1 1. The magnetic field and its features While the magnetic field has similarities to the other fields, it also has some unique features that distinguish it from the others. Gravitational field —gravitational force caused by mass Electric field —electric force caused by electric charge Magnetic field--? gmF r r = G EqF rr = e §20.1 The magnetic field and its application 2 §20.1 The magnetic field and its application 1Magnets and magnetic poles: Unlike poles attract each other, like poles repel each other. The magnetic forces of attraction and repulsion of magnetic poles on each other are similar to the electrical force interactions between electric charges, but they are not at all the same thing. With magnets, no magnetic monopoles ever have been discovered. It is impossible to isolated either a north or a south magnetic pole; hence we say that magnets and the magnetic field always are dipolar. Magnetic poles always occur in north-south pairs that produce a dipolar magnetic field in the surrounding space. B r §20.1 The magnetic field and its application 3 2magnetic field lines The direction of the magnetic field at any point is tangent to the field line at that point; The number density of the magnetic field lines in a region is a measure of the magnitude of the magnetic field there. §20.1 The magnetic field and its application 3the differences of the magnetic field and the electric field (i)A charge q placed at rest magnetic field experiences zero force; (ii)Move the charge along a magnetic field line, the moving charge again experiences zero force; (iii)If the charge is moved at speed v at angle θ with the direction of a uniform magnetic field, a nonzero magnetic force exists on the charge. The force depends on both the speed and the direction of the velocity. §20.1 The magnetic field and its application 4 (iv)The magnetic force varies with the magnitude of the magnetic field B (as determined from the number of the magnetic field lines); (v)The direction of the force on q depends on the sign of the moving charge. In every case the force is perpendicular to both the velocity of the charge and the field direction. BvqF r r r ×= m The SI unit of magnetic field: 1 tesla(T)=10 4 gauss §20.1 The magnetic field and its application 1 A velocity selector jqEEqF jBqvBvqF ? ? e 0m ?== =×= rr r r r B E v BqvqE FF = = =+ 0 0 em 0 rr em0 em0 , , FFvv FFvv << >> deflect up deflect down depends only on the magnitude of the field; is independent of the identical charge; is independent of the mass of the particle. 0 v 0 v 0 v 2. Applications ++++++ ?????? ? ? ? ? ? q ivv ? 0 = r E r x y ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? B r §20.1 The magnetic field and its application 5 2 A mass spectrometer Bq mv R R v mvBq maF BvqF = = = ×= 2 m r r r If the beam is composed of the ions of different isotopes of the same element, all with the same charge, each isotope moves in a circle with a different radius. Choose the speed by a velocity selector §20.1 The magnetic field and its application 3 The Hall effect kBB ? = r ivv ? >=< r Positive charge ivv ? ><?= r negative charge m F r m F r m F r e F r e F r >< v r >< v r >< v r i ? j ? i ? j ? I II I I I §20.1 The magnetic field and its application 6 (i)negative charge carriers jEqjEqF jBvqkBivqBvqF ? ) ? )(( ?? ) ? )(( e m =??= ><?=×><??=×= r r r r jEqjEqF jBvqkBivqBvqF ?? ?? ) ? ( e m == ><?=×><=×= r r r r (ii)positive charge carriers Equilibrium state: BvE BvqEq >=< ><= §20.1 The magnetic field and its application 4 Hall voltage ElyErEVV l l B A AB ?=??=??=? ∫∫ ? 2 2 j ? dj ? d r r nqA I v vnqAI >=< ><= nqd IB nqA IlB BlvV ==>=< Hall 0,0 0,0 Hall Hall << >> Vq Vq x y z B r ????? ????? ????? B A l d A ? +++++ ????? Iv r q §20.1 The magnetic field and its application BlvElVVV BA >=<=?= Hall let 7 5 discovery of the electron and the ratio of mass and charge of electron —Thomson’s experiment yE LB q m B E v v L m qE y 2 , 2 1 22 2 2 =?== §20.1 The magnetic field and its application 6 magnetic focus, magnetic mirror and magnetic bottle R v mvBqF 2 m == Bq mv R = The radius of the circular motion ××××× ××××× ××××× ××××× R v r m B r §20.1 The magnetic field and its application Bq m v R T ππ 22 == The period of the circular motion It is independent v. 8 φ π cos 2 // v qB m Tvh == The screw--pitch For small φ v qB m h π2 ≈ qB mv qB mv R φsin == ⊥ The radius of the helical motion φ v r // v r ⊥ v r B r h q v r B r r §20.1 The magnetic field and its application The magnetic focus The magnetic mirror §20.1 The magnetic field and its application 9 Magnetic bottle §20.1 The magnetic field and its application The Van Allen radiation belts and aurora §20.1 The magnetic field and its application 10 Aurora 7The cyclotron Bq mv R = Bq m v R T ππ 22 == §20.1 The magnetic field and its application Can this process persist infinitely? 11 §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 1. Magnetic forces on a current-carrying wire Electric currents are charges in motion. The force exerted on the wire is a manifestation of the basic magnetic interaction on the moving charged particle in it. Choose a wire segment of length dl and cross sectional area A ><= vnqAIcurrent The average magnetic force on a charge carrier Bvq r r × The total differential magnetic force on the segment of the wire BvlqnAF r r r ×= dd m ld I m F r v r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 12 The total force on the whole wire BlIBlvnqA BvlqnAF rrrr r r r ×=×= ×= dd dd m from ∫∫ ×== wire mm dd BlIFF rrrr We have In uniform magnetic field BLI BlIF rr rrr ×= ×= ∫ )d( wire m m F r I θ §20.2 magnetic forces on currents and torque on a current loop in a magnetic field a b lI r d L r I θ B r BLIF rrr ×= m θsin m BILF = If the wire is not straight line ∫ = LIlI rr d BlIF rrr ×= ∫ )d( wire m from We will get §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 13 2. Torque on a current loop in a magnetic field ] ? sin ? [cos ] ? sin ? )cos[( 24 22 kill kill θθ θθ += ??= r r jBIl kBkiIl BlIF ? cos ? ] ? sin ? [cos 2 2 22 θ θθ = ×+?= ×= rrr jBIl kBkiIl BlIFF ? cos ? ] ? sin ? [cos 2 2 424 θ θθ ?= ×+= ×=?= rrrr 0 42 =+ FF rr I I 2 l r 4 l r kBB ? = r θ x y z 2 F r 4 F r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field iBIl kBjIl BlIF ? ?? 1 1 11 = ×= ×= rrr iBIl kBjlI BlIF ? ? ) ? ( 1 1 33 ?= ×?= ×= rrr 0 31 =+ FF rr I I jl ? 1 jl ? 1 ? kBB ? = r θ x y z 1 F r 3 F r jl ? 1 jl ? 1 ? kBB ? = r θ x z 2 l 24 ll = ? null kBB ? = r 1 F r 3 F r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 14 ] ? sin ? cos)[2/( ] ? sin ? )[cos2/( 23 21 kilr kilr θθ θθ ??= += r r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field jl ? 1kBB ? = r θ x z 2 l 24 ll = ? null kBB ? = r 1 F r 3 F r 1 r r 3 r r O jl ? 1 ? jBllI iBIlki l Fr jBllI iBIlki l Fr ? sin)2/( ? )(] ? sin ? cos[ 2 ? sin)2/( ? ] ? sin ? [cos 2 21 1 2 333 21 1 2 111 θ θθ τ θ θθ τ = ?×??= ×= = ×+= ×= r rr r rr jIABjBlIl ? )sin( ? )sin( 2131 θθτττ ==+= rrr §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 15 AI kAiAA r r r = +?= μ θθ ? )cos( ? )sin(let is the magnetic dipole moment of the current loop. μ r A r is the area vector that perpendicular to the surface of the loop. A r jl ? 1 jl ? 1 ? kBB ? = r θ x z 2 l 24 ll = ? null kBB ? = r 1 F r 3 F r A r then BBAI r r rr r ×=×= μτ jIAB ? )sin( θτ = r The result is valid for planar current loops of any shape. §20.2 magnetic forces on currents and torque on a current loop in a magnetic field Example 1: find the magnitude of magnetic dipole moment produced by an electron rotating with an angular frequency ω along a circle of radius R. ? e μ r Solution: Equivalent current produced by the moving electron is π ω ν 2 q qI == The equivalent magnetic dipole moment is 22 2 2 Rq R q IA ω π π ω μ === §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 16 ω 1 R o 2 R i ? j ? σ+ rrq d2d ??= πσ rr q I d 2 d d σω π ω == krrkIr ? d ? dd 32 σπωπμ == r 4 )( ? 4 ))(( k ? )( 4 ? dd 2 1 2 2 2 1 2 2 2 1 2 2 4 1 4 2 3 2 1 ωωσπ σω π πωσμμ r rr RRQ k RRRR RRkrr R R + = +? = ?=== ∫∫ Example 2: Find the magnetic dipole moment of the ring plane. Solution: r rd §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 3. The motion of the current loop in magnetic field The elements of an electric motor §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 17 4. The potential energy of a magnetic dipole moment in a magnetic field Electric dipole moment in an electric field EpPE Ep r r r rr ??= ×=τ Analogously, magnetic dipole moment in a magnetic field BPE B r r r rr ??= ×= μ μτ §20.2 magnetic forces on currents and torque on a current loop in a magnetic field Example 3: a ring of radius a carrying current i is put in a magnetic field with magnitude B as shown in figure, the direction of the field makes an angle θ with the vertical direction. Find the magnitude and the direction of the force the field exerted on the ring. §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 18 Solution: Choose an element segment of the ring dl The horizontal component of the force is liBF h dcosd θ= The vertical component of the force is liBF v dsind θ= The horizontal component of the total force is zero due to the symmetry. )2(sin dsin aiB liBF v πθ θ = = ∫ Direction of the force points upward. §20.2 magnetic forces on currents and torque on a current loop in a magnetic field F r d θ Example 4: a wood cylinder of mass m=0.25 kg and length L=0.1 m, with N=10.0 turns of wire wrapped around it longitudinally is shown in figure. What is the least current i through the coil that will prevent the cylinder from rolling down §20.2 magnetic forces on currents and torque on a current loop in a magnetic field a plane inclined at angle θ to the horizontal, in the presence of a vertical, uniform magnetic field of magnitude 0.5 T, if the plane of the coil is parallel to the inclined plane? 19 Solution: Newton’s second law for the center of mass )1(0sin ==? mafmg θ Newton’s second law for rotation about the center of cylinder )2(0==? αμ IBfr The magnetic moment )3(2NirLNiA ==μ gm r f r §20.2 magnetic forces on currents and torque on a current loop in a magnetic field Combine (1), (2) and (3) )A(45.2 5.01.0102 8.925.0 2 2 = ××× × == == NLB mg i NirLBBmgr μ §20.2 magnetic forces on currents and torque on a current loop in a magnetic field 20 同学们好 同学们好 ！ ！ §20.3 The Biot-Savart law 1. The relationship between a current and the magnetic field it produces Mass M produce Gravitational Field affects Mass m Via gmF r r = g r Charge Q produce Electric Field affects Charge q Via EqF rr = E r Magnetic Field B r affects Charges in Motion(currents) ∫ ×= ×= BlIF BvqF rrr r r r d produce ? 21 Magnetic Field B r affects Other Charges in Motion (currents) produce Charges in Motion: current 2. Biot-Savart law ∫ × = × = × = wire 2 0 3 0 2 0 ? d 4 d 4 ? d 4 d r rlI B r rlI r rlI B r r r rr r π μ π μ π μ lI r d ld I §20.3 The Biot-Savart law lI r d ld I Qd G: E: M: ∫ ∫ ∫ × 2 00 2 00 2 ? Id 44 Id ? d 4 1 4 1 d ? d d r rl l r r Q Q r r M GGM r r π μ π μ πεπε 3. The similarities of the gravitational, electric and magnetic fields μ 0 =4 π×10 -7 permeability of free space ε 0 permittivity of free space §20.3 The Biot-Savart law 22 4. Applications of Biot-Savart law k zR RI k r R r RI k r lI B r rlI B ? )( )2( 4 ? )2( 4 ? cos sin90d 4 ? d 4 2322 2 0 2 0 wire 2 0 axis wire 2 0 + = = = × = ∫ ∫ π π μ π π μ θ π μ π μ o r r r §20.3 The Biot-Savart law I z P 1Find the magnetic field at a point P located a distance z from the center along the axis of a circular current loop of radius R and current I. ? R lI r d rlI ? d × r r r θcos) ? d( rlI × r θsin) ? d( rlI × r z θ θ r ? 3 0 3 2 0 0 2 0 C 2 ? 2 )( , ? 2 ? )2( 4 ,0 z k z RI BRz k R I k R RI Bz μ π μ π πμ μπ π μ r r r ==>> === For two special points: I I k R I k R RI B ? 4 ? 2 )2( 4 0 2 0 C φ π μ π φπ π μ == r §20.3 The Biot-Savart law 23 Exercise : find the magnetic fields at point C and P in Fig. (a)and (b), respectively. §20.3 The Biot-Savart law Try to draw the magnetic field lines on the straight-cut plane of a current loop . §20.3 The Biot-Savart law 24 k z R R NIR B P ? ]) 2 ([(2 2 3 22 2 0 ++ = μ r k z R R NIR ? ]) 2 ([(2 2 3 22 2 0 ?+ + μ 72.0 0 0 R NI B μ = 68.0 0 0201 R NI BB μ == 2N turns Helmholtz coil of radius R with current I, Find the magnetic field between O 1 and O 2 on axis. k zR RI B ? )( )2( 4 2322 2 0 axis + = π π μ r ?? R R 1 o 2 o O P R z I I §20.3 The Biot-Savart law 1 B 2 o 1 o 2 B O z The Helmholtz coil is an equipment provide an uniform magnetic field in physics laboratory. ?? R R 1 o 2 oO P R z I I §20.3 The Biot-Savart law 25 o R θ ω σ+ θ ?d =B ?d =I?d =q rrq dd θσ= π ω 2 d d q I = r I B 2 d d 0 μ = ∫∫ == R rBB 0 0 d 4 d π ωθμ Rσθωμ π 0 4 1 = Exercise : A sector of radius R with a surface charge density σ rotates about its center O at angular speed ω , find the magnetic field at the center of a sector . Solution: r ? B r §20.3 The Biot-Savart law x y d O P 3Find the magnetic field at a point P located a distance d from a long(infinite)wire carrying a current I. θ lI r d x r r null ) ? sin ? (cos jirr θθ += r kxI jirixIrlI ? sind ) ? sin ? (cos ? dd θ θθ = +×=× r r §20.3 The Biot-Savart law 26 3 0 2 0 d 4 ? d 4 d r rlI r rlI B r rr r × = × = π μ π μ k d I k xdd xId k xd xId k r d r xI k r xI r rlI B ? 2 ? ][ 4 ? )( d 4 ? d 4 ? sind 4 ? d 4 0 222 0 2322 0 wire 2 0 wire 2 0 wire 2 0 π μ π μ π μ π μ θ π μ π μ = ∞? ∞ + = + == = × = ∫∫ ∫∫ ∞ ∞? r r §20.3 The Biot-Savart law §20.3 The Biot-Savart law 27 π θ θ π d dd I R R I I =?= R I R I B 2 00 2 d 2 d d π θμ π μ == Exercise 1: An infinite semicylindrical surface of radius R carrying current I shown in figure, Find the magnetic field at a point on the axis. I P R Solution: Id R P ? ? θ dθ y x B r d B′ r d θ §20.3 The Biot-Savart law i R I i R I iBiBB x ?? 2 dsin ? sind- ? 2 0 0 2 0 π μ π θθμ θ π ?=?= == ∫ ∫ r R I R I B 2 00 2 d 2 d d π θμ π μ == 0d == ∫ yy BB Because of the symmetry Id R P ? ? θ dθ y x B r d B′ r d θ §20.3 The Biot-Savart law 28 Exercise 2: Find the magnetic field at point P. The total current is I. §20.3 The Biot-Savart law Solution: ? x x )1ln( 2 d 2 d 2 d 2 d d 00 00 d w w I x x w I BB xw xI x I B wd d PP P +=== == ∫∫ + π μ π μ π μ π μ P B r Points upward. 5. Forces of parallel currents on each other and the definition of the Ampere j d lII jlBI BilIF k d I B ? 2 ? ? ? 2 210 12 1221 10 1 π μ π μ ?= ?= ×= = rr r 1The currents is parallel wire 2 wire 1 x y 1 I 2 I d l l §20.3 The Biot-Savart law 29 j d lII jlBI BilIF k d I B ? 2 ? ? ? 2 210 21 2112 20 2 π μ π μ = = ×= ?= rr r The forces that parallel wires exerted on each other are attractive if the currents are parallel. If the current are antiparallel, the forces on the wires will be repulsive. wire 2 wire 1 x y 1 I 2 I d l l §20.3 The Biot-Savart law d lII F π μ 2 210 = The magnitude of the force Let I 1 =I 2 =I, d=1 m, l=1m, adjust the current I, until the force equal to 2×10 -7 N, the current is defined to be 1 Ampere. §20.3 The Biot-Savart law 30 §20.4 The magnetic flux and Gauss’ law 1. The magnetic flux The magnetic field lines: a d 2 l 1 l c b l I ∫∫ ?== SS mm SBΦ rr ddΦ The magnetic flux: B r S r d rl r I SBΦ m d 2 dd 1 0 π μ =?= rr Example 1: r × × × × × × × × × × × × l llIl r r Il ΦΦ ll l mm 210 10 ln 2 d 2 d 2 + = == ∫∫ + π μ π μ Solution: §20.4 The magnetic flux and Gauss’ law 31 0d =? ∫ S SB rr 2. Gauss’s law for the magnetic field The flux of the magnetic field through any closed surface must always be zero. Gauss’s law for magnetic field is, in essence, a statement about the nonexistence of magnetic monopoles. §20.4 The magnetic flux and Gauss’ law S r 3. Magnetic poles and current loops The source of the magnetic field is the moving charges. §20.4 The magnetic flux and Gauss’ law 32 The magnet is really nothing more than a current loop consisting of an electron orbiting a nucleus. A macroscopic permanent magnet is a superposition of many small, generally aligned, submicroscopic current loops represented by the electronic orbital motions inside atoms. The idea of magnetic poles is nothing more than an extension of the idea of the two sides of a current loop. §20.4 The magnetic flux and Gauss’ law §20.5 Ampere’s law and its applications 1. Ampere’s law Electric field is conservative field: 0d pathclsd =? ∫ rE r r What about the integral of the magnetic field around a closed path? ?d pathclsd =? ∫ rB r r Ir r I r r I rB r LL 0 2 0 00 d 2 d 2 d μ π μ π μ π ===? ∫∫∫ r r B r I r r d 1A circular path around the wire 33 I r r I r r I rB r LL 0 2 0 0 0 d 2 d 2 d μ π μ π μ π ?= ?= ?=? ∫ ∫∫ r r B r I r r d 2Any closed path around the wire I I r r I rBrB LLL 0 2 0 0 0 d 2 d 2 dcosd μ? π μ ? π μ θ π == ==? ∫ ∫∫∫ r r B r θ ?d r r d r r I §20.5 Ampere’s law and its applications I I r r I rBrBrB LLLL 0 2 0 0 0 d 2 d 2 dcosdcosd μ? π μ ? π μ αθ π ?=?= ?=?==? ∫ ∫∫∫∫ r r B r θ r r d r r ?d? α I Note: The the current threading the path means that the current pierces any surface that has the contour of the path as a boundary, like a soap bubble surfaces. The current must pierce the hat-shaped surface an odd number of times. Current threading the path §20.5 Ampere’s law and its applications 34 0 d 2 d 2 ddd 0 0 0 0 21 = += ?+?=? ∫∫ ∫∫∫ ? ? ? π μ ? π μ II rBrBrB LLL r r r r r r ∑ ∫∫∫ ∫∫ = ?++?+?= ?+++=? )paththeinside( 0 21 21 ddd d)(d i L n LL n LL I rBrBrB rBBBrB μ r r L r r r r r r L rr r r 3A path not threaded by the current ? 1 L 2 L r r d B r r r d B r 4If multiple currents are present simultaneously §20.5 Ampere’s law and its applications ∑ ∫ =? )paththeinside( 0 d i L IrB μ r r Ampere’s law: Note: If you wrap the fingers of your right hand around the closed path in the direction of the path integration, those currents directed through the path in the direction of your thumb appear on the right-hand side of Ampere’s law with a plus sign. Those in opposite direction appear with minus sign. 1 I+ 2 I? §20.5 Ampere’s law and its applications 35 ∑ ∫ =? )paththeinside( 0 d i L IrB μ r r The current must pierce the hat-shaped surface an odd number of times. §20.5 Ampere’s law and its applications I L IIII i 23 ?=?= ∑ 321 IIII i ?+= ∑ Exercise 1: §20.5 Ampere’s law and its applications 1 I L 4 I I 2 I 3 36 Exercise 2: Eight wires cut the page perpendicularly at the points shown in figure. A wire labeled with integer k carries the current ki. For odd k, the current is out of the page; for even k it is into the page. Evaluate the §20.5 Ampere’s law and its applications ∫ ? lB rr d Along the closed path in the direction shown. 2. Applications of Ampere’s law 1The magnetic field outside and inside a long cylindrical wire of radius R with current i (i)outside the wire r r dr B r d §20.5 Ampere’s law and its applications 37 irBrBrBrB r 0 2 2ddd μπ π ====? ∫∫∫ r r r i B π μ 2 0 = (ii)inside the wire 2 2 0 2 2 ddd r R i rB rBrBrB r π π μπ π == ==? ∫∫∫ r r r R i B 2 0 2π μ = §20.5 Ampere’s law and its applications r r d 2Magnetic field inside a solenoid carrying current i §20.5 Ampere’s law and its applications 38 3/222 2 0 ])(2[ )d( d z-dR Rzni B + = μ ) )/2( 2/ )/2( 2/ ( 2 ])(2[ d 2 dB 2222 0 2/ 2/ 3/222 2 0 -dLR dL dLR dLni z-dR zniR B L L + ? + ++ + = + == ∫∫ ? μ μ ? ?? ? ? ??? ??? ?? ??? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? y z O P B r d z d §20.5 Ampere’s law and its applications 12 B′ r B′? r B r P z z' ? ?? ? ? ??? ??? ?? ??? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? o o' The magnetic field inside the solenoid is parallel to the axis of the solenoid. §20.5 Ampere’s law and its applications 39 )(d ddddd 0 nliBlrB rBrBrBrBrB b a a d d c c b b a μ=== ?+?+?+?=? ∫ ∫∫∫∫∫ r r r r r r r r r r niB 0 μ= l §20.5 Ampere’s law and its applications 3Magnetic field of a toroid NirB rBrBrB r 0 2 2 ddd μπ π == ==? ∫∫∫ r r r Ni B π μ 2 0 = The magnetic field on the cross section of the toroid is not uniform. §20.5 Ampere’s law and its applications 40 z x o nullnullnullnullnullnullnullnullnull B r d B′ r d xjI dd = ) ? sin ? (cos 2 d d 0 ki r I B θθ π μ += r 0d == ∫ zz BB Exercise 3: Find the magnetic field at a distance r from an infinite plane carrying current j in unit width. Solution 1: superposition method choose z B′ r d r Id I′d r′ B r d θ θ x §20.5 Ampere’s law and its applications i zx xzj ? d 2 22 0 ∫ ∞ ∞? + = π μ i j z x z zj ? 2 arctg 1 2 00 μ π μ = ∞? ∞ ?= xz +> : 0 xz ?< : 0 2 0 jμ 2 0 jμ ? x B o z x o nullnullnullnullnullnullnullnullnull B r d B′ r d z B′ r d r Id I′d r′ B r d θ θ x i r z r xj BiBB x ? 2 d i ? cosd ? d 0 ?=== ∫∫∫ π μ θ r §20.5 Ampere’s law and its applications 41 2 0 j B μ = abjBablB lBlBlBlBlB b a b a a d d c c bL 0 2d2 ddddd μ==?= ?+?+?+?=? ∫ ∫∫∫∫∫ rr rrrrrrrrrr z x o nullnullnullnullnullnullnullnullnull ab cd ∑ ∫ =? i L IlB 0 d μ rr Same as result of solution 1. §20.5 Ampere’s law and its applications Solution 2: Ampere’s law Exercise : Find the magnetic field at point b. §20.5 Ampere’s law and its applications )( 2 22 2 0 ba I A I j R Ir B ? == = π π μ )(2 22 22 0 2 2 0 2 0 ba Id aj a d a Id B ? = == π μ π π μ π μ I 42 BvqF r r r ×= m ∫∫ ×== wire mm dd BlIFF rrrr The torque on a current loop with the magnetic dipole moment μ r B r rr ×= μτ Lorentz force of a moving charge in a magnetic field The magnetic force of a current- carrying wire in a magnetic field Review Mass M produce Gravitational Field affects Mass m Via gmF r r = g r Charge Q produce Electric Field affects Charge q Via EqF rr = E r Magnetic Field B r affects Charges in Motion(currents) ∫ ×= ×= BlIF BvqF rrr r r r d produce ? Review 43 ∑ ∫ =? )paththeinside( 0 d i L IrB μ r r Ampere’s law: Biot-Savart law ∫ × = × = × = wire 2 0 3 0 2 0 ? d 4 d 4 ? d 4 d r rlI B r rlI r rlI B r r r rr r π μ π μ π μ Review 3. The displacement current and the Ampere-Maxwell law 1 2 ?+ K 2 S 1 S + For time dependent electric field, what will happen? 2 S 1 SCurrent thread No current thread : 1 S IrB L 0 d μ=? ∫ r r 0d =? ∫ L rB r r : 2 S For For ? §20.5 Ampere’s law and its applications 44 D I tt EA t Q t E A t Ed d A t V C t Q CVQ V Q C ≡== === == d d d )d( d d d d d )(d d d d d elec 00 0 0 Φ εε ε ε Displacement current Ampere-Maxwell law LD L IIrB threading0 )(d +=? ∫ μ r r §20.5 Ampere’s law and its applications The crux of Maxwell’ s argument was the realization that magnetic fields are produced via two distinct mechanisms: 1By electric charges in motion(conduction current); 2 By time-varying electric fields(via the displacement current). 4. The magnetic field produced by the displacement current §20.5 Ampere’s law and its applications 45 + + + + + + + ------- A′ E r B r D L IrB 0 d μ=? ∫ r r if 0 d d > t E t E rB t E rrB t E r t EA I rBrB D L d d 2 d d )2( d d d )(d )2(d 00 2 00 2 00000 εμ πεμπ πεμεμμ π = = = ′ = =? ∫ r r The magnetic field induced by the displacement current is perpendicular to the electric field that causes it. Note: §20.5 Ampere’s law and its applications r The magnetic field is perpendicular to the changing electric field that causes it. §20.5 Ampere’s law and its applications