Chapter 5 Magnetostatics The Magnetic Field 5.1 The .
Chapter 5 MagnetostaticsThe Magnetic Field5.1 The Lorentz Force Law 5.1.1 Magnetic FieldsOutside a magnetic the lines emerge from the north poleand enter the south pole; within the magnet they aredirected from the south pole to the north pole. The dotsrepresents the tip of an arrow coming toward you. Thecross represents the tail of an arrow moving away.By analogy with electrostatics, why don’t we studymagnetostatics first? Due to lack of magnetic monopole.If one try to isolate the poles by cutting the magnet, a curiousthing happens: One obtains two magnets. No matter howthinly the magnet is sliced, each fragment always have twopoles. Even down to the atomic level, no one has found anisolated magnetic pole, called a monopole. Thus magneticfield lines form closed loops.How a current-carry wireproduces a magnetic field?12The Magnetic Field of a Bar MagnetDefinition of the Magnetic FieldWhen iron filings are sprinkled around a bar magnet, theyform a characteristic pattern that shows how the influence ofthe magnet spreads to the surrounding space.When defining of the electric field, the electric field strengthcan be derived from the following relation: E F/q. Since anisolated pole is not available, the definition of the magneticfield is not as simple.Instead, we examine how an electric charge is affected by amagnetic field.F qvB sin θ qv BThe magnetic field, B, at a point along the tangent to a fieldline. The direction of B is that of the force on the north pole ofa bar magnet, or the direction in which a compass needlepoints. The strength of the field is proportional to the numberof lines passing through a unit area normal to the field (fluxdensity).3Since F is always perpendicular to v, a magnetic force doesno work on a particle and cannot be used to change itskinetic energy.The SI unit of magnetic field is the Tesla (T). 1 T 104 G4
The Lorentz Force LawForce on a Current-Carrying ConductorWhen a particle is subject to both electric and magneticfields in the same region, what is the total force on it?When a current flows in a magnetic field, the electrons as awhole acquire a slow drift speed, vd, and experience amagnetic force, which is then transmitted to the wire.F q (E v B )F qvB sin θ (nA e)vd B (nAevd ) B I BThis is called the Lorentz force law. This axiom is found inexperiments.The magnetic force do no work.F I BF I B sin θdWmag Fmag dl q ( v B) vdt 0Really? But, how do you explain a magnetic crane lifts acontainer?n: is the number of the conductor per unit volume.: is defined to be in the direction in which the current is flowing.56Example:The Magnetic Force on a Semicircular LoopForce on a Current-Carrying ConductorA wire is bent into a semicircular loop of radius R. It carries acurrent I, and its plane is perpendicular to a uniform magneticfield B, as shown below. Find the force on the loop.The force on an infinitesimal current element isdF Id BSolution:dF Id BdFy IRB sin θ dθπFy IRB sin θ dθ0 2 IRB I (2 R) BThe force on a wire is the vector sum (integral) of the forceson all current elements.The x-components of the forces on such elements willcancel in pairs.The net force on any close current-carrying loop is zero.78
The Motion of Charged Particles in Magnetic FieldsCyclotron MotionHow does a charged particle move with an initial velocity vperpendicular to a uniform magnetic field B?What are the frequency and the period? Are they independentof the speed of the particle? Yes.Since v and B are perpendicular, the particle experiences aforce F qvB of constant magnitude directed perpendicular.Under the action of such a force, the particle will move in acircular path at constant speed. From Newton’s second law,F ma, we haveThe period of the orbit ismvmv 2qvB r qBr2πr 2πm m 2π vqB q B1qB q B fc T 2πm m 2πT f c B 2.8 MHz GaussThe radius of the orbit is directly proportional to the linearmomentum of the particle and inversely proportional to themagnetic field strength.The frequency is called the cyclotron frequency.9All particles with the same charge-to-mass ratio, q/m, havethe same period and cyclotron frequency.10Helical MotionExample: CyclotronA cyclotron is used to accelerate protons from rest. It has aradius of 60 cm and a magnetic field of 0.8 T. The potentialdifference across the dees is 75 kV. Find: (a) the frequencyof the alternating potential difference; (b) the maximumkinetic energy; (c) the number of revolutions made by theprotons.What happens if the charged particle’s velocity has not only aperpendicular component v but also a parallel component v//?Helical Motion.Solution:qB(a) f c 2πm 12 MHz(qrmax B) 2 1.76 10 12 J 11 MeV2m K 2qV 150 keV(b) K max (c)K max / K 11000 / 150 73.5 revs.How to determine the maximum kinetic energy?11The perpendicular component v gives rise to a force qv Bthat produces circular motion, but the parallel component v//is not affected. The result is the superposition of a uniformcircular motion normal to the lines and a constant motionalong the lines.12
Example: Cycloid MotionMagnetic Bottle/MirrorSuppose, for instance, that B points in the x-direction, and Ein the z-direction. A particle initially at rest is released from theorigin; what path will it follow?What happens if the magnetic field is not uniform? Energytransfer between the perpendicular and parallel components.Solution:1. Write down the equation of motion.In a nonuniform field, the particle experiences a force thatpoints toward the region of week field. As a result, thecomponent of the velocity along the B lines is not constant.2. Solve the coupled differential equations.3. Determine the constants using the initial conditions.13Velocity SelectorIf the particle is moving toward the region of stronger field,as some point it may be stopped and made to reverse thedirection of its travel.14Mass SpectrometerA mass spectrometer is a device that separates chargedparticles, usually ions, according to their charge-to-massratios. E Ej E v i viB B Bk Only those particles with speed v E/B pass through thecrossed fields undeflected. This provides a convenient wayof either measuring or selecting the velocities of chargedparticles.15v EB1r mvmE qB2 qB1 B2m B1 B2 rqE16
Example: Mass SpectrometerCurrent and Surface CurrentIn a mass spectrometer shown below, two isotopes of anelements with mass m1 and m2 are accelerated from rest bya potential difference V. They then enter a uniform B normalto the magnetic field lines. What is the ratio of the radii oftheir paths?Solution:v r The current in a wire is the charge per unit time passing agiven point.Current is measured in coulombs-per-second, or amperes (A).1 A 1 C/sThe surface current density, K, is defined as follows:consider a ”ribbon” of infinitesimal width d , running parallelto the flow. Then,dI2qVmK mv2mV qBqB 2then r1 / r2 (m1 / m2 )dIK d 17) ρdtd)) σ d // σvdt18SIn particular, the total charge per unit time leaving a volumeV isI dQ J nˆ da ( J )dτ dtSdIJ da //I J nˆ daIn words, J is the current per unit area-perpendicular-to-flow.)d (σThe current crossing a surface S can be written asThe volume current density, J, is defined as follows:consider a ”tube” of infinitesimal cross section da , runningparallel to the flow. Then,dIJ da //d(Conservation of ChargeVolume Current Densityd ( ρ a dtda In words, K is the current per unit width-perpendicular-to-flow.Note1: How particle is accelerated by a potential difference?d(dVdQ dwhere ρ dτ .dt dt V ddρ V ( J )dτ dt V ρ dτ J dtd // ρvdt19continuity equation20
5.2 The Biot-Savart Law 5.2.1 Steady Currents5.2.2 The Magnetic Field of a Steady CurrentStationary charges produce electric fields that are constantin time. Steady currents produce magnetic fields that arealso constant in time.The Biot-Sarvart law:B(r ) µ0 I rˆµ I dl ′ rˆdl ′ 0 2 4π r4πr2Stationary charges constant electric fields; electrostatics.Steady currents constant magnetic fields; magnetostatics.The integration is along the currentpath, in the direction of the flow.µ0: the permeability of free space.Steady current means that a continuous flow that goes onforever without change and without charge piling upanywhere. They represent suitable approximations as longas the fluctuations are reasonably slow.Definition of magnetic field B: newtons per ampere-meter1 T 1 N/(A m)or tesla (T).The Biot-Sarvart law plays a role analogous to Coulomb’slaw in electrostatics. J 021Example 5.5 Find the magnetic field a distance s from a longstraight wire carrying a steady current I.22Con'tπµ I dl′ rˆ µ IB(r ) 0 2 04π4πr 2 π cos 2 θ scos θ dθs 2 cos 2 θ2πµ0 IµIIsin θ 2π 0 ( 2 10 7 Tesla) 4π s2π ss2What is the force between two parallel current-carrying wires?Sol :B(r ) µ0 I rˆµ I dl′ rˆdl ′ 0 2 4π r4πr2Then, determine the suitable coordinate : cylindrical coordinate ( s, φ , z ).In the diagram, (dl′ rˆ ) points out of page and have the magnitudedl ′ sin α dl ′ cos θl ′ s tan θ dl ′ s sec 2 θdθ and1 cos θ sr23dF Idl BµIµIIdF I 2 0 1 dl 0 1 2 dl2π d2π ddF µ0 I1 I 2 dl2π d(attractive force per unit length, why?)24
The Bio-Sarvart Lawfor the Surface and Volume Currentµ I rˆµ I dl ′ rˆThe Biot-Sarvart law: B(r ) 0 2 dl ′ 0 4π r4πr2Example 5.6 Find the magnetic field adistance z above the center of a circularloop of radius R, which carries a steadycurrent I.Sol : B(r ) µ0 I dl′ rˆ4π r 2Choose cylindrical coordinate (s, φ , z ).In the diagram, (dl′ rˆ ) sweeps around the z axis,thus only the z -component survives.z -component of (dl′ rˆ ) dl ′ cos θ R cos θ dφ11R 2 2 and sin θ 2 2 1/ 22r(R z )(R z )For surface current:B(r ) µ0 K rˆda ′4π r 2For volume current:B(r ) µ0 J rˆdτ ′4π r 2For a moving charge:B(r ) 25Wrong, why?µ0 J rˆµ0 qvδ (r r′) rˆµ0 qv rˆ′′ττdd 224π r4π r4π r 2A point charge does not constitute a steady current.26SolenoidThe Magnetic Field of SolenoidProblem 5.11 A solenoid of length Land radius a has N turns of wire andcarries a current I. Find the fieldstrength at a point along the axis.Solution:Sine the solenoid is a series of closelypacked loops, we may divided into currentloops of width dz, each of which containsndz turns, where n N/L is the number ofturns per unit length.The current within such a loop is (ndz)I.2728
Solenoid (II)Homework #9Con’tz a tan θ dz a sec θ dθnIdz nIa sec 2 θ dθ2µ0 a 2Problems: 9, 10, 11, 39, 49nIa sec dθ2(a 2 a tan 2 θ )3/ 21 µ0 nI cos θ dθ2θ2 1B µ nI cos θ dθθ1 2 01 µ0 nI (sin θ 2 sin θ1 )2dBaxis 2B µ 0 nI (infinite long solenoid)29The Differential Form of B5.3 The Divergence and Curl of B5.3.1 Straight-Line CurrentsSuppose we have a bundle of straight wires. Only wiresthat pass through the loop contribute µ0I.The line integration then beThe magnetic field of an infinite straight wire:B(r ) 30µ0 Iφˆ2πs B(r) dl µ I0 encI enc J daThe total current enclosedby the integration loop.The integral of B around a circular path of radius s, centeredat the wire, is:µI B(r) dl 2π0 s φˆ φˆ sdφ µ0 IIn fact for any loop that encloses the wire would give thesame answer.Really? B dl ( B) da µ J da0 B µ0 JDoes this differential equation apply to any shape of thecurrent loop?Yes, to be prove next.3132
5.3.2 The Divergence and Curl of BThe Divergence of BThe divergence of B:The Biot-Sarvart Law for thegeneral case of a volumecurrent: B(r ) ( (B(r ) µ0 J (r′) rˆdτ ′4π r 2µ0 J (r′) rˆµJ (r′) rˆ)dτ ′dτ ′) 0 (2 4π4πrr2rˆrˆJ (r′) rˆ) 2 ( J ) J ( 2 )2rrr ( A B) B ( A) A ( B)0rˆ1 ( ) (Prob. 1.13)2rr0J (r′) rˆrˆrˆ) 2 ( J ) J ( 2 ) (2rrr B 0 The divergence of a magnetic field is zero.The integration is over the primed coordinates.The divergence and the curl are to be taken withrespective to the unprimed coordinates.3334The Curl of BA Special TechniqueThe curl of B:rˆ ( J ) rµJ (r′) rˆ B 0 ()dτ ′4π 0r20 ( A B) (B ) A ( A )B A( B) B( A)unprimed primedprimed unprimedLet’s prove that thisintegration is zero.dτ ′ 0rˆrˆ (J ′) 2 ,2rrwhere r (r r ′)( J )special technique ( fA) f A f ( A)0 to be seen nextJ (r′) rˆrˆrˆ () J ( 2 ) (J ) 2 )2rrrJ (r ′) rˆrˆ) J ( 2 ) J 4πδ 3 (r ) (See 1.5.3) (2rrµ04π J (r ′)δ 3 (r )dτ ′ µ0 J (r )4π B µ0 J The curl of B equals µ0 times J.2Using the above rule, the x component is:0, for steady currentx-x′x-x′x-x′′ (J) ( ′ J )r3r3r30, since J(r’@ ) 0rˆx x′x x′ (J ) r x2 dτ ′ ′ ( r 3 J)dτ ′ S ( r 3 J) da′ 0(J ′) B 35What happens if J(r’) 036
5.3.3 Applications of Ampere’s Law B µ0 J ( B) da Applications of Ampere’s LawAmpere's law in differential formLike Gauss’s law, ampere’s law is always true (for steadycurrents), but is not always useful. B dl µ J da µ I00 encOnly when the symmetry of the problem enables youto pull B outside the integral can you calculate themagnetic field from the Ampere’s law.amperian loop B dl µ IAmpere's law in integral form0 encamperian loopThese symmetries are:Just as the Biot-Savart law plays a role in magnetostaticsthat coulomb’s law assumed in electrostatics, soAmpere’s play the role of Gauss’s.1. Infinite straight lines2. Infinite planes (Ex. 5.8)Electrostatics:CoulombÆ Gauss,3. Infinite solenoids (Ex. 5.9)Magnetostatics:Biot-SavartÆ Ampere.4. Toroids (Ex. 5.10)3738Infinite Straight WireInfinite PlanesExample An infinite straight wire of radius R carries a current I.Find the magnetic field at a distance r from the center of thewire for (a) r R, and (b) r R. Assume that the current isuniformly distributed across the cross section of the wire.Solution:(a) B dB (b)(r R ) B d B 2πr µ0B Solution: B 2πr µ 0 Iµ 0I2πrµ 0Ir2πR 2Example 5.8 Find the magnetic fieldof an infinite uniform surfacecurrent K Kxˆ , flowing over the xyplane. B dB πr 2IπR 2 B 2l µ 0 Klµ 0K2for z 0 µ K / 2yˆB 0 µ 0K / 2yˆ for z 0(r R )3940
SolenoidToroidExample 5.9 An ideal infinite solenoid has n turns per unitlength and carries a current I. Find its magnetic field inside.Solution: B d B d bcabdacdB d B d B dExample 5.10 A toroidal coil (shaped like a doughnut) istightly wound with N turns and carries a current I. Weassume that it has a rectangular cross section, as shownbelow. Find the field strength within the toroid.Solution: B db B dB aBLab µ 0nLab I B d µ 0 NIµ 0NI2πrThe field is not uniform; it variesas 1/r. The toroidal fields areused in research on fusion power.B µ 0nI415.3.4 Comparison of Magnetostatics andElectrostaticsρ Gauss's law E ε0 E 0 5.4 Magnetic Vector Potential5.4.1 The Vector Potentialρρ E 0 E V and E 2V ε0ε0no name (Faraday's law) B 0 B A and B µ0 J ( A) 2 A µ0 JGauss's law for magnetic field B 0 B µ0 J Ampere's law (Ampere-Maxwell law)F q (E v B )42Is it possible for us to set A 0 equals zero? Yes.Coulomb gaugeProof: If A 0 0, let A A 0 λ B A 0 AIf A 0, then 2 λ A 0 similiar to Poisson' s equationρ1 2 V ρ / ε 0 V 4πε r dτ ′0 A012 λ A 0 λ dτ ′ r4πLorentz' s force law43It is always possible to make the vector potential divergenceless.44
Example 5.11 A spherical shell, of radius R, carrying auniform surface charge σ, is set spinning at angular velocityω. Find the vector potential it produce at point r.The Vector Potential and Scalar PotentialUsing the Coulomb gauge, we obtain : 2 A µ0 Jµ J (r′)A 0 dτ ′4πrFor line and surface current,µ IA 0 dl ′4π rA µ0 K (r′)da′4π rSol :First, let the observer is in the z axis and ω is tilted at an angle ψµ K (r′)Vector potential is A(r ) 0 da′4πrThe surface current density K (r′) σ v′What happens when the curl of B vanishes?Magnetostatic scalar potential. B 0 B U 2U 0 (similiar to Laplace's equation)45µ 0 Rω ( cosψ sin θ ′ sin φ ′) xˆ 2R sin θ ′dθ ′d φ ′4π r 2 R 2 2 rR cos θ ′µ Rω (cosψ sin θ ′ cos φ ′ sinψ cos θ ′) yˆ 2 0 R sin θ ′dθ ′d φ ′4πr 2 R 2 2 rR cos θ ′µ Rω (sin ψ sin θ ′ sin φ ′) zˆ 2 0 R sin θ ′dθ ′d φ ′4πr 2 R 2 2 rR cos θ ′46A (r ) A(r ) R 3σω sinψµ0 yˆcos θ ′sin θ ′dθ ′dφ ′ 24πr R 2 2rR cos θ ′π R 3σω sinψµ0 yˆ cos θ ′ (2π ) d cos θ ′24πr R 2 2rR cos θ ′0A(r ) µ0 R 3σω sinψyˆ ( R 2 r 2 Rr ) R r ( R 2 r 2 Rr )( R r )( )23R 2 r 2 µ0 Rσ 2 (ω r ) insideA(r ) 4 µ0 R σ (ω r ) outside 2r 3Reverting to the “natural” coordinate, we have µ0 R 3σω sinψ yˆudu 222 1 r R 2rRu1 47Surprisingly, the field inside the spherical shell is uniform.48
5.4.2 Summary; Magnetostatic BoundaryConditionsExample 5.12 Find the vector potentialof an infinite solenoid with n turns perunit length, radius R, and current I.Sol :We have derived five formulasinterrelating three fundamentalquantities: J, A and B.A cute method that does the job. B da Φ ( A) da A dlwhere Φ is the flux of Β through the loop in question. B dl µ I 0 enc A dl ΦComments:Using a circular " amperian loop" at a radius inside the solenoid.µ nI2 A dl A2πs B da µ0 nI (πs ) A 02 sφˆ for s R There is one “missing link” in the diagram.Using a circular " amperian loop" at a radius s outside the solenoid.µ0 nIR 2 ˆ2 d As d nIR A Al2πBaµ(π)φ for s R0 492sMagnetostatic Boundary Conditions: Normal These three variables, J, A, and B, are all vectors. It isrelatively difficult to deal with.50Magnetostatic Boundary Conditions: TangentialThe magnetic field is notcontinuous at a surface withsurface density K.The tangential component of B isdiscontinuous.What is the physical picture?Consider a thin rectangular loop. Thecurl of the Ampere’s law states thatConsider a wafer-thin pillbox. Gauss’s law states that B da 0SThe sides of the pillbox contribute nothing to the flux, in thelimit as the thickness ε goes to zero. B dP µ 0 I encThe ends gives nothing (as ε 0), and the sides give////( Babove Bbelow) µ0 K//// Babove Bbelow µ0 KIn short, B above B below µ 0 K nˆ , where nˆ points " upward." ( Babove Bbelow) A 0 Babove Bbelow51How about the vector potential A?52
5.4.3 Multipole Expansion of the Vector PotentialBoundary Conditions in Terms of Vector Potential11 22r r′(r r ′ 2rr ′ cos θ ′)1r′r′ (1 ( ) cos θ ′ ( ) 2 ((3 cos 2 θ ′ 1) / 2) )rrr ′1r ( ) P (cosθ ′)r 0 rLike the scalar potential in electrostatics, thevector potential is continuous any boundary:A above A below A 0 Aabove Abelow//// A B A dl B da Φ Aabove AbelowThe vector potential of a current loopµIµI 11dl ′ 0 n 1 (r ′) n Pn (cos θ ′)dl ′A 0 4π4π n 0 rr- r′µ I 111 0 dl ′ 2 r ′ cos θ ′dl ′ 3 r ′2 P2 (cos θ ′)dl ′ 4π rrrmonopole531 r′ cos θ ′dl′ r 3 r′ P (cosθ ′)dl′ 22Part I ThenA dip 54 ( v) da v dlS ( fA) f A f ( A)P ( cT ) da ( T c T ( c)) da c T da T da Tdl cT dl c TdlSµ0 I 1µI 1r ′ cosθ ′dl′ 0 2 (rˆ r′)dl′2 4π r4π r (rˆ r′)dl′ rˆ da′Recalling Stokes’ theoremLet v cTmagnetic monopole term is always zero.A dip quadrupoleA Special TechniqueMultipole Expansionµ I 11A 0 dl′ 2r4π rdipole SSSP(Eq. 1.108, to be shown later)PP Part IIS ′T ′ da′ T ′dl′, let T ′ rˆ r′P ( A B) A ( B) B ( A ) ( A )B (B ) Aµ0 1µ m rˆrˆ ( I da′) 024π r4π r 2 ′(rˆ r′) rˆ ( ′ r′) r′ ( ′ rˆ ) (rˆ ′)r′ (r′ ′)rˆ (rˆ ′)r′ rˆwhere m I da′ is the magnetic dipole moment.55 Srˆ da′ (rˆ r′)dl′ rˆ da′P＃56
The Magnetic Field of a DipoleHomework #10A dip µ0 mzˆ rˆ µ0 m sin θ ˆ φ4π r 24π r 2B dip A Problems: 15, 16, 24, 46, 58µ0 m(2 cos θrˆ sin θθˆ )4πr 35758
isolated magnetic pole, called a monopole. Thus magnetic field lines form closed loops. 2 The Magnetic Field Outside a magnetic the lines emerge from the north pole and enter the south pole; within the magnet they are directed from the south pole to the north pole. The d
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