Chapter 6. Static Magnetic Fields

Transcription

1 Chapter 6. Static Magnetic Fields

2 Magnetism

3 Magnetism & EM force Magnetism Discovered when pieces of magnetic loadestone were found to exhibit a mysterious attractive force. Found near the ancient Greek city called Magnesia A magnetic field Caused by a permanent magnet, by moving charges (current) Electromagnetic force : electric force + magnetic force Electric force : F e = qe (N) Magnetic force : F m = q u B (N), B : magnetic flux density [T] Total electromagnetic force on a charge q : F = q (E + u B) Lorentz s force Equation Note! Magnetic forces do no work! dw F dl q( u B) udt! mag mag

4 Magnetostatics Stationary charges constant electric fields Electrostatics Steady current constant magnetic fields Magnetostatics Magnetostatics Steady current J t Biot-Savart Law Coulomb s law B ' I dl a T=N/(A m) 4 C 7 permeability of free space 4 1 N/A B( xyz,, ) I( x, y, z)

5 Divergence of B: B B I dl a T=N/(A m) 4 C Idl JSdl Jd J a B d ' 4 a B J 4 B( xyz,, ) d ' a J a ( J) J a J( x, y, z) Divergence and curl are to be taken with respect to (x,y,z) coordinate. J is a function of (x,y,z ) B J S a a a d ds ds sinddr a

6 Curl of B: B J B J a d ' 4 a B 4 J d ' a J J a a ( J) a ( ) ( J a ) d ' 3 B J( ')4 ( ') d ' J( ) B J B B Called Ampere s law da dl J da B dl I enc a a ( J) ( J') ' ' ' J ( ' ) 3 3 J ' ' 3 J ' ' ' J d ' Jda' 3 3 J on the boundary (all current is safely inside) a ( J) d Integral version of Ampere s law

7 Two postulates of magnetostatics in free space The divergence and the curl of B in nonmagnetic media B B J. (in nonmagnetic media) 7 permeability of free space 4 1 H/m No magnetic flow sources (no isolated magnetic charges) V B d Bds Divergence of the curl of any vector field = B J for steady currents S

8 Comparison of Magnetostatics and Electrostatics Coulomb s law E E : Gauss s law Biot-Savart law B B J : Ampere s law Lorenz force law F qe ( B)

9 Ampere s circuital law Example 6-1. An infinite long, straight, solid, nonmagnetic conductor with a circular cross section of radius b carries a steady current I. Determine the magnetic flux density both inside and outside the conductor. a) Inside the conductor B a B, C dl a B rd, dl B rd rb 1 1 The current through the area enclosed by C r r I 1 B 1 a I I I r, r b b b b 1 1

10 Ampere s circuital law Example 5-1. b) Outside the conductor: B C a dl a B B, rd, dl B rd rb B a a I 1, r B r b

11 Ampere s circuital law Example 6-. Determine B inside a closely wound toroidal coil with an air core having N turns of coil and carrying a current I. The toroid has a mean radius b, and the radius of each turn is a. Cylindrical symmetry B has only a φ component and is constant along any circular path about the axis of the toroid. A cicular contour C with radius r. B d l r B NI, b-a r ba C NI BaB a, b-ar ba r B for r ba and r ab.

12 Ampere s circuital law Example 6.3. Infinitely long solenoid with air core Direct application of Ampere s law BL nli B ni A special case of toroid (b) NI N r b B I ni

13 6-3. Vector magnetic potential B : divergence-free B E E V (V: scalar potential) B BA T A : vector magnetic potential (Wb/m) not sufficient definition. Its divergence is needed. B J A J A A A A A J Aa A a A a A o x x y y z z Simplifying it, we choose A= (Coulomb gauge). A J Ampere s law : vector Poisson's equation

14 Vector potential: A? Suppose that A satisfying B A is not divergenceless ( A ) If we add A the gradient of any scalar, A A B A is also satisfied. The new divergence is AA If a function can be found that satisfies the Poisson's equation, A, It is always possible to make A divergenceless. A Amp ere law, B J, becomes vector Poisson's equation, A J ( three Poisson's equations, one for each Cartesian component)

15 Finding A from current density Mathematical analogy with the scalar Poisson s eqn. If goes to zero at infinite, x, y, z 1 V V x y z d,, 4 V ( xx) ( y y) ( zz) The solution of the vector Poisson s eqn. by assuming J goes to zero at infinite, Ampere's law in vector potential, For bulk current: A For line current: A For surface current: A x yz A J, gives x, y, z xyz,, d Wb/m 4 J V I x, y, z I 1 x y z dl dl 4 4,, ' ' K x, y, z,, da' 4

16 Magnetic flux Magnetic flux Φ through a given area S s S (Wb) Bds A ds Adl C

17 Biot-Savart Law For thin wire with cross sectional area S, d equal to Sdl Jd JSdl Idl 1 J I A d 4 V 4 Magnetic flux from the vector potential C dl (Wb/m) Pxyz (,, ) J( x, y, z) I dl I dl BA 4 4 C C The unprimed curl operation implies differentiations with respect to the space coordinates of the field point, and the integral operation is with respect to the primed source coordinate. S

18 Biot-Savart Law Magnetic flux from the vector potential (cont d) Using fg fgf G I 4 BA dl + C 1 1 dl 1 1/ ( x x) ( y y ) ( z z ) ax ay az x y z ax( xx) ay( y y) az( zz) a = ( xx) ( y y) ( zz) 3/ 3

19 Biot-Savart Law Biot-Savart Law I 1 I dl a B dl 4 4 C C (T) I dl a I dl B dbdb or db C

20 A current-carrying straight line Example 6-4. Find magnetic flux density B from A dl a z dz a) z r L I L dz I ln L z z z r L Aaz a 4 z r 4 I L r L A= az ln 4 1 Az BAar a r L r L A r z I L r L IL Ba ln a r 4 L r L r L r I When r L, B a (infinitely long wire carrying current I ) r

21 A current-carrying straight line Example 6-4. dl a z dz b) a ra z r z dl a z a r a z a rdz z r z I dl I rdz IL B db 4 a 4 a z r r L r L 3 L 3/

22 A square loop (example 6-5) Find B at the center of a planar square loop, with side w carrying a direct current I. The magnetic flux density at the center of the square loop is equal to four times that caused by a single side of length w Lr w/ I Baz 4 a w I w z Exercise 8(cm) 6 (cm) rectangular conducting loop and 5 (A) direct current B at the center of the loop?

23 A circular loop (example 6.6) Find B at a point on the axis dla bd, a z a b, z b dl bd zz rb rbzd zb d a r component is canceled by the contribution of the element located diametrically opposite to z a a a a a r dl I bd Ib B az a 3/ z T 3/ 4 z b z b

24 A circular loop (example 6.7) Uniform magnetic field Helmholtz coil MI Magnet

25 The magnetic dipole Example 6.7 I A= 4 C dl 1 cos sinsin a at dl is not the same as a at the P. In fact, a at P is -a dl a sina cos ' bd x 1 y I bsin Ib A= ax or a 4 / d / 1 1 b bcos cos sincos 9 sinsin sin d x x b bcos b bsinsin 1 b 1 sinsin 1 1 b 1 1/

26 The magnetic dipole Example 6.7 When b, b / can be neglected in comparison with b 1 sinsin 1 1 1/ 1 1 b p 1 sin sin 1 x 1 px when x 1. Ib / = sin 1 sin sin / A a A, = a Ib sin 4 b d B Ib A 3 a cos a sin 4

27 The magnetic dipole Similarity between electric dipole and magnetic dipole at distant points. q q d p E 3 a cos a sin, pqd 4 Ib 4 B 3 a cos a sin Simplest form of magnetic dipole Ib Aa ma a a 4 ma pa A sin A m zib zis zm Wb/m V V 4 4 m I

28 Magnetization All matters are composed of atoms, each with a positive charged nucleus and a number of orbiting electrons. In addition, both electrons and the nucleus of an atom rotate (spin) on their own axes with certain magnetic dipole moments. m m because M M and nucleus electron nucleus electron nucleus electron m e- m 1 q M L In the absence of an external magnetic field, the magnetic dipoles of the atoms of most materials (excepts permanent magnets) have random orientations, resulting in no net magnetic moment. The application of an external magnetic fields cause both an alignment of the magnetic moments of spinning electrons and an induced magnetic moment due to a change in orbital motion of electrons

29 Magnetization vector, M To obtain a formula for determining the change in the magnetic flux density caused by the presence of a magnetic material, we let m k be the magnetic dipole moment of an atom. The magnetic dipole moment dm of dv is dm =Mdv will produce a vector potential. The total A is the volume integral of da and the contribution of magnetization to B is A 1 1 lim A/m, # of atoms within n k k n M m 4 d d M a A 4 V d M a B A

30 Equivalent current densities The total A Ma Ma 1 A d d d V 4 4 V 4 M V 1 1 M Using the vector identity, M M M M A d d 4 V 4V Using vector identity, A V M d 4 V 4 F d Fds current S S M a n ds 1 a M J Ma J n m ms A/m A/m Magnetization current density

31 Magnetization current density Magnetization surface current density M : out of paper J Ma ms n A/m P b Pn Jms Jms Magnetization volume current density J m M P A/m P On the surface of the material, Jms Man. If M is uniform (space invariant) inside, the current of the neighboring atomic dipoles will cancelled averywhere. No net current in the interior ( Jms ), or, M Jm if M is space invariant.

32 Physical interpretation of Magnetic current For uniformly magnetized material, all the internal currents cancel. J m M However, at the edge there is a current. mm at & m Ia I Mt surface current: J I / t M ms J Ma ms n

33 Physical interpretation of Magnetic current For nonuniformly magnetized material, the internal currents no longer cancel. I Mt I M ( y dy) M ( y) dz I x z z x J M y m x z dydz M y z J m x M y z I Mt Ix M y( zdz) M y( z) dy M y Ix dzdy z M y J m x z M z y J m M A/m

34 Magnetization current density Example 6.8 cylindrical magnet bar Axial magnetization : M a M z L b magnet Constant magnetization no magnetic volume current M J m J a M a a ms z r M To find B at P,, z, We consider a differential length dz with a current a M dz Ib B a dba z b and use Eq. (6-38) or z to obain 3/ d z B B a M z z L z b zl b z Mbdz zz b 3/

35 Magnetization current density Example 6.8 cylindrical magnet bar M z z L Ba z z b zl b

36 6-7. B&H and Ampere s law Because the application of an external magnetic field causes both an alignment of the internal dipole moments and an induced magnetic moment in a magnetic material, the resultant magnetic flux density in the presence of a magnetic material will be different from its value in free space. B ext B ind B total = B ext + B ind B JJ m 1 BJM H(magnetic field intensity) H B B M J Ampere's law H J A/m M A/m H dl I enc (free current)

37 B&H and Permeability When the magnetic properties of the medium is linear and isotropic, B&H M H, : magnetic susceptibility m m 1 m r Wb/m B HM H H H B H, 1 r m Absolute permeability elative permeability Permeability The permeability of most materials is very close to that of free space, For ferromagnetic materials such as iron, nickel and cobalt, 1 r

38 Analogous elation Electrostatics and Magnetostatics Electrostatics Magnetostatics E B D H 1/ P - M J V A

39 6-9. Magnetic Materials Magnetic materials Diamagnetic, if r 1 m ~ -1-5 the orbital motion of the electrons Copper, germanium, silver, gold Paramagnetic, if r 1 m ~ 1-5 Magnetic dipole moments of the spinning electrons Aluminum, magnesium, titanium and tungsten Ferromagnetic, if r >>1 m >> 1 ( 1~ 1,) Magnetized domains (strong coupling forces between the magnetic dipole moments of the atoms Nickel, cobalt, iron (pure), mumetal

40 Magnetic Materials

41 Ferromagnetic material Hysteresis loops H B I V esidual flux density (permanent magnets) Coercive field intensity M

42 Magnetic recording

43 Ferromagnetic material Hysteresis loss per unit volume per cycle The area of the hysteresis loop The energy lost in the form of heat in overcoming the friction encountered during the domain-wall motion and domain rotation Soft material with tall and narrow hysteresis. Curie temperature Demagnetization temperature (1 ~ 77 C) Ferrites Ceramic like compounds with very low conductivity Low eddy current losses at HF Extensive uses HF and microwave application such as FM antennas, HF transformers and phase shifters Computer magnetic core and disk memory devices and MAM

44 6-1. Boundary conditions The normal component of B B B B 1n n 1 1n n T For linear and isotropic media, B H and B H H = H D D 1n D n s (C/m ) The tangential component H J Hdl Jds abcda H dl H wh w Jwh i 1 H H lim Jh J J 1t t sn h sn A/m n 1 C S surface current density normal to the contour a H H J s abcda E E1 t Et

45 6-11. Inductances and Inductors Mutual flux: B B 1 S 1 d s I I Wb In case C has N turns, the flux linkage due to is 1 1 N Wb L I Mutual inductance between two circuits: L B d s H; henry N 1 1 I S 1 I Self-inductance of loop C 1 : L B d s H N I S1 1 I1 Inductor: a conductor arranged in an appropriate shape (such as a conducting wire wound as a coil) to supply a certain amount of self-inductance

46 Procedure for determining self-inductance 1. choose an appropriate coordinate system for the given geometry. Assume a current I in the conducting wire 3. Find B from I by Ampere s circuital law if symmetry exists; if not, Biot-Savart law must be used. 4. Find the flux linkage with each turn, Φ from B by integration 5. Find the flux linkage Λ by multiplying Φ by the number of turns 6. Find L by taking the ratio L =Λ /I. (Example 6-14) Find the self-inductance of a toroid. Cylindrical coordinate system: For ar b, Ba B and dl a rd; NI By Ampere's law: Bdl C Jds S B rd rb NI B r b NI NIh dr NIh b Bds hdr ln S Sa r a a r a N Ih b N ln a Nh b L ln I a

47 Air Coaxial Cable (Example 6-16) Determine the inductance per unit length of the line. Because of the cylindrical symmetry, B has only a φ-component and the current I is assumed to be uniformly distributed over the cross section of the inner conductor. b a) Inside the inner conductor r a : B a B a 1 1 ri a b) Between the inner and outer conductors ar b : B a B a I r dr r a Consider an annular ring in the inner conductor between r and r dr. The current in a unit length of this ring is linked by the flux a b I a I bdr I I b d B 1 ln r dr B a dr rdr a r a r a r 4a a rdr rdr The flux linkage for this annular ring is d d a fraction of the currents a a a I 1 a b a The total flux per unit length is d a r rdr ln rdr a a a I I b ln 8 a b Inductance L ln H/m I I 8 a

48 Two wire transmission line Example 6-17 Calculate the internal and external inductances per unit length. ri a) Inside the wire conductor r a: B1 ab1 a a a I 1 a I The total flux per unit length of the wire is d a r rdr a a 8 Self inductance of the two wire is L i 8 4 To get the mutual inductance per unit length, B I I, B x d x y1 y The flux linkage per unit length d-a da I 1 1 I d a I d By1Bydx dx ln ln a a x d x a a d L e ln H/m I a Total inductance per unit length of the two-wire line L L L i e 1 d ln H/m 4 a

49 A solenoid with two windings Example 6-18 N 1 1 a I1 l1 N N N a I l1 L N N a I1 l1

50 A conducting rectangular loop Example Determine the mutual inductance between a conducting rectangular loop and a very long straight wire I B a I r B ds where ds a zdr 1 S o tan 6 3 z r d b r d b 3I db r d b dr 3I b 1 ln 1 b d b d r d The mutual inductance 1 3 b L1 b d b I d ln 1 H

51 6-1. Magnetic Energy Work needs to be expended in sending currents into conducting loops and it will be stored as magnetic energy A current generator is connected to the loop, which increases the current i 1 from zero to I 1. The work required to overcome emf is d di I1 1 1 emf : L W11 1idt 1 L1 idi 1 1 LI 1 1 1I1 dt dt Consider two closed loops C 1 and C carrying current i 1 and i respectively. We keep i 1 at I 1 and increase i from zero to I. Because of mutual coupling, some of the magnetic flux due to i will link with loop C 1, giving rise to an induced emf that must be overcome by a voltage in order to keep i 1 constant at its value I 1, the work involved is I W I dt L I di L I I W must be done in loop C in order to counteract the induced emf as i is increased from to I. I 1 W idt L i di L I Total amount of work done 1 1 Wt LI 1 1L1I1I LI

52 Magnetic Energy in generalized form Generalizing magnetic energy to a system of N loops carrying I 1, I, I 3,,,.I n, we obtain N N 1 1 W L I I I, L I m jk j k k k k jk j j k k 1 j 1 For a current I flowing in a single inductor with inductance L, the stored magnetic energy is Wm 1 LI

53 Magnetic energy & Fields Magnetic energy in terms of field quantities Electrostatic Energy & Fields 1 ue E Magnetic energy W 1 I j is the total flux through C from all sources including itself. m j j j j S n j C j j Ba ds j Adl j N 1 1 Wm I j Adlj C j j j j j j AJdI dl Ja dl J V j 1 Wm d J AJ V Desirable to express the magnetic energy in terms of field quantities B and H instead of current density J and vector potential A.

54 Magnetic energy & Fields Making use of the vector identity, A H H A A H A H H A A H 1 Wm d J AJ HB A H AJ V W 1 1 m n HB d V ds A Ha S 1 1 Magnetic energy in B & H W m 1 V HB d B 1 Wm wmd J wm or H V A, H, S Self-inductance from stored magnetic energy W L m I

55 Magnetic energy & Fields Example 6- By using stored magnetic energy, determine the inductance per unit length of an air coaxial transmission line. The magnetic energy per unit length stored in the inner conductor 1 I I W B rdr r dr J/m m a a 3 4a The magnetic energy per unit length stored in the region between the inner and outer 1 b I b1 I b W m B rdr dr ln J/m r a a 4 a 4 Therefore, L W m1w m I 8 b a ln H/m conductor is Exercise 6.14 A current I flows in the N turn toroidal coil in Fig. 6.3 Obtain an expression for the stored magnetic energy Determine its self-inductance and check your result with Eq

56 6-13. Hall Effect In the steady state, the net force on the charge carrier is zero: F q E ub E ub q h h For n-type semiconductors (electron carriers) ua u E E h y a x d V E dx u B d h u B x y z u B 1 Ex Nq J B h NqE Nqu B J B x y z y z y z : Hall coefficient This effect can be used for measuring the magnetic field and determining the sign of the predominant charge carriers ( n type or p type).

57 Magnetic forces F qub m F m I C dl B N

58 Magnetic forces Magnetic forces F qub m Let us consider an element of conductor dl with a cross-sectional area S. If there are N charge carriers (electrons) per unit volume moving with a velocity u in the direction of dl, df NeS dl ubnes udl BIdl B m Magnetic force on a complete (closed) circuit of contour C. F m I C dlb N I When we have two circuit carrying current I 1 and I, respectively, the force F 1 on circuit C. I dl 1 1 a 1 F1 I dl B1, B C C N 1

59 Magnetic Forces Ampere s law of force between two current-carrying circuits. F II dl C C1 1 dl a 1 N F F 1 1 (Example 6-1) Determine the force per unit length F I a B 1 z 1 where B, the magnetic flux density at wire, set up by the current I 1 1 Using cylindrical sysmetry, I1 II 1 B1 ax F 1 ay N/m d d The force between two wires carrying in the same direction is one of attraction

60 Magnetic Forces ail gun

61 Magnetic Torques Small circular loop of radius b and carrying a current I in uniform B. B B B F dl B B dl B dl B dl a, B Ba Fdl B a z r, no net force F F Magnetic moment: x dt a df bsin a Idl B sin bsin 1 x 1 1 df df df, dl dl dl bd 1 1 dt axib B sin d The total torque acting on the loop dt torque rdf, df IdlB x T dta Ib B sin d a I b B x m a I b n T a I b B mb x

62 Magnetic Torques Example 6- The force and torque on the loop B a z B B a B a z, B x x y y F dl B sum of (1)~(4) forces, all directed toward center is zero. no torgue is produced. F Ib a a B a B a Ib B F 1 1 x x x y y z 1 y 3 F Ib a a B a B a Ib B F y x x y y z x 4 Net force on the loop, F F F 1 b T F, F a Ib B a T F, F a Ib b B x 1 y x 1 y 4 4 y 1 The total torque on the rectangular loop x F Ib b B, however, they result in a net torque. T= T13 T4 Ib1b axby aybx Nm TmaxBx ayby

63 Magnetic Torques Direct current motors TmB

64 Electrostatics and magnetostatics