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Transcript
Chapter 28: Magnetic Field and Magnetic Forces Iron ore found near Magnesia Compass needles align N-S: magnetic Poles North (South) Poles attracted to geographic North (South) Like Poles repel, Opposites Attract No Magnetic Monopoles Magnetic Field Lines = direction of compass deflection. Electric Currents produce deflections in compass direction. =>Unification of Electricity and Magnetism in Maxwell’s Equations. p212c28: 1 Magnetic Fields in analogy with Electric Fields Electric Field: – Distribution of charge creates an electric field E(r) in the surrounding space. – Field exerts a force F=q E(r) on a charge q at r Magnetic Field: – Moving charge or current creates a magnetic field B(r) in the surrounding space. – Field exerts a force F on a charge moving q at r – (emphasis this chapter is on force law) p212c28: 2 Magnetic Fields and Magnetic Forces Magnetic Force on a moving charge – proportional to electric charge – perpendicular to velocity v – proportional to speed v (for a given geometry) – perpendicular to Magnetic Field B – proportional to field strength B (for a given geometry) F=qvB p212c28: 3 F=qvB F = |q| v B sinq = |q| v B (v ^ B) F B + v F F=qvB F = |q| v^ B B + v^ v F F=qvB F = |q| v B^ + B^ v B p212c28: 4 Magnetic Fields Units of Magnetic Field Strength: [B] = [F]/([q][v]) = N/(C m s-1) = Tesla Defined in terms of force on standard current CGS Unit 1 Gauss = 10-4 Tesla Earth's field strength ~ 1 Gauss Direction = direction of velocity which generates no force Electromagnetic Force: F=q(E+ vB) = Lorentz Force Law p212c28: 5 Magnetic Field Lines and Magnetic Flux Magnetic Field Lines Mapped out with compass Are not lines of force (F is not parallel to B) Field Lines never intersect Magnetic Flux dFB = B . dA dF B B dA F B B dA B dA 0 no magneti c charge! (no monopo les) p212c28: 6 • SI Unit of Flux: – 1Weber = 1Tesla x 1 m2 – for a small area B = dFB /dA^ – B = “Magnetic Flux Density” Flux through an open surface will play an important role p212c28: 7 Motion of Charged Particles in a Magnetic Field Charged Particle moving perpendicular to the Magnetic Field – Circular Motion! – (simulations) F + v F + v p212c28: 8 Charged Particle moving perpendicular to a uniform Magnetic Field v R + mv 2 F | q| vB R mv R | q| B v | q| B R m cyclotron frequency + v p212c28: 9 In a non-uniform field: Magnetic Mirror v F B Net component of force away from concentration of field lines. Magnetic Bottle Van Allen Radiation Belts p212c28: 10 Work done by the Magnetic Field on a free particle: dW F dx qv B vdt 0! => no change in Kinetic Energy! Motion of a free charged particle in any magnetic field has constant speed. p212c28: 11 Applications of Charged Particle Motion in a Magnetic Field + Recall: Charged Particle moving perpendicular to a uniform Magnetic Field mv 2 F | q| vB R v R mv R | q| B + v p212c28: 12 Velocity Selector makes use of crossed E and B to provide opposing forces + + - - + + - - + + + - - - E upwards F=qvB downwards F = qE No net deflection => forces exactly cancel: |q| v B=|q| E v = E/B p212c28: 13 J. J. Thomson’s Measurement of e/m Electron Gun + + - - + + - - + + + - - - E V and velocity selector: E v B 1 2 mv eV 2 e E2 m 2VB 2 e/m = 1.76x1011 C/kg with Millikan’s measurement of e => mass of electron p212c28: 14 Example: Using an accelerating Potential of 150 V and a transverse Electric Field of 6x106 N/C. Determine a) the speed of the electrons, b) the magnetic field magnitude required for no net deflection p212c28: 15 Mass Spectrometer R2 + + + + + + R1 - - - - - - One method velocity selector + circular trajectory E E v B mv R qB RqB RqB 2 m v E p212c28: 16 Example: Vacuum System Leak Detector uses Helium atoms. Ionized helium atoms (He +) are detected with a mass spectrometer with a magnetic field strength of .1 T. With a velocity selector tuned to 1x105 m/s, where must the detector be placed to detect 4He + ions? p212c28: 17 Magnetic Force on a Current Carrying Wire B I A vd dl Fi F Fi qi v i B Nqv d B n volume qv d B nAdlqv d B JAdl B Idl B ( RHR ) p212c28: 18 Example: A 1-m bar carries 50 A from west to east in a 1.2 T field directed 45° North of East. What is the B magnetic force on the bar? I dl Force will be directed upwards (out of the plane of the page) F ILB sin q 50 A 1m 12 . T sin 45 42.4 N p212c28: 19 Torque on a Current Loop (from F = I l x B ) Rectangular loop in a magnetic field (directed along z axis) short side length a, long side length b, tilted with short sides at an angle with respect to B, long sides still perpendicular to B. B Fb Fa Fb Fa Forces on short sides cancel: no net force or torque. Forces on long sides cancel for no net force but there is a net torque. p212c28: 20 Torque calculation: Side view moment arm a/2 sin q q Fb = IBb Fb t = Fb a/2 sin q + Fb a/2 sin q Iab B sin q = I A B sin q magnetic moment m q B I A B m B Magnetic Dipole ~ Electric Dipole U = - m .B Switch current direction every 1/2 rotation => DC motor p212c28: 21 y Hall Effect Bz Conductor in a uniform magnetic field x z + Jx Magnetic force on charge carriers F = q vd B Fz = qvdB Charge accumulates on edges + + - - - - - - - - - - + + + + + + + + + + - - Jx p212c28: 22 Equilibrium: Magnetic Force = Electric Force on bulk charge carriers Bz Ey - - - - - - - - - - - w t + + + + + + + + + + + + Jx Charge accumulates on edges Fz = 0 = qvdBy + q Ez Ez vd By Ez J x nqv d - nq By nq - J x By Ez Hall EMF V H E z w I J x tw nq - IB y VH t p212c28: 23 Negative Charge carriers: velocity in negative x direction magnetic force in positive z direction => resulting electric field has reversed polarity Bz + - - + - + - + - + -Ey + + - + - + - + - + - + Jx p212c28: 24 Example: A ribbon of copper 2.0 mm thick and 1.5 cm wide carries a 75 A current in a .40 T magnetic Field. The resulting Hall emf is .81 mV. What is the density of charge carrying electrons? p212c28: 25 p212c28: 26