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Chapter 10 Magnetic Field of a Steady Current in Vacuum § 10-1 Magnetic Phenomena Ampere’s Hypothesis § 10-2 Magnetic Field Gauss’law in Magnetic Field §10-3 Boit-Savart Law & Its Application § 10-4 Ampere’s Law & Its Application § 10-5 Motion of Charged Particles in Magnetic § 10-6 Magnetic Force on Current-carrying Conductors § 10-7 The Hall Effect § 10-7 Magnetic Torque on a Current Loop § 10-1 Magnetic Phenomena Ampere’s Hypothesis 1. Magnetic Phenomena (1) the earliest magnetic phenomena that human knew: the permanent magnet (Fe3O4) has N , S poles. Same poles repel each other and different poles attract each other. N pole N pole S pole S pole Magnetic monopole? Never be seen! (2) The magnetic field surrounding the earth (3)The interaction between current and magnet N S I A N N I S B S F N I S attraction repellen t The motion of electron in M-field 2. Ampere’s Hypothesis Each molecule of the matter can be equated with a closed current –called molecular current. When the molecular currents arrange in same direction, the matter appears magnetism in a macroscopic size. S N N All Magnetic phenomena result from the motion of the charge. S 3. Magnetic field magne t curren t Moving charge Mfield Mfield magnet current Moving charge § 10-2 Magnetic Field Gauss’law in Magnetic field 1. Magnetic field Take a moving charge( v and q) as a test charge. The characters of the force on the moving charge by the magnetic field: The direction of M-field at this point each point in the M-field has a special direction. when the q moves along this direction( or opposite the direction), no force acts on it. The direction of the M-force acting on q always perpendicular to v and M-field direction. M-force depends on q,v and angle between and M-field direction. v F qv sin Definition B F --the magnitude qv sin of M-field F or B max tesla(T) qv B along the direction of Fmax v Superposition principle of M-field B Bi i 2. Magnetic field line ( B line) tangential direction of B line--M-field direction. the magnitude of B dN B dS B -line is different with B line: B -lines are always closed lines linked with electric current. They have neither origin nor termination. 3. Magnetic flux and Gauss’ Law in magnetics M-flux:The number of B-lines through a given surface. d B BdS B dS B S B dS n dS unit:weber(Wb)T·m2 For any closed surface S , S SB dS 0 ---- Gauss’ Law in magnetics ----M-field is non-source field B §10-3 Calculation of the magnetic field set up by a current 1. Boit-Savart Law Idl--current element The magnetic field set up by Idl at point P is dB P r Idl I 0 Idl r dB --B-S Law 3 4 r 0 4 10 N A --permeability of 7 2 For any long current, B dB l 0 Idl r 3 l 4 r dB P r Idl I -- superposition principle of M-field 2. Application of B-S Law [Example 1]Calculate the M-field of a straight wire segment carrying a current I. 2 solution Choose any Idl Idl set up dB at P : 0 Idl r dB 3 4 r I l r a 1 direction: all dB set up by all Idl have same direction P 0 I dl sin B dB 2 L L 4 r a r l a ctg sin a dl 2 d sin 0 I sin d B 4a 2 1 0 I (cos 1 cos 2 ) 4a 2 I l r a 1 P discussion for infinite long current 1 0 0 I B (cos 1 cos 2 ) 4a 2 2 0 I B 2a I l a semi-infinite current 0 I B 4a on the prolong line of current B0 r 1 P [Example 2] Calculate the M-field on the axis of a circle with radius R and carrying current I. choose any Idl Solution: Idl I R 0 r x dB P 0 Idl dB 4 r 2 dB dB dB dB// dB// x dB 0 IR 0 B dB// dB sin dl 3 L L 4r 0 IR 2r 2 3 0 IR B I 2 2( R x ) 2 ---- r 2 3 R 0 x dB dB dB// P 2 and I satisfy Right Hand Rule x discussion at the center,x =0 B 0 IR 2( R x ) 2 B0 0 I 2 2 2R the magnetic moment of the circular current 2 pm ISn IR n B 0 pm 2 ( R x ) 2 2 3 2 3 2 [Example 3] Calculate the M-field on the axis of a solenoid with radius R. The number of turns per length of solenoid is n, its carrying current I. Solution l dl Take dl along axis and its distance to P is l. 1 r A2 A1 2 The number of R P turns on length dl , dN ndl the current on dl, dI IdN dB 0 InR dl 2 2( R l ) 2 2 3 direction: 2 Rd dl 2 sin l R ctg l 2 R R l 2 sin 2 2 B dB 0 2 1 A1 P R L 2 0 1 2 nI sin d dl r 2 A2 nI (cos 2 cos 1 ) 0 Discussion B nI (cos 2 cos 1 ) 2 Solenoid “infinite long”: 1 2 0 B 0 nI l 1 A1 R dl r 2 P B ’s direction: satisfy right-hand rule with I. -- the M-field on the axis of a solenoid with infinite length. A2 [Example 4]A long straight plate of width L carrying I uniformly. P and plate current are at same plane. Find B=? of P. Solution I dI Straight line current I dI dx a 0dI dB 2x 0 I dx 2a x L P I dI d x dx All dB have same direction. B dB 0 I L d dx 2a L x 0 I L d ln 2a L L P d x I dI dx Direction : [Example 5]A half ring with radius R, uniform charge Q and angular speed . Find B ? at O. Solution Take an any dl , It charge, dQ dl Q Rd R Q d r dl R x O When dQ is rotating, it equates with 2dQ Q dI 2 d T dB R 0 r dI 2 2( r x ) 2 r x R 2 r dl dI set up M-field at O, 2 2 2 x O 3 2 0 Q r d 3 2 2 ( r 2 x 2 ) 2 2 r R sin All dB have same direction B dB r dl L 2 0 0 Q r d 3 2 2 ( r 2 x 2 ) 2 2 R x O 0 Q 2 2 0 Q sin d 2 2 R 0 8R Direction : 3. M-field set up by moving charge Take Idl,it set up 0 dB 4 0 4 IS Idl r 3 r ( qnSv )dl r 3 r v n dl The number of moving charges in dl, dN nSdl Idl , v same direction 0 q dN v r dB 3 4 r B set up by each moving charge ( q , v ): dB 0 qv r B 3 dN 4 r §10-4 Ampere’s Law 1. Ampere’s Law Question: B dl ? B L Special example, infinite straight line current I 0 I B 2 r I Choose a circle L is just along B-line. B dl Bdl B dl L L B I L B 2 r 0 I L L --does not depend on r choose is any closed line L surrounding I and in the plane perpendicular to I B d l I 0 L Any L surrounding I B dl B (dl// dl ) L L Bdl L 0 I dl// dl dl dl L does not surround I 0 I B2 dl2 B2dl2 cos 2 B2r2d d 2 B1 dl1 B1dl1 cos1 B1r1d B1 0 I d 2 B dl I L L1 B dl1 B dl2 0 L 2 dl1 r1 d L1 a B2 dl2 r2 L2 b L Conclusion B dl 0 I --Ampere’s Law L Notes: Amperian loop I is the algebraic sum of all currents closed by L. I >0 when it satisfy right-hand rule with L. otherwise, I <0. I has not contribution to B dl if it is outside L l Set up by all I (inside or outside L) B is non-conservative field. 2. Application of Ampere’s Law [Example1]A long straight wire with R,uniform I. Find B=? inside and outside it.. Solution Analyze the symmetry of B --Axis symmetry I R B r L Take L to be a circle with r, same direction with B, r>R: L B dr B 2 r LB dl Bdl L B 2 r 0 I R 0 I B 2 r L r<R:the current closed by L, I 2 I R 2 r B 2r 0 I 0 Ir R 2 2 0 Ir B 2 2R B 0 R r [Example 2] Find the M-field of a infinite solenoid. The number of turns per length of it is n, its carrying current I. Solution:analyze the distribution of B Choose a rectangular loop abcda LB dl Bbc Bda 0 a b d c B B 0 nI --uniform field exterior: B 0 R [Example 3] A straight cylinder conductor with R. A hole with radius a is far b from the central axis of cylinder. The conductor has current I,Find B=? at point P. Solution Assume a current I in conductor Current density: I j 2 2 (R a ) Compensatory R a O b method : Imagine there are j and j in the hole. P The B set by the conductor with a hole = the B1 set up by one without hole + the B2 set up by the hole’s negative current B1 R a O b 0 2 B1 (a b) j 2 ( a b) 0 (a b) j 2 Direction :see Fig. P B1 For hole’s -j : 0 2 B2 a j 2a 0a j 2 R O P a b B2 Direction: see Fig. 0bI a b a 0 0 BP B1 B2 j j 2 2 2 ( R a ) 2 2 Direction: [Example 4] A conductor flat carries current The current density is j per unit length along the direction of perpendicular to j. Find the distribution of B outside the flat. j B dl1 dl2 dB1 P B dB dB2 Solution Take a rectangle path abcda B dl B dl B dl 2 Bl 0 jl bc L B 0 j 2 At the two side of the flat, M-field has same magnitude and opposite direction. da B a b d l c B §10-5 Motion of Charged Particles in M-field 1. Lorentz force --Magnetic force acting on the moving charge. Fm qv B v Fm v// B v Notes Fm v Fm does not do work to q. --Change v ’s direction, don’t change v’s magnitude. there are E-field +M-field in the space, a moving charge q sustains: F Fe Fm q( E v B) 2. Moving charge in uniform M-field Let q goes into M-field with initial velocity v v // B: q --straight line motion with uniform velocity. B v v B: --Circle motion with uniform speed in the plane of B F qvB m v R O R 2 mv R qB 2R 2 m period T qB v Application: mass spectrometer (质谱仪) A charged particle from S is speeded U up by U 1 2 qU mv (1) 2 S2 Enter M-field 2R B mv R (2) qB Combine (1) and (2) q 2U 2 2 m B R Application: cyclotron (回旋加速器) 2 m T qB do not depend on R E: speed up q B: change the velocity direction of q v and B with any angle v// v cos ---- // B uniform, straight line v v sin ---- B uniform, circle v v v v // v v// B Moving path---helix mv Revolving radius R qB helical distance B h 2 mv // h v//T qB Application: magnetic focusing (磁聚焦) B A· · A h The particles have same v// same h They focus on point A again 3. Moving charge in non-uniform M-field mv As R qB 2 mv // h qB R, h are different when B is not constant. plasma Magnetic restraint ---Magnetic bottle M-field of the earth Van Allen radiation belts beautiful aurora §10-6 Magnetic force on a currentcarrying conductor B 1. Ampere’s Law Idl is in M-field The force acting on each electron is IS n v Fm ev B The numbers of electron in Idl is dN nSdl The resultant force acting on the dN electrons is dF Fm dN ( ev B)nSdl enSvdl B vdl vdl I dF Idl B --Ampere’s Law of M-force for any shape of current-carrying wire, F dF Idl B L L 2. The application of Ampere’s Law [Example 1] A straight wire with length L carrying I is in a uniform B . Find F =? Take Idl force dF BIdl sin direction: I B Idl L The M-force acting on L is F dF BI sin dl BIL sin L l 0 [example 2] A curved wire segment with I is in the plane which B . Suppose AB=L is known.Find F =? Set up a coordinate system, take any Idl dF Idl B dFx dF cos dF sin IBdl sin IBdy I y A dFy dF dF x Idl B x Similar, dFy dF sin IBdx Fx dFx IB l yB yA dy 0 Fy dFy IB dx IBL l xA Vector express: F IBLj xB Same as the straight wire from A to B. I A B Conclusion in a uniform B , the M-force acting on any shape wire = the M-force acting on the equivalent straight wire . for a closed wire, F=0 in uniform B [Example 3] I1 I2 . AB=L . Find F =? acting on AB. L dl the force acting on dl is dF I1 A d x I2 L dF IBdl 0 I1 B 2x F dF I 2 Bdl B L 0 I1 I 2 2 dL d dx 0 I 1 I 2 d L ln x 2 d 3. The interaction between two parallel currents 1 set up B1at 2 , d 0 I1 B1 2d dF21 The M-force acting on I 2 dl , dF21 I 2dl B1 I1 I2 I 2 dl B1 I1 I 2 0 Magnitude dF21 I 2 B1dl dl 2d direction:21 I I 0 1 2 Similarly F12 dl 2 d , dF21 , dF12 have opposite d B2 dF21 I 2 dl B1 dF12 direction. The force for per unit length wire, dF 0 I1 I 2 dl 2d I1 I2 §10-7 The Hall effect 1. The Hall effect --there is an electric potential difference on the direction of B when a current-carrying plate is put in M-field. Experiment result, B d b 1 I VH H IB d V H:Hall coefficient. 2 Depends on the material. 2. Theoretical explanation Let the velocity of free electrons is v , number density is n Fm ev B In equilibrium state, ev B eEH 0 EH v B Fe eEH B 1 Fm I E Hv F e 2 V Hall E-P-difference, VH U1 U 2 2 1 2 EH dl 1 ( v B) dl 2 vBdl vBh B 1 1 I nevbh 1 IB VH ne b 1 And H ne EH I V 2 For moving positive charges, VH U1 U 2 ( v B) dl 2 1 B 1 FL I v EH Fe 2 2 vBdl vBh 1 1 IB VH nq b 1 H nq V Notes: n has large magnitude in conductors (~1029/m3). The Hall effect is not obvious. The Hall effect is obvious in semiconductor n type semiconductor:electron conduction. p type semiconductor:hole conduction. Positive charge to measure H(or VH) can judge the moving charges and find n. §10-8 Magnetic torque on a current loop 1. The torque acting on a loop by M-field The normal direction of loop n : Satisfy right hand rule with I Fad a l1 l2 I b d B I n c Fbc Fad Il 2 B sin( ) Il 2 B sin Fbc Il 2 B sin Same magnitude, opposite direction, locate on a line. Fad Fbc 0 Fcd B d (c) Fab Fcd Il1 B Do not locate a line. Set up a torque l2 l2 M Fab sin Fcd sin 2 2 Il1l2 B sin ISB sin direction: n a(b) Fab For the loop with N turns, M NISB sin Vector express: Define M NISn B pm NISn --M-moment of a current loop M pm B Fcd B d (c) a (b) Fab --can be used for any shape plane loop in uniform M-field n Discussion : M M max pm B 2 =0: M=0 --stable equilibrium position. = : M =0 -- unstable equilibrium position When suffers disruption, it turns =0 The resultant force acting on loop=0 in uniform M-field. But the torque 0 --only rotation, not translation In non-uniform M-field, M0, F 0. --rotation and translation 2. Potential energy of current loop A current loop has pm ISn It suffers:M pm B sin M ’s direction: n and B n B I M makes decreasing Increase 1 to 2,external force does work: 2 2 1 1 W Md pm B sind pm Bcos1 cos 2 = The increment of potential energy of the loop in B Wm 2 Wm1 pm is put in B, the A loop with M-moment potential energy of the system ( loop + M-field) is Wm pm B cos pm B