole toward the south pole of a magnet and magnetic field linescontinue inside a magnet. The field round bar is ―non-uniform‖ strength and direction varyfrom place to place. The earth‘s field locally, however, is uniform Fig 33 (c) 45
  46. 46. Fig 33(a) Fig 33(b) --- the number of lines per unit area is proportional to the strength of the magneticfield Fig 33 c Fig 34 The Earths magnetic field is shown in Fig. 34.The pattern of field lines is as if therewere an (imaginary) bar magnet inside the Earth.. The Earths magnetic poles do not coincidewith the geographic poles, which are on the Earths axis of rotation. The north magnetic pole,for example, is in northern Canada, about 1300 km from the geographic north pole, or "truenorth." This must be taken into account when using a compass. The angular differencebetween magnetic north, as indicated by a compass, and true (geographical) north, is calledthe magnetic declination. In the U.S. it varies from 0° to perhaps 25°, depending on location. 46
  47. 47. Electric Currents Produce Magnetism During the eighteenth century, many scientists sought to find a connection betweenelectricity and magnetism. A stationary electric charge and a magnet wereshown not to haveany influence on each other. But in 1820, Oersted (1777-1851) found that when a compassneedle is placed near an electric wire, the needle deflects as soon as the wire is connected to abattery and a current flows. As we have seen, a compass needle can be deflected by amagnetic field.What Oersted found was that an electric current produces a magnetic field.He had found a connection between electricity and magnetism. Fig 35 Fig 36 The Magnetic field lines produced by a current in a straight wire are in the form ofcircles with the wire at their center. Fig. 35(a). There is a simple way to remember thedirection of the magnetic field lines in this case. It is called a . "right-handrule: you grasp thewire with your right hand so that your thumb points in the direction of the conventionalcurrent; then your other fingers will encircle the wire in the direction of the magnetic field.Fig. 35(b). The magnetic field lines due to a circular loop of current-carrying wire can bedetermined in a similar way using a compass. The result is shown in Fig. 36.Force on an Electric Current in a Magnetic Field;Definition of B In previous section we saw that an electric current exerts a force on a magnet, such as 47
  48. 48. a compass needle. By Newton‘s third law, we might expect the reverse to be true as well; weshould expect that a magnet exerts a force on a current-carrying wire. Experiments indeedconfirm this effect, and it was also first observed by Oersted. Fig 37 Fig 37( c)Let us look at the force exerted on a current-carrying wire in detail. Suppose straight wire isplaced between the poles of a horseshoe magnet as shown in Fig. 37. When a current flows inthe wire, a force is exerted on the wire. But this force is not toward one or the other pole ofthe magnet. Instead, the force is directed at right angles to the magnetic field direction,downward in Fig.37(a). If the current is reversed in direction, the force is in the oppositedirection, Fig. 37(b) It is found that the direction of the force is always perpendicular to thedirection of the current and also perpendicular to the direction of the magnetic field, B. Thisstatement does not completely describe the direction, however: the force could be either up ordown in Fig. 37(b) and still be perpendicular to both the current and to B. Experimentally, thedirection of the force is given by another right-hand rule, as illustrated in Fig.37( c). Orientyour right hand so that outstretched fingers can point in the direction of the current, and whenyou bend your fingers they point in the direction of the magnetic field lines. Then your thumbpoints in the direction of the force on the wire. This describes the direction of the force. What about its magnitude? It is foundexperimentally that the magnitude of the force is directly proportional to the current I inthe wire, and to the length ℓof wire in the magnetic field (assumed uniform). Furthermore, ifthe magnetic field is made stronger, the force is proportionally greater. The force also dependson the angle θbetween the current direction and the magnetic field (Fig.38). When the currentis perpendicular to the field lines the force is strongest. When the wire is parallel to the 48
  49. 49. Fig 38magnetic field lines, there in no force at all. At other angles, the force is proportional to sinθ(Fig.38). Thus for a current I in a wire, with length ℓin a uniform magnetic field B, we have F IℓBsin Up to now we have not defined the magnetic field strength precisely. In fact, themagnetic field B can be conveniently defined in terms of the above proportion so that theproportionality constant is precisely 1.Thus we have F=IℓBsinIf the direction of the current is perpendicular to the field (θ = 90°), then the force is F=IℓB I B If the current is parallel to the field (θ = 0°), the force is zero. The magnitude of B F maxcan then be defined as B = where Fmax is the magnitude of the force on a straight Ilength ℓ of wire carrying a current I when the wire is perpendicular to B. The relation between the force F on a wire carrying current I, and the magnetic field Bthat causes the force, is vector equation. F=IℓBhere, ℓ is a vector whose magnitude is the length of the wire and its direction is along the wire 49
  50. 50. (assumed straight) in the direction of the conventional current. The above discussion applies if the magnetic field is uniform and the wire is straight.If B is not uniform, or if the wire does not everywhere make the same angle with B. thenabove equation can be written dF=IdℓB . , |where dF is the infinitesimal force acting on a differential length dℓof the wire. The total forceon the wire is then found by integrating.F= dF = IdxB Above equations can serve as a practical definition of B. An equivalent way to defineB. in terms of the force on a moving electric charge, is discussed in the next Section. The SI unit for magnetic field B is the tesla (T). From above equations,it is clear that1T = 1 N/A. m. An older name for the tesla is the "weber per meter square" (1Wb/m2 = 1 T).Another unit commonly used to specify magnetic field is a cgs unit, the gauss (G):1G=10-4T.Torque on a Current Loop; Magnetic Dipole Moment When an electric current flows in a closed loop of wire placed in a magnetic field, asshown in Fig. 39. the magnetic force on the current can produce a torque. This is the basicprinciple behind a number of important practical devices, including voltmeters, ammeters, andmotors. The interaction between a current and a magnetic field is important in other areasaswell, including atomic physics. Fig 39 50
  51. 51. When current flows through the loop in Fig. 39(a), whose face we assume is parallel toB and is rectangular, the magnetic field exerts a force on both vertical actions of wire asshown, F1 and F2 (see also lop view. Fig. 39(b) Notice that, by the right-hand rule, thedirection of the force F1 on the upward current on the left is in the oppositedirection from the equal magnitude (F1=F2=F) force F2 on the descending, current on theright. These forces give rise to a net torque that tends to rotate the coil about its verticalaxis. Let us calculate the magnitude of this torque. From Eq. F=IℓB, the force hasmagnitude F = IaB. where a is the length of the vertical arm of the coil. The lever arm foreach force is b/2, where b is the width of the coil and the "axis" is at the midpoint. The totaltorque is the sum of the torques due to each of the forces, so τ = IaB. b + IaB. b =IabB= IAB 2 2where A = ab is the area of the coil. If the coil consists of Nloops of wire, the torque on ,Nwires becomesτ =NIAB If the coil makes an angle θ with the magnetic field, as shown in Fig. 39( c). the forcesare unchanged, but each lever arm is reduced from ½ b to ½ sin θ. Note that the angle θ ischosen to be the angle between B and the perpendicular to the face of the coil. Fig. 39( c). So the torque becomesτ =NIABsin The formula, derived here for a rectangular coil is valid for any shape of flat coil. The quantity NIA is called the magnetic dipole moment of the coil and is considereda vector: =NIAwhere the direction of A (and therefore of μ.) is perpendicular to the plane coil I the greenarrow in Fig. 39( c) consistent with the right-hand rule (cup . right hand so your fingers wraparound the loop in the direction ot current flow then your thumb points in the direction of 51
  52. 52. (μand A). With this definition of μ, we can rewrite τ =NIABin vector form:or τ= Bwhich gives the correct magnitude and direction for τ . Electric Charge and the Magnetic Field From what we already know we can predict the force on a single moving electriccharge.If N such particles of charge q pass by a given point in time t, they constitute a currentI = Nq/t. We let t be the time for a charge q to travel a distance ℓ in a magnetic field B; then ℓ= v t where v is the velocity of the particle. Thus, the force on these N particles is. byEquation F=IℓB= (Nq/t)(vt)B=Nq v B The force on one of the N particles is then F=q v B This gives the magnitude of the force on a particle of charge q moving with velocity vat a point where the magnetic field has magnitude B. The angle between v and B is θ. Theforce is greatest when the particle moves perpendicular to B (θ = 900): F=IℓB I B The force is zero if the particle moves parallel to the field lines (θ = 00). Thedirection of the force is perpendicular to the magnetic field B and to the velocity v of theparticle. In other words, the magnetic force is always at right angels both to the velocityv of the charge and to the magnetic field B.It is given again by a right-hand rule. Orientyour right hand so that your outstretched fingers point along the direction of motion of theparticle ( v ) and when you bend your fingers they point along the direction of B( Fig 39).Thenyour thumb will point in the direction of the force. 52
  53. 53. Fig 40Force on an Electric Charge Moving in a Magnetic Field We have seen that a current-carrying wire experiences a force when placed in amagnetic field. Since a current in a wire consists of moving electric charges, we might expectthat freely moving charged particles (not in a wire) would also experience a force whenpassing through a magnetic field. The magnetic force changes only direction of theparticle’s velocity not its magnitude.Indeed, this is the case under the influence of themagnetic field a particle of charge q with velocity v follows circular path or circular motion.The force exerted on the particle has a magnitude F=q v B The same force is keeping the charge on circular path in other words this is centripetal v2force .Newton‘s second law F=m a we have a centripetal acceleration a = r mv 2 Of a particle= =q v B r mv Radius of circular path r= qB e mv 2 of the electron is very important . For electron F=e v B= m r 53
  54. 54. e = m B.r The time T required for a particle of charge q moving with constant speed v to make 2 .rone circular revolution in a uniform magnetic field B( v ) is T= where 2 r is thecircumference of its circular path. CEN mv 2 .r 2 .m Radius of circular pathr= so T= = qB qB 1 qBSince T is the period of rotation, the frequency of rotation is f= = This is T 2 .moften called the cyclotron frequencyof a particlein the field because this is the rotationfrequency of particles in a cyclotronMTB-EEE Sources of Magnetic Field We will now see how magnetic field strengths are determined for some simplesituations, and discuss some general relations between magnetic fields and their sources. Webegin with the simplest case, the magnetic field created by a long straight wire carrying asteady electric current. We then look at how such a field, created by one wire, exerts a forceon a second current-carrying wire. Interestingly enough, this interaction is used for the precisedefinitions of both the units of electric current and electric charge, the ampere and thecoulomb. Then we develop an elegant general approach to finding the connection betweencurrent and magnetic field known as Amperes law, one of the fundamental equations ofphysics. We also examine a second technique for determining the magnetic field due to acurrent, known as the Biot-Savart law; which does allow us to solve problems more readily inmany cases than Amperes law. Finally, we try to understand about iron and other magnetic materials and how they 54
  55. 55. produce magnetic fields.Magnetic Field Due to a Straight You might expect that the field strength at a given point would be greater if the currentflowing in the wire were greater; and that the field would be less at points farther from thewire. This is indeed the case. Careful experiments show that the magnetic field B at a pointnear a long straight wire is directly proportional to the current I in the wire and inverselyproportional to the distance r from the wire: I B r This relation is valid as long as r, the perpendicular distance to the wire, is much lessthan the distance to the ends of the wire (i.e., the wire is long). The proportionality constant is Iwritten as μ0/2π; thus.B= 0. [outside a long straight wire] 2 .rThe value of the constant μ0, which is called the permeability of free space, isμ0 = 4π xl0-7 T.m/A Force Between Two Parallel Wires Consider two long parallel wires separated by a distance d as in Fig 41 They carrycurrents I 1 and I 2 respectively. Each current produces a magnetic field so that each mustexert a force on the other, as Ampère first pointed out. The magnitude of the the magneticfield B1 produced by I 1 at the location of the second wire. 0.I 1 B1 = 2 .d 55
  56. 56. Fig 41 Operational Definitions of the Ampère The ampere, the unit of current, is now defined in terms of the magnetic field B itproduces using the defined value of μ0. In particular, we use the force between two parallel current-carrying wires, to definethe ampère precisely. If I1=I2= lA exactly, and the two wires are exactly 1 m apart, thenF II (4 x10 7 T .m / A)(1A)(1A) = 0 1 2= =2x 10 7 N / m 2 d 2 (1m) Thus, one ampère is defined as that current flowing in each of two long parallelconductors 1 m apart, which results in a force of exactly 2 X 10-7 N/m of length of eachconductor. The amount of current in a wire can be varied accurately and continuously (by puttinga variable resistor in a circuit).Thus the force between two current-carrying conductors is fareasier to measure precisely. The magnetic field in terms of the force per unit length on acurrent-carrying wire, via equation F=IℓB. Ampères Law I 0. Equation B= gives the relation between the current in a long straight wire and 2 .rthe magnetic field it produces. This equation is valid only for a long straight wire. The Frenchscientist Andre Marie Ampère(1775-1836)proposed general relation between a current in awire of any shape and the magnetic field around it. Heconsidered an arbitrary closed patharound a current by using compasses in sequence as shown in Fig. 42, and imagined this pathas being made up of short segments(needle of compasses) each of length  . First, we take 56
  57. 57. the product of the length of each segment times the component of B parallel to that segment(call this component Bll). If we now sum all these terms, according to Ampère, the result willbe equal to μ0 times the net current Iencl that passes through the surface enclosed by the path: Fig 42 B11  = I 0 encl The lengths  are chosen so that Bll is essentially constant along each length.Thesum must be made over a closed path; and Iencl, is the net current passing through thesurface bounded by this closed path. In the limit  → 0, this relation becomes B d = I 0 encl Ampère Lawwhere d is an infinitesimal length vector and the vector dot product assures that the parallelcomponent of B is taken. Above equation is known as Amperes law. The integral in aboveequation is taken around a closed path. To understand Amperes law better, let us apply it to the simple case of a long straightwire carrying a current I which weve already examined, and which served as an inspirationfor Ampere himself. Suppose we want to find the magnitude of B at some point A which is adistance r from the wire (Fig. 43). Fig. 43 We know the magnetic field lines are circles with the wire at their center. So to applyequation B d = I 0 encl . We choose as our path of integration a circle of radius r. The choiceof path is ours, so we choose one that will be convenient: at any point on this circular path, 57
  58. 58. Bwill be tangent to the circle. Furthermore, since all points on the path are the same distancefrom the wire, by symmetry we expect B to have the same magnitude at each point. Thus forany short segment of the circle (Fig. 43), B will beparallel to that segment, and hence 0 I = B d =B d =B(2 r) Ampère Lawwhere d = 2πr, the circumference of the circle, and Iencl=I.. We solve for B and obtain I 0..B= 2 .r This is exactly the same equation for the field near a long straight wire as discussedearlier. Its importance is that it relates the magnetic field to the current in a direct andmathematically elegant way. Amperes law is thus considered one of the basic laws ofelectricity and magnetism. It is valid for any situation where the currents and fields are steadyand. not changing in time, and no magnetic materials are present. CEN Magnetic Field of a Solenoid and a Toroid A long coil of wire consisting of many loops is called a solenoid. Each loop producesa magnetic field as was shown in Fig. 44. A toroid, whichis essentially a long solenoid bentinto the shape of a circle (see Fig. 45). In Fig. 44a, we see the field due to a solenoid when thecoils are far apart. Toward the center of the solenoid, the fields add up to give a field that canbe fairly large and fairly uniform. For a long solenoid, with closely packed coils, the field isnearly uniform and parallel to the solenoid axes within the entire cross section, as shown inFig. 44b. The field outside the solenoid is very small compared to the field inside, except nearthe ends. Note that the same number of magnetic field lines that are concentrated inside thesolenoid, spread out into the vast open space outside. Fig44a Fig44b Fig45 We now use Amperes law to determine the magnetic field inside a very long (ideally, 58
  59. 59. infinitely long) closely packed solenoid. We choose the path a,b,c,d shown in Fig.46.Far fromeither end, for applying Amperes law. We will consider this path as made up of foursegments, the sides of the rectangle: ab, bc, cd, da. Then the Amperes law, becomes b c d a B d = B d + B d + B d B d a b c d The field outside the solenoid is so small as to be negligible compared to the fieldinside. Thus the first term in this sum will be zero. Furthermore, B is perpendicular to thesegments bc and da inside the solenoid, and is nearly zero between and outside the coils, sothese terms too are zero. Therefore we have reduced the integral to the segment cd where B isthe nearly uniform field inside the solenoid, and is parallel to d , so d B d = B d =Bℓ cwhere is the length cd. Now we determine the current enclosed by this loop for [the right sideof Amperes law, equation B d = I 0 encl . If a current I flows in the wire of the solenoid,the total current enclosed by our path abcd is NI where N is the number of loops our path Fig 46encircles (five in Fig.45‖‖). Thus Amperes law gives us Bℓ = 0 NI If we let n = N/ℓbe the number of loops per unit length, then B= 0 nI This is the magnitude of the magnetic field within a solenoid. Note that B depends 59