Chapter 5: Magnetostatics, Faraday s Law, Quasi-Static Fields 5.1 Introduction and Definitions We begin with the law of conservation of charge:

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1 Chapter 5: agnetostatics, Faraday s Law, Quasi-Static Fields 5. Introduction and Definitions We begin with the law of conservation of charge: v d d Q J J a v t t d J t conservation da (5.) of charge, J B arbitrary volume agnetostatics is applicable under the static condition. Hence, and (5.) gives [for magnetoststics] (5.) t J Assuming a magnetic force F B is eperienced by charge q moving at velocity v, we define the magnetic induction B by the relation: F B q v B, which is consistent with the definition in (5.).

2 5. Biot and Savart Law The Biot-Savart law states that the differential magnetic field db B at point P (see figure) due to a differential current element d in d loop p loop is given by d B I (5.4) 4 I Thus, the total field at P due to I in I P d linear superposition, an eperimental fact d loop d loop is: B () 4 F Integrating the I 4 force on I in loop due to I in loop, we obtain d B d( d) II ( ) d d 4 II I d ( d d ) d d ( d( ) (5.7)

3 5. Differential Equations of agnetostatics and Ampere slaw B Gauss Law of agnetism : d d Rewrite (): B I 4 cross section of wire I Change to, to, and let Id Jdad Jd, we obtain B ( ) J ( ) d J d ( ) 4 4 B a aa ( J) J ( ) [ J ( )] d 4 ( ) J (5.6) 4 d operates on B [Gauss law of magnetism] (5.7)

4 5. Differential Equations of agnetostatics and Ampere s Law (continued) Ampere'sLaw: Rewrite (5.6): J ( ) B ( ) 4 d B ( ) ( ) J 4 d J ( ) d ( ) J J ( ) d [ J ( ) J ( ) ] d J ( ) d J ( ) J ( ) 4 [ d ] ( ) ( ) a a a ( )d d 4 ( ) n B( ) J( ) arbitrary (5.) da Bn da Jn da loop B d I (through the loop) d Ampere'slaw law, amuchmoreelaborate more elaborate B d I (5.5) representation of the Biot-Savart law 4

5 5.4 Vector Potential J ( ) Vector Potential : Rewrite (5.6): B( ) d 4 B A, A (5.7) where the vector potential A is given by A 4 J ( ) d, (5.8) which shows that A may be freely transformed (without changing B) according to AA (gauge transformation) (5.9) We may eploit this freedom by choosing a so that t A (Coulomb gauge) (5.) See proof on previous page. (5.8) ( ) d A J, 4 Coulomb gauge requires everywhere and hence const. 5

6 B J Rewrite: B A A J ( A ) A J 5.4 Vector Potential (continued) Choose the Coulomb gauge ( A ) A J (5.) J ( ) A d 4 (5.) Note: (5.) is valid in unbounded (infinite) space, i.e. the volume of integration must include all currents. If there is a boundary surface, the currents on the boundary must be accounted for by application of boundary conditions (See eample in Sec. 5..) 6

7 5.4 Vector Potential (continued) A Comparison of Electrostatics and agnetostatics : Electrostatics agnetostatics Definition of E: Definition of B: FE qe FB qvb Coulomb's law: Biot-Savart law: ( )( ) E ( ) d B ( ) J ( ) d 4 4 E E B B J E daq Ed Bda Bd I Gauss's law Gauss's Law Ampere's law of electrostatics of magnetism 7 J

8 5.6 agnetic Field of Localized Current Distribution, agnetic oment agnetic (Dipole) oment : A ( BC) B( A C) C( A B) J ( ) d A 4 [Eq. (5), Ch. 4] [ J ( ) d ( ) d ] J 4 J [ ( )]d Proved on net page. J Proved on p.85 under the [ J( )] d. J is localized 8 within volume conditions: m If is far of integration from source. (5.55) 4. J where m J( ) d [magnetic (dipole) moment] (5.54) Note : In (554) (5.54), m is defined dwith respect tto a point of reference. Here, it coincides with the origin of the coordinates ( ). 8

9 5.6 agnetic Field of Localized Current Distribution, agnetic oment (continued) Problem : Prove the relation J ( ) d under the conditions: J and J is localized within volume of integration. Proof : Since J outside the volume of integration, we may etend the volume of integration to without changing the integral value. J ( ) d d dy dz(j e J yey J zez) Consider the -component first: e J ( ) d dy dz J d J dy dz d J J y ( J J dy dz z y z ) d J d J d The insertion of these terms will not Similarly, the y- and change the value of the integral because z-components also vanish. J y J z Thus, J ( ) d ( ) dy Jy & ( ) dz Jz y z 9

10 5.6 agnetic Field of Localized Current Distribution, agnetic oment (continued) Anti - symmetric unit tensor ( ): (used on p.85 and p.88), if two or more indices are equal ijk, if i, j, k is an even permutation ti of f,,, if i, j, k is an odd permutation of,, Eamples :,,, Levi-Civita symbol ijk ( AB) AB, ( A) ( AB) i ijk j k i ijk k jk jk j ijk ijk ijk i ijk AB j k Aj ijk B k ijk A j i Aj kijbk jik A j B( A) A( B) i B B k i k i A ()

11 5.6 agnetic Field of Localized Current Distribution, agnetic oment (continued) Eample of magnetic moment: : plane loop I m ( ) I J d d d ( area ) da I( area) (5.57) m m is normal (by rig ht hand rule) to the plane of the loop. Eample of magnetic moment: a number of charged particles in motion angular momentum J qivi ( i) L v i ( ) i d q i i i i i i i i i i i q m J v L (5.58) e L if q / e/ for all particles. i i L: total angular momentum (5.59)

12 5.6 agnetic Field of Localized Current Distribution, agnetic oment (continued) Dipole Field : (valid far from the source) Rewrite (5.55) : m A (5.55) 4 BA m 5 4 ( m is a constant.) [ 4 m m m m ] m m y m ( ) ( ) ( ) z 4 y z A( B) B( A) e m 4 5 y z nnm ( ) m magnetic dipole n (5.56) 56) 4 field A B B A A B

13 5.6 agnetic Field of Localized Current Distribution, agnetic oment (continued) As in the case of the electric dipole moment, here we characterize a localized current distribution by a constant quantity, the magnetic moment m, which turns an otherwise complicated field calculation (see, for eample, Sec. 55)i 5.5) into a simple one (with ithlimited validity.) Consider, for eample, a circular loop carrying current I. Using (5.57), 57) we have m I a ez (see figure). Hence, the dipo le field ldis z nn ( m) m B n er (5.56) 4 e m I a e r z r er( er ez) e I a z 4 r a ( ez er cos e sin ) I cos er sin e I a (5.4) 4 r When r a, the dipole field is a good approimation of fthe total t field [see Jackson (5.4).]

14 5.7 Forces and Torque on and Energy of a Localized Current Distribution in an Eternal agnetic Induction agnetic Force in Eternal Field : F J ( ) B ( ) d (5.) Epanding B :[see lecture notes, Ch. 4, Appendi A, Eq. (A.4)] B( ) B() ( ) B() This implies "After differention of B ( ), set in results to," i.e. ( ) B() ( ) y ( ) ( ) y z B B z B [ ] F J( ) d B() J( ) ( ) B() d (proved in Sec. 5.6) J ( ) ( ) B() d ( mb ) U,, See derivation on pp (5.69) where U mb potential energy. (5.7) 4

15 5.7 Forces and Torque (continued) agnetic Torque in Eternal Field: N f( ) d [ J( )] [ J( ) B( )] d (5.) B d B() ( ) B() B () J( ) J( ) J ( ) [ J ( ) B()] d [ B () ] J ( ) d B () J ( ) d = [ B() ] J ( ) d B() [ J ( )] d () [ ( )]d [Using the formula at the bottom of p.85, replacing with B().] ] B J ( ) d s J a J is localized. J on surface mb () (5.7) 5

16 5.7 Forces and Torque (continued) A Comparison between Electric and agnetic Potential Energy, Force, and Torque in Eternal Field : Potential energy Force Torque U pe (4.4) FU NpE U mb (5.7) F U N m B p + E Both p and m tend to orient along the positive field direction under the actionofof the torque (see figures on the right). This I results in a state of minimum potential m B energy. In this state, p reduces E, whereas I m enhances B. Questions: () How does a permanent magnet attract a piece of iron? () How does it attract another permanent magnet? 6

17 5.7 Forces and Torque (continued) Force in Self -Consistent Field; agnetic Pressure and Tension : A self-consistent field is the combined field generated by the source J under consideration and the eternal source (if present). Thus, using B J, we may epress the magnetic force density f (force / unit volume) in terms of B. -field f JB ( B) B a solenoid B lines ( ab ) ( a) b+b ( ) a+a( b)+ b( a) a uniform electron beam B ( B ) B [see p. )] () magnetic tension force density, magnetic pressure as if a curved B-field line tended force density B-field to become a straight line. B lines In regions where J, we have f [pressure and tension force densities cancel out]. uniform current 7

18 5.8 acroscopic Equations, Boundary Conditions on B and H acroscopic Equations : To be more general, we move the point of reference for m from to and write ( ) ( ) [See Sec. 4.] Sub. this relation into ( ) A J 4 d How about J free? [(5.)] we obtain vanishes ih only if J is localized li J within the volume of integration. ( ) d ( ) ( ) J m A, (4) 4 4 where m ( ) ( ) J ( ) d. J 8

19 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) To proceed, we consider the orbital motion of atomic/molecular electrons, which can collectively give rise to a permanent or induced magnetization (total magnetic moment / unit volume) given by ( ) N m (5.76) volume density of type i molecules i i i magnetic moment per type i molecule averaged over a small volume As will be shown in (5.79), a current density ( J ) is associated with. In addition, there is also a current density due to the flow of free charges, which we denote by J free (Jackson denotes it by J in Sec. 5.8). By the principle of linear superposition, we may write A ( ) A free ( ) A ( ), where A free and A are due to J free and J, respectively. J ( ) free Obviously, A free( ) d 4 9

20 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) For A, we have the epression for, but not yet for J. So we approimate A by the dipole term in (4). J ( ) d A ( ) m ( ) ( ) 4 4 d J, where we have set J( ) because J is formed of current loops of atomic dimensions ( volume of integration). Under this condition, m is independent of the poin t of reference because d m ( ) ( ) J ( ) (5.54) J ( ) d J ( ) d m(). To represent A by the dipole term, we must have the dimension of the dipole. So, we divide the source into infinistesimal volumes. In each small volume V, the dipole moment is V, which generates a small A at given by

21 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) ( ) ( ) V A ( ) 4 where we have replaced the notation with. This gives A ( ) 4 ( ) ( ) d ( ) d Volume of integration 4 includes all sources. ( ) ( ) d d 4 4 a aa ( ) s n da v A d s n A da ( on S) ( ) d What if on S? 4 Question: Does this relation still hold as? V

22 Griffith /4 6. The Field of a agnetized Object 6B 6.. Bound dcurrents Suppose we have a piece of magnetized material (i.e. is given). What field does this object produce? The vector potential of a single dipole m is A( r) m rˆr 4 r In the magnetized object, each volume element carries a dipole moment d, so the total vector potential is A( r) ( r ) rˆ 4 r d

23 Griffith /4 Vector potential and Bound Currents Can the equation be epressed in a more illuminating form, as in the electrical case? Yes! By eploiting the identity, rˆ r r ( ˆ yˆ zˆ ) y z ( ) ( y y ) ( z z ) ˆ( ) yˆ( y y) zˆ( zz) rˆ / (( ) ( y y) ( zz) ) r The vector potential is A( r) ( r) ( )d 4 r Using the product rule ( f A ) f A f ( A ) and integrating by part, we have A r) [ ( r)] d r ) ( r ) ( [ ] d 4 r d [ ( r)] [ ( r) ˆ] da 4 r 4 r n How? See net page.

24 Griffith /4 Gauss's law ( E) d E da v S ( ( vc)) d c( v) d v v Let E v c, ( ) d d v c a c v a S S Since c is a constant vector, so ( v) d vda v S 4

25 Griffith 4/4 Vector potential and Bound Currents A( r) [ ( r)] d [ ( r) nˆ ] da 4 r 4 r J With these definitions, A r) b (r) volume current K bound currents Jb K d b da 4 v r 4 S r ( b ( r) nˆ surface current The electrical analogy volume charge density b P surface charge density σ ˆ b Pn 5

26 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) Thus, A( ) A ( ) A ( ) free J ( ) ( ) free d 4 (5.78) For comparison, in Sec. 5., we have A J, (5.) whic J ( ) h has the solution: A ( ) d (5.) 4 In (5.) and (5.), J represents the current due to both free and bound (atomic) electrons, whereas in (5.78) contributions from free and bound electrons are separated into two terms. Comparing (5.78) and (5.), we find that the bound electrons contribute to A ( ) through a magnetization current density ( J ) given by J, (5.79) which is the macroscopic ehibition of the atomic currents. 6

27 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) Hence, by separating free and bound electrons, B ( ) J ( ) [(5.)] can be written B ( J ) (5.8) Defining a new quantity called the magnetic field H : Effects of the atomic H B, (5.8) currents are implicit in H. we obtain from (5.8) the macroscopic version of (5.): H J free (5.8) Question: Does H have a physical meaning? Diamagnetic, Paramagnetic, and FerromagneticSubstances : The counterpart of (5.8) in electrostatics is D E P [(4.4)]. In Sec. 4., it is shown that, for small displacement of the bound electrons, we have the linear relations: P e E (4.6) D E, with ( e) (4.7), (4.8) 7 free

28 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) However, the magnetic properties of materials are such that is not always proportional to B, depending on the type of the material. We summarize, without derivation, possible relations between B and H.. For diamagnetic and paramagnetic substances, is proportional to B and we epress the linear relation as B, paramagnetic B B, diamagnetic (5) Substituting into H B, we get the linear relation: B H, (5.84) where is called the magnetic permeability. Question : The plasma is diamagnetic. Why? B. For the ferromagnetic substance we have a nonlinear relation (see figure): H B F ( H ), (5.85) which ehibits the hysteresis phenomenon shown in the figure. 8

29 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) Boundary Conditions : n : unit normal pointing (i) BvBd s Bda from region into region ( B B ) n B B B (5.86) n (ii) H J free region pillbo of infinitesimal H d a J free d a region B thickness ( LHS) H d (see lower figure) n d ( H H) ( n n) L K free rectangular n[ n( H n loop of H)] L infinitesimal a( bc) b( ca) d height L ( RHS) J free nda K free nl K free : surface current of n( H H) K free free charges (unit: A/m) (5.87) Special case : K H H (6) free t t t: tangential to surface 9

30 5.8 acroscopic Equations, Boundary Conditions on B and H (continued) Application to a Solenoid : i i l n turns unit length permeable material H B / in in B out Approimate the manetic field by that of an infinite solenoid. So, = constant. H in B-field lines H dl I free Hinl nil Hin ni Bin Hin ni B B ni out in Question :" Bout ni" implies that filling the solenoid core with material (while keeping i at a constant value) can greatly enhance B. Why? out

31 5.9 ethods of Solving Boundary-Value Problems in agnetostatics We put the basic equations : B and H J free (5.9) in forms suitable for types of boundary - value problems. Type : Linear medium with const (in each region). (a) Equation for vector potential (with or without ) H A [ ( A) A] J free B H AH A J free A J use Coulomb gauge, A (7) free (b) Equat ion for scalar potential (only for J free ) B H and H H (5.9) (8) Typically, we use (7) or (8) to solve for A or in each uniform region and find the coefficients by applying conditions (5.86) and (5.87) on the boundary. An eample will be provided in Sec. 5..

32 J 5.9 ethods of Solving Boundary-Value Problems in agnetostatis (continued) Discussion: In a vacuum medium, we have A J I free (5.) m B In a uniform- medium, we have I A J. (7) free Hence, the effect of medium is to increase the ability of to produce B by a factor of / (see figure above). free In electrostatics, we have free E (vacuum medium) p + (.) and free (uniform dielectric medium) (4.9) Hence, an medium reduces the ability of free to produce E by a factor of / (see figure above).

33 5.9 ethods of Solving Boundary-Value Problems in agnetostatis (continued) Type : Hard ferromagnets (permanent magnet, given, J free = ) (a) Vector potential B H ( ) real current B J, where J [see (5.79)] BABA( A) A J ( ) A d (5.) J J A( ) 4, (b) Scalar potential is a mathematical H H tool, not real charge. B ( H ) H (5.95) where (effective magnetic charge density) (5.96) ( ) ( ) 4 4 d d ( ) ( ) d 4 4 aa+a d (5.98)

34 5.9 ethods of Solving Boundary-Value Problems in agnetostatis (continued) Effective magnetic surface charge density : Rewrite (5.96): (5.96) vd s da v d (see pillbo below) n surface area A ( ) na A n (5.99) a mathematical tool thickness Surface current density K due to magnetization : In Sec. 5.8, Here by the same algebra, HJ J free real current n real current K n( H H ) K n( ) n K free free H H K n 4

35 5. Uniformly agnetized Sphere Consider a permanent magnet with magnetization : ez, r a B r P, r a vanishes everywhere z a ecept on the surface. ( ) n( ) d 4 4 s da by (5.95) (.7) 4 = 4 r Y (, ) a r r d cos r a 4 cos r [ Y (, ) Y (, ) cos, r Inside: H in,, Y discontinuous! H rcos z, r a Y (, ) (, ) (5.4) a r a 4 cos Y(, ) Y(, )] Bin Hin ( H ) in Bin Outside: dipole field with dipole moment 4 a m. 5

36 5. agnetic Shielding, Spherical Shell of Permeable aterial in a Uniform Field Consider a spherical -shell in an eternal B. B l m im a r r P (cos ) l e l m im b r Q (cos ) Eq. (8) l e l Hrcos l Pl (cos ), r b (5.7) l r Hr cos l r P (cos ar b gives the ( (5.8) l l l) l ), l r eternal B. l lrpl(cos ), r a (5.9) l b boundary conditions b H (5.9) Ht Ht (6) (outside) + a a B H B B (5.86) (inside) r B H r b b The shell is assumed to be a linear medium. r a r a 6

37 5. agnetic Shielding, Spherical Shell of Permeable aterial in a Uniform Field (continued) Boundary conditions result in solutions for the coefficients: B if l l l l l ( )( )( ) b a H ( b H )( ) ( a b ) 9 9 H H ( )( ) ( ) a ( a ) b b in as, implying that B out B (5.) & 9 (5.) materials tend to "absorb" B-field lines and thereby provide the shielding effect. High-materials can have / as high as. 6 When, B reduces to B everywhere, i.e. a static megnetic field penetrates into the shell as if there were no shell present (even if the shell is made of good conductor, such as copper). + dipole field B in : uniform field ( / if ) 7

38 5.5 Faraday s Law of Induction The Biot-Savart (or Ampere's) law relates the magnetic field to electrical current. Faraday then discovered eperimantally that time-varying magnetic flu through an electrical circuit could induce an electric field around the circuit. This not only provided the first link between electric and magnetic fields, but also led to a new mechanism to generate the E-field, i.e. a time-varying i B-field. Referring to the figure, let loop C loop C be an electrical circuit (as in Faradays Faraday's B (, t ) original eperiment) or any closed path d n in space (a generalization of the original da observation with immense consequences). Faraday's law states S: an arbitrary surface c d B da, E s n (5.4) t bounded by loop C where E is the electric field at d in the frame in which h d is at rest, and B is the magnetic induction in the lab frame. 8

39 Rewrite (5.4): B n c 5.5 Faraday s Law of Induction (continued) loop C E d s da t d da n B (,) t (5.4) Assume loop C is at rest in the lab frame, then E E (electric field in the lab frame) and (5.4) becomes E d B da integral form of Faraday's law c s n (9) () t where both E and B are lab-frame quantities. (9) can be written (by Stokes's theorem: c E d s En da ) s Enda B s nda t Thus, E B differential form of Faraday's law) (5.4) t t 9

40 5.6 Energy in the agnetic Field To find the energy associated with amagneticfield field, weevaluateevaluate the work needed to establish the current J ( ), which produces the magnetic field. We break up J ( ) into a network of thin loops. In the build-up process, an E field will be induced by B/ t. The rate of work done by E within each loop is is the cross section of the loop (same as integration over the area Jackson's ). J & may vary along d. encircled by the loop dw loop n J Ed a thin s J E nda dt B / t loop J B s n da t Stokes's thm. S Work done within each loop to generate B : d ( d J) W loop sj n Bda sj Anda : cross section of loop A A d J J d J d J d AJd d () loop 4

41 5.6 Energy in the agnetic Field (continued) loop loop AJd As shown in W [()], the work done within each loop is an integral over the volume of the loop. Thus, an integration over all space gives the total work done to generate B : Assume the rate of build-up H obeys the static law H J. Otherwise, the static law breaks down and there will be radiation loss. W A J d A ( H ) d (5.44) H( A) d ( HA) d For this integral B ( ) s H A da to vanish, the volume of integ- HBd ( HB) d ration must be. Assume linear medium: B H or Bμ H Total work done to bring the field up from to the final value B : By conservation of energy, this is W ( ) d H B (5.48) the total magnetic field energy. w H B [field energy density] () Note: w HB ( H j) ( Bj) ( H j Bj) j j j 4

42 5.7 Energy and Self- and utual Inductances Assume linear relation between J and A for nonpermeable W AJd ( AJ) d (μ ) medium (5.44) d A ( ) (5.) (5.49) W A J ( ) ( d d JJ ) 8 ( ) 4 J d J ( ) J ( ) 8 for N current- (5.5) carrying circuits N N i j N N N d i d j LI i i ijiii j i j ij i i j i, (5.5) where self-inductance for a thin wire L J ( ) ( ) i i Ji didi 4 I d (5.54) Ci i d Ci i 4 C i Ci i i i i i J ( i) J ( j) d idj ij d (5.55) i d 4II C i C j j 4 Ci C i j i j j i j mutual inductance ( ij = ji ) for thin wires 4

43 5.7 Energy and Self- and utual Inductances (continued) ( ) A( ) d (5.) 4 J Vector potential at circuit i due to current in circuit j: A J ( ) j ij( i) 4 C j d j i () j From () and (5.55), we obtain II A ( ) J( ) d i j C i Assume J flows along wire d of negligible cross section da J ( ) d i i Jdad Id i B ij ij ij i i i magnetic flu from circuit j passing through circuit i ( ) ij I A (5.56) Ci ij d si ij ij j I A nda j I F j d d ij Fij ij I ij : induced voltage in circuit i due dt dt j to current variation in circuit j. The sign implies that t the induced d ij tends to drive a current tin circuit i to inhibit the flu change caused by circuit j (Lenz s law). 4

44 Homework of Chap. 5 Problems:,, 6,,, 8, 9,,, 44