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Chapter 28 Magnetic Field and Magnetic Forces Physicist: Hans Christian Orsted Born: 14 August 1777 Died: 9 March 1851 Dane What Orsted saw... Magnetical Field Intensity of Magnetic field Biot-Savart law Magnetic Force To Moving Circular law with Magnetic material To Coil Particle Guass’s law Interaction Hall Effect Key words permanent magnet(永磁体) magnetic monopole(磁单极) magnetic field(磁场) tesla(特斯拉) magnetic field line(磁场线) gauss(高斯) magnetic flux(磁通量) weber(韦伯) magnetic flux density(磁通密 度) Key words cyclotron frequency(回旋频率) mass spectronmeter(质谱仪) isotope(同位素) magnetic dipole(磁偶极子) Hall effect(霍耳效应) magnetic moment(磁矩) solenoid(螺线管) magnetic dipole moment(磁偶极矩) 28.1 Magnetism • Permanent magnets. • Magnetic poles: a north pole(N-pole) and a south pole(S-pole). • Compass. • Opposite poles attract each, and like poles repel each other. • There are not magnetic monopoles. 28.2 Magnetic Field • A moving charge or a current creates a magnetic field in the surrounding space(in addition to its electric field). • The magnetic field exerts a force F on any other moving charge or current that is present in the field. Magnetic field (cont) • Magnetic field (B) is a vector field. • At any position the direction of B is defined as that in which the north pole of a compass needle tends to point. • For any magnet, B points out of its north pole and into its south pole. 28.3 Magnetic Force on a Moving Particle • Consider a particle of charge q and velocity v at a point in a magnetic field with induction B. • The magnetic force exerted on the charge is given by the definition of B, F q B q B sin F qv B • Units are those of force N 1 1 NA m Tesla (T ). 1 charge velocity C (ms ) 1G 10 4 T Magnetic force on a moving particle (cont) Orders of Magnitude • • • • • • • Magnetic field at surface of Earth 10-4T. Bar magnet 10-2T. Write head for hard disk 10-1T. Lab electromagnet 1T. Superconducting magnet 101T. Interior of atoms, up to 101T. Surface of a pulsar(脉冲星) 108T. 28.4 Magnetic Field Lines Magnetic flux • Basic concept is as electric field lines but: – they have the direction a compass needle would point to at each location; – tangent to the field lines gives the direction of B at a point; – they never cross and – they form closed loops - a continuous field. Magnetic flux d B B dA B cos dA B dA B B cos dA B dA • Magnetic flux is a scalar quantity. • This unit is the Weber (1Wb) 1Wb=1T· m2=1N· m/A 28.5 Gauss’s Law for Magnetism • The total magnetic flux through a closed surface is always zero. B d A 0 d B B dA (magnetic flux through any closed surface). Magnetic field B is sometimes called magnetic flux density. 28.6 Moving Charges in B Fields • • • • Uniform field of B. Non-uniform field of B. Combined E and B fields. Force on a current carrying conductor. Basic Motion in Magnetic Field • The velocity of a charged particle can be divided into two components one parallel and one perpendicular to the field. • The perpendicular component gives circular motion in a plane perpendicular to B. • The parallel component gives constant linear motion along B (no force). • Combination is a helical path. Uniform Field of B F B F + O + S F + P Uniform Field of B (cont) • If v is perpendicular to B – F=qvB the force is perpendicular to the velocity giving circular motion. – Modulus of force is qvB=mv2/r r=(mv)/(qB). – Define cyclotron frequency 1 v qB fc . T 2r 2m • Independent of speed and radius of orbit, particles of same q/m have same fc. Uniform Field of B (cont) Uniform Field of B (cont) • If v is not perpendicular to B. F q B q per par B q per B • So we have circular motion from the perpendicular component. • The parallel component gives constant linear motion along B (no force). • Combination is the helix. Non-Uniform field of B • Effect can be to confine particles • Examples magnetic bottles and van Allen belts. Charged Particles from Sun Enter Earth’s Magnetic Field Example: Helical particle motion q 1.60 1019 C m 1.67 1027 kg B 0.500T At t=0, the proton has velocity components vx=1.50x105m/s, vy=0, and vz=2.00x105m/s • At t=0, find the force on the proton ant its acceleration. • Find the radius of helical path, the angular speed of the proton, and the pitch of the helix (the distance traveled along the helix axis per revolution). Example: Helical particle motion (cont) • The magnetic force on the proton is F q B q x iˆ z kˆ Biˆ q z Bˆj 1.60 10 19 C 2.00 105 m s 0.500T ˆj 14 1.60 10 N ˆj • The acceleration is F 1.60 1014 N ˆ 12 2 ˆ a j 9 . 58 10 m s j 27 m 1.67 10 kg Example: Helical particle motion (cont) • The force is perpendicular to the velocity, so the speed of the proton does not change. • The radius of the helical trajectory is given by m z 1.67 10 27 kg 2.00 105 m s 3 R 4 . 18 10 m 19 qB 1.60 10 C 0.500T The angular speed is qB m 1.60 10 C 0.500T 7 4 . 79 10 rad s 27 1.67 10 kg 19 Example: Helical particle motion (cont) • The time required for one revolution (the period) is 2 2 7 T 1 . 31 10 s 7 1 4.79 10 s • The pitch is the distance traveled along the xaxis during this time, so d xT 1.50 105 m s 1.31107 s 0.0197m 28.7 Combined E and B Fields F q E B • The velocity selector. – Uniform, perpendicular B and E; v perpendicular to both. – Balance when q[E+(vB)]=0 v=E/B – Undeflected particles have been velocity selected. Combined E and B fields (cont) • e.g. Thomson's Experiment. – First accelerate particles so ½mv2=eV. – Then force straight motion so E/B=v=(2eV/m)½. – Thus gives e/m, which is constant! Combined E and B fields (cont) • e.g. The Bainbridge Mass Spectrometer. – First a velocity selector. – Then uniform magnetic field as r=(mv)/(qB) m/q. 28.8 Magnetic Force on Current Carrying Conductor Magnetic force on current carrying conductor (cont) F q B so dF ed B • Where vd is the drift velocity. If n conduction electrons per unit volume, force summed over all the electrons is F nAled B •Since I=nAvde, F Il B. •For variable magnetic fields, or curved wires dF Idl B Example • The conductor has a straight segment with length L perpendicular to the plane of the figure on the right, with the current opposite to B. • Find the total magnetic field force on these three segments of wire. Example (cont) • There is no force on the segment on the right perpendicular to the plane of the figure on the right. • For the straight segment on the left, L points to the left, perpendicular to B . The force has magnitude F=ILB, and its direction is up. • Choose a segment dl with length dl Rd on the semicircle, so magnitude dF of the force on the segment is dF I ( Rd ) B Example (cont) • The components of the force on segment are dFx IRd cos dFy IRd sin • To Find the component of the total force, we integral these expression. 0 0 Fx IRB cosd 0 Fy IRB sin d 2IRB • Adding the forces on the straight and semicircular segments, we find the total force: Fx 0 Fy IB L 2 R or F IBL 2 R ˆj 28.9 Torque on a Current Loop Torque on a current loop (cont) • Consider a square current loop. Each side – parallel to axis produces a force IaB, – perpendicular to axis produces a force IbBcos(). • Forces balance but parallel sides yield a torque =2(½bIaBsin())=IABsin()=|IAB|. • So define = B. • is called the magnetic dipole moment. Units? Torque on a current loop (cont) IA NIA NIAB sin 28.9 Hall Effect Hall Effect • Charges in a wire within a B field will be pushed to one side of wire. • Charge will build up until E and B forces match i.e. FB=qvdB=FE=qE=qVH/h. (h is the height of the bar.) • Recall that I=nqvdht, so VH=(IB)/(nqt). (t is the thickness of the bar, so ht is the cross sectional area; vd is the drift velocity, n the carrier density.) • Measures field or charge carrier density. • Sign of Hall voltage gives sign of charge carrier. Electrostatic vs Magnetic Fields (I) • Electrostatic Field • Electric monopoles exist (charges, +q, -q) • Created by charges at rest through E field • The field exerts a force (F=qE) on a static charge. • Magnetic field • Magnetic monopoles do not exist • Created by moving charge (or current) through B field. • The field exerts a force (F=qvB) on a charge moving at velocity v. Electrostatic vs Magnetic Fields (II) • E field is a vector (F is parallel to E). • The electric force does work on a charged particle. • The energy of a charged particle changes. • B field is a vector (F is perpendicular to v and B). • The magnetic force does no work on a charged particle. • The energy of the particle does not change. Electrostatic vs Magnetic Fields (III) • E field lines start at +ve • B field lines are and end on -ve charges. continuous (form loops). • Gauss's law tells us • Since B lines have no end points E dA q0 / 0 . B dA 0. • Since E dl 0 the E field is conservative, • Wait and see… hence definition of V. Chapter 29 Sources of Magnetic Andre Marie Ampere Born: 20 January 1775 Died: 10 June 1836 French C B dl 0 I enclosed Key words principle of superposition of magnetic fields(磁场叠加原理) law of Biot and Savart (毕奥-萨伐 定律) source point(源点) field point (场点) ampere (安培) Ampere’s law (安培定律) toroidal solenoid(螺线管) Bohr magneton(波尔磁子) Key words paramagnetic(顺磁性的) relative permeability(相对渗透性) permeability(渗透性) magnetic susceptibility(磁化率) diamagnetic(抗磁性的) ferromagnetic(铁磁体) magnetic domain(磁畴) hysteresis(磁滞) magnetization(磁化) displacement current(位移电流) 29.1 Magnetic Field of a Moving Charge ELECTRIC FIELD Proportional to q. Proportional to 1/r2. MAGNETIC FIELD Proportional to q. Proportional to 1/r2. AND to v. AND to sin(). is the angle between and . r Magnetic field of a moving charge (cont) MAGNETIC FIELD ELECTRIC FIELD q moves with velocity v. For a point charge q. The field is circular, i.e in Field is radial, i.e. the direction of E is along r. the direction of vr. Magnetic field of a moving charge (cont) B r rˆ r 0 q sin B 4 r2 The field line directions are given by the right-hand rule. 0 q rˆ 0 q r B 2 4 r 4 r 3 0 • In SI unit the numerical value of 0 is exactly 4 10 7 0 110 7 T m / A 4 c 2 1 00 Example: Forces between two moving protons y • Find the electric and magnetic forces on the upper proton, and determine the ratio of their magnitudes. + q r + q z x Example: Forces between two moving protons (cont) Solution: By Coulomb’s law, the magnitude of electric force is 1 q2 FE 40 r 2 The force on the upper proton is vertically upward. The magnitude of B caused by the lower proton is 0 q B 4 r 2 From the right-hand rule, the direction of B is in the +z direction. Example: Forces between two moving protons (cont) The magnitude of the magnetic force on the upper proton is 0 q 2 2 FB qB 4 r 2 The ratio of the magnitude of the two forces is 2 FB 0 q 2 2 4r 2 2 2 0 0 2 2 FE q 40 r c c 2 1 00 So in non-relativity analysis, FB<<FE, FB can be ignored. 29.2 Magnetic Field of a Current Element B dQ nqAdl I n q d A 0 n q d Adl sin 0 dQ d sin dB 2 4 r2 4 r 0 Idl sin 4 r2 0 Idl r̂ 0 Idl r dB 2 4 r 4 r 3 Law of Biot and Savart 29.3 Magnetic Field of a Straight Current Carrying Conductor y a dl y r P O I a dl dy Ixdy 0 Idl sin 0 dB 2 4 x 2 y 2 3 2 4 r 0 I a xdy 0 I 2a B 3 a 2 2 2 4 x 4 x x 2 a 2 x y dB When a>>x, x2 a2 a 0 I B 2x r x 2 y 2 sin sin x x2 y 2 Magnetic field of a straight current carrying conductor (cont) For a long straight currentcarrying conductor, at all points on a circle of radius r around the conductor, the magnitude B is 0 I B 2r y Btotal Example B2 B1 B1 P1 Wire 1 Btotal I 3d B2 Wire 2 P2 d d I 2d P3 B1 Btotal B2 • Find the magnitude and direction of B at points P1, P2, and P3. • Find the magnitude and direction of B at any point on the x-axis to the right of wire 2 in terms of the xcoordinate of the point. x Example (cont) Solution: a) At pint 1: 0 I 0 I 0 I 0 I 0 I B2 B1 Btotal B1 B2 2 4d 8d 2 2d 4d 8d At pint 2: 0 I B1 B2 2d At pint 3: 0 I 0 I B1 2 3d 6d 0 I Btotal B1 B2 d 0 I B2 2d 0 I Btotal B2 B1 3d Example (cont) Example (cont) b) The magnitudes of the fields due to each wire are B1 0 I B2 2 x d 0 I 2 x d The total field is in the negative y-direction, and has magnitude Btotal B2 B1 0 I 0 I 0 Id 2 x d 2 x d x 2 d 2 At points very far from wires, so that x>>d, and Id Btotal 0 2 x 0 I B 2r ( a long, straight, currentcarrying conductor) 29.3 Force between Parallel Conductors Force between parallel conductors (cont) The magnitude of B field produced by the lower conductor, at the position of the upper conductor, is 0 I B 2r The magnitude of the force exerted on a length L of the upper conductor is 0 II L F I LB 2r Force between parallel conductors (cont) The force per unit length F/L is F 0 II L 2r (two long, parallel, currentcarrying conductors) • Similar result for force on the lower conductor. • Like currents attract, unlike currents repel. Definition of the Ampere • One ampere is that unvarying current that, if present in each of two parallel conductors of infinite length and one meter apart in empty space, cause each conductor to experience a force of exactly 2x10-7 Newtons per meter of length. 29.4 Magnetic field of a circular current loop 0 Idl dB 4 x 2 a 2 0 Idl a dBx dB sin 4 x 2 a 2 x 2 a 2 12 0 Idl x dBy dB cos 4 x 2 a 2 x 2 a 2 12 dl 2a B 4 x Due to rotational symmetry about x-axis, the y-component cancel. Iadl 0 x 2 a 2 3 2 0 Ia 0 Ia 2 (on the axis of a dl 3 3 4 x 2 a 2 2 2 x 2 a 2 2 circular loop) Magnetic field of a circular current loop (cont) • For N loop, Bx N 0 Ia 2 (on the axis of 2 x 2 a 2 2 N circular loop) The maximum value of the field is at x=0, the center of the loop. 3 0 NI (at the center of N Bx 2a circular loops) Magnetic field of a circular current loop (cont) 29.5 Ampere’s Law • Recall Gauss's law vs. Coulomb's Law. • Ampere's law is the alternative to the B-S law. – Use to find B fields from highly symmetric current distributions. – Find current distributions for particular B fields. Ampere's Law (cont) 0 I B dl B// dl B dl 2r 2r 0 I The line integral is independent of the radius of the circle and is equal to 0 multiplied by the current passing through the area bounded by the circle. Ampere's Law (cont) B and dlare antiparallel, so B dl Bdl B dl 0 I The sign depends on the direction of the current relative to the direction of integration. Like directions is a positive sign, unlike direction a negative sign. Ampere's Law (cont) B dl B// dl B1 dl 0 dl B2 dl 0 dl b c d a a b c d 0 I 0 I r1 0 r2 0 2r1 2r2 0 Ampere's Law (cont) B dl Bdl cos dl cos rd Brd 0 I B dl 2r rd 0 I d 2 d 2 0 I Ampere's Law (cont) B dl Bdl cos dl cos rd Brd 0 I B dl 2r rd 0 I d 0 d 2 0 Ampere's Law (cont) • It relates the tangential component of B summed around a closed curve (C) to the current (I) which passes through the curve. B d l I 0 enclosed (Ampere’s Law) • Holds for any curve provided the current is continuous, i.e. it does not begin or end at any point. Ampere's Law (cont) • If B dl 0 , it does not necessarily mean that B 0 everywhere along the path, only that the total current through an area bounded by the path is zero. B dl B dl B I dl B dl dl B I B dl dl B B dl 0 I I 0 Ampere's Law - The recipe • Let geometric closed path pass through the required field point and lie in a plane. • And encompass the current • Ensure B is constant along the path, and that B is tangential to the line or – it is perpendicular to the field lines (zero contrib.) – or it is in a region where B=0. • To achieve this use the symmetry of the field lines. Example: inside a long cylindrical conductor • The current is uniformly distributed over the cross-section area of the conductor. • Find the magnetic field as a function of the distance r from the conductor axis for points inside (r<R) and outside (r>r) the conductor. Example: inside a long cylindrical conductor (cont) Solution: i) r<R Choose integral path a circle with radius r<R. B dl B dl B2r 2 I Ir 2 2 I enclosed J r 2 r 2 R R By Ampere’s law, B dl I 0 I B2r 0 I B 2r 0 enclosed Example: inside a long cylindrical conductor (cont) ii) r>R Choose integral path a circle with radius r>R. B dl B dl B2r I enclosed I By Ampere’s law, Ir 2 B2r 0 2 R B dl 0 I enclosed Example: inside a long cylindrical conductor (cont) 0 I r 2 R 2 , B 0 I , 2r rR rR Example: Field of a solenoid • The solenoid has n turns of wire per unit length and carries a current I. • Use Ampere’s law to find the field at or near the center of the solenoid. Example: Field of a solenoid (cont) Solution: Choose the integration path the rectangle abcd. Side ab, with length L, is parallel to the axis of the solenoid. Side bc and da are taken to be very long so that side cd is far from the solenoid; the the field at side cd is negligibly small. b c d a B dl B dl B dl B dl B dl a abcd b c BL 0 0 0 0 nLI d B 0 nI Example: Field of a solenoid (cont) Example: field of a toroidal(环行) solenoid • A toroid is wound with N turns of wire carrying a current I. • Find the magnetic field at all points. Example: field of a toroidal solenoid (cont) Solution: To find the field using Ampere’s law, we’ll use the integration paths shown as black lines in the right-hand picture. i) Path 1 B dl B2r 0 I enclosed 0 circle1 B0 Example: field of a toroidal solenoid (cont) ii) Path 2 B dl B2r 0 I enclosed 0 NI circle 2 0 NI B 2r iii) Path 3 B dl B2r 0 I enclosed 0 NI 0 NI 0 B 0 circle 3 29.5 Displacement Current B dl 0 I enclosed Where I enclosed ic or 0 ? This is a clear contradiction. As the capacitor charges, the electric field E and electric flux e through the bulged-out surface are increasing. Displacement current (cont) • The capacitor charge q is A q C Ed EA E d • As the capacitor charges, the rate of q is the conduction current. dq d E ic dt dt • We invent a fictitious current iD in the region between the plates, defined as d E iD dt (displacement current) Displacement current (cont) • We include displacement current iD, along with conduction current iC, in Ampere’s law: B dl 0 iC iD enclosed (generalized Ampere’s law) Displacement current (cont) • Apply Ampere’s law to a circle of radius r, iD 2 B dl 2rB 0 R 2 r 2 ic 2 r 0 2 r 0 ic 2 R R For r>R, B is the same as through the wire were continues and plates not present at all. or B 0 r i 2 C 2 R Summary • The magnetic force exerted on the charge is given by B, F qv B F qE B • The force on a straight segment of a conductor current I in a uniformly B is F Il B • For a infinitesimal segment dl the force is dF Idl B • Gauss’s law for magnetism, B dA 0 Summary • Magnetic field of a moving charge • Magnetic field of a current element • Magnetic field of a long straight current • Ampere’s law B dl 0 iC iD enclosed 0 q r B 4 r 3 0 Idl r dB 4 r 3 0 I B 2r 29.6 Magnetic Materials • A extern magnetic field B0 • Put a magnetic material into the magnetic field. • Interaction between the magnetic field and the magnetic material. • The accessional magnetic field by the magnetic material is B¹. • The total magnetic field: B B B 0 Magnetic materials (cont) • Define the relative permeability: B r B0 • Paramagnetic(顺磁性) material: μr>1 • Diamagnetic(反磁性) material: μr<1 Molecule current • The molecule current can produce a magnetic moment. • Define the magnetization of the material: total M V e I • The total magnetic field B in the material is B B0 0 M Molecule current (cont) I a d b c l • Molecule current line density: js p m I s S js Sl p M m js Sl js Sl V M dl M dl M ab Ml jsl I s ab Ampere’s law in material B dl 0 I I s 0 I M dl I s M dl B M dl I 0 B • Define: H M 0 • So: H dl I (Ampere’s law in material) Boundary conditions of magnetic field B1 S 1 2 B2 B dS B2n S B1n S 0 B1n B2 n Boundary conditions of magnetic field (cont) H1 a d 1 2 b c H2 H dl H1t ab H 2t cd H1t H 2t